ML20238F392

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Rev 0 to Calculation 42111C-011, Fragility Derivation for Refueling Storage Tank 11
ML20238F392
Person / Time
Site: Calvert Cliffs  Constellation icon.png
Issue date: 12/21/1994
From:
EQE ENGINEERING CONSULTANTS (FORMERLY EQE ENGINEERING
To:
Shared Package
ML20238F391 List:
References
42111C-011, 42111C-011-R00, 42111C-11, 42111C-11-R, NUDOCS 9809030328
Download: ML20238F392 (63)
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Text

{{#Wiki_filter:- __-__ _ _-___- _ _ l ENCLOSURE (B) 1-l i EQE Engineering, Inc. Calculation No. 42111C-011 l l l l l l l 1 I-l l 9809030328 990831 PDR ADOCK 05000317 P PDR-Baltimore Gas and Electric Company 1 q Calvert Cliffs Nuclear Power Plant 1 h August 31,1998 g 4 'd #' (j M9 ' g __ ]

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i Project: Calculation

Title:

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References:

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EGE g EQEMEf#MnOfW. 'Jj SHEET NO. M Ca lveH d //b ( JOBNo.U///'/0 JOB By M7 to S DATE '0l24/ CALC.No. SOll SUBJECT /2W67 CHKD NM DATE /lb/#, i?e s-pen s e &s laa t/sn (der, L) Se caose Me o set /ar n/rf mo r,,ew I capa ci dy is 6/tg/tt/y in lluen e e d b., Me ky alsod na,., le y fr&SSa/e,s a<Weg en Yh e- +ae4 so m // a + -ts a f/rrse of ind v/ nu nr errr e a f, the re.sfen3 e i / e.<f eg ed map f Va ca /c v la fe d a t a bo a i

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ESE EQE INTERNADONAL fg-N_8 SHEET NO. '27 ~~ de Isted C// By >!!ui 4 DATE Wl2/l, (' Jos NO. 42lll. I 9 job CHK'D NM6r DATE //b/9' CALC. NO. 0 # ll SUBJECT 4W TT ~ i i R es c> o n s e eva I va +/es (eos L. ) Spe en ta. 6hase hv<tst ( 12 el. 4, TD f:> t e 3 - 2,) '.( y

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E EQE INTEftNATIONAL Qy SHEET NO. O['l'I ~

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EQE . g soe m mw o w. j 'kB SHEET NO. ~2 ~7 ( JOB NO. 4 28\\\\. th jag [a(ves+ C l) {hs gy y,,% DATE **I28I' cal.C. NO. (OII SUBJECT C W6r CHKV N M DATEl//N4 4 l Pe pe-q e //e cts ( co d. ) l l $,. - l J. { * * / R ) ~ - l,L f " /ccan) r .os c' @g= Je {% fjQ>kf l Gac. s ) = c.sg i he G. tG (a ll $, ) L c, o a es t ca n a c e.s i.ie made <,s,a n s a, Th <- u n c e rt vin t.7 due -l o v a *va h! !' Li t.m co n v e-c uv7 ( mode ) < e g ven cy oned c/n~yr y will h a-n e:1 +/m er +cd /n

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l M EQE INTERNATIONAL Q9 SHEET NO. #/ ~37 1 (' JOB no, 4 2 IIl l9 JOB $ctlVe'E $ll 6 BY ^^ w A DATE tel*29, CALC. NO. col / SUBJECT /2 vv 6 T CHK'D /M DATE //Nb F <9vency (-( se2+..ca ) "2. c. o s z e. o s )"2 D ?~ y t < $: ha: {. tG 2 2

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EQE NTERNATONAL a ': -;. 4ll.: J@ ~~ SHEET NO. /2 7 A* I'' 4 N' 6 BY M w 5 DATE /d/7#/ [' JOB NO. 42 /// /9 JOB CALC. NO. CON SUBJECT Ew6 / CHK'D M M DATE //h o v aras/ M de s, R - s,,a,, u ./ Anacia ad l/aeselsh<y s 5 -.{s s {/s. l( / I"e 1M (sv) u re J e-(. s C) { l.6o ) { l.6) bl.6){ \\.6 )(l.6) ( 4.G ') { st,ca t ) s 2 A,82 4 lc-kop, = {C.7c*}415*fC.68*t5.01 2 h 'I* 6'. 2 *) e \\ ( , [ 5,74 I r 41a 3 ) ##5 s d.2 7 $*1 ldJ la 4 (ls tc0 pres s a /e Q -l /7/e d/a a R e.g p a,is.$ e (mq.:-d 9 ab Th e e a ys e n., ne -1 v., s-is 6ti 3vy d. p., de t hs C# 't n e hluid f tese oee 'tos e-ca q e. 5 5,c,., 61d v ct,,, A L.er, ste n s/ 4 e el 't h *- 'l <% k cr1 'Ih er 4 i m a. cl mer c i m a.. m a-at. t w-t 4t, e p/cbs>re Oc $ IJ ers y' b E= ^) 'fM 7'O} c SC s I /g HS - 2 45, t I 4: ]M l' ~

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ESE O. * *"o" SHEET NO. ' /Z 7 JOB NO. 42 f fl.19 JOB [a lu wf. dfi 6 BY A w < DATE B/2J/- CALC. NO. Oll SUBJECT EW6T CHK'O NMd DATE //,[//b blold pe css s ia s (cent.)

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l <~ EQE INTERNATIONAL 3 iM 'Jc3 y!T SHEET NO. (' JOB NO, 421li.10 JOB de Ivd* dIl 6 BY A w S DATE M/3'/ ! CALC.NO. CB/I SL9 JECT RW6T ChK'D- [-r _ DATE j Th.e b.s t es + /%49 4/ce4/de u e v$ ht a/ q.s e f en n Jc-a+ -t h e t/en c ol + 1 x,'m o m o J e. o l a # -, i n y m ome A+ is gloe-b y: Vf VI y .I~ 6 A0 As y (W6 *n)(I" 0~40 b) ( cist e u.n) { !- c.40 (%)(c-49)) = =5 ~ ] '5 5, 7 /# l E s % er t e M e d/a r, d.r y a e r Li FoIlewiog TZ e f 4, .fy, a me. gig,, s s.e 4sen s e g n a me., t c.:e,7 et e t c y s-> r a y -p e esuma4ed kom: a x a# Psa c M sr J + me d /c - u.) ne u e /A, e c a e.t 4 0 < - i e 9 mem-n+ et p o e /dy. N,- m e d/er n co a e i d u ent q m e-n.-n t fes onbe. k= m - dia., /n e la 4 e err e<g y nbssspira.; fa < -id' To in / Ve n.< brh ey is > ( hw ' h u e. Y Se

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l l l EQE INTERNATIONAL { l I ~--"' SHEET NO. /C' (~ JOB NO. 43/// /9 JOB d4/e'i d/' 6 BY. M A/6 DATE /d/~W l CALC.NO. Col / SUBJECT 8W 6 Y CHK'O //4 DATE //N/ / 1 i I 7)* f crm in a /?7e aia n davast& l / / l 71, a -fa n k ca77aci 4y is o a e<n ech b /h a y knk &sek //y cayae/L7 en 1he compr ess/e n i S/de gl th e fais k, and /7y 1/ e .ca es Jes t 1 l cl +h e lo//o.arn7 rn Y/r

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'EGE EQEINTERNATIONAL Y3 SHEET NO.

  1. /27

~~~ JOB NO. 47///. /0 JOB $dlVe'4 $l'Y'S BY wA DATE to/2/. 1 CALC.NO. [O// SUBJECT 8W6I CHK'D M//d DATE ///k/ N4 chm en 4 f y< e $en iS the' de n's a W o n af ef % r%k th tt m e m e as -f gmy asMy usr y ar llv/ <I fresss'e. Yb e e ertwm +11 e minim o = lls/d yre-$$ s<e cue S a /C o US ech for a cmse* v Pt'e si si a d Ta d T.( g L PSl} ( Keyj,,, ) f/Opjjp ) { pyj,;,) (g,p, ) P,_ : 8 9. G ~1, P,_; /0 9 3

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DATE #/#//5 ( JOB NO. 42 /// /9 JOB CALC.NO. b0 SUBJECT Ew S T CHK'D NM DITE I//4b I ~ i l he fa 6+ /c $*r <:~ r. era AbSe'p t/er, fatter l <g a 'Tise fa / lstre=" sue:r ele- / deer f/ f/N b ' +1t

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G6.6 0 ret e det /Yy Yh/6 l6 sl 4.5 ( c.sa) z c.46*q. 7he<</e@, Ho / /e a's fron <csynn:se is n e e esso y. o,, s c5 ECE INTET4NATONAL %f "'~ SHEET NO. %7 $d l red dll BY MwA DATE /8/2// I JOB NO. A2/ll./4 JOB CALC.NO. foil SUBJECT 12 A47 CHK'D N//4 DATElll0 Vasta J //H r4 //n.<ie s i y s w.cie y as b <,,,ed Fa sa m e -tr-i e gas e S 10 die.s as .ea e in el -th e $<s y //Hy y ase,ne fe/s to noIc 4u < $ec t on w,e <e s a it rny momen + tesJ/z!/sybo~c/n da b///*7-c:cyae/ly* Th e e.s r- /ma fed e a e^ mer.b/r can tp en be &1) Y Mes Ap h~' : ) e :I, n eases Mel Ea; Ct2 el. 5, at:n 1-%) } f '= 4 l No da6 e Rad.u ilos T. f, C a g T, f, r M4 N y, g e I %e 4s a se,51 ss.VI S.o l 8.4 01 09 .otG 1%@ I.' A 2 tradscedQ 1 % 10 1 11 A.co ~ 14.1t 5 17sdsca) T -1p ossso 1.79 1 x,joca ps,se -ip s 4 n.na e. 0 0, < .c,< ima

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( EGE 1 g eQE PRERfMnOfW. yJ SHEET NO. Ei i JOB NO. 47111. top yog dalueet dllhs g, p CALC. NO. Ul/ SUBJECT /2W6r CHKD N M I//4k DATE f s ~ o<y tn dia, 0, ca,,,a 79'Om e + e r e 6F=4.C Re fes en ce 6,M 64[ &(6 FT(C*Ic): 6,4: rho-em Resp. 24, 6 E4 lt.x

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= ( eet g coeurtamom Q.3 SHEET NO. 27 bo UC'b $ll BY S 'd b DATE!0l23/4 i JOB NO. k 2III* 19 JOB l CALC. NO. [Oll SUBJECT S W 6I CHK'D N/M DATE ///#/ f I ~ IEe ere n ee s " In te</ n Repa<t Cl.'lls fra1,l/fies42/ll-cr-014,lo /. EgE do eresfondenea c/ vert ZPEee'u r on a L pe. 29, ie e 4. lnel Lug, ,,n, 2. 6 4 e vense n j /spo eta + es, Le tfe' -lo s EGE .E n 1c en a +ro ne l, " Tran.sarria f.a / A< c w p /fij n e FIe o - TEesfsiis e Spca fra ly Se le e ie eb t$ a! l ding $ snd 6fraefa'e3 +c< una ~ t.u i e r 1/ie s C a I a c el-t d // ls IPEEE Piz 4, " Gep s. 2, 199 4. de l<J lm Ucn, ^Re lueIry

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q F e s w i. r s... F f f i S ..:...j. l C r. t ...f r e >. 4X.. .f.4 v L. > :... !. - ..s..> 4 t l g ..p a C s 1 4 .3 .. i. : .i. 3 8 0 0 0 0 0 4 3 0 0 0 0 0 gv CO~. 8g4 e 11: 11 - (-o l \\ A_ttachment B - Refuelina Water Storaae Tank Response Calculations P9'7 This MATHCAD template computes the response parameters which are needed in performing a tank evaluation per EPRI TR-103959 methodology for vertical tanks. Inputs required are an I l assumed earthquake, and the necessary tank parameters. Two notwiimenolonal parameters are I also needed for the calculation of the diamond 4xickling capacity, and for the calculation of the compressive buckling capacity (elephant's foot buckling). Base units are feet, sIeconds, and pounds. This calculation presents the response of the refueling water storage tank at Calvert Cliffs. The tank is evaluated against the 10,000 yr retum period uniform hazard spectra. Dedved Units: BY.&fd DATE lo/FS /94 l kip =1000 tbf hzm 1.sec-1 ksis1000 psi CHK'DbATEl'/4 Ih Define Tank Geometry: R :=20.75 A NominailnnerTaiik Radius R :=35 A Dome Radius d H:=39 A Heightto Water Elevation tb := 0.25.in Bottom Plate thickness td:= 0.1875 in Dome thickness ) f f.1g W l Clearance between peak of dome to spring line hd := Rd 1 - cos asin R t 4 L djjj arings := 4 Number of different diameter rings composing the tank shell 'O.3438' O.2813 t := in She!! Thickness at each ring from bottom of the tank to the top. ,0.1875, ' 83 ' 83 Hr := in Height of each ring measured from the bottom of the tank to the top. ,249, 1 Define Anchoraae Details: I I n := 36 Number of equally spaced anchor bolts ( := 1.375 in Anchor bolt diameter Define Material Prooerties: E, := 27.710' psi Young's Modulus for Shell Material Eb := 2910' psi Young's Moduluts for Bolt Material ye := 3710' psi Effective yield stress for shell material, from Table 3-9 of Ref.1 e y g := 62.45 Unit Weight for liquid A' y, := 0.284-Unit Weight for shell material to 8 r g := 3.2510 psi Bulk Modulus of fluid,3.25x10**5 psi for water Ill~b !! P */ Define Earthauske Amplification Parameters: The earthquake input ground motion for the IPEEE evaluation is defined in terms of Uniform Hazard Spectra with a Peak ground acceleration of 0.10g. For IPEEE, a 84% Spectral shape is assumed. For PRA, a 50% Spectral Shape is assumed. The 50% spectral shape is assumed for the purposes of this calculation. The response spectra are given in Reference 6. ) pga := 0.106 g j Assume that the peak vertical spectra values are 2/3 of the horizontal over the entire frequency spectrum. gyMd/DATE '"/'5/* Comoute averace shell thickness. and total shell helaht g ff/ql9q i := 1.nrings H,:=[Hr, H, = 41.5 A i [t, Hr, t, := t, = 0.234 in Define Dimensionless Parameters from Refs. 2 and 4: - Obtain Cwi from Reference 2, Table 7.4: Parameters needed for table 7.4: =1.88 = 9.414 10' R R For H/R = 1.8, t/R=0.0005 : Cw=0.0642 For H/R = 2.0, t/R=0.001: Cw=0.0904 Interpolate to get value for ts/R = 0.000941: ) C wg := 0.0673 1 l 4 ~& ' l Tank Welaht and C.G. Components!

O' /7 Note that the distance to the component C.G. is measured from the bottom of the tank.

(Shell) (Bottom Plate) BY.22LflDATE lo17N9 j := 1.nrings W :=(s R ) tb'7 s 2 p W, := 2 s R y, Hr, t, Wb = 13.83 kip W,:={W, i tb W, = 51.866-kip by t cg :={Hr-(isj)- } Xb = 0.01 A j g i { W,-cg, X,:= I W, i X, = 18.059 A 1 l ~ (Liquid) (Dome) 2 Ww:=xR Hyg h:=(2 xR h ) t67s W dd W, = 3.292 10' kip W =11.491 kip H h .fRI 2 a := asm (Rdj x .19.5 A w t t I 2 sin (a)~ I-,+ Xh := H,+ h - Rd d 3a R t 2-, R Xh = 45.996* A Fluid Hydrostatic Pressure: P,g := y g H Maximum fluid pressure occurs at base of tank Pst = 16.9 psi i l l i l l l l Comnute Horizontal Imnulsive Mode Response: 42 lll. c.c g [ impulsive Mode Frequency: 0.127 Y s BY M DME b/75/94 CLi := C wg-(Reference 7, equation H-2) g g ) CLI 'E '8 s f ; g* l y fg =4.439 hz Compute Spectral Acceleration at this frequency, damping for the impulsive mode may l be taken as about 5% (this is considered to be a conservative damping estimate) Sah := 0.12 g (see Attachment A for response spectra) Compute Weight of fluid effective in the impulsive Mode, and its corresponding C.G.: 1.732-)R l R W :=if "s, ,1.0 - 0A36TI w 3 W @ef. 3, Eqn. C3500-1,-2,-3,4) i y g 1.732 g l @e. 3, Eqn. C3500-1,-2,-3,4) X g := if s,03 75,0.5 - 0.188-H 3 W g =2.52810 kip X g = 15.599 fl Compute impulsive Mode Base Shear and Overtuming Moment: Sah h + W + W l} (Ref. 7, Eqn. H-3) V g := g -(W 3 Sah h+ W X + Wl X ) (Ref. 7, Eqn. H-4) M g := g -(W 'X 3 3 i h V g = 310.987 kip 3 M j = 4.90810 kip fl Estimate hydrodynamic fluid pressure on the tank at the bottom plate Sah I I' g (Ref. 7, eqn. H-8: Note this is conservative at fluid p I._ '~ 2 136 R II depths less than about 0.15*H) P g = 0.766 psi i Compute Horizontal Convective iSloshina) Mode Response: h {} l,[,o l \\' g Convective Mode frequency gYgg,hm teltEly I.5 # ygy (tl4lqt.( f := tanh.1.835.3 @ef.7, eqn H-10) c ,{ R j t' Rj, f =0.269 hz e Compute Spectral Acceleration at this frequency, damping for the convective mode response is primarily fluid controlled and is estimated to be about 0.5%. Sac :=0.043 g Compute Weight of Fluid acting in the convective mode and its C.G. location W O.46 R tanh 1.835 H'.W, (Ref. 7, egn. H-11) 't I f t l . '4 Hj t Rj, c f i cosh 1.835 H,- 1.0 k Xe := 1.0-H (Ref. 7, eqn. H-12) 1.835:H sinh 1.835g I i f tRj t R, 4 W = 804.025 kip c X =28.389 A e Compute Convective Mode Base Shear and Overtuming Moment: S" W (Ref. 7, eqn. H-13) Ve := - e 8 S E Me := WX (Ref. 7, eqn. H-14) e c 8 V =34.573 kip c M = 981.484 A kip e Compute Hydrodynamic Convective Pressure at fluid depth "y" y :=H This maximizes the hydrodynamic convective pressure f H-yi 0.267 W.S w ac ( R' (Ref. 7, eqn. H-16) Pc := g l.835 E f gRH i R3 P =0.021 psi e Compute the fundamental mode fluid slosh height S h, := 0.837 R. ** (Ref. 7, eqn. H-17) 8 h, = 0.747 A fL ~ 8.,p.t,g Compute Vertical Fluid Mode Response: Compute the vertical fluid mode fundamental frequency ByJfM/DATE 0173/** .1 curnWans ///4/6 2 f := 1 71 f 2R 1.I (Ref. 3, eqn. C35,00-13) +- y / / 4H g t, E, r gj ' f1 S,y := 2 I 0.122 g f = 6.184 hz y 34/ Compute the hydrodynamic vertical fluid response mode pressure, based on a tank - on a rigid foundation, note this pressure is also at y=H, which maximizes p.

  • "~I P := 0.8 y g H

'E cos y g 2 i H j, P = 1.1 psi y Combine Individual Mode Responses to net Total Seismic Demand-Base Shear: Vtot:= V1,yc) 2 Vtot =312.902 kip Overturning Moment gog:=fM g + Mc) ~ 2 M 8 l Mgg = 5.005 10 kip ft Fluid Pressures: Note that these are best estimate, parameter variation are required for fragility analyses: Psh := P g, p 2 Total Horizontal Seismic Response 2 c Psh =0.766 psi Maximum and minimum compression zone Pcmax = Pst + Psh + 0 4 P v pressures at the time of maximum base moment, (Ref. 7, egn. H-22) Pcmin := Pst + Psh - 0.4 P y P = 18.106 psi cmax l Pcmin = 17.226 psi Ptmin := Pst - Psh- 0.4 P Minimum tension zone fluid pressure at the time of y maximum base moment (Ref. 7, eqn. H-23) i Ptmin = 15.694

  • psi Minimum average fluid presssure on the base plate P,yg := Pst - 0.4 P V at the time of maximum base shear (Ref. 7. eqn H-14)

P,yg = 16.46 psi Expected minimum total effect weight of the tank shell acting on the base at the time of the maximum moment and base shear: 1 - 0.4 2_,,pga (Ref. 7, eqn. H-26) te ;= (W + W ) \\ W h 3 3 8i Wte = 61.667 kip c_--_____---___________-___ R: fir:nces:

  • P' U/'/
1. Methodology for Developing Seismic fragilities, EPRI TR-103959, j

Electric Power Research Institute, Palo Alto, CA, June,1994. gl

2. A.S. Veletsos, " Seismic Response and Design of Uquid Storage Tanks", Chapter 7, CHK'DhATE II/4!@

j Guidelines for the Seismic Deslan of oil and Gas Pipeline Systems. ASCE,1984.

3. ASCE Standard and Commentary -Seismic Analysis of Safety Related Nuclear Structures, ASCE 4-86, ASCE, September 1986.

l

4. Bucklina of Thin-Walled Circular Cylinders. NASA SP-8007, National Aeronautics and Space Administration, August 1986.

S. Newmark, N.M., and Hall, W.J., Devetooment of Crtteria for Seismic Review of Selected Nuclear Power Plants. NUREG-CR 0098, U.S. Nuclear Regulatory Commission,1978.

6. Stevenson and Associates," Transmittal of Probabilistic Floor Response Spectra for Selected Building and Structures for use with the Calvert Cliffs IPEEE PRA," letter to Paul Baughman, EQE Intemational, September 2,1994.

7 Electric Power Research Institute, A Methodology for the Assessment of Nuclear Power i Plant Seismic Margin (Revision 1), EPRI NP-6041-SL, Revision 1, EPRI, August,1991. j 4 N 2 lLl -C-cq d P.I/3 Attachment C - Maximum P*rmissible Load in Top Plate + Maximum Permissable load in Plate. This template is based on yleid line theory as it applies to k % the top plate bending in bolt chairs. Bolt chairs are typically used in the anchorage oflarge, flat 4 bottomed storage tanks to distribute shears in tanks to the anchor bolt. Large bolt projections S h provide for ductile response under seismic loading. The moment capacity of these tanks is g limited by the compressive buckling capacity on the compression side of the tank, and by the g allowable bolt hold down tensile capacity on the tension side of the tank. The allowable tensile. i capacity is,in tum, limited to the smallest of the: 1) force required to yleid the anchor bolt,2) i force required to pull the bolt from the concrete embedmont (concrete failure),3) force required to d bend the top plate to a maximum allowable deflection, and 4) force required to tear the bolt chair l to tank wall weld. This template computes the force required to collapse the top plate based on top plate bending. I define unit variables kip a1000 lbf ksis1000 psi define plate dimensions: a := 2.844 in dimension of top plate from tank wall to C. Line of bolt hole b := 5.0 in dimension of top plate from tank wall to outer extent l g := 3.0 in dimension of top plate adjoining tank wall f:= 1.344 in dimension of top plate from edge of bolt hole to free edge tpi := 0.50-in top plate thickness 9 :=.50 in vertical (gusset) plate thickness t ttnk :=0.3438 in tankwallthickness R :=0.7188 in radius of the bolt hole h R g:= 1.0313 in effective location of the applied bolt load (may be taken as equal to one-half e of the width of the nut). define plate material properties oink := 37 ksi Yield Stress for tank wall material o,p :=44 ksi Yield Stress for vertical plate material 3 o p=44 ksi Yield Stress for top plate material p o I L f lll - C ol' 2 c,P 3/3 calculate plastic moment capacities for vield lines shown in Fiaure 1. M 11:=.5 ttak '8tak M,13 = 2.187 kip h SYMATEM@f w w 2 M g :=0.25 t g pl DE ^ p p M g =2.75 kip in- / p J m 'o Myp := 0.25 typ yp =2.75 kip g-M yp m 4 Mp3 :=if(M 11>Mpi,M g,Mw,ij) M pi =2.187 kip h w p m yp>M g,M g,Myp) Mp2 :=if(M p2 = 2.75 k,p ;,- p p M i Mp3 := Mp1 f p3 = 2.75 kip *"- M Mp4:=Mp3 in p4 =2.75 kip " M In calculate load corresponding to co!! apse for vield lines shown in Flaure 1. i := 1 90 p,:= (i-1)I deg 24 a-R sin (p;) h a; := atan 78 - R cos(p )jl. h g . (2 a-R sin (p ) h g g', i=(s) 2 Pu; := Mg8 cot (a;) + Mp2 b + Mp3-(l; sin (a;) + f) + Mp41; cos(a;) cot (a;) p - R rj ef C g := min (Pu) C g = 112.179 kip 114.5 !!4 i i 113.5 IN; EP - 113 112.5 f f i t f 112 0 10 20 30 40 50 60 8 1lll-b@\\ f3/3 Tcnk sh;ll stresses calcul;ted cccording to GIP, p. 717: y := 37 a := 5.0 a gy@ ATE i CHKMATh h:=8_ R := 249 t, := 0.3438 tb := 0.25 e :=2.844 1.0 Z := Ref. SQUG GIP, p. 7-17 " Local Membrane Stress" I, f0.177 a t } ftb b +1.0 4 Rt, j (t,j P e 1.32 Z 0.031 ? u ,p 2 R t, fl.43 a h ) +(4ah)i s 2 , ( R t, Z =0.988 o(87.7) = 176.831 guess at ultimate load that will produce yielding in the tank shell (cy=37 ksi for A 204 SS) guess := 80 max :=r ot(c - o(guess), guess) P y P = 18.35 max I i l L.'._.__ Atta:hment D - Medirn Fluid Pre:2ures (RWST) 42.lll-G.Oli b P tiq This MATHCAD template performs e almulation for an assumed pressure distribution. The fluid pressures are cornputed in a separate program. This procedure is recommended for use in calculation the capacity of flat bottom storage tanks in accordance with EPRI TR-103959. 8Y Itid.h> ATE 101731' M ATE Establish inverse Cumulative Normal Function for Simulation: c 0 := 2.515517 d g := 1.432788 c g :=0.802853 d2 := 0.189269 c 2 := 0.010328 d3 := 0.001308 I I t(p) := in ,p -(p50.5) + (1 - p)2 (p>0.5), 2 f o + c g t(p) + c 2 t(p)2 } inonn(p) := t(p)- -((p>0.5)- (p 50.5))

c l + d g t(p)+ d t(p)2+ d t(p)';

l { 2 3 unity := 1 j The purpose of this calculations is to find the Median fluid pressures acting on the tank base plate and tank wall at the base. Each of the input parameters assumed to be constants are defined below: PST := 16.9 psi Static Fluid Pressure PHS := 0.766 psi Fluid pressure from horizontalinput motion P ys := 1.1 psi Fluid pressure from verticalinput motion The random variables are assumed to be lognormally distributed with uncertainty as defined below; f1med ; 4 5 E fl := 0.34 f2med := 0.40 p f2 := 0.57 f und ;= 1.09 p gg := 0.10 l Perform Multi Variable, Latin Hypercube simulation: N := 100 N independent samples 1 i := 1.. N Generate random pressure factors: i-1 + md(unity)Ii f f g := f1med exp p fg inonn i i i N 43 I N f2f2med exp p f2 inonn 1 i t i N j3 f fi-1 + md(unity)" fu :=fMmed exp.S ginonn i i i i N 33 1 b-8l} ), Each of the samples need to be " shuffled"in ord:r to remove corrition: - I gy }71AlA ATE 80178)*f D B, :=rnd(N) generate a vector of N randorn niimbers CHKDhTE IMN C := augment (f g,B) create an array of two columns in order to mix up the original array D := csort(C,2) rearrange the original array by sorting on the random numbex, f g := D,,, redefine the. original array B, := rnd(N) C := augment (f,B) 2 D := coort(C,2) f,:= D,,, 2 B, := rnd(N) C := augment (fu,B) ' D :=csort(C,2) ILA; := D,,, At the location around the circumference where the axial compression stresses are maximum, the fluid pressure, pc,is given by: cmu, := PST+ f,'(IM, P gg3 + f P y3) P 1 2; Pcmin :=PST+ f,'(IM, P gis - f, P y3) 1 2 g Similarly, at the location of maximum tensile uplift, the fluid pressure, pt,is given by: Ptma :=PST + f,' (-fM)iP yg3 + f, P y3 l 2 g IIS-f, P v3] tmin, := PST + f,' (-ILA), P P 2 J 1 Plot values of pressure as a function of the trial 5 1.5 10 i i i g t t t N +g ,+, tt /* ,.. +.+# i io +, ++# * + * ++ +:;,,g +,f u +, *

  • 4+

L ",*i 5 + + j /kCK ,8 gg "s.%[ [%:<bbe d **: m *b n v V C 4k* 9

  • (X P q g

/ ? s P" *A" f"% + + " "....* "l " = $ y I O " Dds O " O g m ap,,, o ao DO o Do j o o o t i f i l 0 i 0 20 40 60 80 100 $20/-C Clf D 7'3,/ orint out median and +1 standard deviation variates: IA(P)ut A(P) := sort (P) p(P) :=In gf g _ gj CHwoI %om llM194 L Pressures on the compression side of the tank, first for the minimum pressure: [ P := Pcmin A(P)3, = 18.565 psi A(P)u = 19.594

  • psi A(P)3, = 17.098
  • psi p(P) = 0.054 l

And then for the maximum pressure: l P := P cmax A(P)3, = 22.82 psi A(P)u =25.678 psi A(P)3,= M. ETsi p(P) = 0.ll8 Pressures on the tension side of the tank, first for the minimum pressure: P := Ptmin A(P)so = 10.974 psi A(P)u = 12.837

  • psi l

A(P),, = 8.096 psi p(P) = 0.157 And then for the maximum pressure: P := P tmax A(P)so = 15.2 psi A(P)u = 16.493 psi A(P),, = M.W psi l p(P) = 0.082

References:

gy g g to/75/94 h/// [ Of[ P4/f q agm /4/44 n

1. Electric Power Research Institute, Methodology for Developing Seismic Fragilities, EPRI TR-103959, June,1994,
2. SQUG, Generic Implementation Procedure (GlP) for Seismic Verification of Nuclear Plant Equipment," Available from USNRC Public Document Room (920918-MS), Washington, D.C.,~

February,1992. .nsr h 1 l i l

Oil l -C-Ol \\ Attachment E - Calculati7n of Fluid Hold D7wn Forces for RSWT E' b 1/ BYkWhDATE D125lef Fluid Hold-Down Forces CHICDhATE Il N For tanks with minimum anchorage, hold-down forces resulting from fluid pressure acting on the sank' bottom will contribute significantly to tthe ovoertuming capacity of the tank. 'EPRI TR-103959, Methodoloav for Develoolna Seismic Fraallities. Section 7,contains procedures for calculating these fluid hold-down forces as a function of tank uplift. This Mathcad template follows the procedure in EPRI TR-103959 for determining the fluid hold <fown forces. Inputs are tank parameters, the output is a plot of the fluid hold <fown tension as a function ~ of both uplift displacement and uplift length. Derived Units: kipa1000 lbf hzs1 sec" ksia1000 psi Define Tank Geometrv Snout): R := 20.75 ft Tank Radius II:= 39 ft Height to Fluid Surface t,:=0.3438 in ShellThickness tb :=0.25 in Baseplate Thickness Amax := 0.02 in Maximum permissible uplift height, this is a function of the failure mode O :=2.5 red Assumed angle to the neutral axis of the tank n 1 Define Tank Material Properties Gnout): E,:= 27.710' psi Young's Modulus o := 0.30 Poisson's Ratio u := 84 ksi Ultimate Strength of Tank Material o y := 30-ksi Minimum Yield Strength of Material o ye := 37 ksi Effective Yield Strength of Material o Define Fluid Pressure Parameters Gnout): Pcmin := 18.57 psi Pcmax := 22.82 psi Ptmin:= 10.97 psi Ptmax := 15.2 psi. Qg$ned oressures to be used in analysis: P := P Pc := Pcmax t tmax - Calculate Additional Parameters needed for analysis: i i tb Ib 'Ib = 8.2810" *ft' ,_ 2) 3 E*t 8 K:= K = 103.08 kip in 12-(1 - u') .i R_,}3,[3_,2) r = 34.593 c := I s

$2lll-C-Ol{ F. -( 274 K, := ,,,,, K, = 28.641 kip R fg _ p (EPRI TR-103959 calls this term Mf/P) BYMATE M/U/# s r := CHK'DhATE //4/fY dl2-(1-9 ) A / 2 i Perform iterative solution for uplift helaht. tank shell hold-down tension end moment. and maximum positive moment in the base plate as a function of the uplift lenoth and fluid pressure: F(L) := 1 + - gL I I K ( 2 E gIbj 2f L' 1 Ki TL p 6(L,P) := - F(L) (72 E gIb 6 j, E gIb ---+ 24 L 1 K gL y)' 2 T (L,P) := + - - - + - -P ~ e 2 F(L) (12 E gIb Lj, p Kgd I I M fL,P) := --- + r F(L) (12 EgIb j L M fL,P) fM fL,P)32 2 max (L,P) := p M + 2 _4 P ( PL j 'MfL,P)" tgr F (L,P) :- -+ P Horizontal force in the tank wall at the 3 P R 2 intersection with the tank bottom 12-[I- (u ){ Hold down tension as a function of uplift length, the term f (b' )3 T fL,P) := T,(L,P) + F (L,P) : Fh(L) is the additional correction which accounts for h ( L / membrane effects EPRI TR-103959 aooroximates the relationshlo between fluid-hold down forces and uolift displacement with a linear expression. This simotification is usefulin the overturning moment capacity evaluation. Solve for the minimum and maximum length of plate effective in resisting uplift (corresponding to zero uplift, and maximum uplift at the extreme outer tensile fiber). The fluid pressure is a function of the distance from the neutral axis of the tank: P(0):= (P - P ).I - N Reference 1, eqn. 7 38 I+P t c t ( 2 Assume an angle to the neutral axis in order to calculate the minimum hold down force, this angle may be changed later. O = 2.5 rad n P(0 ) =22.062 psi j n L L__-____--__-__--_--__-______

9 LIll -C-ol1 E. 7.sj4 l Calculate fluid hold down force at v:rious I: cations about tha circumf r:nce cf ths tank. Note th st tt th3 niutral axic ^ l the fluid hold down forces will be minimized since there is zero tank uplift. At the point of maximum tension the fluid i hold down force will be maximized:- l foos(0)-cos(e )I W M MTE l# M

  • n

/ M M ON ( 1-cos(e ) j n I N:=20 x := 10 in (Initial guess, note that the above derivation it is assumed that delta is less than 2/10 of L ) i := 1.. N f p,:=0 ' i~ i n (N-l j p :=P(p,) g L is the length of plate effective in resisting uplift and is calculated y := r ((A(p )- 6(x,p,)),x g here as a function of the angle beta measured aboQt the tank circumference Tfis the fluid hold-down pressure and is calculated here as a function Tf,:=T (h,p,) of the angle beta measured about the tank circumference. f Print out the length of baseplate effective in resisting uplift and the corresponding fluid hold-down forces at both the neutral axis and at the point of maximum ter.slon in the tank in order to check published solutions: L, = 8.981 *in Point 1 is at the point of maximum tension in the tank Tf, =0.096 5 I" L = 7.355 in g Tf =0.122 h g m EPRI TR-103959 recommends fitting a best-fit linear relationship for the fluid hold-down force as a function of the angle beta: a, := cos(p,) Tfn := intercept (a,TO fn = 0.111 h T m AT := slope (a,Tf) f f -0.014 h 4 AT m y, := AT pcos(p,) + Tfn

$2lll G olI G F 4p't Plot fluid hold <fown f:rce versus tha gngla beta f:r th2 tank: 0.125 i i 3 g m 0.12 ) 0.!!$ T,(KipsAn) Ns y; \\s. g 0.11 N \\ N. 0.105 0.1 s. i e i i e i i i ,,3 -1 -0.8 - 0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 Cos (p) The Uoner limit on this eaustion is based on a fully plastic moment at both ends of the uplifted zone of the base plate and should be limited to the followina values: tb 2Gu+c I pb = 1.031 kip in-y M M pb := 4 ( 3 j m M, := M 4,p,) cWk := max (M) chrk = 0.298 kip h m The maximum moment in the tank shell should be limited to 90% of the fully plastic moment capacity. M gi := 0.90 M pb M.ig = 0.928 kip-(ok, note that check is less than the allowable)

Attachment F: Calvert Cliffs - RWST Capacity Computation N< F. sp. i jb Overturning Moment Caoacity The overtuming moment capacity of anchored tanks is computed in an iterative process. EPRI hd% TR-103959, Project 2722-23, Methodoloav for Devetoolna Seismic Fraallities. June 1994, Section 7, N contains the procedure for calculating the overturning moment capacity. EPRI NP-6041, A D :P i Methodology for Assessment of Nuclear Power Plant Seismic Margin, Appendix H is also used as a 3 D / tcference. This Mathcad temp!ste follows the procedure in EPRI TR-1003959 for determining the overtuming E moment capacity of tanks. This template is intended to be used in conjunction with the template g FLUIDHD. The FLUIDHD template esiculates the fluid hold-down forces which lead to slightly gg increased capacities for marginal tanks. This template may be used with or without consideration of the FLUID hold-down forces. Derived Units: kip s1000 lbf hzul sec-1 ksin1000 psi Define Tank Parameters R := 20.75 ft Tank Radius WTE := 55.7 kip Expected Minimum Total Effective Tank Weight f N:=36 Number of equally spaced anchor bolts 6eo := 0.02 in Permise!ble uplift elongation 1 8 E, := 27.710 ksi Young's Modulus for the Tank Material t,:= 0.3438 in Tank wall thickness at bolt chair location Pe := 18.57 psi Pressure in compression zone at the time of maximum moment ye := 37 ksi Effective yield stress fc r tank material e Define Anchor Bolt Parameters h, := 30. in Depth to embedded anchor bt aring surface (estimated) h := 8.0 in Height of anchor bolt chair above tank foundation (thickness of baseplate) e 2 Ab := 1.4849 in Nominal Area for each anchor 3 Eb := 2910 ksi Anchor modulus of elasticity (ksi) Tbp := 0.0 kip Anchor bolt pre-load (klps) TBC := 18.4 kip Maximum anchor bolt load (force in bolt required to yield tank wall) Obtah <iiamond bucklina coefficient from Reference 4. Fiaure 6: Parameter needed from Figure 6: 2 P iR3 c = 0.352 E (t ; 3 Read-off value for delta-gamma' f f Ay := 0.18 Input Linear Annroximation to fluid hold-down forces (from template FLUiDHD.) l fn := 0.09 h zero delta fluid hold-down force, fluid hold down force at neutral axis T i m 0.015b Slope oflinear approximation ate: - _ _ - _ _ - _ _ _ _ _ _ - _ _ _ _ _ _ _ = _ _ _ _ _

Altil-C-oII V 7 2ja Compute the Bucklina Canicity of the Tsink Shell based on EPR NP-6041 aporoach: axial stress limit at tha onset of elephant's foot buckling: BY.ME.JOATE aJ75/< l Sg:= S g = 1.811 CHK'hATEh / t a ; := 0.05 E -.' classical buckling capacity c 3 R 2 Sg+ f PR3'7 c } ) p := a g 1-(Ref.1, egn. H-27) a e l.12 + S g 5 S 1+I 1 t ('ye :j ( j e = 14.578 ksi o l = 23.139 *ksi p e diamond buckling capacity based on NASA SP-8007: e v=hf y := 1 - 0.731-(1 - e ) d Ay a := 1.25 y+ 0.605) recommended approach to diamond buckling per reference, equation 7-4 4 note that the 1.25 factor accounts forjudged conservatism in the model tests performed to support the NASA SP-8007 approach when the y factor cl is based on a loading thatinduces bending. A -. ' ye OAS 11 := + 0.18 -(651.6)-(A>0.55) + 1.0-( A50.55) Ref 7, eqn. 7-50,7-51 L h b 2 c bd 11 cl e bd = 20.318 ksi Compressive shell capacity: B If('bd<0p bd' 8p) t, C 2 B = 5.012 h Median Buckling Capacity C m Estimate bucklina capacity of tank based on Eu'ropean Convention fo Construction Steetwork (ECCS) approach. Ti approach provides a lower bound code limit stato capacity. The equation uncertainty may then be estimated from ti l l median buckling capacity given above, and the ECCS method. I Calculate imperfection factors for normal construction for the case of axial compression and bending withoutinterni pressure: . = 724.25(ok) Note, that the following factors are only applicable for 212 < R/ts < 1500 per EPRI TR-1039' t, i 0.70 i a AO ;"" 0.1 + 0.01 R I. f [sj 1

k b-G \\ eBO := 0.1887 + 0.8113 a AO

f. p.5)b revise imperfection factor to account for internal pressure. Assume that bending occurs:

BY W## ) ATE 'd/71/* f Ref. 7, egn. 7-67 CHK'DbATE ///4/94 F p := E,(t,j Fp + 0.007 a BO ~ p := Ref. 7, eqn. 7-66 a Fp + 0.007 The buckling capacity of the tank is estimated from the following equations. The reader is referenced to EPRI TR-103959 for a detailed explanation of the procedure. a p(R,) := ol Ref. 7, egn. 7-71 o e a I* A (R,) := Ref. 7, eqn.~7-72 p $ 'P(R,) i cu(R ):=if 1(R,)222,0.75.o (R,),o 0.4123-(A (R,))1.2' Ref. 7, eqn. 7-70 o p p ye p f I ~ o h :=P R Ref. 7, eqn. 7-69 c /sj i i au(R,) := (ocu(R,)2- 0.75 o h- 0.5 o h o To begin the solution scheme, and assumed R, term must be selected. The solution has converged when the computed R,is equal to the assumed starting value. ( := 0.80 R. := root (oau(5)- C eu(C)'5) R, = 0.52 Compute ECCS bucklina capacity: cap 8au(R,) t, Ref. 7, egn. 7-75 ECCS = 3.804 h ECCS cap m l Compute uncertainty on eaustion f Cg i Ref. 7, eqn. 7-60. Note that this is computed at the Ocbc:[I ) minimum fluid pressure acting on the compression (ECCS capj side of the tank. p ege = 0.138

12all c-ou. F. p.g ComeAe Overtumina Moment Cenativ of the tank based on median coini e.idve bucklina caoacity as determined above. - yg 1612319' CommAe dimensionless parameters for solution scheme Note: the angle Beta represents the angle to the neutral axis,it is assumed prior to the start of the algorithm. i(8} W }+ ~ I + #E) C (E) C g(p):= in(p)+ (n-p) cos(p) 1 + cos(p) 2 s ME}- E*E) /I + WE)' c 4(p) := 0-'ISCE)*E) c (p):= in(p)+ (n-p) cos(p) (1 - ms(p)3 1 - ms(p) 3 s Compute Anchor Tension in Each Anchor Bolt as a Function of Location i := 1,2.. N (1-1) 3 deg

a. :=

(N), 6,o A Eb cos(a;)- cos(p)I f b Ty :=Tbp + i h,+h c ( 1 - cos(p) 4 3 =if(TB nBC T,,TBC) T B g T,:=if(T,20.0 lbf,TB,0.0 lb B B 3 AT * =-0.015 kip, Slope is based on best fit from template FLUIDHD in WTE+ T B, C (p)+ AT C (p) C' m(p) := +Tgp 3 e 3 2R 2 2 B; R cos(a;) + T gR 2 sin (p) + AT C (p) R 2 M SC(p):= C'm(p).C (p) R + T e 4 2 i l

$141) - C-o \\\\ F g, c/ Assume a beta value to beain solution: Note th t th3 tngir beta r: presents the cngl3 between the mbmum tension side of the tank and the neutral axis. For a lightly anchored tank the angle beta should approach pl. For anchored tanks, the angle should be between zero and pl. The assumed beta and the calculated angle below should be identical for the final solution, pe2.816 assumed beta j Solve for beta. Note that this is an iterative process. The beta printed below should be equal to the assumed bet for the final solution. Therefore, the user must input a new assumed beta which is equal to the assumed beta for the final run. The plot below shows the difference between the compressive stress in the tank wall and the calculated (:= 1.1.1. 3 buckling stress as a funciton of an assumed angle. The point at which the curves intersect is the correct angle beta, f(q) := C' m(()- CB d*10 ( := 3.0 i i p := root (f((),() 8 gg) 2*10 p = 2.816 g a o 6 I I -2*10 Final Results: Moment Capacity: d D N MSC(p) = 1.50510 kip ft CHXDhATE II/4/9{ Sum of anchor bolt tension (needed for shear capacity): {T, = 370.16 kip B i l I l l

RIIIr:nces:

1. Electric Power Research Institute, A Methodology for the Assessment of Nuclear Power

~ Plant Seismic Margin (Revision 1), EPRI NP-6041-SL, Revision 1, EPRI, August,1991. BYd.GDATE Mh*/' j

2. A.S. Veletsos," Seismic Response and Design of Liquid Storage Tanks", Chapter 7, Guidelines for the Seismic Desian of Oil and Gas Pioeline Systems. ASCE,1984. '

CHKD ATE M

3. ASCE Standard and Commentary -Seismic Analysis of Safety-Related Nuclear Structures, ASCE 4-86, ASCE, September 1986.
4. Sggk[ina of Thin-Wallsd Circular Cylinds, NASA SP-8007, National Aeronautics and Space Administration, August 1986.
5. Newmark, N.M., and Hall, W.J., Development of Criteria for Seismic Review of Selected Nuclear Power Plants. NUREG-CR 0098, U.S. Nuclear Regulatory Commission,1978.
6. Stevenson and Associates,' Transmittal of Probabilistic Floor Response Spectra for Selected Building and Structures for use with the Calve'rt Cliffs IPEEE PRA," letter to Paul Baughman, EQE International, September 2,1994.

7 Methodology for Developing Seismic Fragilities, EPRI TR-103959, ~ Electric Power Research Institute, Palo Alto, CA, June,1994. ) A Y---_.___-_-____-__

N M l l d-O(( g _r= +..--_M G! Calvert Cliffs - RWST Capacity Computation. Tank Wall is at Ultimate Stress i'h/s Qvertumina MomentCapacity The overtuming moment capacity of anchored tanks is computed in an Kerative process. EPRI g TR-103959, Project 2722-23, Methodoloov for Develoolna Seismic Fraallities. June 1994 Section 7, contains the procedure for calculating the overtuming moment capacity. EPRI NP-6041, A 3 Methodology for Assessment of Nuclear Power Plant Seismic Margin, AppendixN is also used as a g reference. i This Mathcad template follows the procedure in EPRI TR.1003959 for determining the overtuming moment capacity of tanks. This template is intended to be used in conjunction with the template FLUIDHD. The FLUIDHD template calculates the fluid hold-down forces which lead to slightly E increased capacities for marginal tanks. This template may be used with or without consideration of Ni the FLUID hold <fown forces. Derived Units: d kips 1000 lbf bzal see ksis1000 psi Define Tank Parameters R := 20.75 ft Tank Radius WTE := 55.7 kip Expected Minimum Total Effective Tank Weight N := 36 Number of equally spaced anchor bolts 8,o := 0.037 in Permissible uplift elongation ~ B,:= 27.710' ksi Young's Modulus for the Tank Material t, := 0.3438 in Tank wall thickness at bolt chair location P := 18.57 psi Pressure in compression zone at the time of maximum moment c ye := 37 ksi Effective yield stress for tank material o Define Anchor Bolt Parameters h, := 30. in Depth to embedded anchor bearing surface (estimated) hc := 8.0 in Height of anchor bolt chair above tank foundation (thickness of baseplate) l 2 Ab := 1.4849 in Nominal Area for each anchor Eb := 2910' ksi Anchor modulus of elastic liy (ksi) Tbp := 0.0 kip Anchor bolt pre-load (kips) TBC := 41.8 kip Maximum anchor bolt load (force in bolt required to yield tank wall) Obtain diamond bucklina coefficient from Reference 4. Flaure 6: Parameter needed from Figure 6: PcI .' = 0.3 52 RI E, (t,; Read-off value for delta-gamma .I' Ay := 0.18 Input Linear Anoroximation to fluid hold-down forces (from template FLUIDHD) -Tfn := 0.09 zero delta fluid hold-down force, fluid hold down force at neutral axis e :=.0.015h ' Slope of linear approximation AT

(f -bO ( l g 7, Zg Compute the Bucklina Capacity of the Tank Shell based on EPR NP-6041 approach: axial stress limit at the onset of elephant's foot buckling: ByMDATEE/ZF/4 1 CHMATE II/d4f S;:=/R_I S; =1.811 (t,) 400 o d := 0.605 E s classical buckling capacity 3 R fPR3'f Sg+ 2 e } 1-1- (Ref.1, eqn. H-27) p := c j-88 S 1+I o c t (cye :j 1.12 + S 3 j o = 14.578 *ksi a l =23.139 ksi p e diamond buckling capacity based on NASA SP-8007: p=.L. R_ j f 16 t, y := 1 - 0.731.(1 - e ) d I Ay a := 1.25. y + 0.605 recommended approach to diamond buckling per reference, equation 7-4 ( 4 note that the 1.25 factor accounts for judged conservatism in the model tests performed to support the NASA SP-8007 approach when the y factor J ' cl is based on a loading that induces bending. } A= ' ye 0.45 q := + 0.18 -( A51.6)-( A>0.55) + 1.0-( A50.55) (Ref. 7, eqn. 7-50,7-51) i \\ A b o bd :"'9'8 cl o bd = 20.318 ksi j l l Compressive shell capacity: B := if(o bd<8p>' bd 8p) t C 3 B=5.012ah Median Buckling Capacity C m Estimate bucklina capacity of tank based on European Convention fo Construction Steelwork (ECCS) sooroach. Th ' approach provides a lower bound code limit state capacity. The equation uncertainty may then be estimated from th median buckling capacity given above, and the ECCS method. Calculate imperfection factors for normal construction for the case of axial compression and bending without interna pressure: Note, that the following factors are only applicable for 212 < ras < 1500 per EPRI TR-10395 = 724.25(ok) t, 0.70 aAO f I 0.1 + 0.01, R t 1 i s3

c l l eBO := 0.1887 + 0.8113.e AO l revise imperfection fact' r to account for intamal pressure. Assume that bendirig occurs: WMDATE Id/73#4 o DhamMM P c.,fR 3 ta p:=~E,}t,j. (Ref. 7, eqn. 7-67) F l l F + 0.007 ABO p p := (Ref.7, eqn. 7-66) a F + 0.007 p The buckling capacity of the tank is estimated from the following equations. The reader is referenced to EPRI TR-103959 for a detailed explanation of the procedure. o (R,) := o1 (Ref. 7, eqn. 7-71) c p I* A (R,) := (Ref. 7, egn. 7-72) p a (R,) p 3 cu(R,) :=if 1(R,)222,0.75 o (R,),o, 1 - 0.4123-(1 (R,))1.2' (Ref. 7, egn. 7-70) o p p y p f ~ oh:=P. R\\ i (Ref. 7, eqn. 7-69) c \\,l t 1 su(R,) :=(acu(R,)'- 0.75 o h ) - 0.5 a h e To begin the solution scheme, and assumed R, term must be selected. The solution has converged when the computed R,is equal to the assumed starting value. ( := 0.80 R, := root (o'au(5)- 5'8cu(5) 5) R = 0.52 a Compute ECCS bucklina capacity: ECCS cap := oau(R,).t, (Ref. 7, egn. 7-75) ECCS,p = 3.804 h c m Compute uncertainty on eaustion ) f C 3 Ref. 7, eqn. 7-60. Note that this is computed at the ,_1 E cbe -{ minimum fluid pressure acting on the compression ECCS**E' side of the tank. 0 cbe = 0.138 5

M il l -C-CD l.\\ d p.4/6 Compute Overtumino Moment Capacity of the tank based on median compressive bucMina caoactiv as determined above. Compute dimensionless parameters for solution scheme ByMWd )ME id/75/4 Note: the angle Beta represents the angle to the neutral axis,it is assumed prior to the CHKDhATE II/9/94 start of the algorithm. 1 I +

  • 8(0)

(} }+ ~ C (p):= in(p)+(n-p) cos(p) C (E);* g 2 s 1 + cos(p) f + ms(p)' C (p):= b Mp)- Eco@) 1 C (p):= 3 4 sin (p)+ (x-p) cos(p) i - cos(p); I - cos(p) 1 Comoute AnchorTension in Each Anchor Bolt as a Function of Location i := 1,2.. N (i-1) 3. deg

a. :=

(N), fcos(a,)- cos(p)I 6 AEb 3 co TB :=Tbp + i h,+h e i 1 - cos(p) j T,:=if(TB;fB C,T,,TBC) B B T, :=if(T 20.0 lbf,T ,,0.0 lb B B B j AT * =-0.015 kip Slope is based on best fit from template FLUIDHD in WTE + T B; C m(p>:= .Typ C,(p> AT.C (p) c 3

2.,

SC(p):= C'm(p).C (p).R'+ {T, R cos(a;) + TgR 2. sin (p) + AT C (p) R 2 2 M 2 B e 4 i

4 2 l1.c.ot t 6,p.s/3 t Assume a beta value to bealn soluggn: Note that tha angle beta r presents the tngla between the maximum tension aide of the tank and the neutral axis. For a lightly anchored tank the angle beta should approach pl. For anchored tanks, the angle should be between zero and pl. The assumed beta and the calculated angle i I below should be idantical for the final solution. p e2.614 assumed beta Solve for beta. Note that this is an Iterative process. The beta printed below should be equal to the assumed bot for the final solution. Therefore, the user must input a new assumed bets which is equal to tho' assumed beta for the final run. The plot below shows the difference between the compressive stress in the tank wall and the calculated (:= 1.1.1 3 buckhng stress as a funciton of an assumed angle. The point at which the curves intersect is the correct angle beta, f(() := C' m(()- CB 8*1 ( := 3.0 i i p := root (f((),6) 8 g(3 s 10 p = 2.614 g __A o -s to', Final Results: Moment Capacity: g M SC(p) =2.61210 kip ft CHichATE.$h'[i 8 Sum of anchor bolt tension (needed for shear capacity): {Tg =711.522 kip i -)

2 fit -c-oll 5. 3l ofb '. R:f:r:nces:

1. Methodology for Devel'oping Seismic Fragilities ', EPRI TR-103959, Byggd0 ATE /d/F//t ElecMc Power Research Institute, Palo Alto, CA, June,1994.

Cwo%EJMeg j

2. A.S. Veletsos," Seismic Response and Design of Liquid Storage Tanks", Chapter 7, Guidelines for the Selsmic Desian of Oil and Gas Pipeline Systems. ASCF.1984.
3. ASCE Standard and Commentary-Seismic Analysis of Safety.Related Nuclear Structures, ASCE 4-86, ASCE, September 1986.
4. Bucklina of Thin-Walled Circular Cvlinders. NASA SP-8007, National Aeronautics and Space Administration, August 1986.
5. Newmark, N.M., and Hall, W.J., Development of Criteria for Seismic Review of Selecid Nuclear Power Plants, NUREG-CR 0098, U.S. Nuclear Regulatory Commission,1978.
6. Stevenson and Associates," Transmittal of Probabilistic Floor Response Spectra for Selected Building and Structures for use with the Calvert Cliffs IPEEE PRA," letter to Paul Baughman, EQE Intemational, September 2,1994.

7 Electric Power Research Instnute, A Methodology for the Assessment of Nuclear Power Plant Seismic Margin (Revision 1), EPRI NP-6041-SL, Revision 1, EPRI, August,1991. j I l I

42III.- c-of) j./ CHECKING CRITERIA CHECKLIST t 1 ) . m n s w e v;: n ; v.: n.

~~c x NN N'M Client Ok Project EI Job No.

N ll l - Calc. No. Revision No. Criteria Yes No N/A Comment No. /

1. Originator followed defined procedures.

[

2. Title, purpose and function of the work checked are adequately described.
3. Work method clearly stated and appropriate.

[

4. Assumptions identified. Open items flagged for subsequent verification where necessary.
5. Technical bases and references current, correctly selected, and incorporated.
6. Technicalinput properly selected and adequately identified. Any specific input to be excluded are adequately identified.
7. Applicable codes, standards and regulatory requirements identified and properly used.
8. Analytical steps can be verified without recourse to originator.
9. Each page of the work identified and trace-able to originator, date and job or equivalent control number.

/ 10.All markings legible and identifiable. [ Page [/ of.3

  • kfjffhhfkffffh!hf!{!f 1

1 i

42/// _ c _ o//y-A +S: CHECKING CRITERIA CHECKLIST ) +T wh;.} W "TX L &

a
-. :.:~::if s+:-;
a>
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Client Oh Y b Project N/#M O'h Job No. Nb' N Calc. No. ~ Revision No. Criteria Yes No N/A Comment No. I

11. Work clearly references any final supporting

[ computer runs. \\ 1 l

12. Final computer runs include input listing and l

output. N L

13. Final computer runs contain unique number Identifier.

l 14.Results consistent with inputs, technical j procedures, and other project criteria. / 15.Results are reasonable.

16. Revisions are clearly documented.

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17. Technical interface requirements in the

/ Project Plan have been satisfied. I 18.All documentation available to checker. / j

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w(.1 HMh+ Checked by: Date: { (2[ Pageg2of 3 ffh h h h h hfff h f h

2 lil. - C - otj, p ll l ( CHECKING CRITERIA CHECKLIST j McW2 WftM21P4%6e?2 2 2Mistrig# Client 0& Y5 Project [Ed6f$ llNl~s f 2--lll. l@ Calc. No. b I Job No. i Revision No. CHECKING CRITERIA CHECKLIST Comment No. Comment Resolved by: Date: l AIOS/6 ~ l t l 1 l i ht h' P*9* f$'5 *I 3 hh*h h'hk$bhhh _. _ _ _... _ _ _ - _ _ _ _ _ _ _ _ _ _ - -}}

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