ML20117K860
| ML20117K860 | |
| Person / Time | |
|---|---|
| Site: | Pilgrim |
| Issue date: | 01/31/1994 |
| From: | Lashkari B, Jeffrey Reed JACK R. BENJAMIN & ASSOCIATES, INC. |
| To: | |
| Shared Package | |
| ML20117K845 | List: |
| References | |
| 91C2672-C018, 91C2672-C018-R00, 91C2672-C18, 91C2672-C18-R, NUDOCS 9609120202 | |
| Download: ML20117K860 (111) | |
Text
,
l Attachment B S&A Calculation No.91C2672-C018, Revision 0 Fragility Analysis of Station Blackout Diesel Generator Pilgrim Nuclear Power Sation l
i i
9609120202 960905 PDR ADOCK 05000286 l
P PDR
Ca(c. No O C 1(,% - C c \\ b (N.o Fragility Analysis of Station Blackout Diesel Generator Pilgrim Nuclear Power Station
- - - - - ~
by Bahman Lashkari John W. Reed 1
Prepared for Stevenson & Associates January 1994
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Estimation of the Median Ductility Scale Factor, Ftsu l
. The median ductility scale factor, Fmu, is obtained by several methods (See Reference 1):
ne following variables are used in the analysis:
f = 7.5856 412 De A8 Bus frequency m
b:= pets b = 0.05 Elastic damping
& ::2.51 Response spectnan knuckle frequency [\\ g AddM sec s = 0.13794 Strain hardening ration.
mu = 1.67125 Ductdityfactor Riddle-Newmark method for calculating FuRn The details of the methodology are given in Reference 1 lets first calculate Fmu at the peak ground acceleration level (zpa)
(
The zpa valueis:
zpa ::0.4 g and the spectral acceleration at the fhaA*-tal frequency, f, and clastic damping, b, is:
i Sa ::SSa(fas,b)
Sa = 0.65961 *g Fmu is estimated as:
Fu4 :: 8* mu "
Fu4 = 1.74485 rpa Then,in the acceleration range of the spectrum, Fmu is Mma'M as:
Fu3 := (2.67 mu-1.673)""
Fu3 = 1.52439 Finally, in the velocity range of the spectrum, Fu is estunated as:
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23 s.
The median ductility scale factor, Fu.RN, using the Riddle Newark method,is obtained as:
i Ful = 1.52439 Ful :s Fu3 (Fu3 <Fu4) + Fud-(Fu3 2Fu4)
Using the ratio of ultimate static capacity to yield static capacity, R MW as:
R = 1.09259 R := 1 + s-(mu-1)
Fu RN := (Ful-(Fu2<Ful) + Fu2-(Fu22Ful))
FuRN = 1.3952 R
Modified Riddle-Newmark method for calculating TuMRN Because the Ridhe Newmark method does not account for second slope of the foros<leformation curve, the e slope is accounted for by modtfymg the ductdity ratio. Derefore, the marhw ductdity ratio is:
j l
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+
)+
mu ; := 0.5 + (""-
mu 3 = 1.50718 2 R*
At the peak rpa level.Fu is:
' Fum4 :=b mu 3 Fum4 = 1.72513 zpa in the acceleration range of the spectrum, Fmu is:
Fum3 = 1.42101 Fum3 := (2.67 mu 3 - 1.673)"d" l
In the velocity range of the spectrum, Fmu is:
Fum2 = 0.52402 Fum2 := (2.24 mu 3 - 1.24)"'" Cf where Cf has been defined previously.
The median ductility scale factor, Fu.MRN, using the modi 5ed Riddle-Newmark method is:
Fumi := Fum3 (Fum3 <Fum4) + Fum4-(Fum32Fum4)
Fumi = 1.42101 Fu MRN := Fuml-(Fum2<Fuml) + Fum2 (Fum22Fuml) Fu g = 1.42101 l
i 1
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7-S N Effective Riddle-Newmark method for Calculating FuERN l
Because the modified Riddle Newmark method only accounts for second slope of the force-deformation curve, the last correction to perform is to account for ground motion duration.
He factor to account for earthquake duration is:
CD:=1.0 d $g ).0 6p'W kMA 4% (M h and the median ductility scale factor, FuERN, usin effective Riddle-Newmark method is:
Fug.= 1 + CD-(Fu g - 1)
Fu ERN = 1.42101 l
Effective Spectral method for Calculating FuSA In this method, Fmu is calculated using the effective frequency and damping ratio of the structure First, the ratio of secant stiffness to clastic shWnee is #=='M as:
Ks g := U Ks g = 0.65376 mu Next, the ratio of the secant frequency to clastic frequency iWma'M o be:
t
(.
fs f:= jKs g fs f = 0.80855 Compare to:
fs =6.13335 Hz f,, = 6.13377 Hz f
fs := fs f,
The ratio of the effective frequency to elastic frequency is calculated as follows (See Reference 1):
Coefficient to account for short duration motion
(
cf := 1.9 Al := cf (1 - fs f)
A := Al.( At 50.85) + 0.85-( A!>0.85) fe f:=(1 - A) + A-(fs f) fe f = 0.93036
?
ne effective damping ratio, be,is calcu:sted as follows:
Cn := 0.15 Coefficient to account for short duration motion l
bh.= Cn. (1 - fs f)
Hysteretic energy dissipation damping bh = 0.02872 ifs \\
- f be :=
- (b + bb) be = 0.05945 b'fi
25' M '
The median ductility scale factor, Fu SA, is estimated as follows:
_ -- effective frequency, fe,%
1 f
fe = 7.05734 %
fe := fe f ns 1
Sa, := SSa(fe,be)
Sa, = 0.59505.g Sa.e is the spectral acceleration at the effective frequency, fe, and the effective damping, be ffe f)2 Sa Fu SA = 1.46763 l
.Fu SA := I* f Se,
(
t y,:, -
l Final Estimate of the median ductility scale factor i
Different methods have been used to estimate the median ductility scale factor.
Based on a recent study, the median ductility scale factor, Fu, is taken as the l
average of the ductility scale factors found using the Effective Riddle-Newmark i
method, Fu.ERM, and the FEcetive Spectral Method, Fu.SA.
i l
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Fmuma, = >.m32 4. i.mm.
'" sa"l '" s^
Fme m,, :=
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38.03 br Fmu := 0.4-[0.06 + 0.03-(Fu SA R-1)]br Fmu = 0.03124 ne uncertamty due to story driA==Ma'~4 with failure is calculated, assummg that the 10' elastic deflection is at 1.0 standard deviation from the mean, to be 1.25673 using the same equation as above. Derefore:
I fFmumedian bu Fmu - ( l.25673 #
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= 8.16 bFmu = 0.14935 bFmu := br Fmu + bu Fmu + bu Fmum 1
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i w I a l l j i I ANALYSIS AND DESIGN OF STRUCTURAL CONNECTIONS: Rein breed Concrete and Steel i io.. I l of London-e j M. HOLMES, BSc., DSc, CEng., F.I.CE, FStructI l Professor and Head of Department of Civil Engineering i University of Aston in Birmingham { ad Technology l 1 L. H. M ARTIN, BSc., PhD. j Reader in Structural Engineering w Department of Civil Engineering .rsity University of Aston in Birmingham s
- loo, Ontario.
i ! Enginects. l ELLIS HORWOOD LIMITED l Publishers Chichester Halsted Press: a division of JOlIN WILEY & SONS New York Brisbane Chichester Toronto l ondon. I 1 } e . ~.... .,s. ...,c,
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a 7 g,. _ w 6 i (Ch.4 Sec.4.13) Elastic neory for Prying Forces 145 the tee shown graphically in Fig. 4.10. The tensile forcesin the bolts adjacent to the iGange tension flange of the beam are approximately equal,ne preloaded bolts adjacent itubs ne to the compression Gange eventually resist part of the bending moment. De l Prying force in the elastic stage of behaviour may be greater than at ultimate l .mn con. J enental load and depends on the number of contact areas. f of those lange are 4.13 ELASTIC THEORY FOR PRYING FORCES nen the
- rythin, The bolts must be designed to resist the external forces plus the prying forces, and it is therefore necessary to develop a theory to calculate the magnitude of t contact
- nts by the prying force. For an allowable stress method of design at service load condi-tions, the prymg force Qd, is related to the extemal force F., assuming linear ic stage i
attom of elastic behaviour of the components. l j De theoretical model in Fig. 4.ll shows an end plate of thickness t, and of cantilever length (a, + b ). The extremity of the plate is in contact with a j -orce in p column nange, for example, and this introduces the prying force Qd.. The i urtion in extemal force F is assumed to be balanced by the prying force Os, and an axial j strength force F, in the bolt.The axial force in the bolt produces an extension of the j 3 j bolt of 6. s b fe i I rl 1 3' I 3/ o o 1 / W i a origin t ~ y j _____.__._b.q_____ bt* f
- O e F
e b o I s o o i e Fig. 4.11 - Forces acting in the elastic stage for prying force theory. ( Applying McCaulay's method for the deucction of a beam with the origin of O and the deflection positive downwards. day El p = - Q3.x t (F, + Ot.) [x -a ) (4.29) p 16. Integrating
- nt # and
+ (F, + 03.) [x -a,] : 2 d x El y = - Od, 2 +A (4.30) mn con-dx 2 ,,m,- ion J, i l .t ~, , w c:a,, y s up,, .s.
l I 1 3 4 l 1 146 Bolted Connections ~ (Ch.4 3,,, 4,3 g } Integrating Obu "- + (F. + On.) [x -a,]3 3 x + Ax + B (4.31) Ely = - On, 6 i 6 where F, = l l i whenx = a, + b,, dy/dr = 0 and therefore from equation (430) Destrainins I [4.20], and the 59 (a, + b )2 b 2 0 = - On, + (F. + Q3,) +A-values of eg, andl p l j thickness plus wmt Ihe width of: l Rearranging 6a Lth jj 13 defined and is ad j j A = Ob.a (a, + s b,) - F. f-I
- column connectic
[
- h. Nhf[5,
{ tions in a colum p 2 2 i theory (4.22], a when x = 0,y = 0, and therefore from equation (4.31) B = 0 whenx = a the whichever is the t extension of the bolty = -6,and from equation (431) b -EI 6 e = - Qu. + a, On. (a, + 2b,) - F. b . 4.15 CONNECT 1 Some connectios E F. - 2EI6 /a,b ' diagram is genest b p ) 2(a,/b,)+ (2/3)(a /b )* * } beam, and colus p p i are applied to th 1 The extension of the bolt originally preloaded with a force Fe, is stiff bearing W l force H are re@ -Fw)sp/A En = (F. + Qe. -Fw)sp/AnEn. (4.33) fasteners, e.g. be I) 3b " (Fbt b j Substuting equation (4J3) in (432) and rearranging i F,(1 - k,) + k, F, (434) 1
- 06. = 2(a,/b,) + (2/3)(a,/b )2 + k, p
j i w, t,*g /64,b,' Ag En. (435) i where
- k. = Ep p
This form of the equation is applicable in the linear elastic range of behaviour and relates the prying force Q3, to the extemal applied force F. ,{ 4.14 ULTIMATE LOAD THEORY FOR PRYING FORCES j At ultimate load when the bolt fractures it is' preferable to relate the prying ) i force Qw to the ultimate tensile strength of the bolt F,. Equation (4.34)can i i be expressed in terms of the bolt force and strain in the bolt, when combined with equation (4.28). At ultimate load the equation would be j l i 1 l t i l ? - w.,Jr. ..s ,,,m r, ,,..,w y
J t <~. o 1 ;;! 2 1 0 no,.crsJA Nm-I e, m ~ o. -r- : o ) cumcr Re s~ A A~ = " -,- co.ov o,1. , w, x, s t L fSSI.MJ i O. at N-WM M -' @N M e -- - 2/A l aon vw ( it. 3=o ) no u og rv tcy+s c; W 4- % p sri d', _ Og * (fc. %1(x fl E I C N, CG d 6-Ac E%i :-4k&(Ic+%l(d, A E I d._m. : 434 c_ GN L Lv bl = Ch .dFed&)(l'd,bb ETg = -CA N_. + Cx + 0 3, c o e i N ~~ f lC $ ): Yu - A aj + A =. - Qse. af c ~ => A=C = 2-2 + bop +0:- +kO 4hrp b O' C Jock R. Benjamin & Associates,Inc. I) Consulting Engineers 9h W
sy DATE PROJECT PACE cF CHKD. 3Y DATE CUCJECT JOD NO. j UOW Rq +.Q dy, d,% O = Cb_ (*f ( EiC )M + A R + .2 A = he a (bg + u p ) ~ Fe_ d g l '2 g d cdr xco %=o =7 D = 0 4 6 =- o (.%~ ~oaun-)ank 4 % k A,i Q % k A Oob O M OV CEM.A W G 7 \\ is NO 1Xd Ch E d % 1 ('^ k ( M i . a 2+ a y +c at,A I O b Y b Qg) \\' f c'O M Q d% i % tbg)+ 4+ e, y Q. t b( A = c-4 G = D =0 d ko OkOws bh KIN s -k" i t( W +o o ss-GA %doA4 l l Jack R. Benjamin & Associates,Inc. k { Consulting Engineers 4
~. CY DATE PROJECT PACE gr CHKD. SY DATE SUBJECT JOB NO. A.Sc S: Of+bg d% El 9 e = O e I k + ( Ee.+ @ge'. [ + A o ~7. - 3 A = EI9 + OQ a t N + 7_ b ) -- Fe. M C e e v r, h4 -Q/X W bok g : -- $9 kG %O69 / + t+ Ap [4[t1 bp,) p a.e(Ft9_ae_ trit h E Of f A 9 t 2.bf g be 2. ~ _1 2.
- g
^ d ( f. 2.f5%_. __ 2 El$ t, Te, ' br a.esf .4 i % 2.4 .] Ap ' V Ip, ~36'7 Fe_.-YCG'+k) gbe = _ '~), Y b( Now b~bU%/o %s L = (Ft +%D se As E, e - t Jock R. Benjamin & Associates,Inc. Consulting Engineers 9
CY 3 ATE P2OJECT PAGE OF CHKD.SY DAT3 SUOJECT JOD NO. i sthstr:vk m% h 1 4 ( F. +Qgdog ' q _ 2Cp {ge 4 gbe. _- be Ag Eu ap i 2.@/h ) (2/3)(Af/be)*- 7:7----t g 2rrbe* ZEI B P ft. - y gk 4 Kbf A6 Eb a.,p k ~ 21 0f [ 2E r.9e, Y Ab4"f t 6f Q Se., - g4 2.E Io 7 6tA E p 3 b S'f E pw '. b AP f \\? 1 Op f 0f W f kbb^{ b Gf f kb b h b i 1Y t I - kt) ~ b 5-j g t-e f 2NAD+ (%X4/W 4 % .,I C Jock R. Benjamin & Associates,Inc. Consulting Engineers 9 }(
SY DATE PRCJECT Pact Y CF CHKO. SY DATE EU3 JECT JO3 NO. p()' l 1) I U[ ~ t bb of1/u Sg(_r) M do A a Nf C s m c2 Q g w M b v v.^oe u dQ eA c A M M( =0 -u - f%.ic) ec A A =. o % M + M - (Fe 'Q,d bg=o 7 Qu (a +9 - 9) = Fdr e Q" R* R( st'S h idu Gw m
- e. 4 4Sdvi~3 hoc F u,. k )
StTbt bP Fh 4 1 ~ "f yc + w.c =. k ( k'.)-(k+kL)bthO l y ae = Y }\\ -Ye (\\ + ke[' - y h e %( (Af b c. 2.Er./6p' Jock R. Benjamin & Associates,Inc. b Consulting Engineers [
(* OY DATE PROJECT PACE OF CHKD.DY DATE sucjgCT JO3 NO. ( hk INb I*' 'S \\ c a %J to v42 c u os p), t 4 4
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e DATE PROJECT E l '^ pAgg OF CY ^' ~ JOD No.US 3 CHKD.CY DATE SUCJECT I i .J,, y e,; = 0 <2:.r u _~ n 63% Fe 4 v \\ / =r 07.~.} a 33. QTwr.u \\) ika ciuna .L .s 4 i w.y,.,. wuvan fxsw yns,
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BY DATE PROJ2CT PAGE OF CHKD. GY DATE SUOJECT JGO NO. 5% -Fem ~ 6 % ( v w 30, sk ,~ us i.e. 3 pm.M A) l l 19 (.6C FS l, 6 3 'c CE \\/ = +- =
- 2. 4 3 y
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- MO
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. - - -_~ .. ~ Cy DATE PROJECT PACE b cF CH KD. DY DATE EU3 JECT JDC NO. $w w-U$ l J W l r s, -J r 'f c S = 0. 5 5 b Fe_. + 1. M L w .r J 'N i f'm 6d V -4 ("
- ( - %
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- 0. C oarcti 5 93 (d A P I 1.-
,,,, I rn Jacy = l CHKD. CY DATE SUCJECT JO3 NO. a 2 E# ', hMNN Lv. s 1 & C kdif7 ( ) < Mut WM l O 9 b ^? Y *1 C.) fN k O. ; 4 LEE d ~bi. @ fi.-@+A (x+( (E I d., ~ x g dd E._ .E 4 ,f- ] EI d h ._.fg"v'- Ob 3 [ =. (9f e hk-l.)tOkj(d + A r - M' CL L g3 1 7 t-Slk, Mk As.,.0 EIx -._4 j 4 g gg 1 G l os ssag' de u = l be l 3 = p oaw) e N I~O f Di ' $ t. h, = ) 4 6 Bro 4=C l Jock R. Seri)omin & Associates,Inc. I) ConsuHing Engineers
i CY DATg PAOJ2CT PACg QF CHKO.3Y DATE SUCJECT JO3 NO. vom 6 x=eeb da _ O ( 3,so~l N. Do 3 9 y* - (o ll n;t-J h we,,J O=C. Ef1jd + h k-3) +%' g,4 hz diCdi+z_L3df<-3-)W A = Q a._ e 2-2. 4 d x c o Su = O =7= g= o 4 Sc o W"d M 15 )b &#A N A No~ a < +% a u a pp 4 2 ~J a a p +e A A 3 c) i Mc 6Nw (14 h dm = o %J Ax L de = C%(9+bg)+(Fe+L)L \\/ t 7 g \\/ = 5 % r + h al 'A e = _C% Cag +6 )+[jFe - 3) + GA_)r M tit g L4 A=c 4 b=Dzo a h cr A 4v Po bv NN d Nkh4 d y c+< % satoit"J.'.,7' *'"'" '"*' di W +0 ass ~
CY DATE P'1oJECT PAgg or CHKD. CY DATE SUDJECT JO3 NO. d x=2rbe d, _gC ds E I G2g (4i{b'.) p 5- ))-b +A t f (Q[t 2 () - ~ IJL ~ O -1. J bu v) 2 - 9 h d NC p 1 l -tTS
- -% Y 4 4p EIQ 4 % $(ap24)_((%-5)@ l 3
b L t ~ t N j'e 3 = Op y<-3) __ gyg_ g 6[ f 2 g: , t =1 h(_3) 2NV _ F'E "9 bf O g _ hP - _3lT 6p' - sf bf 4 ~ ~ CQ w t ~ g vs,9 g( w y F& bol+ %/o pwAsvi ~ 5 _- ( Fe +A.) o f 8 A'3 E s E i Jock R. Bonjomin & Associates,Inc. Consulting Engineers S
., ~ _. GY DATE PROJECT PAgg gy CHKS.GY DATE SUDJECT JO3 No. o - ) W} } U' /r \\9 = 'a: ) O ( $ >g} C k 4-f A :.g e e a,be. - -r P dp) 4 2g f ap/g 2. W k = 'p . [6 7 ( a nstsyJ ~-: m K pt {o _ A Er r _,2W __ 2x r%. _1 a 'r
- G)g, =
br te rne e, kf A E d ef 9 N E= h Me n_ E M 4f $? We 2E-I$F n bi-A Ep q 6qQAE 3 g F< (%'A) z'Ib 1 G2o L U + ke l Jock R. Ben}omin & Associates,Inc. Consulting Engineers
sY DATE PROJECT PAGE OF l CHKD.CY DATE SU3 JECT JOD NO. l [ dolt. d \\s b b ).' < b
- g b
.wNauf d ka 5\\.) ~ '2%rA O ^n R % ez Q J\\ % \\~4 uw Da wK Ml% V M. WM-h '[d. 2 3 t F~J Sc A Mo =0 f fe 9 L) Y bt. fY { C-3 + 00 ~~ CA(.a4 '4<) = W b =A br e ( = (Q Fe-e,~) 6 p I ba6 fh 0Dl] b M 1} 1 L 0 x F4 Ot'- b ) 3 " * ' - 3 W~
- (4h.-3)M l
< + x< e = [0,,,, '-' t Y lf C* $ -} l tog ' 49 k ~3j- 't [t, ~ i 2rr /qs I Jock R. Benjamin & Associates,Inc. Consulting Engineers D J
CY DATE P3OJECT Pact or CHKO.3Y DATE SUCJECT J33 NO. f\\wb i ' ft W V4 hQg O E'4A JN ( l5 uw &=o 7 O be. (^9 + 'of) - ('k ft 7 l1'NQ 6f c; Fe 6g = Qx Ca p-% -\\p) + 3 g k Fi=Q*"f+[3h I l O I Jock R. Ben)omin & Associates,Inc. Consulting Engineers D
i CY DATE PROJECT Pact or CHKO.DY DATE SUOJECT JOB NO. 9 _q 4 \\ ~ f i /k A \\ i g L' t \\) f lV ~ ~; 4 & q' { f ~~ g = I y .y - g ' f -1
- n C, y J i I
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I .b f1 Ii b ~ '" b Dart anoJECT Fact or cv I~ 'Sh
- ^^"'N J33 NO.
'C N b CHK QY DATE SUOJECT l ftl) ) hj()jhya ~ JVU w .Lvu n s ^ C { Q Q i'S l I h) {&.M i h f t, 7 w l I b bh< SS gd' % ALM) __g 7.; r l Rssvm ( A g 1 a a t - m 1~a. + - b h '~g g d m) t = P[ w +0A f,wi F V Av4 VN) " h A M @ H + OiN f f0M 4 l M ~ AMc= A v. h t. + b d r _ P~s +u PW g = ~ C ' J 5,e) 3 b i ,55,(Es,Qid F S ks = d ,,3 ho 4 cag e. 2. C2.trfu)* l M $hk ( $ws, z S L (fe.,ht)F (kg 4, tei 4 NS /
- g 8 Jack R. Benjamin & Associales,Inc.
I Consulting Engineers 9 l l
? b DY DATE PROJECT PACE CF CHKD. SY DATE SUOJECT JCD NO. l b,5w b I SW e E.W \\ 'l - 4 l kS=b v__ 6d M r i I 3wl Sew L.. %.6 kMT khL L b g::: su y n ,3 l tar.% -J
- o. + Pw k
l .h = 4 (Aw.At). c n[dA hS
- l 63 Le :o L
Ne, d) t 0,4 f v _ 4 3 b tN ~~/_ k hwiSM 2 i'l fd'd l 1 e tw i y qq l I Jock R 8enjamin & Associates,Inc. D Consuming Engineers
l l l l l CY DATE PROJECT PACE OF CHKD. DY DATE SUDJECT JO3 NO. I l i 1 I l l l 1 i l \\ r ~ -.. u: - 1 l I l H r l I Jock R. Ben)omin & Associates,Inc. D ~ ~ ~ ' Consulting Engineers 1
C:\\WINMCAD\\ FILES \\A8 COMPS 2.MCD 12/17/93 kip.: 1000 Ibf i ksi := 1000-psi fgig :=IOHz Fixed base frequency of cabinet f =7.5856 Hz f, = 9.848 Hz peta := 0.05 Frequency and damping of the entire cabinet g e 'O.375290~ ' 2.5 ' i 0.60 5 1 f SA := g F := Hz Groundinput response spectrum 0.702350 10 0.723530 ,25, SSa(f,p):=exp'(linterp(in F .,ln' SA,ln f Q0.05 f f f f g-Values of Sa at different frequencies Hzt (g3 (Hzi 4 i (f3 and damping valuca l .g i Wt := IOkip Weight ofcabinet Spectral acceleration Spectral accelerataco ' Spectral acceleration l in n/sdrection in e/w direction in vertical directaan Se,:= SSa(fns,Seta) Se, := SSa(f,,,$ cts) Savert := 0.40 g a e .,a, =0.65961 *g \\ Sa,, = 0.69991 *g} Savert =0.26667 g 3 h:= 109 in Height ground to top of cabinet 1, := 91 in Base lever enn for north / south direction axial force at support 1,,g := 96in Base lever ann for east / west direction axial force at support at first bolt Iew2 := 19.2 in Base lever ann for east / west direction axial force at support at second bolt -Wt hli Sa t) Sa Sa, Wt vert h ew n - -Wt t g (21 g g P,:= p = p o 2 1,,g + I ( ewtj P = 1.97519
- kip P,, = 1.91031 kip Pvert = 0.66667 kip as Wt Wt Sa 24 3: 4 ' ns Y g := Pns + 0.4-(P,, + Pvest)
Y T g i T 3 = 3.00598 kip Y 2 = 2.5 kip Y3 = 1.64901
- kip l
i T 3 c g =0.54858 c 3 := Yl Y 'T3 2 e 2 := c 2 = 1.37144 kip Yi
2 STOP1 l l ELASTIC PODif ANALYSIS This program determines the allowable capacity of the plate bolt anchorage system for the 4160V Swithchgear A8 at the l Pilgrim Station l Plate and steel properties E ;= 2910' psi Modulus of elasticity for steel E 0:= Modulus of rigidity for steel i 2-(1 + 0.3) p := 0.25 in 'Iluckness bold down plate t }wp := 2.25 in } Vrxithhold down plate db := 0.625 in Diameter bolt holein plate Fa{:= 255"iei (I,angih piste from,and to bok6iinxinium value) bp= 1.375 in thig6iblaite'fhun boh~ts'siiiiiiert yp := 44 kai Medianyield capacity plate e Bolt properties 8 Eb := 2910 psi Modulus ofelasticity for bolt b := 0.196 in' Area bolt (Gross area for stretching) A gp := 2.5 in Effective length bolt for stretching ( Channel properties C6x8.2 (AISC 9th Ed.) d := 6.00in Hetsht ofchannel b g := 1.92 in Width channelflange tf:= 0.343 in
- lhickness channel flange t, := 0.200 in Thickness channelweb i
o := 0.599 in Distance outside edge channel web to shear center e K := 0.08 in' Torsion constant C,;= 4.72 in' Warping constant ! x := 13.1 in' Moment ofinertia about x-axis I := 0.692 in' Moment ofinertia about y-axis y L := 38.375 in Channel span between assumed supports L a := - Distance of anchorage to support along channel 2 l q := 0.25 in Distance from edge of flange where point of contact betwcen plate and channel flange occurs l I
O General properties c1 and c2 are used for the case where the shear force c I =0.54858 on the bolt is proportional to the the F on the e c 2 = 1.37144 kip anchorage. V=c g F, + c 2 pp:=t + 1-(l6 assumed ;l) Offset in plate which causes moment in plate from t f e shear force on the bolt (equal to thickness of channel flanSe P us deflectim upward) l Calculated properties t EE T =0.85807 Variable reduction factor on the force Fe due to shear t := 1 - c j b on bolt p t (:= BEE c 2 ( = 0.35482 kip Castant reductim factor on the force Fe due to shear on the bolt p t W !,= P P 1 = 0.00293 in' Moment ofinertia plate 12 h ;, (wP-d )'l h P Z g = 0.02539 in' Plastic section modulus plate at hole Z 4 I Mh = 1.11719 kip in Plastic moment capacity plate at hole Mh ;* Z yp h CASE 1 - MOMENT BETWEEN PLATE AND CHANNEL FLANGE IS ZERO This calculatim finds the value of s such that the slope at the end of the plate is zero and the moment at the j p J attachment to the channel (i.e., at x = a + b )is also zero. This is the limiting case where the channel offers no p p resistence to rotation See derivation for definition of terms and theoretical basis (Reed i1/22S3) Qbe(a).= Allowable prying force correspondmg to plastic P a moment capacity plate at hole p 2 E1gP k (a ) := e p 2 A E p b b b p k(a ) := 2 b + 2_, p b l p 3 (b pj 4 5 'a \\ F,(a ) := -Q be(a ) +( p p l
- -~_ 4 Constram A to equal zero, which corresponds to zero slope at the end of the plate Find ap 2' { Q be(a ) a -(a p + 2 b )- (F (a ) T-(). P b p - p p e p ,, p j p.= root a f bf b b i i F,(a). T-k,(a)- 1 + T- -k(a)t- - ( (k(a p) + k,(a ))- p p p p i Pl. P. p. l 2 2 b p l a =0.75511 in p I i Find limiting value of 6, ~ I b I' b - k(a ) Y-[b - ( (k(a ) + k,(a }) F,(a ). T-k,(a). 1 + T-p p p p p i Pt. P. P. 6, = 2EI P l l 0, = -0.014004148 Tension force on anchorage at edge of channel flange F,(a ) =1.3604 kip p Q e(a ) = 1.4795 kip l Prying action force on plate b p F (a ) =2.8399 kip F (* p) := F,(a ) + Q e(a ) b p Bolt force b p b p Ybolt(* p) = 2.11773 kip bolt (s ):=c g F,(a ) + c 2 Shear force V p p l Plot deflected shape, slope, shear and moment of the plate Qbe(a ) a -(ap + 2 b )- (F,(a ) T-()- + E I 0, p A(a ).= p p p p f 3 1 (x-a )3 y(x,a ) := -Q be(a )-+ A(a ) x + (F,(a ) T-(+ Q be(a ))- ' P) x p p p p p 6 (x-a }2 f 2 ) r(x a ) = h f de(* g)7 + ^(* P),+ (F,(a ) T-(+ Q be(a ))-
- P) x g
p p 2 v(x.a,) .Q w(a,). (F,(a,) + Q a(a,)). (x>a,) u(x.a ):= Qw(a ).x+ ;(F (a ) T-()+Q e(a )}(x-a )-(x>a ) g p e p b p g g p,a ) A E P si,(a ) :- (s -y P b b e 1t(a,) =2.8399 kig so g 8P
I s P* P n = 100 6x. = 6x = 0.0213 *in n i.=l..n+I x; =(i-1)e def;:= y(x;,a ) slope;.= r(x;,a ) shear; = V(x;,a ) mom; = M(x;,a ) p p p p y(a,a ) =-0.00125 in y(a p + b,ap) =-0.01636 in y(0.in,a ) = 0 in p p p p 0 i~ i i i U! I in M i Y -0.01 8 -u2, J, 1, 8 l 4 h ~ ~'- p + b,a ) =-0.014004148 r(0.in,a ) =7.06720925910" r(a,a ) =-0.004%3919 r(a p p p p p 0.0t a i i j 9 w M } l -9.01 - 1 I I i 0 0.5 1 1.3 2 2.5 'i:-. Pbolt(*p) =2.8399 kip V(ap + b,a ) = 1.3604 kip v(0.in,a,) =-1.4795 kip p p l 2 i i in M i l _kir 0 i 0 I I i 1 '0 0.5 1 1.$ 2 15 s. t M(a,a ) =-1.11719 kip.in M(a p + b,a ) = 0 kip.in M(0.in,a ) =0 kip in p p p p p 0 g g i i , song ""0.$ .[ 0 - -1.$ 0 0.$ 1 1.5 2 2.5 i = t
k CASE 2 ROTATION AT CONNECTION BETWEEN PLATE AND CHANNEL FLANGE IS SET TO. 8, For thlS case a value of 9 between zero and the limiting value 0 = -0.014 is assumed and the plate response is e e calculated. Again, the slope at the end of the plate is set equal to zero. 0, = 0assumedei Assumed value of 0, 2 EI0* obc (a )-(k(a )+ k (a ))+ +c g p e g ,P F,(a ) = p +-t(a) c g a -(ap + 2 b )- (F,(a ) T-(), p + E I 8, A(a p) := p p p y(x,a ) := -Q be(a )-+ A(a ) x + (F,(a ) Y-(+ Q be(a )) (x-a,)8,[x>, p) I 3 I ~ 1 x p p p p p (x-a ")' ~~ 2 i r(x.a ):=h;fo be(a )i + ^(a ); + (r (a ) v-c+ o be(a ))- I* A g g e g g 2 p l V(x,a ) :=-Q be(a ) + (F,(a ) + Q be(a ))-(x>a ) p p p p p M(x,a ) ':-Q be(a ) x+((F,(a ) T-() + Q be(a )].(x-a )-(x>a ) p p p p p p bolt (a ) := c g F,(a ) + c 2 V p p Given Constrain slope to equal zero at end of plate r(0 in,a )=0 p a = 0.87422 *in p := find (a ) a p p 1 i F,(a ) = 2.53015
- kip p
Q be(a ) = 1.27793
- kip p
F (a ) = 3.80808 kip F (a ) := F,(a ) + Q 3,(a ) b p b p p p p.a ) A.Eb bolt (a ),,-y(a p b pbolt(s ) =3.80808 kip P p p.- 8p y (a ) = 2.75942 kip p
DY DATE PROJECT pAgE OF CHKD.QY DATE SUDJECT JOD NO. l l l E Jack R. Benjamin & Associates,Inc. Consuming Engineers 9 I
-= li 1 0.091 3hkw h b N-2.5 0.375 5 0.8 ~ \\ x := 10 y := 0.702 25 0.724 33 0.4 t l , 50 , 0.4 i := 1. 7 g i i ~ ~ git \\ ya 2.s b ~ 0.1 Y_i i 0.01 1 10 100 xg l l l
7* i := 1. 7 l l I 1 '0.091 2.5 0.375 8 0.. x := 10 y := 0.702 j 25 0.724 33 0.4 Yi ,50 , 0.4 zg.- -
- i I
Y W 1 i j gg 2. s 14, i 0.1 Z_i V 0.01 I 0.001 l 1 10 100 I l
l 7 P' P n :100 Ax : Ax = 0.02249 *tn n i : 1.. n + 1 x, :(i - 1 ) ax def; :y(x,,a ) slope, :r(x;,a ) shear; : V(x; a p) mom, :M(x,a) p p i p p,a ) =-0.0016749 *in y(a p+b p,a p) =-0 01275 *in y(0-in, ap) = 0 in y(a p 0 g i
- E Wi i
j i m -g.01 - 9.. l f f t -'O.02 0 0.5 1 1.5 2 2.5 is r(a,a ) =-0.005747713 r(a p + b,a ) = -0.00362 r(0 in,a ) =0 p p p p p l 0 i i
- E ini 4
-0.005 - 9. -0.01 0 0.5 1 1.5 2 2.5 3i 5 p + b,a ) = 2.53015
- kip V(0 in,a ) =-1.27793
- kip Pbolt(a ) =3.80808 kip V(a p p p
p 4 I i l i i "f. is sheer; 2 0 0 i i f I I I I 0 0.$ 1 1.5 2 2.$ "i 5 p + b,a ) = 1.38013
- kip in p,a ) =-l.11719
- kip in M(a M(0 in,a ) =0* kip in M(a p p p
p 2 i
- E i
i. i _ in 0 W 0 -g - I l' I I n e l n 0 0.5 1 1.5 2 2.5 i 1 5 i
T Calculate rotation of channel due applied shear and moment and compare to assumed rotation Also, cateulate the total deflection of anchorage system and channel. Rotation and deflection of Channel Flange. Assume moment ofinenia is equal to sum of:
- 1. Plate and channel flange composite for plate width wp tf P (t f
- p fp = 0.0391 *in' I I fp.:
- 2. Channel flange width equal to twice distance from edge of flange to inside web surface 3
2-(b-t,)tf I = 0.01157 *in, f I f: f 12 Distance from ground to center of area of plate channel flange composite A := 2-((b - tw) t ] fp ::w -(tp + t y) A f f f p t t p+tf + Ar f A fp-c:= c = 0.23784 *in Afp,Af Total moment ofinenia 32 I t 32 f tp+tf f t f +Af c-- I = 0MS *in, IIfp + I + Afp' 8 - t 2 j Calculate rotation and venical deflection of flange / plate composite at edge of flange F(a)-(bg-t w-ri)3 1 e p 0 g,EI 2 t f i' -(b - t w-i)) p, b,a ) + Vbolt(a ) t f+ -c M(a + f 0 g =-0 00325 p p p F,(a )-(b - t,- 71)3 p f 1 aven g := E 1 3 t f t \\' -(b - t w-ri)2 p+b,a)+Vbolt( p) tf+ -c M(a f p p +. Aven i = -O 00277 *in bolt (a p) = 2.75942 kip V
4 Rotation of channel due to twisting and deflection at edge of11ange. Torsion moment at shear center p) (b - q+ e )) V bolt (a p)- -tg2 p, b ,a ) + F,(a 0 :[M(a T f o p p i T o = 0.13417 kip in At distance "a" from the exterior channel intersection to anchorage a = 19.1875 in i Rotation of channel The following fonnulation is taken from Roark 6th Ed (as given in Mathcad Roark's I Handbook) nis file corresponds to Table 21, Case 1, and Table 22, Cases le-1g,in Roark's Formulasfor Stress and Strehn. Concentrated latermediate torque ag {~ 8
- T, T
(\\ a T, L Case f Left end free to warp but not twist, right end fixed (no twist or warp)
- <f0 NOTE THAT THE "LEFT" SUPPORT CORRESPONDS TO EXTERIOR CONNECTION AND THE "RIGHT" l
SUPPORT CORRESPONDS TO AN i INTERIOR CONNECTION ON THE A! To SWITCHGEAR CHANNEL BASE FRA) 1
Io i p = f KO Y p - 0.08074.1 in E/ (C w F g(x) = cosh (p x) F (x) = sinh (p x) 2 F (x) = cosh (p x)- 1 3 l F (x).= sinh (p x)- p x ( 4 i l F,g(x) =(x>a) cosh (p-(x-a)) l F 3(x) :=(x>a)-(cosh (p-(x-a))- 1) .,,; ~' F,4(x) := (x>a) sinb(p-(x-a))- (x>a)-(x-a) p C g '= cosh (p L) C 2.= sinh (p.L) C 3 = cosh ($ L)- 1 C 4 := sinh (p L)- p L C 3 = cosh (p-(L-a))- 1 C 4 := sinb(D-(L-a))- p-(L-a) 1 Vector ofend constrsints 0 for this case. Recall: 'C C,4-C C,3). _A l To 3 4 'O C E 2(C C 4-C C 3j deg 9A g 2 w 0.02022 I:= gA 1 0 I= o IC g C,4 - C 2 C,3) g t' A -48.22618^ .To C C 4-C C 3 ; lbfin T A ( g 2 l i. 8 fa {iT F (x)+ glbfin To I 2 g C,E p, F 4(x) F (x) + F (x) + 0(1,x) = 1, deg + 4 3 2 D p C,Ep Rotation at point of anchorage attachment to channel 02 =-3.51099 10 rad 02 =- 0(1,a)
ll x = 0 ft,1.. L Range of x-values: 100 Torsion Torsion Finned end Fixed end 0.03 g g g Angle of twist,0 0.02 41.x),,, 0(1,0 ft) = 0 tad _ des 0(1,a) = 3.51099 10 rad j o " -c 0 t 0 1 2 3 0(1,L) = 0 tad A l l Venical deflection of channel at contact point between plate and flange due to channel rotation l .my Awrt2 =-7.96644*10 ' in ~ 2 :=0 -(((b + e )-in Avert 2 f o Vertical deflection of channel between aoss channels - assume fixed / pinned span as efective boundary conditions 3 -7 F,(a ) L p Avert 3 := Aven 3 =-0.00343 *in 768 E 1 x Horizontal deflection of channel between cross channels - assume fixed / pinned span as effective boundary conditions 3 -7 Vbolt(a ) L p Ahonz := Aboriz =-0.07083 *in 768 E l l y l Total rotation from channel support system and deflection from concrete to switchgear i 0Tel=-0.0036003 red 0Tel:= 0 g + 02 l I p+ b,a ) =-0.00362 rad f Compare to calculated rotation at end r(a p p of plate (connection to channel flange) Totaldisplacement Aboriz =-0.07083 in AvertT := Averti t Avert 2 + Avert 3 Aven T = -0.01974 *in +y(sp+b,a ) p p Cantilever flange Avert ; =-0.00277 in 2 =-7.9664410 ' ain ~ Channeltwisting Avert Channelbendmg Aven 3 = 400343 in p+ b,a ) p,a ) =401275 in Aelplate:=y(a i Plate bendmg y(ap+b p p p ,(a ) = 2.53015 ki Force at ends of plate Q be(8p) = 1.27793 kip p ~ Force in bolt Pbolt( p) =3.80808 kip bolt (a ) =2.75942 kip Abar,g.= y(a p+ b,a ) V p p p T
STOP2 12., Calculate scale factor F,(a)+t3 _.T FS = \\ FS = 1.67338 \\ p I Tg Calculate frequency of cabinet in East / West direction l /0.4 Pew) AvertT SS*(Iew, peta) FS h A g = 0.1181 in ew2) l\\ + a,= e 2 (2 n ffaed) (l ewi-I T 1 3 SpAcc g := SSa(few peta) FS SpAcc,g = 1.17121 g e S Acc,g P f_new,,. = i - f_new,, = 9.84806 Hz Calculated frequency j 2z Act 3 f, = 9.848 Hz Assumed frequency e Calculate frequency of cabinet in North / South direction h SS*(fns, peta) FS \\ fPns + 0,4 Pvert ahonz + j AvertT + Ael = 0.18759 *in ag= (2s.fg)2 e I ns( Y l j SpAcc,; :SSa(fns, peta) FS SpAcc g = 1.10377 g e S Acc,j P I I D'* ns = 7.58572 *Hz Calculated frequency fnew ns 2s a,i 3 f = 7.5856 Hz Assumed frequency ns l t 3 i 1 a I
~ - _. l O STOP3 SECANT POINT ANALYSIS l Spectral acceleration Spectral acceleration Spectral acceleration m in n/s direction in e/w direction in venicaldirection i SSa(fnsp, peta) Sa,:= SSa(f , peta) Sa,n = 0.40 g Sa y ns Sa = 0 62852 g Sa,, = 0.67716 g Sa,n = 0.26667 g y ns 1 t Sa Sa, Wt. ' Sa 'hi e h ven ns.Wt-) Wt 7 g (2t g (2 g pvert, p. p i ew ~ f 2 4
- as,2 I ew2) ns 2 1,,g + l
-L=e ewtj l P = 1.88211 kip P,, = 1.84823 kip P =0.66667 kip j vert ns Wt Wt Sa ns t g := Pas + 0.4-(P,, e P n) T T 2*7 3*7 g j y t ; = 2.88807 kip T2 = 2.5 kip t 3 = 1.5713 kip 1 T3 c g := T c g =0.54407 I l T 'T3 2 c 3.= c 2 = 1.36017 kip T i l l t l I l
_...... = .- - -.. - ~ -..- N l General propemes Cl an C2 are War k cm wke se sedom c g = 0.54407 on the bolt is proportional to the the F on the e c ; = 136017 kip anchorage: V=c 3 F, + c 2 t pp ' : t f+ 20(l6bar ih Offset in plate which causes moment in pir.e from e shear force on the bolt (equal to thickness of channel hge plus deflection upward) Calculated properties t P_E t = 0.76341 Variable reduction factor on the force Fe due to shear T '= 1 - c g b m Wit p ('= b(EE c 2 ( = 0.59147
- kip can.aant reduction factor on the force Fe due to shear onh bolt p
3 t w 1:= P P 1 = 0.00293 *in' Mnmant ofinenia plate l 12 2 h * (* p-d )t b p Z h =0.02539 in, Plastic section modulus plate at hole Z 4 M = 1.I1719 kip in Plastic moment capacity plate at hole Mh :Z '8h yp h 1 CASE 1 - MOMENT BETWEEr4 PLATE AND CHANNEL FLANGE IS ZERO l l ) This calculation finds the value of a such that the slope at the end of the plate is zero and the moment at the p attachment to the channel (i.e., at x = a + b )is also zero. This is the limiting case where the channel offers no p p resistence to rotation See derivation for defmition of tenns and theoretical basis (Reed i1/22/93) l Ms Allowable prying force corresponding to plastic Q be(* p) := a p: moment capacity plate at hole 2 EIgP k (a ) := e p b,AEa p b b p k(a ) = 2- + - p P k P) Ia ) F,(a ) := -Q be(8p) + C p l
~,. - -.. _.. l N P+b E a n : 100 Ax : 8x = 0.0218 *tn l n i :1..n + 1 i l x, : (: - 1 ) Ax j def; y(x,,a ) slope; :r(x,,a ) shear; = V(x;,a p) mom, :M(x,,a) p p p l y(0 in,a ) = 0 *in y(a p,a p) = 400142 *in y(ap+b p,e p) = -O 01698 *in p 0 01 i i i i h l def 0 o -o oi - l I I - 1 02 0 0.5 1 1.$ 2 2.5 h is l r(0 in,a ) = 1.465303015*10' r(a p,a ) =4005291601 r(ap + b,a ) =-0 014331831 l p p p p 0 01 i i i i i nW 0
- i l
t -o.oi - t i l 0 0 0.5 1 1.5 2 2.5 8i ~ e i,(a,) = 3.22678 kip v(a,. b,,a,) = l.83908 kip v(0 in a,) =-i.3827 tip so 2 i
- F i-i
-lup o 0 i i i I t I I 0 0.5 1 1.5 2 2.5 8i p,a ) =-1.11719
- kip in M(a p+b p,a ) = 0 kip in M(0 in,a ) =0 kip in M(a p
p p u i i i ( o i @m - N 0 -t - i I -1.5 l o 0.5 1 1.5 2 2.5 =4
11 CASE 2 ROTATION AT CONNECTION BETWEEN PLATE AND CRONNEL FLANGE IS SET TO. 8, For thlS case a value of 9, between zero and the tuniting s alue 0, = 401433 is assumed and the plate response is calculated. Again, the slope at the end of the plate is set equal to zero. Assumed value of 9, 6, = 0assumedp 2E10* p)-(k(a)+k(a))+ +( Qbe(a p e p P F (a ) : t-k,(a) ] p p a ) a -(ap + 2 b )- (F (a ) t- ()-+ E 10, be p A(a ) : p p e p p (x-a )3 I 3 I -Q be(* p)y + A(a ) x + (F,(a ) T-(+ Q e(* p))' P [x>,p) 3 x y(x,a).= p p b p 6 -Q e(8p) + A(a ) + (F,(a ) Y-(+ Q be(a )). - a )'-(x>a) ) (x f 2 P r(x,a ) = b p p p p p v(x,a,) =.o 3,(a,). (F,(a,) + o s(a,)).(x,a,) u(x,a ) :=-o s(a,).x+((F (a ).,-() + o be(a,)}(x-a,)-(x>a ) g e p g bolt (a ).=c 3 F,(a ) + c 2 V p p Given Constrain slope to equal zero at end of plate r(0 in,a )=0 p p.= find (a ) a =0.92366 in a p p i F,(a ) =3.04146 kip p Q e(a ) = 1.20953 kip b p l F (a ) = 4.25099 kip F (a ) :F (a ) + Q e(a ) b p b p e p b p l p,a ) A -y(a p b Eb Pbolt(a ) =4.25099 kip bolt (* p) :: P p P V bolt (a ) = 3.01492 kip p
N P+ I n :100 tsx = ex = 0.02299 *tn n i 1..n+1 l x, = (i - 1 ) Ax def; :y(x,a ) slope; =r(x;,a ) shea, e V(x;,a ) mom; = M(x;,a ) g p p p p y(0 in,a ) = 0 in y(a p,a ) =-0.0018697 in y(a rb p,a ) =-0 01383 in p p p p 0 i in.
- def, in -g o g 0
I I i -0.02 ( 0 0.5 1 1.5 2 2.$ l sg g l p,a ) =-0.0060727o9 r(a t b,a ) =-0.0049 I r(0 in,a ) =0 r(a p p p p p 0 ,g i
- 5 ini
-0.005-I -$.01 f l 4 I -0.01$ l 0 0.5 1 1.5 2 2.5 . i. im p + b,a ) =3.04146 kip V(0 in,a } =-1.20953 kip Pbolt(*p) =4.25099 kip V(a p p p 4 i ii I i
- 2 w
almar; 2 ? 0 0 l l ! t I I , ^ 0 0.5 1 1.5 2 2.5
- i as p + b,a ) = 1.26212 kip in M(a,a ) =-l.11719 kip in M(a M(0.in,a ) =0 kip in p p p p p
2 i
- i i
i
- E wl i
l i i u,, 0 -~ - 0 .g I I I I O 0.5 1 1.5 2 2.5 s; ~~ In
I N l Calculcte rotation of channel due applied shear and moment and compare to assumed rotation Also, calculate the l total deflection of anchorage system and channel. l Rotation and deflection of Channel Flange. 1r Assume moment ofinertia is equal to sum of:
- 1. Plate and channel flange composite for plate width wp
- p- (t p + t )3 f
l fp. = fp = 0.0391 in, 1 I
- 2. Channel flange width equal to twice distance from edge of flange to inside web surface 3
l 2-(b - t w)tf f I := I g = 0.01157 in, f 12 Ihstance from ground to center of area of plate channel flange composite - ~ - - - - -- - A :=2-((b - t,) t ] fp := w -(tp+t ) A f f f p f t fp p+t t f+A f A f
- =
c = 0.23784 in Afp + Af Total moment ofinertia P+tg*+ A rs - g\\
- f t
I t I =0.06045 in' t I + ^ fp' 8 - I ;*I f t fp t l Calculate rotation and vertical deflection of flange / plate composite at edge of flange F,(a )-(b g-tw - 11)* 1 p 0 ;.= E1 2 t y(b-t - vi) M(a tb,a)+Vbolt(a ) t + p p p p f + - f w 0 g =-0.00351 l l F,(a )-(b g-t,- il)' I .I p avert j := E 1 3 t I t \\- -(b - t - 11)* M(a p + b,a ) + Vbolt(a p)tf+ -c f w p p +- Avert j =-0.00304 *in i 1 4 y bott(* p) = 3.01492 kip
l 2.0 Rotanen of channel due to twisting and deflecnon at edge ofilange. l Torsion moment at shear center p + b,a ) + F,(ap)-(b g-a+ e ))- V bolt (a p) 0 =(M(a -t T p p o f-l T0 =0.5294 kip in At distance "a" from the exterior channel intersection to anchorage a = 19.1875 in 1 Rotation of channel I The following formulation is taken from Roark 6th Ed (as given in Mathcad Roark's 1 Handbook) m This file corresponds to Table 21, Case 1, and Table 22, Cases le-Ig,in l Roark's Formulasfor Suess and Swain. l Concentrated latermediate torque s y 1, T, fN I L s l l Case f Left end free to warp but not twist, right end fixed (no twist or warp) l $of NOTE THAT THE "LEFT" SUPPORT e CORRESPONDS TO EXTERIOR CONNECTION AND THE "RIGHT" SUPPORT CORRESPONDS TO AN INTERIOR CONNECTION ON THE A8 Io SWITCHGEAR CHANNEL BASE FRAND
~. 21 1 f 2 p =; K G 1 = 0.08074 *h (C E m w j F g(x) = cosh (p x) F (x) = sinh (D-x) 2 F (x) = cosh (p x)- 1 3 F (x) = sinb(p x)- p x - 4 F.g(x) :=(x>a) cosh ($-(x-a)) F.3(x) := (x>a)-(cosh (p-(x-a))- 1) ' F,4(x) := (x>a) sinh ($-(x-a))- (x>a)-(x-a) p C 3 '= cosh (p L) C 2 := sinh ( L) C 3 = cosh (p L)- 1 C 4 := sinb(p L)- p L t' C,3 := cosh (p (L-a))- 1 C 4 := sinh (p-(L-a))- p (L-a) Vector of end constraints 0 for this case. Recall: IC C,4 - C C,3) g To 3 4 2 BA C,E5 ( C g C 4-C C 3 ; deg 2 I= g 0.07977 0 A l= != 0 fC i C,4-C C 3) g 0"A 2 -T- -190.28083 0 CjC 4-C C 3 ; lbfin T 2 A If I ibfin T0 g F (x) + F,4(x) F (x) + F (x)+ 0(1,x) := 1 deg + 4 3 2 3 3 3 p p C,E-p C,Ep Rotation at point of anchorage attachment to chacncl 4 i 02 :=-0(1,a) 02 =-0.00139 rad l l l l
- 2 *t x = 0 ft,1.. L
. Range of x-values: 100 Torsion Torsion Pmned end Fixed end Angle of twht,0 o.05 a r, ) 0(I,0 ft) = 0 rad .28 l 0(1,a) =0.00139 rad -o.05 o i 2 3 0(I.L) = 0 tad l E l Vertical deflection of channel at contact point between plate and flange due to channel rotation 2 :=0 }(b + e )-q]) ~2 =4.00314 in ' Avert Amt 2 f o Vertical deflection of channel between cross channels - assume fixed / pinned span as effective boundary conditions 3 -7 F,(a ) L p l Aven,'= Avert 3 =4.00412 in i 768-E I x l Honzontal deflection of channel between cross channels assume fixed / pinned span as effective boundary l conditions 3 -7 Vbolt(a ).L p Abortz := Abonz =-0.07738 *in 768 E I y l Total rotation frora channel support system and deflection from concrete to switchgear 0Tp =-0.0049 red 0Tp := 0 + 02 Compare to calculated rotation at end r(a p + b,a ) =-0.0049 rad 1 p p of plate (connection to channel flange) Total displacement Abortz =-0.07738 in i l AvertT =-0.26523 in l AvertT = Avert 3 + Avert 2 + Ave:t3 +(t-t pp) f Cantilever flange avert 3 =-0.00304 in Channel twisting Avert 2 =-0.00314 in Channelbendmg Avert 3 =-0.00412 in =-0.25493 in Plate bending t f-tpp Force at ends of plate Q be(a ) = 1.20953 kip F,(a ) = 3.04146 kip p p I Force in bolt P bolt (* p) = 4.25099 kip bolt (a ) = 3.01492 kip V p t
-_.~-._ 23 STOP4 Calculate scale factor c( P)
- 2 FSS :=
1.91g74 Y l Calculate frequency of cabinet in East / West direction f.4 P,,) AvertT S88(Iewp, eta) FSS 0 b A =0.17525 in AE := (1,wj-1,w2)( 2 (2.x ffixed) T 3 j SpAcc := SSa(fewp peta) FSS SpAcc, = 1.2993 g SpAcc" (newp, := 1. f_newp e, = 8.51512 Hz Calculated frequency l ~ ~ ~ Assumed frequency f,,, = 8.51538 Hz Calculate frequency cabinet in North / South directa h I SS*(Insp, peta) FSS T j IPns + 0.4 Pvert \\A = 0.31351 m, \\ AP := Ahonz + AvertT + p 1 as( T 1 j (2 n fw) ~ SpAccp := Ssa(fns, peta) FSS S Acc =1.20597 g P p i i i l SpAcc " f_newpa := 2 n f_D'*H ns = 6.13345 Hz Calculated frequency A i i f
6.13377 Hz Assumed frequency nsg 0g 0 in S Acc l AA *
A,g P e SPACC := SA A , P cc p p, i := 1. 3 Force / Deflection Diagram l t.$ i i i i f0 ) f0 ) sm' SPACC = 1.10377 g AA = 0.18759 in (l.20597j (0.31351) 0.5 l l t i t 0 0.1 0.2 0.3 0.4 l I Calculate ductility factor and slope l SpAcc,- SpAcc g ~~ e A" A-Ag 4 e mu= -- mu = 1.67125 s := s = 0.13794 IS Acc g) i P e A,g 1 A el
-.-- =. 2% l STOP6 Estimation of the Median Ductility scale Factor, Fmu i The median ductility scale factor, Fmu, is obtained by several methods (See Reference 1): l l The followtng variables are used in the analysis: f = 7.5856 Hz The A8 Bus frequency ns 1 b : peta b = 0.05 Elastic damping fk = 2.5 1 Response spectmm knuckle frequency i see. s = 0.13794 Strainhardening ratxal mu = 1.67125 Ductility factor i i Riddle-Newmark method for calculating FuRn l The details of the methodology are given in Reference i Lets first calculate Fmu at the peak ground acceleration level (zpa) The zpa value is: zpa.= 0.4 g l and the spectral acceleration at the fundamental frequency, f, and elastic damping, b, is: l Sa.: SSa(fas,b) Sa =0.65961 g Fmu is estimated as: Fu4 : mu " Fu4 = 1.74485 zpa Then, in the acceleration range of the spectrum, Fmu is sima'~4 as: Fu3.:(2.67 mu-1.673)" Fu3 = 1.52439 l Finally, in the velocity range of the spectrum, Fu is estimated ar-1 \\ I \\ I & \\ f & <l.0 + k 21.0 Cf = 0.32957 Cf - [nsj[ns j [ns j f Fu2 :(2.24 mu-1.24)*'" Cf Fu2 = 0.5774 1 i
._~ 25 l The median ductihty scale factor, Fu RN, using the R2ddle-NewTnark methed, is obtained as-Ful Fu3 (Fu3<Fu4) Fu4 (Fu3 2Fu4) Ful = 152439 i Using the ratio of ultimate stanc capacity to yield static capacity, R defined as: R = 1 + s (mu-1) R = 1.09259 FuRN = (Ful-(Fu2<Ful) + Fu2-(Fu22Ful))Fu RN = 1.3952 R Modified Riddle-Newmark method for Calculating FuMRN l l Because the Riddle-Newmark method does not account for second slope of the force-deformation curve, the effect of second slope is accounted for by modifymg the ductility ratio. Therefore, the modified ductility ratio is: r l )+I mu j = 0.5 + I""- I I'I + mu 3 = 1.50718 2R At the peak zpa level, Fu is: Fum4 = 5*.mu g3 3 Fum4 = 1.72513 zpa In the acceleration range of the spectrum, Fmu is: Fum3 = (2.67 mu j - 1.673)uti Fum3 = 1.42101 In the velocity range of the spectrum, Fmu is: Fum2 := (2.24 mu j - 1.24)"33 Cf Fum2 = 0.52402 where Cf has been defmed presiously. De median ductility scale factor Fu.MRN, using the modified Riddle-Newmark method is: Fumi : Fum3-(Fum3 <Fum4) + Fum4.(Fum3 2Fum4) Fumi = 1.42101 l FuMRN :Fuml-(Fum2<Fuml)+ Fum2-(Fum22Fuml) FuMRN = 1.42101 i i l I
Y1 Effective Riddle-Newmark method for Calculating FuERN Because the modtfied Riddle-Newmark method only accounts for second slope of the force-deformation cun e, the last correction to perform is to' account for ground motion duration. l The factor to account for earthquake duration is: l CD : 1.0 and the median duculity scale factor, FuERN, using the effective Riddle-Newmark method is: Fu ERN : 14. CD-(Fu g 1) FuERN = 1.42101 f 1 l Effective Spectral method for Calculating ruSA In this method, Fmu is calculated using the effective frequency and dampeg ratio of the structure First, the ratio of secant stiffness to elastic stiffness is estimated as: 2 Ks g = U + Ks g = 0.65376 mu Next, the ratio of the secant frequency to elastic frequency is estunated to be: fs g.: fKs g fs f = 0.80855 Compare to: fs = fs f as f: =6.13335 Hz f,p =6.13377 Hz f a The ratio of the effective frequency to elastic frequency is calculated as follows (See Reference 1): cf : 1.9 Coefficient to account for short duraban motion Al :cf-(1 - fs f) A :: A1-( A150.85) + 0.85-( Al>0.85) fe r =(1 - A) + A-(fs g) fe f = 0.93036 The effective damping ratio, be, is calculated as follows: Cn = 0.15 Coefficient to account for short dura.' ion motion l bh : Cn (1 - fs f) Hysteretic energy dissipation damping bh = 0.02872 I Iis gY I be:: -(b + bb) be = 0.05945 (f*f) -n-
I 21 The median ductility scale factor. Fu SA, is estimated as follows: effective frequency, fe,h fe.: fe rf fe =7.05734 Hz o3 Sa,.= SSa(fe,be) Sa, = 0.59505 g Sa.e is the spectral acceleration at the effective frequency, fe, and the effective damping, be Ife g\\
- s, 1
Fu SA = 1.46763 Fu SA,= fs t; Se l ( e Final Estimate of the median ductility scale factor i Different methods have been used to estunate the median ductility scale factor. Based on a recent study, the median ductdity scale factor, Fu, is taken as the average of the ductility scale factors found using the Effective Riddle-Newinark method, Fu.ERM, and the Effective Spectral Method, Fu.SA. l l Fu ERN + Fu SA Fmumedian = 1.44432 Fmumedian := 2 l l l i r l l [ l l t 5 1 l
~_ __-.-.=. -.... - 2T I**I I*"I " Assumed valuca Calculated values Elastic point f n 7.5856 Hz f_new, = 7.58572 % m FS = 1.67338 f,,a 9.848 Hz f_new,, = 9 84806 % =401275 m eja-0.01274 in act plate Fmumedian = 1.44432 Gassumedeis 0.00362 OTel = 40036 I-FS Fmu g = 2.416897743 l f,ps6.13377 Hz f_newp = 613345 % g l f,,pn8.51538 Hz f_newn,w = 8 51512 % i F_' ' Tp " SN I fFS Fmumedim ~in BETA.=3 In BETA = 2.57973 10 l ( 2.416897743 j i \\ I l l l l I I t 4
M f Log::rithic standard Deviatien of tha Ductility Scalo rector (' L i l h variability due to random scatter of time history. computed Fmu versus predicted l Fmu values using approxunate methods (for example, the spectral averagmg method) is: l I I br Fmu : 0.4-[0.06 + 0.03-(Fu SA R-1)]br Fmu = 0.03124 The uncertamty due to story drift associated with failure is calculated, assummg that the 10' elastic deflection is at 1.0 *=viard deviation from the mean, to be 1.25673 using the same equation as above. Iberefore: I Fmumedian A bu Fmu = 0.13913 bu ya := 1.0 i I Ibe uncertamty due to inelastic energy absorption model is: r bt:Fmum :=0.1-(Famedian-3) bu Fmum = 0.04443 l The combined variabilityis: b Fmu = 0.14935 bFmu := br Fmu + bu Fmu + bu Fmum 7 A.' l a r I
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