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
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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
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by Bahman Lashkari John W. Reed i
1 Prepared for Stevenson & Associates January 1994
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l Estimation of the Median Ductility Scale Factor, Ftsu l l
. The median ductility scale factor, Fmu, is obtained by several methods (See Reference 1):
l ne following variables are used in the analysis:
De A8 Bus frequency fm= 7.5856 412 b = 0.05 Elastic damping b:= pets
& ::2.51 Response spectnan knuckle frequency [\ g AddM sec Strain hardening ration.
s = 0.13794 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: 1 zpa ::0.4 g and the spectral acceleration at the fhaA*-tal frequency, f, and clastic damping, b, is:
i Sa = 0.65961 *g Sa ::SSa(fas,b)
Fmu is estimated as:
Fu4 = 1.74485 Fu4 :: 8* mu "
rpa Then,in the acceleration range of the spectrum, Fmu is Mma'M as:
Fu3 = 1.52439 Fu3 := (2.67 mu- 1.673)""
Finally, in the velocity range of the spectrum, Fu is estunated as:
\
l I & i Cf = 0.32957 Cf:= 1 & \ t h <l.0) + 21.0 (f as (I ns j (I ns j i
Fu2 =0.5774 Fu2.:(2.24 mu- 1.24)"II Cf f
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) l Using the ratio of ultimate static capacity to yield static capacity, R MW as: l R = 1.09259 R := 1 + s-(mu- 1)
FuRN = 1.3952 Fu RN := (Ful-(Fu2<Ful) R + Fu2-(Fu22Ful))
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 j
slope is accounted for by modtfymg the ductdity ratio. Derefore, the marhw ductdity ratio is:
l
} + )+ mu 3 = 1.50718 mu ; := 0.5 + (""-
2 R*
At the peak rpa level.Fu is:
"" Fum4 = 1.72513
' Fum4 zpa :=b mu 3 in the acceleration range of the spectrum, Fmu is:
Fum3 = 1.42101 Fum3 := (2.67 mu 3 - 1.673)"d" l l
In the velocity range of the spectrum, Fmu is:
l 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 = 1.42101 Fumi := Fum3 (Fum3 <Fum4) + Fum4-(Fum32Fum4)
Fu g = 1.42101 Fu MRN := Fuml-(Fum2<Fuml) + Fum2 (Fum22Fuml) 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 = 0.65376 Ks g := U mu Next, the ratio of the secant frequency to clastic frequency iWma'M to be:
(. fs f = 0.80855 Compare to:
fs f:= jKs g fs := fsf f , fs =6.13335 Hz f,, = 6.13377 Hz The ratio of the effective frequency to elastic frequency is calculated as follows (See Reference 1):
cf := 1.9 Coefficient to account for short duration motion
(
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 Hysteretic energy dissipation damping bh = 0.02872 l bh .= Cn. (1 - fs f) 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 fe = 7.05734 %
1 fe := fe ffns Sa , = 0.59505.g Sa.e is the spectral acceleration at the effective frequency, fe, ,
Sa , := SSa(fe,be) -
and the effective damping, be I
I ffe f)2 Sa l .Fu SA := Fu SA = 1.46763 Se, (I* f t y,: , -
l Final Estimate of the median ductility scale factor i
Different methods have been used to estimate the median ductility scale factor. l l
Based on a recent study, the median ductility scale factor, Fu, is taken as the average of the ductility scale factors found using the Effective Riddle-Newmark l
i method, Fu.ERM, and the FEcetive Spectral Method, Fu.SA.
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P Logarithic Standard Devictica of tho "x tility f r:1 : rt-- M M ne variability due to random scaner of time history-computed Fmu versus predicted Fmu values using appronmate methods (for example, the spectral averaging method) is:
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 # bu Fmu = 0.13913 n&.t<) 1.0 L@ ne uncertainty due to inelastic energy absorption model in: -
bu Fmum :=0.1-(Fmumedian- I) bu Fmum = 0.04443 The combined variabilityis:
= 8.16 bFmu := br Fmu + bu Fmu + bu Fmum bFmu = 0.14935 l
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' 8 l l I% bk d l l I i i r D Jock R. Ben)omin & Associates,Inc. Consulting Engineers 1 l
i , w I a l l j i I ANALYSIS AND DESIGN : ; OF STRUCTURAL
- CONNECTIONS:
i io.. Rein breed Concrete and Steel 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 w Reader in Structural Engineering
- 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.
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.mn con. Prying force in the elastic stage of behaviour may be greater than at ultimate l 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, t contact and it is therefore necessary to develop a theory to calculate the magnitude of
- nts by the prying force. For an allowable stress method of design at service load condi-
! ic stage tions, the prymg force Qd, is related to the extemal force F., assuming linear elastic behaviour of the components. i attom of l j De theoretical model in Fig. 4.ll shows an end plate of thickness t, and j -orce in of cantilever length (a, + b p). The extremity of the plate is in contact with a i urtion in column nange, for example, and this introduces the prying force Qd.. The j extemal force F is assumed to be balanced by the prying force Os, and an axial strength j force F3, in the bolt.The axial force in the bolt produces an extension of the j . bolt of 6b . s fe
' i I *
, rl 1 3' o o I 1 t 3// W i a origin
~
y j _____.__._b.q_____ - F bt* " f e* O be oo I so e i 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 p) (4.29) 16. Integrating '
- nt # and d x 2 mn con- El dx y = - Od, 2 + (F, + 03.) [x -a,] : +A (4.30) .
2 !
,,m,- ion
- J, i l
~, ., . . .
. .. , w c:a , , y . . - - . .t s up,, .s. , ., .._ . ,. ,
l I 1 3 _ _ _ _ 4 l 1 146 Bolted Connections ~ (Ch.4 3,,, 4,3 g } Integrating Obu "- x3
+ (F. + On.) [x -a,]3+ Ax + B (4.31) i Ely = - On, 6 6 where F, = l i ' l I
whenx = a, + b,, dy/dr = 0 and therefore from equation (430) Destrainins b2 [4.20], and the 59
- (a, + b p)2 values of eg, andl 0 = - On, + (F. + Q3,) +A-j thickness plus wmt l ,
l Rearranging Ihe width of: j I j 6a Lth jj 13 defined and is ad A = Ob.a p (a, + s b,) - F. f-
- column connectic
{ i 2 2 [ when x = 0,y = 0, and therefore from equation (4.31) B = 0 whenx = a the
- h. Nhf[5, tions in a colum theory (4.22], a whichever is the t extension of the bolty = -6 b,and from equation (431)
-EI 6 e = - Qu. + a, On. (a, + 2b,) - F. .
b . 4.15 CONNECT 1
*'* E Some connectios F. - 2EI6b /a,bp ' diagram is genest
) beam, and colus 2(a,/b,)+ (2/3)(ap/bp )* * , }
1 i are applied to th
' The extension of the bolt originally preloaded with a force Fe, is stiff bearing W l force H are re@
3b " (Fbt -Fw)sp/A En = (F. + Qe. -Fw)sp/AnEn . (4.33) i j I) b Substuting equation (4J3) in (432) and rearranging fasteners, e.g. be
- F,(1 - k,) + k, F ,
1 . (434)
- 06. = 2(a,/b,) + (2/3)(a,/b )2 p + k, j -
i where k. = Ep w, t,*gp/64,b,' Ag En . (435) i 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 ! i i force Qw to the ultimate tensile strength of the bolt F ,. Equation (4.34)can 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
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i DATg PAOJ2CT PACg QF CY CHKO.3Y DATE SUCJECT JO3 NO. vom 6 x=eeb 3 9 da _- O ( 3,so~l Do N. y* (o ll n;t-J h we,,J O=C. hz Ef1jd + h k- 3) +%' g ,4 _ ._ A= Q a._ 2- diCdi+z_L3df<-3-)W e 2. 4d x c o Su = O =7= g= o 4 Sc o No~ &#A N A W"d M 15 )b a < +% a u a pp 4 2 ~J a a p +e A A 3 c) i Mc 6Nw (14 h dm Ax = o %J L de = C%(9+bg)+(Fe+L)L 7
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sY DATE PROJECT PAGE OF l CHKD.CY DATE SU3 JECT JOD NO. l [ dolt. d \s b b ).' < b
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C:\WINMCAD\ FILES \A8 COMPS 2.MCD 12/17/93 i kip .: 1000 Ibf ksi := 1000-psi Fixed base frequency of cabinet fgig :=IOHz f g=7.5856 Hz f e, = 9.848 Hz peta := 0.05 Frequency and damping of the entire cabinet
'O.375290~ ' 2.5 '
i 0.60 5 1 SA := g F := Hz Groundinput response spectrum f 0.702350 10 0.723530 ,25, f f SA f f f Q0.05 Values of Sa at different frequencies
.,ln' ,ln g-SSa(f,p):=exp'(linterp(in (g3 F(Hzi l
Hzt 4 i (f3 and damping valuca
*' .g i :
Wt := IOkip Weight ofcabinet Spectral acceleration Spectral accelerataco ' Spectral acceleration l in n/sdrection in e/w direction in vertical directaan See, := SSa(f,,,$ cts) Savert := 0.40 g Sea,:= SSa(fns,Seta)
.,a , =0.65961 *g \ Sa ,, = 0.69991 *g} Savert =0.26667 g 3 '
h:= 109 in Height ground to top of cabinet Base lever enn for north / south direction axial force at support 1, := 91 in Base lever ann for east / west direction axial force at support at first bolt 1,,g := 96in Base lever ann for east / west direction axial force at support at second bolt
- Iew2
- = 19.2 in Sa ew Sa vert t
San, -Wt t)h
-Wt hli, -Wt g g g (21 p = p .,
Po,:= 2 1,,g + I ( ewtj Pas = 1.97519
- kip P ,, = 1.91031 kip Pvert = 0.66667 kip Wt Wt Sa Y T Y g := Pns + 0.4-(P ,, + Pvest) 24 3: 4 ' ns g i
T 3 = 3.00598 kip Y 2 = 2.5 kip Y3 = 1.64901
- kip l ,
i ! T3 c g =0.54858 c 3 := Y l Y2'T3 c 2 = 1.37144 kip e 2 := 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 Pilgrim Station l l Plate and steel properties E ;= 2910' psi Modulus of elasticity for steel E Modulus of rigidity for steel 0:= i 2-(1 + 0.3) t p := 0.25 in 'Iluckness bold down plate
}wp := 2.25 in } Vrxithhold down plate Diameter bolt holein plate db := 0.625 in
! (I,angih piste from,and to bok6iinxinium value) Fa{:= 255"iei thig6iblaite'fhun boh~ts'siiiiiiert
- bp= 1.375 in e yp
- = 44 kai Medianyield capacity plate Bolt properties 8
Eb := 2910 psi Modulus ofelasticity for bolt A b := 0.196 in' Area bolt (Gross area for stretching) Effective length bolt for stretching gp := 2.5 in Channel properties C6x8.2 (AISC 9th Ed.) ( d := 6.00in Hetsht ofchannel b g := 1.92 in Width channelflange
*lhickness channel flange tf:= 0.343 in t , := 0.200 in Thickness channelweb ,
i ! e Distance outside edge channel web to shear center o := 0.599 in K := 0.08 in' Torsion constant C ,;= 4.72 in' Warping constant Moment ofinertia about x-axis
! x := 13.1 in' I y:= 0.692 in' Moment ofinertia about y-axis L := 38.375 in Channel span between assumed supports L Distance of anchorage to support along channel a := -
2 q := 0.25 in Distance from edge of flange where point of contact l ' 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 Fe on the c 2 = 1.37144 kip anchorage. V=c g F , + c 2 t pp:=t f+ 1-(l6 assumed Offset in plate which causes moment in plate from e ;l) shear force on the bolt (equal to thickness of channel flanSe Pl us deflectim upward) 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 I t ( = 0.35482 kip Castant reductim factor on the force Fe due to shear (:= BEE p c2 on the bolt t W
! ,= P P 1 = 0.00293 in' Moment ofinertia plate 12 Zh ;, (wP-hd P)'l Z g = 0.02539 in' Plastic section modulus plate at hole 4
I Mh = 1.11719 kip in Plastic moment capacity plate at hole Mh ;* Zh yp , CASE 1 - MOMENT BETWEEN PLATE AND CHANNEL FLANGE IS ZERO This calculatim finds the value of s psuch that the slope at the end of the plate is zero and the moment at the j J attachment to the channel (i.e., at x = a p + bp)is also zero. This is the limiting case where the channel offers no resistence to rotation ! See derivation for definition of terms and theoretical basis (Reed i1/22S3) Allowable prying force correspondmg to plastic Qbe(a).= P a moment capacity plate at hole p 2 E1gP k e(ap) := bp 2 A bE b p k(ap ) := 2 b + 2_, p3 b (b pj l 4 5 'a \ F,(a p) := -Q be(ap ) +( l
._ .-. - - --. _ _ . - - -~_ . . -.
4 Constram A to equal zero, which corresponds to zero slope at the end of the plate Find a p b 2' l { j a p .= root Q be(a p - p) a -(a pp + 2 e b )- p (F (a ) T-(). P ,
,, p f bf b b i i
F,(a). p T-k,(a)- p 1 + T-
-k(a)t-p - ( (k(a p) + k,(ap ))-
i Pl. P. p. l ' 2 2 bp l ap=0.75511 in I i Find limiting value of 6,
~
I b I' b F,(ap). T-k,(a). p 1 + T-
- k(a p - ( (k(ap ) + k,(ap })
) P.Y-[b i Pt. P.
6, = 2EI ! P l l 0, = -0.014004148 Tension force on anchorage at edge of channel flange F,(a )p =1.3604 kip l Prying action force on plate Q be(a )p = 1.4795 kip Bolt force Fb(* p) := F,(ap) + Qbe(ap) F b(a p
) =2.8399 kip Shear force V bolt (sp ):=c g F,(a p) + c 2 Ybolt(* p) = 2.11773 kip l
Plot deflected shape, slope, shear and moment of the plate p A(ap) .= Qbe(a )p a -(ap + p 2 b )-p (F,(a ) T-()- + E I 0, f x 3 1 (x- a )3 y(x,ap ) := p )-+ A(ap ) x + (F,(ap) T-(+ Q be(ap))- ' P)
-Q be(a 6 2 )
f x (x- a }2
+ (F,(ap) T-(+ Q be(ap))- ** P) r(x a )g
= fhde(* g)7 + ^(* P), 2 v(x.a ,) .Q w(a ,) . (F ,(a ,) + Q a(a ,)). (x>a ,)
u(x.ag):= Qw(ap).x+ ;(Fe (ap ) T-()+Q be(ap)}(x- a )-(x>a g g)
-y p,aP ) AbEb P si,(a g ) :- (s e so 1t(a,) =2.8399 kig 8P
I s n = 100 6x . = P* P 6x = 0.0213 *in n i.=l..n+I x; =(i- 1)e def;:= y(x;,ap) slope; .= r(x;,ap ) shear; = V(x;,a p ) mom; = M(x;,ap) y(0.in,ap) = 0 in y(a p,a p) =-0.00125 in y(a p + bp,ap) =-0.01636 in 0 i~ i i i ! U! I in Mi ! Y -0.01 - 8
-u2 , J, ', ,',, ', 1, 84 l '- ~ ~'-
h r(0.in,a p) =7.06720925910" r(a p ,a )p=-0.004%3919 r(ap + bp,a p) =-0.014004148 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:-
v(0.in,a,) =-1.4795 kip Pbolt(*p) =2.8399 kip V(ap + b p ,a )p= 1.3604 kip l 2 i i in M i _kir 0 l i 0
, I I i 1
'0 0.5 1 1.$ 2 15 s.
t M(0.in,ap) =0 kip in M(a p,a p ) =-1.11719 kip.in M(a p + bp,ap) = 0 kip.in 0 g g i i
, song ""0.$ .[ -
0 -
-1.$ 2 2.5 0 0.$ 1 1.5 i
=
t _ -
k l CASE 2 ROTATION AT CONNECTION BETWEEN PLATE AND CHANNEL FLANGE IS SET TO. 8, For thlS case a value of 9 ebetween zero and the limiting value 0 =e -0.014 is assumed and the plate response is calculated. Again, the slope at the end of the plate is set equal to zero. 0, = 0assumedei Assumed value of 0, 2 EI0* ! obc (ag )-(k(ap )+ ke (ag ))+ ,
+c
,P F,(ap) =
+-t(a) c g l
- l A(a p) := ap-(ap + 2pb )- (F,(a p ) T-(), p + E I 8,
~
I 3 I x y(x,ap) := 1
-Q be(a p )-+ A(ap) x
+ (F,(a p ) Y-(+ Qp be(a )) (x- a,)8,[x>, p)
~~
i 2 (x- a ")' l r(x.a ):=h;f o be(ag )i + ^(ag); + (r e(ag) v- c+ o be(ag ))- p 2 I* A V(x,a p) :=-Q be(ap ) + (F ,(ap) + Q be(a p ))-(x>ap ) M(x,ap ) ':-Q be(ap ) x+((F,(a p) T-() + Q be(ap )].(x- ap)-(x>ap ) V bolt (ap ) := c g F,(ap) + c 2 Given Constrain slope to equal zero at end of plate r(0 in,ap)=0 a p := find (ap ) a p= 0.87422 *in 1 i F,(a )p= 2.53015
- kip l Q be(a p) = 1.27793
- kip Fb(a p) := F,(a p) + Q 3,(ap) F b(a p
) = 3.80808 kip p.ap) A b.Eb P bolt p(a.- ) ,,-y(a pbolt(s p) =3.80808 kip 8p y (a p) = 2.75942 kip
DATE PROJECT pAgE OF DY CHKD.QY DATE SUDJECT JOD NO. l l l l Jack R. Benjamin & Associates,Inc. E I Consuming Engineers 9
. . - . . . _ . ...-. . . - . _ - _ _ _ ._ . . -= . . _ . - . - . - _ - - . . .
li 1 0.091 2.5 0.375 3hkw h b N-0.8
~ \
5 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 Y 0.1 _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 l 8 0.. x := 10 y := 0.702 l j 25 0.724 l , 33 0.4 ;
,50 , 0.4 Yi zg .- -
*i I
W Y 1 i j gg . 2 . s 14 , i 0.1 Z _i V 0.01 - I 0.001 1 10 100 l I l
l 7 n :100 Ax : P' P Ax = 0.02249 *tn n i : 1.. n + 1 x, :(i - 1 ) ax def; :y(x,,ap) slope, :r(x;,ap ) shear; : V(x; a p) mom, :M(x,a) i p y(0-in, ap) = 0 in y(a p,a p) =-0.0016749 *in y(a p+b p,a p) =-0 01275 *in 0 , ,
; g i
*E Wi i j ! -
i m -g.01 - 9.. l f f ! t
-'O.02 0.5 1 1.5 2 2.5 0
is r(0 in,ap) =0 r(a p,a p ) =-0.005747713 r(a p + bp,a p ) = -0.00362 l 0 i , , , i
*E .
ini 4 -
-0.005 -
9.
-0.01 2.5 0 0.5 1 1.5 2 3
i 5 V(0 in,a )p =-1.27793
- kip Pbolt(ap) =3.80808 kip V(ap + b p,a )p= 2.53015
- kip 4 I i l i i "f.
is sheer; 2 i 0 0 i f I I I I 0 0.$ 1 1.5 2 2.$
"i 5
M(0 in,a p) =0* kip in M(ap,a p) =-l.11719
- kip in M(a p + b p,a )p= 1.38013
- kip in 2 i , , ,
*E .
i - i. i _Win 0 0 -
- -g -
I l
' e ,
I I n l n 1.5 2 2.5 0 0.5 1 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 P- (tp f tf I fp = 0.0391 *in' I fp .:
- 2. Channel flange width equal to twice distance from edge of flange to inside web surface 2-(b-t,)tf f
3 I = 0.01157 *in, I f: 12 f Distance from ground to center of area of plate channel flange composite A fp ::w p-(tp + t y) Af:= 2-((bf - tw) tf ] t p+tf t A fp- + Ar f c:= c = 0.23784 *in Afp ,Af Total moment ofinenia f tp+tf32 I t 32 IIfp + I +f Afp' 8- +Af c-- f I t= 0MS *in, t 2 j Calculate rotation and venical deflection of flange / plate composite at edge of flange 1 F(a)-(bg-t e p w- ri)3 0 g ,EI 2 t f i'
+ -c M(a p, b p ,a p) + Vbolt(ap) t f+ -(bf - t w- i)) 0 g =-0 00325 1 F,(a p)-(b -f t ,- 71)3 aven g := E 1 3 t
f t \' M(a p+b,a)+Vbolt( p) tf+ -c p p -(bf - t w- ri)2
+.
Aven i = -O 00277 *in V bolt (a p) = 2.75942 kip
4 Rotation of channel due to twisting and deflection at edge of11ange . Torsion moment at shear center T 0 :[M(a p , b p,ap) + F ,(a p) (b f- q+ e )) o V bolt (a p)- -tg2 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 {~ g a 8
- T, T (\
a T, L Case f Left end free to warp but not twist, right end fixed (no twist or warp) l
# <f0 NOTE THAT THE "LEFT" SUPPORT CORRESPONDS TO EXTERIOR l
CONNECTION AND THE "RIGHT" SUPPORT CORRESPONDS TO AN i INTERIOR CONNECTION ON THE A! To SWITCHGEAR CHANNEL BASE FRA) 1
Io , i 1 1 p = f KO Y p - 0.08074.1 in (C w E/ F g(x) = cosh (p x) F 2(x) = sinh (p x) F 3(x) = cosh (p x)- 1 l ( F 4(x) .= sinh (p x)- p x i l 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) 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: l To 'C 3 C ,4- C 4C ,3). _A
'O CwE 2(C C 4-C 2C 3j g deg 9A I:= gA 0.02022 0 1 I= o IC g C ,4 - C 2 C ,3) g t' A
-48.22618^
.To lbfin
( C gC 4-C 2C 3 ; TA l i. 8 fa 0(1,x) = 1, deg + {iT F2(x) + g2 I F3(x)+ glbfin F 4(x) + To D p C,Ep C ,E p, F 4(x) ! Rotation at point of anchorage attachment to channel 02 =- 0(1,a) 02 =-3.51099 10 rad
ll Range of x-values: x = 0 ft,1.. L 100 Torsion Torsion Finned end Fixed end 0.03 g g g Angle of twist,0 0.02 0(1,0 ft) = 0 tad 41.x) , ,, _ _ _ des o " 0(1,a) = 3.51099 10 rad j
-c 0 t 0 1 2 3 0(1,L) = 0 tad A
l Venical deflection of channel at contact point between plate and flange due to channel rotation l _
.my l
' ~ Avert 2 :=02 -(((bf + oe )-in Awrt2 =-7.96644*10 ' in Vertical deflection of channel between aoss channels - assume fixed / pinned span as efective boundary conditions 3
-7 F,(ap) L 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 Ahonz :=
-7 Vbolt(ap) L Aboriz =-0.07083 *in 768 E l y l
l Total rotation from channel support system and deflection from concrete to switchgear i 0Tel:= 0 g + 02 0Tel=-0.0036003 red l I f Compare to calculated rotation at end r(ap+ bp,a p) =-0.00362 rad ! 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+bp,ap)
Cantilever flange Avert ; =-0.00277 in
~
Channeltwisting Avert 2 =-7.9664410 ' ain . Channelbendmg Aven 3 = 400343 in i Plate bendmg y(ap+b p,ap) =401275 in Aelplate:=y(a p+ b p,ap)
' Force at ends of plate Q be(8p) = 1.27793 kip ,(a )p = 2.53015 ki
~
Force in bolt Pbolt( p) =3.80808 kip V bolt (ap ) =2.75942 kip Abar ,g .= y(a p+ bp,ap)
12., STOP2 Calculate scale factor I F,(a)+t3 p
- _.T FS = \ FS = 1.67338 \ -
Tg Calculate frequency of cabinet in East / West direction l l l 1 h /0.4 Pew) AvertT SS*(Iew, peta) FS a,= - 2
+ A eg = 0.1181 in (l ewi-I ew2) Tl\1 3 (2 n ffaed)
SpAcce g := SSa(few peta) FS _ SpAcc,g = 1.17121 g SPAcc ,g f_new ,, . = i - f_new ,, = 9.84806 Hz Calculated frequency j 2z 3 Act i f e, = 9.848 Hz Assumed frequency 1 I Calculate frequency of cabinet in North / South direction 1 ! h _ fPns + 0,4 Pvert \ SS*(fns, peta) FS , ag= ahonz + j AvertT + Ael = 0.18759 *in e I Yl j (2s.fg)2 ns( SpAcc,; :SSa(fns, peta) FS SpAcc eg = 1.10377 g I SPAcc ,j Calculated frequency ! fnew ns *' I D'* ns = 7.58572 *Hz 2s 3 a ,i f ns
= 7.5856 Hz Assumed frequency l
t 3 i 1 a I
._ . .- - -. . . ~ - _ . . ._ ..- - - - . - - - - - . - . - . . . - _
l O STOP3 SECANT POINT ANALYSIS l m Spectral acceleration Spectral acceleration Spectral acceleration in n/s direction in e/w direction in venicaldirection i Sa ns SSa(fnsp, peta) Sa ,:= SSa(f , peta) Say ,n = 0.40 g Sa ns = 0 62852 g Sa ,, = 0.67716 g Say,n = 0.26667 g 1 Sa t Sa ven
'hi Sae, Wt. ' h Wt 7
ns.Wt-) -. - - g (2t g (2 g
- p. i p
ns ew ~ pvert , i f 2 4
; as,2 I ew2) 2 1,,g + l
-L=e ewtj l 1
P ns= 1.88211 kip P ,, = 1.84823 kip P vert =0.66667 kip j l ' ( Wt Wt Sa ns T t g := Pas + 0.4-(P ,, e P y n) T 2*7 3*7 g j t ; = 2.88807 kip T2 = 2.5 kip t 3 = 1.5713 kip 1 T3 c g =0.54407 c g := T I I T 2'T3 c 2 = 1.36017 kip l . c3*
.=
Ti l l l t l 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 Fe on the c ; = 136017 kip anchorage: V=c 3 F , + c 2 Offset in plate which causes moment in pir.e from t pp ' : t f+ 20(l6bare ih 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 ( = 0.59147
- kip can.aant reduction factor on the force Fe due to shear
('= pb(EE c 2 onh bolt 3 t w Mnmant ofinenia plate P 1 = 0.00293 *in' l 1:= P 12 2 b p Z h * (* p-d )t Z h =0.02539 in, Plastic section modulus plate at hole 4 Mh :Z '8h yp M h= 1.I1719 kip in Plastic moment capacity plate at hole 1 l CASE 1 - MOMENT BETWEEr4 PLATE AND CHANNEL FLANGE IS ZERO l This calculation finds the value of a psuch that the slope at the end of the plate is zero and the moment at the ) ! attachment to the channel (i.e., at x = ap + b p)is also zero. This is the limiting case where the channel offers no 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 ke(ap) := b,AEa p b b p
+ -
k(ap) = 2-P k P) Ia ) F ,(ap) := -Q be(8p) + C l . _ .
. . _ - . -. . - _ . _ - . ~ , . - - . . _ . .
a N l n : 100 Ax : P+b E 8x = 0.0218 *tn l n i i :1..n + 1 l x, : (: - 1 ) Ax j def; y(x,,ap) slope; :r(x,,ap ) shear; = V(x;,a p) mom, :M(x,,a) p l y(0 in,a p) = 0 *in y(a p,a p) = 400142 *in y(ap+b p,e p) = -O 01698 *in 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 l l r(0 in,a p) = 1.465303015*10' r(a p,a p) =4005291601 r(ap + b p,a )p=-0 014331831 0 01 i i i i n i 0 W
*i l t -o.oi -
t i l 0 0 0.5 1 1.5 2 2.5 8
~
i v(0 in a,) =-i.3827 tip e so i,(a ,) = 3.22678 kip v(a , . b ,,a ,) = l.83908 kip 2 i , . . ! *F ! i-i
-lup o -
0 i
! i i
I t I I 0 0.5 1 1.5 2 2.5 8 i M(0 in,ap) =0 kip in M(ap,a p) =-1.11719
- kip in M(a p+b p,a p) = 0 kip in u , , i i i
( o . i
@m N 0
- - i
, -t I ' ' ' '
-1.5 o 0.5 1 1.5 2 2.5 l
=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. 6, = 0assumedp Assumed value of 9, , 2E10* Qbe(ap)-(k(a)+k(a))+ p e p +( P F (ap ) : t-k,(a)p ] A(ap ) : be ap ) ap -(ap + p2 b )-e (F p (a ) t- ()-+ E 10, I 3 I 3 x (x- aP)3[x>,p) y(x,a).= -Q be(* p)y + A(ap) x + (F,(ap) T-(+ Q e(* b p))' p 6 f ) 2
+ (F,(a ) Y-(+ Qp be(a (x P)). - a )'-(x>a) r(x,ap) = -
-Q be(8p) + A(ap ) p p v(x,a ,) =.o 3,(a ,) . (F ,(a ,) + o s(a ,)).(x,a ,)
u(x,ag) :=-o s(a,).x+((Fe(ap).,-() + o be(a,)}(x- a,)-(x>ag ) ! V bolt (ap ) .=c 3 F ,(a p) + c 2 l Given Constrain slope to equal zero at end of plate r(0 in,ap)=0 a p .= find (ap ) a p=0.92366 in i F,(a )p =3.04146 kip Q be(a )p = 1.20953 kip l e )p + Q be(ap) F b(a p
) = 4.25099 kip Fb(ap) :F (a l
-y(a p,ap) A b Eb P bolt (* p) :: Pbolt(a p) =4.25099 kip P V bolt (ap) = 3.01492 kip
1 N P+ I n :100 tsx = ex = 0.02299 *tn n i 1..n+1 l x, = (i - 1 ) Ax def; :y(xg,ap) slope; =r(x;,ap ) shea , e V(x;,a p ) mom; = M(x;,ap ) y(0 in,ap) = 0 in y(a p,ap) =-0.0018697 in y(aprb p,a )p=-0 01383 in i 0 , in . def, in -g o g - 0 I I ' ' i -0.02 0 0.5 1 1.5 2 2.$ ( l sg g - l l I r(0 in,ap) =0 r(ap,a p) =-0.0060727o9 r(ap t b p,a )p=-0.0049 0 , ,g , i
*5.
ini
-0.005-I f !*"".. -$.01 -
l 4
' ' ' I
-0.01$ i 0 0.5 1 1.5 2 2.5 l
. i.
im V(0 in,a p} =-1.20953 kip Pbolt(*p) =4.25099 kip V(ap + bp,a p) =3.04146 kip 4 i I i ii
*2 w
almar; 2
? -
0 0 -- l
, l ! t I I
^
0 0.5 1 1.5 2 2.5
*i as M(0.in,ap) =0 kip in M(ap,a p) =-l.11719 kip in M(a p + bp,a p ) = 1.26212 kip in 2 i ;i i i
*E.
- i -
wl i i l -
- u,, 0 . . . . . . ..
-~ - 0
.g I I I I 0.5 1 1.5 2 2.5 O
s;
~~
In
I l N 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 w p
- p- (t p + tf)3 1 I fp = 0.0391 in, l fp . =
- 2. Channel flange width equal to twice distance from edge of flange to inside web surface 3
I f:= 2-(bf - t w)tf l I g = 0.01157 in, 12 Ihstance from ground to center of area of plate channel flange composite - ~ - - - - -- - A fp := w p -(tp+t f) Af:=2-((bf - t,) t f] t tf A fp p+t f+A f
- = c = 0.23784 in Afp + Af
, Total moment ofinertia I t f t
> I ;*I t fp t I f+ ^ fp' 8- P+tg*+ A rs - g\
- I t=0.06045 in' l
Calculate rotation and vertical deflection of flange / plate composite at edge of flange 1 F,(a )-(b p g- tw - 11)* 0 ; .= E1 t 2
+ M(ap tb,a)+Vbolt(a p p p ) t f+
y(b-t
- f -wvi) 0 g =-0.00351 l
l I .I F,(ap)-(b g- t ,- il)' avert j := E 1 3 t I t \- M(a p + b p,ap) + Vbolt(a p)tf+ -c
-(bf - tw- 11)*
+- -
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 . Torsion moment at shear center l
-t f-T 0 =(M(ap + bp,ap) + F ,(ap)-(b g- a+ e o))- V bolt (a p) 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 Is L l l Case f Left end free to warp but not twist, l right end fixed (no twist or warp)
# $of e
NOTE THAT THE "LEFT" SUPPORT CORRESPONDS TO EXTERIOR CONNECTION AND THE "RIGHT" SUPPORT CORRESPONDS TO AN INTERIOR CONNECTION ON THE A8 Io SWITCHGEAR CHANNEL BASE FRAND
_ _ ~. _ . _ . 21 1 f 2 KG1 p =; = 0.08074 *h m (CwEj F g(x) = cosh (p x) F 2(x) = sinh (D-x) F 3(x) = cosh (p x)- 1 F 4(x) = sinb(p x)- p x - F.g(x) :=(x>a) cosh ($-(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: To IC 3C ,4 - C 4C ,3) g C,E5 (2 C g C 4- C 2C 3 ; deg BA I= g 0.07977 0 A l=
!= 0 fC i C ,4- C2 C 3) g 0"A
-T- -190.28083 0
CjC 4-C 2C 3 ; lbfin TA If Igibfin T0 0(1,x) := 13 deg + F2(x)+ F 3(x) + 3 F4(x) + 3 F ,4(x) p p C ,E-p C,Ep 4 Rotation at point of anchorage attachment to chacncl i 02 :=-0(1,a) 02 =-0.00139 rad l l l l
*2 *t . Range of x-values:
x = 0 ft,1.. L 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
' Avert2 :=0 2 }(b f+ eo )-q]) Amt ~2 =4.00314 in Vertical deflection of channel between cross channels - assume fixed / pinned span as effective boundary conditions 3
! -7 F,(ap) L 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 Abortz :=
-7 Vbolt(ap).L Abonz =-0.07738 *in 768 E I y Total rotation frora channel support system and deflection from concrete to switchgear l
0Tp := 0 + 02 0Tp =-0.0049 red 1 Compare to calculated rotation at end r(a p + bp,ap) =-0.0049 rad of plate (connection to channel flange) Total displacement i Abortz =-0.07738 in l l AvertT = Avert 3 + Avert 2 + Ave:t3 AvertT =-0.26523 in
+(t-t f pp)
Cantilever flange avert 3 =-0.00304 in Channel twisting Avert 2 =-0.00314 in ! Channelbendmg Avert 3 =-0.00412 in Plate bending t f-tpp =-0.25493 in Force at ends of plate Q be(ap ) = 1.20953 kip F,(a )p = 3.04146 kip I Force in bolt P bolt (* p) = 4.25099 kip V bolt (ap) = 3.01492 kip t
- - - . . . . - . - . ..-.._ - -_.~-._ _ - - .
23 STOP4 Calculate scale factor FSS := c( P)
- 2 1.91g74 Yl Calculate frequency of cabinet in East / West direction b f0.4 P,,) AvertT S88(Iewp, eta) FSS A =0.17525 in AE := 2 (1,wj-1,w2)( T 3 j (2.x ffixed)
SpAcc := SSa(fewp peta) FSS SpAcc , = 1.2993 g SpAcc" (newp , := 1. f_newp e, = 8.51512 Hz Calculated frequency
~ ~ ~
f,,, = 8.51538 Hz Assumed frequency l Calculate frequency cabinet in North / South directa h j IPns + 0.4 PvertI SS*(Insp, peta) FSS T > Ahonz + - AvertT + = 0.31351 m, \ AP := \A p ~ T1 j 1 as( (2 n fw) SpAccp := Ssa(fns , peta) FSS SP Acc p=1.20597 g i i i l SpAcc " Calculated frequency f_D'*H ns = 6.13345 Hz i f_newpa := 2 n A i f nsg =6.13377 Hz Assumed frequency 0g 0 in SP Accel AA * = A ,g SPACC :=
, SA P cc p A p,
i := 1. 3 Force / Deflection Diagram l t.$ i i i i _ - f0 ) f0 ) sm' SPACC = 1.10377 g AA = 0.18759 in 0.5 - - (l.20597j (0.31351) l t i t l , 0 0.1 0.2 0.3 0.4 l I Calculate ductility factor and slope
~~
l SpAcc ,- SpAcc eg A" A-Ag 4 e mu = 1.67125 s := s = 0.13794 mu= -- i A ,g ISPAcc eg) 1 A el
- .-- = . -_. - .- - - - __
2% l ! STOP6 i Estimation of the Median Ductility scale Factor, Fmu 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 ns
= 7.5856 Hz The A8 Bus frequency 1
b : peta b = 0.05 Elastic damping fk = 2.5 1 Response spectmm knuckle frequency i see . s = 0.13794 Strainhardening ratxal I mu = 1.67125 Ductility factor l 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 and the spectral acceleration at the fundamental frequency, f, and elastic damping, b, is: l l Sa .: SSa(fas,b) Sa =0.65961 g Fmu is estimated as:
- Fu4 = 1.74485 Fu4 : mu "
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 \
Cf - I & \ f & <l.0 + k 21.0 Cf = 0.32957 [nsj[ns j [ns j Fu2 :(2.24 mu- 1.24)*'" Cf Fu2 = 0.5774 f 1 i
_ _ ._~ 25 The median ductihty scale factor, Fu RN, using the R2ddle-NewTnark methed, is obtained as-l 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) R + Fu2-(Fu22Ful))Fu RN = 1.3952 Modified Riddle-Newmark method for Calculating FuMRN l Because the Riddle-Newmark method does not account for second slope of the force-deformation curve, the effect of second l slope is accounted for by modifymg the ductility ratio. Therefore, the modified ductility ratio is: r l mu j = 0.5 + I""- 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 l In the acceleration range of the spectrum, Fmu is: Fum3 = 1.42101 Fum3 = (2.67 mu j - 1.673)uti In the velocity range of the spectrum, Fmu is: Cf Fum2 = 0.52402 Fum2 := (2.24 mu j - 1.24)"33 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 FuMRN :Fuml-(Fum2<Fuml)+ Fum2-(Fum22Fuml) FuMRN = 1.42101 l 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. The factor to account for earthquake duration is: l l CD : 1.0 and the median duculity scale factor, FuERN, using the effective Riddle-Newmark method is: l l 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: Compare to: fs g.: fKs g
, fs f = 0.80855 f: =6.13335 Hz f a,p =6.13377 Hz fs = fs ffas 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 Hysteretic energy dissipation damping bh = 0.02872 bh : Cn (1 - fs f) 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 =7.05734 Hz fe .: fe rf 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 1 Ife g\
- s, Fu SA ,= Fu SA = 1.46763 Se e l (fs t; .. - ._ _
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 := Fmumedian = 1.44432 l 2 l l i r l l [ l l t 5 1 l
-- . .. ~_ __-.- .= . - . . . . - - - . . . . . . _ -
2T I**I I*"I " Assumed valuca Calculated values - Elastic point fm n 7.5856 Hz f_new , = 7.58572 % FS = 1.67338 f,,a 9.848 Hz f_new ,, = 9 84806 % eja-0.01274 in act plate =401275 m Fmumedian = 1.44432 Gassumedeis 0.00362 OTel = 40036 I-FS Fmu g = 2.416897743 f ,ps6.13377 Hz f_newp g = 613345 % l f,,pn8.51538 Hz f_newn,w = 8 51512 % l F_' ' Tp " SN i _. _____ . _ _ 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 ya := bu Fmu = 0.13913 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: bFmu := br Fmu + bu Fmu + bu Fmum b Fmu = 0.14935 7 A .' l l l l l 1 a r I
. . . . . .. _ _ . . _ . . . . _ _. . .._ ._ _~.. - _ _
PAOJECT PACE OF DY DATE JOD NO, CHMO.CY DATE SUOJECT I l
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