Text

Rev 0 to Calculation 91C2672-C-018, Fragility Analysis of Station Blackout DG Pnps
	ML20247D364
Person / Time
Site:	Pilgrim
Issue date:	01/31/1994
From:	Lkashkari B, Jeffrey Reed; STEVENSON & ASSOCIATES
To:	;
Shared Package
ML20247D359	List: ML20247D355; ML20247D364; ML20247D429;
References
	91C2672-C-018, 91C2672-C-018-R00, 91C2672-C-18, 91C2672-C-18-R, NUDOCS 9805150034
	Download: ML20247D364 (114)
	v • d • e

- _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ - _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

l I.

I i

I

\\

AC Bus A8 Calculation 91C2672-C-018 l

l 1

l l

l I

I i

i I

9005150034 990507 ADOCK 05000293 PDR.

PDR P

- < t r.

i

~

CaECH c ton 2-UTE (4/. o Fragility Analysis of Station Blackout Diesel Generator Pilgrim Nuclear Power Station l

Bahman Lashkari John W. Reed l

Prepared for Stevenson & Associates f

January 1994 l

OY DATE PROJECT PAgr or CHKO.BY DATE SUBP"*:T JOB NO.

4 r

6 1

se mawa a up %%~% SL l

h I

bnMan LAthKhR1 do\\W % RfED 9

wk lvu ob nk M &

Uw

'NSB 8

Jock R. servamin & Associates,Inc.

Consulting Engineers 9

CH K3. CY BATE CUOJECT J33 ND, AM. 690 6%0 DAh>(*d A OrW_

0 tii k

5b(

l.0 6bMdM kb i

\\.I &

\\

~

1.1 b\\

2.

l.3 A

(

S

~fd k T a

1.sg gy $Q{4 hg c

2.0 M 8

4.0 Db bb Sils0 V bd bT 9

6.1 k

9 4.\\.\\ Ad

%M

\\\\

T 6.g.t hM gww

(

12.

a.\\.s A ! b.' L it is i

s.1 6%% A L n6 it.o ( -

A w

d 2--5bYA.d.b ltf 8). I e

6 4.t.s sLp 4,(,

a g 4 4,4 m Ad & A A

2o 6.).S Mk 2I#

o

-I.

Jock R. Benjamin & Associates,Inc.

Consulting Engineers E

CY 9 ATE PAOJECT PAfr SF CHK3. GY 9 ATE CU3 JECT JDD NO.

h Ndk 23 9.%

e 91.1 bM hk 29 4.1.1.1 fa d e k p S 6% k M 11, N

YA4, 21 4.2.i.a R A Q d, p k at

8. A. A 80 diab d bMb 37 WA. '6k. A 82.2 as 4a.% MM a3 g

4.2.s.s mA g a3 9.)A.A M O ih 39 4.aA.sup 39 4.1 A.S ' h,-

36 4.1A.5 [

66 4.1.Sh h 6 3b h[-

Ad se 5.o Qu Ayk A _ SRT WAdeva V4,%

MLast.wysd q M Rdd 2 b4kg 6 BM

~~i Aki a_ w.ka gqa ha

% A,J s M& W L h A nad 3_ wAAA

~

g E

Jack R. Benlomin & Associates,Inc.

Consulting Engineers

CHK3.CY BATE CUOJECT J33 NO.. D6 o40 bM DIfA 0_, 4x&

OW~

I u

Q 6tah 3 M rde M hMM phb

&~

LA 79 9Ah em. a g c pA A Ag omtalLL,a g gSkwi M,4A M b

Y N 0^L W

5 N

S SN

. 0$ 4 O MhdbL ba 6H ht)

,kM :

'\\

d M,Y

.R bh Mkbbp d EA A A

c,aK p 4 g L M ty,d A sect. At A

--a a nau asu y a u QANW d h 6 d.A Mbk p gA %.

O%

b M b*f K W~ W

^

L w % f A uLL + L dy ofb~

wo y sdp As.

O Jack R. Ben lomin & Associales,Inc.

Consulting Engineers 9

CY 9 ATE PR3 JECT PAgr er CHK2.DY MATE SUOJECT J33 NO.

bh\\.0 D

L S&

6 8 0 % >t't 3 Y 0

d A wy&. %% zo LA L M y (qu-

), A A wh (pa8), A s.o g tL c.

hk w r; 6(4T As al u p a L. W d&

d w% s.o w % L 6

' d s. b% s.o5x e a -m. A 44 u

d Qaa.A.

m g & sar 4 7

o a At A da L

w. 4 dAsw2. 7 - o 9 q L caan a l

1 Y

i I

Jock R. Benlomin & Associates,Inc.

i Consulting Engineers D

j l

{

I

CY

ATE PR3 JECT PAge 1

gr 1

cHKD. 3Y nATE CUOJECT JO3 NO.

O

%3N DM b g,g

\\.\\ Mbb

\\

Ob 3 M kb la A, M1 T' h M M N

kM b4 41 AM b

l

$4 Q. O Mh E M d

Lv A p @ hyLA,da.%vSwirp M.

l

%bbMbb b

M4 d

bb kMd b

M I

1M M MM $ M db.

l Nb M M Mb bh f b v

4 e A 9a e A,g um.sa ~atc.1~,

x 4W & 4 g Auh. p g h abb,J l

l.,

1 l

l f

l e

I Jack R. Benkmin & Associales,Inc.

Consulting Engineers D

1 CY MATE PDCJECT PAf;r g y-CHK3. CY HATE EU3 JECT J33 N3.

\\.1 6~&, _ hw U

h. % up e Q A o %A htA x A p. L 1A 4 p % d g,ti G k d 12 y L p b L q ).

% d e k w e 4 s m ( A g.aj 6. J A m q L %s d.

l AudE % 4 AA AL 4L.abwA1.eL g

AL q.% dbu AkA g g h a dia.a 4 td y a4 L Ak A 4JJ g a. %,1 ad n-E L.p g h to p.h,w;2fs L h.t> y.b L l

eed a q A,L.A.L g ocJ a4. A q fL. a.%

w g g o iL~.

hd n L e q p% b.47 a.J eg L

w. b 4g bw 'r hmo41, L pA Ap A -

N Mib

%#Y YhTO m L w g b h, s A, 3 1 4J. A h, &. $ ~ 4, 6 L k. e i

$ g hs p ti.Aeb d, % L Q L.=3 %

% % C p o 4,Q,b % 5 p

% L L.

M A 6 p L A M p L % d;A a s' d % % A m n b. cam, %.eo 4 W,m,w.es.

K e 4 L % (pg. cins,N.5e) Am A A LE w \\15 + #d LL & y hw.

~ khkJ s cs a gA & u ~J & M L p M x p.

wa n ne $

e g> aoen.wmin u.

w.ine.

-"a **-

l w y yx,

O CF j

CY DATE PROJECT PAge SU. JECT M N D.

7 CHK3.0y

_nArg I

b b

g; i

M. h@

)1 UNb\\\\D f

CA%2, %.so

.)

. U hbt\\b M.f,@

g

.2

. Uk b \\\\

b. 60

. 23. %s, w. n

3. wvse, %.e

.2

NV ONY &. $Q g

O I* b M U bhN.O M.S 4

f

\\

l

~3 g.

hb h B 9* f-0 j

l Vg.

htah t 0 2 j tb.5p

. Dg.

t% h 2 9 \\, b.EO

\\.

L 1

~

Jack R. Beniamin & Associates,Inc.

S.)

ConsuNing Enginews 3

f' t

b OY -

CATE PROJECT PA%8' C F' CHK3. Gy nATE CUOJECT J20 NO.

Dbpb bk b hk g,3 v

iV h% M<d' b

O d 0-20,000 l]

hbh Md&

p a u w g a g aau n

M M (l-O h h a. M Ab M h OMddM L

M M

& 0 4; hurw Q R bbk M k.0M b l

% 9-l 1

h 6AL dM d 0, p S t.t.A n A y st. s g - & a g a v ~,ut L a 9.,1. 6, b p,.

, q u w a a q p a _

N-db b

b %

h amb h

M t.T M %0g. b D

, h aM OL Nf

&f jW&

f $-

(A3 w.s - a p.L y 6 g

% atg.

GM'6, b. 60

.h

. Gu@y %. 60

. bg t-34 3g Jock R. Benlomin & Associates. Inc.

I.

Consulting Engineers D

CY-

" TATE P;22 JECT PACE.b gr CHK3. CY nATE CUOJZCT JO3 NO.

l

\\*

0y 544 W V

h kT b M do 0..D 9V M. M M 4h Ot%u q s aa. a w ess 6. n a un au d A g n%o W b c~;mg q v' W.

w t.L& o d g tti c p 4 A LJ 6 4

.b& &.

&&a 4 \\0%

l ys sk k u 6 % d 4. t r a q f

Y O yA k

34 hhA AM A M b b M b.k bb e

N4 4 6 f k %w A TdA g

ca a. L a q

_ g sor tm w u p y ks ~ tu gtt n (a n a %

m) ad A u m t.L A.

a L s.1 6.e W L A u paA a E% a ~%4-g+uap a.a d L 5-4 4 g p'( Att.

N M*0 Y y

LM, % y i %%

d aL sh.

7 1

Jock R. Benlomin & Assoclofes,Inc.

E Consulting Engineers B

l l

h CY

" ATE PR OJECTo_

PAtr gr i

CHKD. 3Y nATE CUEJECT JO3 NO.

MMM tmNu

[w 4 M-(atk b

& mL wt#g L h p kbA +4. AL

$@ @'9 b A T

9t w.p a At a ~t h g (m m 6 m)

Lp %ubtApAMLb4xq %d4 p -gu m Lga.

%M ts b ENe% b d #sRT dN.TNI Ibn Ad M J wy 54Adpa g ahdJL A ub5 4 h abY\\LLMd,h w

[kbtM b.1 4 h<%p,o)q o

A ku N

wM O

AM w

i M

=

l bd M

LM b,K

' NA

+L 4 M 4 % wh, mp p e 4~u g 4 E Q,J 1 Ap.

4 L u pqt s a n g s a L

N g ( U ^ ts $

W&N h"

4 w a m. L m@,x +

u-w l

g uT u p g. 6 A a,o

> AL 1 p.,va6-@t9 e q u c4 o p%aeW-Anum,anW~

l Jock R. Benlomin & Associates,Inc.

Consulting Engineers 9

l cv

Art encerCT 1

PArr c r --

CHKD. CY MATE SUOJECT-JO3 HD.

I

'W Q V

p%N.4b%, M d d wI, b 8 8 % 3 M ($30 l

NN kA NNDN 0$

b% &

bdb db M b d

b 8bD N hwhv.

h.cW-A wAf q A sp d44eL A L.4 A 4A a S %

Dea.kmMAAL l

i

& p ay.

i l

r l

i l

l l

I Jock R. Beniamin & Associates,Inc.

Consulting Enginoots B

l

CY SATE PMBJECT PAf E" Y

CF CHK3. CY MATE CUSJECT J23 ND.

7.0 dh Imd

(.Mu-

)

k T p 6 % S wA p p d. a 5l, L u

b ke+db (

(4-1

$b M

V.

t.p mA AtaAaL46**""

A. uia. y $ Att tA LL & L 4 6 mfd a;, rx,g4 gte a 9.a4.4.

i 1.00 i

i i i

,,,il i

i i ii,iig i

i,ii,ii 5% Damping i

Ground

)

0.75 -

)

S 1

,e l

0.50 -

l i

8 l

}

m 0.25 -

l

~

\\

1 l

0.00 -

i i i i i i; i iiiiij i

i i

i iiiiii 0.1 1.0 10.0 100.0 Frequency (Hz)

N Jack R. Beniamin & Associates, Inc.

Consulting Engineers D

L___------------_----------------------

k CY ATE P%CJECT PAtr CF CHK3. CY nATE CUOJECT J33 NDe o.o h ath e d %ov6 stb M u

s v

k poA d 4 L A4p u g a AbM s.

L nJap teedt 4 s c& b % A % A & L E W b M-uLJtApatpp.4 Ak1 u Aim S.

);&a AA 4 4AAAm 9. L Ag s.\\

u

_.% @ A cu nws A p d m,,

9 in m

s 9

l l

l

%N w

-I "I_M

\\-

/

)

,u.s, e, _,,,,

I fx, L Q g ng N=qu-i 1

/

--amma w iai

{

4 1

.A Ro.utJJ 4 y L rLS LL.

l

__g l

bbMA bb*

bb CAI' f

N b int-

- Eb s.a %% a W i-Jack R. Benlomin & Associates,Inc.

Consulting Engineers D

l j

CY NATE P2OJECT PAgr 10 gy CHK3. CY nATE CUCJECT J3O NS, s

DM B OTi M A 3\\ #f b:

3 f

j amtsn s x

. u%s

l I2' bnwA, b d' N

yMAd U

^

a Ap. % > % >

qlo y

4 42.

W = 10 C l

1 I

k w A, ID 'db b

AN dA sj % 'G.

T l

i Jack R. Beniamin & Associales,Inc.

Consulting Engineers B

i

CY 4 ATE PRCJECT PAgr 11 GF l

CHK3. CY nATE CUOJECT JO3 NO.

l S.11 kbafS. NM l

u A died 4a A AA d d 2.. A A 4 2 Aa e AA g aup A u A d W.

l A we.W g%

A; pgA p w,

.L was Atai h 6 L A A p g 7sa a

s h A. m usn. 4 7 gaa es w a s s.

BA AA A p a g h w&O y u b)e l

sa w p a&2.d ataa4 g2./a sche n A L AA

%bxA p

un f f wh \\.o z...n-:-4 L p k k2.d

4. b 4 A % d at L kw p A.

4&AJ ev b h4

.J a 4) Lp pAb) L b A 4.

i DL bNE i f

4.aak k% g g%caaoaAA1.ssasaL ch147 d

s.es % a L. d. 4 W L.

l O

Jock R. Benlomin & Associates,Inc.

Consulting Engineers S

i

CY DATE POCJECT PAgr 17-CF CHK3. CY MATE CUOJECT JOO N O.

i c.s.2 HAd b %t (c4)

L wR & 4 }.L MAEm p a L a db d% Ao.L.

a.

Mb

(Af s.x, I

k 1

(% bbT A

3, P

4 E

Td ds M d M N g.%i saft.( u 1 & 2.2d A

. sttAA. LA 4

t Td Mwu

- o.N b39 QhtL Sj $

Im & ndLtd % aj$ f '

A wiuo,

i.u, ou AJ..

oA p uA p

$m &"

Y NOY&$

LL.Sa

- p r m y A k auw) % a a; 4JJ Aun et A wgM.

~

w O

Jock R. Beniamin & Associates,Inc.

ConsuNing Engineers B

l

l0 CY-

) ATE PROJECT PAgr g p-CHK2. CY DATE CUOJECT J33 NO:

a.t.s A A %!L

% A La M f %" 4 a ha, ce m1.4Ja&,

sen i

p. (+ wAcd w.w L (w. Ed.4 3,A g Ar puu, 4

dp 4 % 4 WT a.

T T%

s.vos k

c V.

A 2.ua k

c i

u d.0.....

v k

h.soi.Mt a.M % % - Un ha. %.%. LL 1

kit m k o.tv. il yl $ kit.% Q,.L ab dL i kk t \\45-4.

D

/

v M

T oA5 A = oAy, teno.\\% t.a.

k A

g v

A k e 0.bl A " #* i#Ll tI'\\ E 8' t " s'u k

b OS 6-Q

.$, Y $

Y s ni n L,1.c a %(

g

.f

~

+.~

. \\.0

( O

\\. i /

l

.,,, 7, g (

0 Ma T,,

b Yv -

=

Jock R. Benkmin & Associates,Inc.

l I

Consulting Engineers S

Cy NATE PR$ JECT PAf' EF CHK3.CY "ATg CUOJECT JO*) N D-A e

z e

t 9 16S 3 IJ 10T 1T 3:e 1.0 snap.

[g,1.lf 4

e.e /

\\

3.t i

4 g m ; g 3 $ A tt p c. 13s o w. A v r -

o g

y e \\.tn.

L.4dAe%m & teLk, M G.9.....

I A A A p.4 p s hthA & kJ(yac4 Qu-Y h O=Y

/

b b 6

y

, 000 6

u u

1 ia

/

J 4

4 O.

bI*b c s,,, _ <.

& s. a u

/r wps 2 tabli y

_Ay 0,.(L-Q deAb W

h f = db i bDM O

con.umne main

OY OATE PRDJECT PAfr CF CHK3.CY nATE CU2 JECT JO3 N2.

D w 9. D f

i

~

.0. % n u.s. 13.s f hM M4

= %oo xt.i- %foo pd 1%, '

t~

A 4 i.2 %A p sa g(pa), A

.J L &

LA A p

.a i.s m.,n. a,3.t 7 (L us.vas e A, 4 % %yv 2. My.A ah&.

bA @ A,

k. aist,3 a.~ 4 m A 9 m. Tib x A, li xo.s xt 5.s 2i.2.

L d

a r3 an.te x ai.x. seu b

in Cah11 A A4 4 k

h A

y (6 e.aL T

"g v

A h 3.ypy.2.T, {,

4 7,,,y,gg s.vas Ob UN b l bl Acals. !'

d Ali, D,

& W f M.

E Jack R. Benjamin & Associates,Inc.

Consulting Engineers 9

CY

ATE PMDJECT PAGE

SF CHKD. CY nATE EUDJECT JO3 NO.

a.2 6 L L,J, L ty d t

9g S W 1 % 4 L A d d. p A g L Ax, A aldi. q le m+ % Q L cJmL,%< %,

.h.

L.

wk Q. %. % L uA Ra. %LL f i2 ag+

C=~ak W (3Aw<ad b 9 %,ka. w p 4 p.w).

g TL /A JJ.k aAd M @ Ahe~ p aa, 4t) 4. &%

W gL. M ~1ty-bf.a.cAA4 =g t

la L emb a e.u d.Ait ( M L d=.).

Ss a w a c4L,, A h

.E_x 6.Sw x4

)

mb u

a 2-U A-M&t43 m mJs (pu. Aht4 3,T9 i

6s.

. o.1og (W% % 4, oAg TM)

% g a 4. is i a

w. i, A x e.1x !!i. i6.idt, k.l.

m%e _L8 W$WWh L

T.c. N 2d A

f,= ss.ais' d A %

~4, E A a 2 p.

O Jack R. Benlomin & Associales,Inc.

ConsuNing Engineers 9

1

CHK3.CY MATE CU JECT JC0 NO.

I g

T. c..

Inn,,1.h51 4 Msv.315 Lbu a - p a q4 (ad &

Apm v, Es1.o.sezeA

.e

k. son M.L k. us k.

W

Ie v w, N & LL Att m.

.w A*. q,,L t,1t d6th t

k )'

4.

V v

l l

g. o.92. g A,.,.bzx k s xe.17. a.es 4 04 A t= t.JLo A ru_a. K,q, g 4. & 7

{. s.ow. c.us e 5b2.

I i

M dd b b l kl A4[4.

MM /w s

l i

i b

b

}

j i

@ Nd A 0.551 M di M W w "...%

}w bbW MA k m. k b b M Cow., W (t A : L A A A t m ). % e, % A 6 a a

.b.

~

1

\\.\\el s..

d?='0A k 4&

1 l

Jock R.Benlomin & As E

l Consulting Engineers

CY DATE PROJICT PACE N

CF CHK3. cY nATE CUCJECT JO3 NO-t l

k kdk M.o v

9 a

I kk b %D N OM kT 6 M6M h

h kW W

'& b N

N

)

y

(% g u a g-4 a w e.

I h b "N d

M b

1. y u p s a u g a & J g 1. 6a s As a hu a &

p p y g && u.~L.1

-)

pN& w -

uA.

2.

s A g

.a p A L q u e, A (yoA). (p (p o.9 A e.9 6 A p A w A 4 4 & p a u A. pN k M NW(4 A A

hd le & $a M h Uh-e o lutL 4,A q 4 4 h uJ u b Ju4 A W

P A f%& A4 A 4-VJ s dA.i4 W l

h&

h_

u h

Eb M k

>$k84

%Q %WM L " %MW-I Jack R. Benjamin & Associates. Inc.

Consulting Engineers 9

N CY

" ATE PRDJECT PASr

'J F CHKD.Cy nATE SUCJECT JO3 N3.

Lf AN i

s.\\.t k Mf fu%,&, o a% M uk & Q 1

- L u& L ~ wus

%.n a L. A L d

AA A sh z.o d s va a o.3.

'p4AAJLkh Lg b@b p L L q=J (LA) p. t M

lA (fkb bdd 2MS, %

f tj O

b'To\\N\\%)

f,.s. i,(3 a.s Ay 0)

A wLtd g.- % 4 L (A1& L a6JA 0 a K 7g w L A d a p p % Jr, q q L p t. E n-

- & p ua A p. ut qu u w a s s.w.,. $ N NY9 s. % %S 09uLLSaysua g qLp4 W u 4 M GL o 4p (p kg are d Ae,i ut), p%(pt) d P y&w % a (um 4

(<. A c e a # J % 2.sso.I,a. L ;

~ l'r" 2.sso,s..*ka.s k.i.u13 1.....

a db N hA M, Mdd w ik. dd bM

~

cus;ay + g g.s a K s w w su a,,yz).

m E

Jack R. BetSomin & Associates,Inc.

Consulting Engineers S

w

CY

'5 ATE P2CJECT PACE' CF l

CHK3. CY nATE SUOJECT J33 NO.

i h

I t' q,g,,

vy om %a % e aA,at 4 4 p a;L a.

el Q g ( Add h cAkta.

+

~1 d &&. A, 2o h A Lt b p w h i @( % W L t J L L dd pk M %d.

p 3.6 e.eins ( Aa )dt,uld 3,Ty2e,Mp).L;,

h,#v#. 2,x s.oi2.s o.asi644 cm 9 b ndL caA L. (A,u Abddo,Tegn),[( uso cads 1J 1.0,e.

(. 1.42.

AN k ckSL f 4 L L.uk.t.L f

L w p 1. p t & A 4e & M L p t.

Au p aop)p, r

= l- @

b "#'hN51

$'T

\\'*

M d

Q g,s & k L r*tJf.* &,

NM b Mb b b bk s Ud

/WM

~

b W sft,uA 4 s,Ty n.)

h,,,.ussi ad (.s.n d

I Jack R. Bankmin & Associates,Inc.

Consulting Engineers S

l

M CY DATE P33 JECT PAgr SF CHK3. CY nATE CUOJECT J33 NO.

h t, E t b % 4. A 4 - g g a A J <afL p 6 a

,.e e t, its J L A 1 a.a 0.isstie.. %, L.

e

% AL u.

6 4. (ht

,b.ssv,31) 161x o.69o. t. tot D 6

Au,f, (Y s% (6.invt). i.qtx e.wy, t.2,i, D fL b M1 bid YdM k

,5:

6

\\.14 1.%s.,

l I

i I

8 I

i l

I I

f.it?

V.31 c

e

$1(A.)

o.t 0.7.

AS A+

nu y " 8.814 ~ }'(t) d P *~4 5 *k*

0.tte l

l*4k W1 Mp g3 q b e.34 s.\\W 6.

--, 8.lS gd d 3, s.wz 8.Wii' E

Jack R. Beniomin & Associates,Inc.

Consulting Engineers D

2t M

Estimation of the Median Ductility Scale Factor, Fmu

. The median ductility scale factor, Fmu, is obtained by several methods (See Reference 1):

The following variables are used in the analysis:

fas = 7.5856 *Hz The A8 Bus frequency b:= peta b = 0.05 Elastic damping Add M 6)

&:= 2.51 Response spectrum knuckle frequency g

sec s = 0.13794 Strain hardemng ration mu = 1.67125 Ducedityfactor i

i Riddle-Newmark method for Calculating FuRn i

j The detads of the methodology are given in Reference 1 12 first calculate Fmu at the peak ground acceleration level (zpa)

The zpa valueis:

zpa :=0.4 g and the spectral acceleration at the fundamanf at frequency, f, and clastic damping, b,is:

Sa := SSa(fas,b)

Sa = 0.65%1 g Fmuis estunated as:

Fu4 := S.mu' l' Fu4 = 1.74485 rpa Then, in the acceleration range of the sp-amu, Fmu is a='ima'aA as:

Fu3 :=(2.67 mu-1.' 73)'#"

Fu3 = 1.52439 l

6 l

Finally,in the velocity range of the spectrum, Fu is a=+ima'~I as:

i f &

i l

f & tfm <l.0 +iI 21.0 Cf = 0.32957 Cf -

1 f

(Ins; i ns 4

t ns j

Fu2 :=(2.24 mu-1.24)'#38 Cf Fu2 = 0.5774 i

23 5!

The median ductility scale factor, FuRN, using the Riddle-Newmark method, is obtained as:

Fui := Fu3-(Fu3 <Fu4) + Fu4-(Fu3 2Fu4)

Ful = 1.52439 Using the ratio of ultimate static capacity to yield static capacity, R. Mned as:

R := 1 + s-(mu-1)

R = 1.09259

,(Ful-(Fu2<Ful) + Fu2-(Fu22Ful))

FuRN = 1.3952 p,

R Modified Riddle-Newmark method for Calculating FuMRN Because the Riddle-Newmark method does not account for second slope of the force-deformation curve, the effeet of second slope is accounted for by modifymg the ductility ratio. Therefore, the moddied ductdity ratio is:

mu g := 0.5 + I""~ I)'I * ) + I mu g = 1.50718 2R l

At the peak zpalevel,Fuis:

c.t1 Fum4 :=. *.mu 3 Fum4 = 1.72513 i

i in the acceleration range of the spectrum, Fmu is:

Fum3 := (2.67 mu 3 - 1.673)'d!'

Fum3 = 1.42101 la the velo.;ity range of the.W um, Fmu is:

Fum2 := (2.24 mu g - 1.24)831.Cf Fum2 = 0.52402 where Cfhas been defined previously.

l The median ductility scale factor, Fu.MRN, using the moddied Riddle-Newmark method is:

l l

Fum! := Fum3-(Fum3 <Fum4) + Fum4-(Fum32Fum4)

Fumi = 1.42101 Fu MRN := Fuml.(Fum2<Fuml) + Fum2-(Fum22Fuml)

Fug = 1.42101

'A% e Effective Riddle-Newmark method for Calculating FuERN Because the modAfied Riddle-Newmark method only accounts for second slope of the force-deformation curve, the last correction to perform is to account for ground motion duraticm.

The factor to account for earthquake duration is:

CD := 1.0 d $t 1.0 Op'W lo A E4 % ( M b and the median ductility scale factor, FuERN, usin e effective Riddle-Newmark method is:

FuERN := 1 + CD-(Fug-1)

FuERN = 1.42101 Effective Spectral method for Calculating FuSA In this method, Fmu is e* lenta'~i using the effectzve frequency and damping ratio of the structure. First, the ratio of secant stiffness to clastic stiffbess is Wa'~i as:

Ks g := U Ks g =0.65376 mu Next, the ratio of the secant frequency to clastic frequency is eshmated to be:

fs f:=}Ks g fs f = 0.80855 Compare to:

fs := fs f as fs =6.13335 Hz f

=6.13377 Hz f

nsp ne ratio of the efrective frcquency to clastic frequency is calculated as follows (See Reference 1):

cf := 1.9 Coefficient to account fo, -hxt duration motion Al := cf.(1 - fs f)

A := Al-( Al 50.85) + 0M(A1>0.85) fe f:=(1 - A) + A-(fs f) fe f =0.93036 ne effective damping ratio, be, is calculated as fo? lows:

Cn := 0.15 Coefficient to account for short duration motion bh := Cn-(1 - fs f)

Hysteretic energy dissipation damping bh = 0.02872 fs t:

)

be:=

f

-(b + bh) be =0.05945 I* fj I

27 M i

Tbc median ductility scale factor, Fu.SA, is wi='M as follows:

G2-

. _ ' _ ^ efrective frequency, fe,%

fe := fe f,,

fe -7,03734.Hz f

l l

Sae := SSa(fe,be)

Sa, =0.59505 g Sa.e is the spectral acceleration at the effective frequency, fe, l

and the effective damping,be Ieg)* Sa f

l Fu SA :=

Fu SA = 1.46763 t f)

Sa, i

Final Estimate of the snedian ductility scale factor 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 i

average of the ductility scale factors found using the Effective Riddle-Newmark method, Fu.ERM, and the E*.fective Spectral Method, Fu.SA.

(v

[p = l MD1 j

FuERN + Fu SA Fmumedian = 1.44432 Fmumedian "

2 l

i 1

)

l J

l

CY SATE PROJECT PAte' Ab cr CHK3. OY nATE CUOJECT J33 ND.

n.\\.s 6 M L b.p 1

l A wo fM&W % u 4AAL %

L u.LL%.% as 9.w. m a k JA Ah,A Js aJ w y tr p muJe =. % A % Ltd e &

l A qL.tA.cAEd @.

MLd, L' 3.e.tz(Aw(

. as L.), cAa.1 s

s

( a ( p v (-n), % 2' 39.a La.

l g

/* N L u A @ GA v.as.or. L.

b r,. i. wit w 4. i.m

/5 A thisxMY\\. o.os A

l kh %\\.%3 \\

I f.

L% L wLA L hafl f \\o %

y

. A <bh Aco3 au~.~d,.e at M, 4 0 A 4. J f

j U1 h cm. (.-\\t).

/mw.At.4 % @ A

( ( a), t.as1.

~

I Jack R. Benlamin & Associates,Inc.

Consulting Engineers 9

V9 Al M

Logarithis ctandard m.victica af the

- '-~

The variability due to random scatta of time historymw Fmuversuspredicted Fmu valu:s using yrh methods (for example, the spectral averaging method) is:

l brFmu := 0.4-(0.06 + 0.03-(Fu g R-1)],

brFmu = 0.03124 7 = 8. 0 3 that the 10' elastic deSection is at 1.0 standant deviation from the mean, to be 1.25673 using'the same equation as above. Therefore:

l fFmumedian In

( l.25673 4 buFau :*

Fam4N 3

1.0 b@

l The uncertainty due to inelastic energy absorption model is:

l bu Fmum :=01-(Fmuw,,n-1) buFmum =0.04443 i

The combinedvariabilityis:

t

8.16 bFmu :

brFmu +buFmu +buFmum bFmu =0.14935 i

.i e

h

CHK3. QY CATE CUOJECT JOD ND.

hhtt A.

3. 0.\\b b

~

hb

.N,M A

. 0.\\5 Nk,

3 0.L5 pb *M n Nu l

i t

Jack R. Benlomin & Associates,Inc.

Consulting Engineers S

CY DATE P;iCJECT PAgr 7_4 SF CHKD. QY 9 ATE CUOJECT JOO NO.

>\\.s %,,. 1 % A ~ b h J b

% 43 qu w e&L M 4:

(

.sML

\\

q

,wAwkh

. 44.k yA 6Rh 9

. c,a su is 4 a ka %.

l 64 &

o.1 64 ~%u sh5k, LLk

. ail p % h % k g w. E ga W Ap

. QTA

' M Y Y..! '

A& A N W A kH d %LTd w kJ 3111 n

t i

i i

G i

d-b N

Ykf f W Shk

, W $0 i

% ko.aa d s.m,

. h &Q, a (L. o.2o E.

A& u an,\\.O.

Jack R. Ber@ min & Associates,Inc.

Consulting Engineers D

OO cy nATE

'G3 JECT PAfr CF CHKD. OY nATE surJECT J;O N3.

6A e o gg m.sw(tk L,.sn 4) d W L wx

& k 2D-(* w e A A. w adL 4 3,...xs 4 t.o.x,ca s A

\\

Q b

l

[3,. o.za

/ m m % d % @ ( A b k 4 s') J g 6,. g e t t J "

(s. s.3 3 a

4. i.m l

I M:

1 p, g 1 I15X L %%% _, o.22.

t

['

\\' sos x w tA A...u NMM 3,. l. AS

(

3 ose s o.22.

o.nA l

o.6 s

l l

Jack R. Benlomin & Associates,Inc.

Consulting Engineers B

O!

CY DATE PRCJECT PAcr Cr CHK3. CY MATE CUOJECT J30 NO-N% N3d dd kew d

.o.Al, M d ( d W

U

\\

[ e M e ta.h b ~. h ( M a d d %) M 6,

,4. M

[

3 J s,_, & A Q l" >

(,i.ssu 4 (. i.ev I

i b

(b, L }'UbK

. o. ole l

t.sse x t.%t, i

I i

ll."A.\\.K Nt b e d M NAd

~

\\

V kMb dk d

4 h

$3 b

NW W

U W s'o a. 0.0\\ &

u A p wa b L t.s AA $ m y o.ss.

le n&W tb O (AE J 9 GJL 6 Ad 1,.ss t,.

{ < tab %

= \\.%1 j

g,g

\\4% At.4%%

< 0 03

. L.5 % xt. % T i

I Jack R. Benlomin & Associates,Inc.

Consulting Engineers 9

CY

" ATE PMCJ2CT PA;r dl gr CHK2. CY nATE CUOJECT JD3 NS:

i 4.41 Idd bed LowMb

\\

l

% unho.Ao A e & L k k u o. & 9 8

%N N. 00$ f & &

n 100 \\t8 t00 A Cn y

(W W %.k h o L A., M.& N 3.0 T l

da, s u LA # A 2 7, L A A L). (+ nJu d Mh 4 (c0, loo b0 d ;

4 I

0.6%\\0 O {p

\\.%\\0 ga I3, 1 ( %D7.M%%,,,1y G

o.4%te M.Elo hM A I' D I

l l

l Jock R. Benlomin & Associates,Inc.

Consulting Engineers D

l l

CY DATE PR OJECT PAge SS g y.

(

C H K D. C Y_.

nATE SUOJECT J33 NO, l

li.'2. 3 k Wk%'6 iw h(

A w<., L M L kkh h- ~A, a $ p h q A ag 4.tk

.1 h by% d N 4A bd A.

. e. 0 5 A 4 50WN.h &

.4 \\. 0.

4.2%

h %%%M 4 % ph :

Vd w-n.,R_ _,b

h a

m egr aa s%

i 31.u M A s h L%

Ay L L.L Q q d g o p

\\5

NY 0.05 h 6.t$, $~f: %&

l b

d AA b

.W 4

>b,e.ese d.% 4 4 \\.e.

I Jack R. Beniamin & Assoclofes,Inc.

Consulting Engineers D

J

b CY

" ATE PROJECT PACE CF CHKD. CY nATE CU3 JECT J33 NC.

'o A 6 W [ 4., h 4,1,s.2

% -h.s h t g A 4 a.s A A w.a.

I y

% sk. L L s Q se p k%%

aAe. a M a. g - o.3 A ei tr.

L w,E Q % @ ( W A 4 Q 4 g.o.e5

(,.s.wn. a (.i.3ss AJ s

A i.umtms...n

~

\\.u n.ms

%uk Ea \\'O.

g.2.s.s M 1

0 l

l M&

L&

&p

& w N b ]>e-bb IAbph+

bh Ay IO

~~~~-~% a dAtia % AJ x g qts L aaigr.

h&

A b

IP N9' s

b to x \\o,, 2. lb N6 Q

tb

[+ NLt ) % @ (ALculd9 GE a.

eg IS % d

._ _. A hh4 T ID Of-)

,3

L 6 6 2. d s 1.'b65' Jack R. Benlomin & Associates, be.

Consulting Engineers D

CY

" ATE P3OJECT PAgr bb CF CH K3. 3Y DATE CU:: JECT -

J33 NO.

A4 r3, g i.tsu x1.vm

,,,iq

\\.0bu \\.ss 5 k&

do\\.0 4.2.8.3 AA u l

% cad ^& 9 '

, r s'j, LQ w.o w J y L.

adip. A A L y.

a $m A. t A. e b b ddl

.ht x a.3/, b a d tr.

D

!A% DOld b @ ( Ad ad.d i) d A 8.5 /, h

(= l.Haf ad (, l.SYi kd A

wn n.w 4

,,, i3

\\.%T.x\\.%fg h k f 410.

hi.S km i

u 1

bM l

/

% h d hr A 10 ll$. k A b b ens Ay i

l LL g ca ~ s4 x m

.w ge 6.x.

Jack R. senlamin & Associates,Inc.

Consulting Engineers S

l

CY

, SATE PROJECT PAgr Ob gr CHK3. CY nATE CU' JECT J33 NO.

l Ohh dM S) d A,

, b

+ -

2.

D U'k )

4 f, \\hlte M

, \\.515 3

LA 2,L

\\.u m l. w \\

, o.ov 1.m x1.se 4

% & f la 10

.N 1.4.b bb N

'\\

)

h idtd O.tw tdd

' b d Ackdd L y A g a &~ x u p m m 44km p.J A L w.L 4 A % d e.h pt k dfb'k D)

A b

l Ct.

i b

l

/

.- e U = T xb

.N % a.

p I

Jack R. Benjamin & Associates,Inc.

Consulting Engineers 9

CY

" ATE PROJECT-PAge 01 gy CHK3. OY nATE SUOJECT JO3 NO.

h N

WM S

Wh & &

w em u ~ A p _\\r.

I

$ tsb W 0 $ M &

W

)

a\\\\%

a 1.408 g

AJ l

3. t i.v a xi. m o.n

~

\\ Yh \\.10$

E.

Jack R. Benlomin & Associates,Inc.

Consulting Engineers 9

i

CY MATE PM3 JECT PAgr OY gr CHKU. CY HATE SU3 JECT JD3 NO-uu

+Q A &L& dud

.&. A J ys y am a ww w L & u tAaM &. a mt s 2 A mh A ODD b bb kal w&

4

/!

C#+dh

.s b

'h C.O $

hkk lM o.os e.I6 O

l.o c.ts bM %b l.0 0.3 c.lf MTd%

l.0 0 010 YN I.0 0.03 M

l.o e.l@

MdbbM 1.0 0.05 wm

' 6.R.

10 1>?'A

\\.o e.\\S

~

W.

g.o

c. tS\\

M l.0

o. ll b

lO

$.ll MM h 10 e.oS

%M pg o.19 0.31 Jack R. Beniamin & Associates, Inc.

Consulting Engineers

M Cy DATE P!ICJECT PAgr cr cHK3. CY DATE CUOJECT J 33 N O.

(* L w (%kh4dLpb, A W A oAg,

N b

c1

(%h) = 0.S%%M = 0.910

=

oh) p = 0.b1

>0.JA.

1 a

b4 b bV3Atk b = D Cli f (YG h = 0.% E B e n f t v )

o.st g h p,1 M M @ L% lb b 4 aa pe44 6 e daeyl M.4zy; a >J au(q.aps)%L L u A.

o,A-aL 9,g n w a s a -

p d

l I

f l

I Jack R. Beniamin & Associates,Inc.

Consulting Engineers B

CHK3. OY nATC CLCJECT J33 NO.

5.0 $%

\\

I

%L G %

Q s k w hhiuaL % 4L %A

\\.

f A k R. h p 't M& bgay Stb swh b A S, add A k 19, me.

2. " b A

) %i q sb b i Cwu 1 m Q<AiwJ Q.stut," g n. A.J L. ud, ek; kub \\M e

O.

_MS *Is Sa 6 d," A

% e g 6, **.7. x 4

.w. ud hb,% W, cA, odA M93.

sata 9 4 abp 4 04/ Mg,.4d.,

9 W A L a w, A u s, w g m s.

l l

l I

Jock R. Benlomin & Associates,Inc.

Consulting Engineers 9

CY

" ATE PRCJECT PACE

, _ cr CHK3. SY DATE CUDJ ECT J3s NC.

i i

i l

b S9ct NJabum %,b l

I I

i l

j j

l 1

I Jack R. Ben lomin & Associates,Inc.

Consulting Engineers B

l

__.___._________.____._.______________J

%l v Mh

}

4 Pictures of Pilgrim Station A8 bus amMMMustand 7

c f.

t' g.

.(.

},~

Picture i Picture 2 y

.s.o_

l s

~

?

^

N

]

p 4

l Picture 3 Picture 4 1

1 Page1 L----- ------

Pictures of Pilgrim Station A8 bus -

t.

?.

1.

4 I

e 2

~

t 4

\\.

3 l

1 Picture 5 Picture 6

\\,(i

\\'

.k; 4 J:,.

i i

i

,. l (I)h

.(j i il l a. ?

t l7 -

g, l

..l.

m

- r,,.

J l

i.

1 p

p e

a

,e g

er 3

Picture 7 Picture 8 r

i l

Page 2

Pictures of Pilgrim Station A8 bus ~~

\\

j -

k,..'(,

e

.{,y e

=>

1.

6

.P 1.

g-l '. >

-i l

ji l

k i -

,I e

p

(\\,-

i J.

i l

8

}

X-

, p,.

i :

y.

e t.

'j I

.-. 4 [j,.

t

. t

- i Picture 9 Picture 10 i

n

. i i -

_ i..

-..,.v,.

I g

},

[

~.

/;.

....i t

-1:.

gg -

l g,,

- intifli, '

a l

I Picture 11 Picture 12 I

Page 3

i Pictures of Pilgrim Station A8 bus aushemmengd j

l

~

9

\\

m.

.'~

l

,t i

j fff c....

a_

.p'1ffy a I/},

n..

I.'

.s

{

ma:n

-),.

)

Picture 13 Picture 14

. q),

.d,

1. t.

lh'-

j l

'J il I...

.:l f,) '

- )

i

.j I t I '.

~

l j

8,..

l e

4 L'-

se L' f I i

,l i

l l..

l f,

'),

s I

'I i,

j f.1

!, I 1

Ji i.

. g

[.'

?

l.

z.

1

Pictures of Pilgrim Station A8 bus WSommmel

! (l j'

i l'.

ffh l.I I

s o

..i

i

i l

I ? ".;]

I, i

I 1,.

t:,

\\;;)

t '-

A '

3

?

i~.. 2:

I.'

i.

'lM

)

l i 'w $

4)

{1

.1

.i 1 ii

.. j u

^

g)3 i;

'" n i

,_l 1-

~

4 i

[ l _.

l.

Picture 17 -

Picture 18 (1

~

iy 11 l

Picture 19 Picture 20 y

l4 y

Page 5

CY MATE POCJECT PAgr gr CHK3.QY f* ATE CUOJECT J33 NO.

\\

Mb0 Y bdk "^' kT bd b

m Jock R. Benjamin & Associates,Inc.

E Consulting Engineers D

a p

Ii.

,e 0-n

-n a

c!

l ipg _E lll lI l1E Jg

.. i

! j! ;

l

" -H a

q j

..'F1 ii i i

t ii i i' 8! I

. e g

4 s

a

\\ L u. M4sl ese leh j .4 8 544= M G54 e e U U e 7,_

h=ff i

j. g r:

==

===;

g-

,,l 1

5.l i

i 1

s ei 1

l

(*

i_= = = = = = = = =.==:

by I

3 1,

1

]

l o

as

ggl lG

~

g- ((ll[l[{ [

,.',D N

.h n

{

1 f'

M.N lp. ; (tI l,'

g ga g,,g

$ _Q m.

M U.

[

F:3:

1 1.%

,.Ip 3

vi6 i-i la h

IQ.hf A

k l$

,,...)

.~.

..t l

( ' s a. ?,,.

"~g m_

_ m _._

'.l.

y l

t 3

s j '..

t l

H a.

\\

E E [,r...,

l f '4h,l 1

I I I l

$8' 3 Y ~~T g

'l t

~

lI l l l

s y

i s e M $i :[ ]

' NM.hl' q'. :--

f (

I ll i i s

i tie z '@g 5 Se

.h E

<a.

E N

N A y y'y p y l

i,.

's s

[E

.w 3

's x

n.g g

g m s-

. o o

g Im A

g '6F i

=

_d ] l "n

i 1

H 5

ut M

1 ia._.

~'

e li ______ __-- I I

T g.

sgg

[L e

g.

e l e

g. ima i=i c i 6

g 88G g

i-itblic; i

J

..s i

o o

,I d

.l

\\

Q M

~

i l

g M

9

==

O e

'd : h.

lIN e

a m-

.L e

.~

ie

. W... ;. '-* '..d.Rp..'.;r

% - @ d__m_f__

g gu gg g,[ji

. f.;,,.;:'pg" g; s. 'i+.

P T

7 g.*

4;,,

a ggg x..,... t 5.

-e i e.- a"..,.s.2 t.

]

~[

g '){

E= 1, ;i 8llEij 4

I MG

[t S

I g*[ l l.$ >. $

[

Al t

l ". r'4,;..'

%gh w ',

g 4<

7.x e

.7

~t ug s!

r k

uL.

.s. a u

c.

M.

-..4, gem

.M _q l.,p.,,. y;,f.pf,

.c_,_-

=

. --- e

..p

. thy J.r,.4 [::.g.jl e' ~ ~ --

W a.-

CY MATE PT4CJECT PAgr gy j

CHK3. QY

" ATE CUOJECT JO3 NO.

l l

d1 l

l b

U l

l i

I l

l.

E Jack R. Beniamin & Associates. Inc.

Consulting Engineers D

l l

e ANALYSIS AND DESIGN li.. g

' OF STRUCTURAL

"'2 neogy, Uni-CONNECTIONS:

Reinforcec Concrete and Stee:

etiendon.

M. HOLMES, BSc., DSc, CEng., F.lCE, FStmetE Professor and Head of Department of Civil Engineering University of Aston in Birmingham and L. H. MARTIN, BSc, Ph.D.

Readerin Structural Engineering Department of CivD Engineering vdty University of Aston in Birmingham 8

rloo.ontasio.

i Engineers.

ELLIS HORWOOD LIMITED Publishers a Chichester Halsted Press: a division of i

JOHN WILEY & SONS andon.

New York Brisbane.Chichester Toronto i

O

' _.5, ~ ? v'.E i ~ Fi f [, 2

M i,; g.f -Ensi ~

~~ ' '

%'.- r

_hTNKGd. _:~. A!?*. -

M__

L (Ch.4 Sec.4.13],

Elastic neory for. Prying Forces

145, the tee shown graphically in Fig. 4.10..ne tensile forces 1n the bolta adjacent to the

Sange tension flange of the beam sre approximately equal.ne preloaded boltsadjacent.

tubs are to the compression flange eventually resist part of the bending moment.De in con-Prying force in the elastic stage of behaviour may be greater than at ultimate snental load and depends on the number of contact areas.

of those 4.13 ELASTIC '11IEORY FOR PRYING FORCES ry thk, ne bolts must be designed to resist the external forces plus the prying forces, e contact and it is therefore necessary to develop = thcory to calculate the magnitude of ents by the prying force. For an allowable stress method of design at service load condi.

k stage tions, the prying force Qb. is related to the extemal force F., assuming linear attom.cf clastic behavim of the components.

ne theoretical:rodel in Fig. 4.11 shows an end plate of thichiess t, and

' cree in of cantiever length (a, + b,).ne extremity of the r*ste is la contact with a e in column flange, for example, and this introduces the prying force 06.. De strength extemal force F is assumed to be balanced by the prying force Gw and an axial force F in the bolt.The axial force in the bolt produces an extension of the b

bolt of 8. s b

Fe f

l /

o O

/

W

" ~--Oor\\(n 9

t y

Tbt" Fe + O e l b

b

p p

\\

Fig. 4.11 - Forces acting la the elastic stage for prying fores @ary.

Applying McCaulay's method for the deflection of a beam with the origin of O and the 6eflection positive downwards.

day Elp = - Q3.x,p (F. + On.) [x -a,].

(4.29) sb.

Integrating cen-El

= - Qbe

+ (Fe +Ges)

+A.

(4.30) l

_ine-don de 2

2

146 Bolted Connections ~

[Ch.4 sec.4.15]

Integrating Obu *

+ (F. + Qw)-[x - a la 8

x p

+ Ax + B (431)

Ely = - Qe, 6

,o where F, =

whenx = a, + b,4v/dr = 0 and therefore from equalion (430)

. The strain in p

(a + b )S b8

[4.20], and the s 0 = - Obe

+ (F. + 0be) P- + A -

values of enu p

p anc 2

thickness plus was Rearranging The width o:

b:

Ad Y3) 13 defined and is al A = On,-a,(a, + de),b,) - F. P-column connectk

[4 g h pA 2

f tions in a colun 2

f.

when x = 0,y = 0, and thsrefore from equation (431) B = 0 whenx = a the theory [4.22]. s f

whichever is the I extension of the bolty = -8. and from equation (431) b b = -Qw I + a, Qw (a, + 2b,) - F.

-EI6 6

. 4.15 CONNECT Rearran@g

. Some connection F. - 2EI8 /a b diagram is gener.

8 6 p p W,and d n 2(a /b,)+(2/3)(a /b,):

p p

Th. cxtension of the bolt originally preloaded with a force Fe, is

'?

are applied to th stHT bearing nes:

force H are resis 6 = (F ~Fw)fp/A Ed = (F. + On,-Fw)fp/A Ee. (433) 8 bt b

b fasteners, e.g. bo Substuting equation (433) in (432) and rearranging F.(1 -k.) + k. Fe, 2(a /b,) + (2/3)(a,/b,)2 + k.

p where

k. = E, w, tp t,/6a,b,8AeEb.

(4.35) s This form of the equation is appilcable in the linear clastic range of behaviour i

and relates the prying force One to the external applied force F..

4.14 ULTIMATE LOAD THEORY FOR PRYING FORCES

,I At ultimate load when the bolt fractures it is' preferable to relate the prying force Ge,. to the ultimate tensile strength of the bolt Feu. Equation (434)can be expressed in terms of the bolt force and strain in the bolt, when combined 1

with equation (4.28). At ultimate load the equation would be l.

i i

I 1

1 lm. t.....

., _.. n.. _ __, _.u s.

w/ det.-e

f' J33 N3.1

-0 M b I

CHK3. CY nATE CUOJECT F%)w 6

l l

-aks b o.Wvt gp pos51N.wtr p

u

]O)

M VW I t.

WO

(

Qp $%

9g + b ON A

Y L

srIf. = _Og * (Fc. Hal(X o )

Er AN G2p r

i c1#'

dc

{

1 t.::.dM

&(Ect%)(X*[l 4

Cg=

4 C cLs s

La s

b J. 3, = Ch deh)(ped,Ag.6 ETs = 4

+ Cx + 0 C

b f

b\\= %

Y~

~

t v

- A af + A =. - b Q c

.e 2):

A=t 1

2

+Aap+0=A

+ Aar 4 p c ep G :: O 6

l O

Jack R. Benjamin & Associates,Inc.

Consulting Engineers E

h l

CY MATE P.7"JJECT PAfr gr C HK3. CY nATE CUEJECT J33 N3-w 03 x = mg + h d y =c

___is l

o=_A h+%f + ( R4Qw)'of + A A =

Cdkc " f *f + 'd ) ' E N C

f l

z_.

l 4

cd xco

%=o

=7 D = 0 4 6 =- o

(.hn~_. No m f fuA { 14cb 4 MA M

\\ovN O I

Idow M ceuu M

(nd m) ts Th.)

b d

%i (5 N 14d CkN M

N M

YO Nt )k 4

'd 0

b Y Q S

5"g)

\\'

' c'O C4H4.

dx Me =

Ca<.( 9 +tp)+(h.+ %)b y A =- c_.

4 B = D =. 0 l

d hv

'6kows NW kN l-{-

i< W

+v a s s a~.

@s % 40 Au Jock 't. Benlomin & Associates,Inc.

Consulting Engineers 9

1

CY RATE PROJECT PAtt ggr CHKD. CY nATE CUOJECT J30 NO.

d'

% = Ge + h 4p; e A EI @ e = - %e f ' d 4- ( Fe. + GAQ

+A A = - EI9

+ 9_b-a t ap + 7L b )-- Fe 9 C

e c

L z,

\\ M bok

'j " - kg N O Ey V

/

+k 9+

Ap(4(t?. b ) p,1

~

4_

p be

~

Ak 2 hf j~

t

~

~~

~

~~

~

a-a

( f.

? IDEA _ 2EI$g

'(.

be*

o.g97 g

4 3

4M 7/.

~5 st sr

}.y^

s Sb Ep }

~ 2 E.I bC Y

Qe=_

b 2("f/b ) + $ ("%rP f

y0%.)

N %

0

s,=. ( F 1 + c A a a e O

Jack R. Beniamin & Associales,Inc.

Consulting Engineers 9

CY 9 ATE PRCJECT PAgr gr CHKD, CY nATE CUEJECT J20 NO.

sdsMudy h%

2C1 I 4 R +Qb< joe

~

b=

b' "ge

~

A Euaq J

b p

2@f/br)(2/3)(Af/brP-w K

I 2er 2U0 f It. ~ y f

4 g

be K be Abet, g g

~~

Df

_ QE l~.he, ff Ah%*r.,

y Q

z w-k+

2 C 0f 6

Ag E b*f p

i px

I x *e' %

\\2 I

2N p Wh Of

)

~

67AA^t (o ae br-As ers l

Fe_. (',1 - ke) ~ 1NNf j

62-]

g f

- be _--

2( AP/b )+ ('dS@e/b )* 4 ke_,

f f

E l

('

Jock R. Beniamin & Associates,Inc.

C I

Consulting Engineers D

I L

i

CY CATE PROJECT PAcr T tr CHKD. OY nATE CUCJECT J 33 N D.

NU {

l l$

+ O<-

vu s4p3 & +c a

sne A u b -

1d % A me M

M Mt =O

%)

em A A =. o M + bd - (R 'Gb_.) log =- o G2 P

A A(YN N) = Mr Q"

C. *

~(

sMhh idu 4m

e. 4 4Sdvi~3 hoc i

l F (t-%)

aew.

bP Ft. h 4

_~

"f l

yc, + w.c i

P E M g = g(g_ g} _(g,q q q hg eg-ar

= Fe Ti-te (t+}.eJ _ x L {

_e y

me be.

2.ET /Lg -

S N

Jack e. ser*nnin a Associates,inc.

Consulting Engineers i

________-_________-______L

CY 2 ATE PR2 JECT PAge gp CHKD. Dy nAyg CUCJECT J33 NO.

hk inh I*'

\\

R., kt L

hJ to A cw~s4 te %3A a

oh b h

(

V

)$

s I YD M y

kPb G

L k

q t

k t,

"f 2 Ekbf F.am ld p a p3 Y G L ((cc + t ) + 2

N '

l bc g

gi

(-DL ft 4

l l

l I

Jack R.% min & Associates,Inc.

Consulting Engineers 9

1 I >

tave N

eng;cc7 h

PA2r

__cr cy Af bn k m un. LLT - 0N O j

~

cmo. cv nave suasect i

b\\)e!ca?

/ q{

Y wim o b

Fe 4 v

j F OM 4 3 dA (b

\\/

uV

/wdlu-u v vM v

mf ON L d-,sM www towue ydh 4, be)

MSS M b NS Sh Y

Sh

( No ythw) q Usa vs, u% o 7, r N, (M l&)

c 3

Q

%== o & c

%,3 3

\\ f%

%\\"

%. * / (,'l o ) s V4 Y N*

+

3 s

~

/

qt; mL I

Te Fm (V=10N hN-s (C.6 $)(10)(ld9 / 2.)

2.

g

~

b 'Mb (91)(2) e.z'

[*

Pe m (0 6O(lo)(109 /L) a m

2(% + u a<.z) tu I

h vd b+7~7)[lO)(l[t)

OibT)

T

=

t

~

.\\.u + cm (1.TNo.al F5 L-k 1

4

% A tsA 4-

~ 2.43

~

E Jock R. Benkmin & Associates,Inc.

Consulting Engineers B

i J

L CY

ATE PR2 JECT PAfr cp CHK2. CY nATE CU' JECT J30 N D-b'h EN M ho \\h

( Ub<.

d%

05) 1,f,

i~

W W)

. V

= 7 (. 6d FS m

1, 6 3 E T S e

1

=

2. e 3 w

~

_,_, s k.

1.63

-=

055 W Q,.+(,Mk_

/,

S ) 0,8g' G.

(

lione d 5t; Adc) -h Fedd- ( w q v) ae he 7

~

s Ah V34 l__t\\'

't 7

l d4 i

n

~

n O <--

b

.... y

- - Fe + Q g (Qssw hnot %adin m mod M 4 9 49+Gg

( w re ) h) c

/

b fA NDO Q

'="

b f

min u..ociai. ine.

Joeu. seine-q =t + %

r con.omo e

e i o bM 0i_. N*Ytw,4bdic

b CY 2 ATE P23 JECT PAcr gr CH K3. 3Y nATE CUOJECT JO3 N3.

i Ja p&

P<

w&

L

{-

\\

6p Ju 3 = o, 5 6 6 Fe + 1. M L Go v4 @

WA Fe.

1-5" M - # (1.38 4 h) 4 f

c>lo h d

N G]) & N gno md d' w:

\\

A b4

^$ '

O h Q(4

%7.f.6)'.M[w 13 n

be_(Al k by (13%

l o,65%_g' f

/

q i

l E

Jack R. Benlomin & Associates,Inc.

Consulting Engineers 9

chMw oATcll D 9

A racJacT-PAtr gr

\\

. CHK3. QY MATE CUOJECT--

JIO N D.Dbd ',

I

(

A t vC kwb 5

SM

)

cy 6 Y be+%

o S st Emi 1

kX b

A.+ k% +

Di =

c. -

uQ u

i = M +-

R.-$)NR,

+A dT h

bE M

k O f pq

-[x g

4%

E Q T ~= M

+fft-3)4[(Pf f)n{3 EIg -.4 Y 4 G 9D e

(,

7 u c.

1 oissb+d bg 3 = p (1,3%)

f DIG P Di ~ 9 t.

hi =

)

4 4

4 f

6=o

&=L l

l

{

l l

I Jock R. Beniamin & Associates, Inc.

Consulting Engineers E

b CY OATE PR3 JECT PAf 8' CF C H K ?>. O Y

' ATE CU2 JECT JO3 NO.

4_o

( 3 ( d o <.; d ruov d

X=9<+b3 (o11 nat~u~d h c6J 2

A=Q g q (q +7_b b' t-t) $

e y 1

e d

'A c o S u = O

=7= O= o g 6c o

~

ruo~ csd om A

J{m<)d u M )%

a s+% cm o a p p 4 a m! at a g +o &, &

e) i Mc 6 W u (a b dm=o %)

u I

I L,

% = __Cg (q + be ) + (Fe + %.)L

\\/t 7

g A

V=.5*Fe + h 31

i f

(q+bt+

Fe. - 3) + C&_)r

4 A = c-4 b=Dzo 4 cnab 4v Qob~ HmEM4 M du I

em m e:*J,,ma^"*'**'" J w +o ass ~

l L.

on A a

l

CY DATE PR3 JECT PAtr gyr C H K3. CY

", ATE CUCJECT J 2D N 3. -

l

%d x = ag + bp dn, _ gL

~'

E I& =

CQg (9-t bd Fe-) }A

,4 A= cr% 4 g3_. ap(q,,g)_aq._pg

) L buh 7 ~k d & cQ

-dJ p

-tT% = -4 Y og cIq4% q(%)(UFe3)1kk 4

b L

f f-t-3)

E}if-) - E s.1 = O 1

f

p s

.g

(~g_3 )._ 2na_

zus.

bg' ap g hg=

hp

.1 9 L,g sg 3 ys 5

Co (4 % -3) - k

( C9 M l

M[g )

} 3 :.

49 h,%

2.,

9 t

vf c)

[%

A.

O v

3

= ( fE +A.) o 3

r

%e s O

. lock R. Ben)omin & Associates,Inc.

Consulting Engineers 9

GY_

SATE PAOJECT PAgr g y-CHKD, 3Y

.gyg SU2 JECT 233 NO.

b m

b (qs-g _ zu rq A+%he' EV S>Ep 29 4

t y =. -

2 (S /sp) 4 2/3 (#f[Nb P

(

~

k k

,4 2Eh f

_ MFe 3) 2

{Q +nu =s:q pe f

khf-As ts ra a

K wr% _

F h'>

b b*

bb "I _

f f

l l

g=

f ze coe wa E

f b

b Af k

E=

h Wp Il f

Y Of

'Q b % E *P 6 or $ AuE f

f 9

F4(T-4) z*p 3

g f

K + ke O

Jack R. Beniamin & Associates,Inc.

Consulting Engineers 3

CY MATE PLCJECT PACE OF CHKD. OY nATE SUOJECT JD3 ND.

t N 1s N

&x

$\\.)

&'d w

l s~a a s% m va em 1ww w v% my At = a l

icimJ 8<

A M o = 0 (fe p.t) l 1

l Ne_ = t% @ f +6g )-

4Fe-3) 44] bp t o fE f

ff~

bf c

l kbe

I

~

f' k

& SU M)

G b 6

'l

']

1 M

b 0

X 1

{

bil_ Q 4_$

V

! 4 R-3) h.p

\\

vc + Re "f

l l

l 2r

% = Fe. (T-4) - (R+Q(9 c-5) 3 F

h-t, j'

~

f 2e r / q'-

Jack R. Benjamin & Ass Consulting Engineers 9

CY DATE PR3 JECT PAfr OF CHKD.CY n ATE - ---

CUOJECT JO' N O.

I C

\\NC) f bedhg de n~ M J h h

is p M

OD 1

l W +9) d(U Ft-5) 6Qd 6f L' be b{ -. C g ( A p +

.f 43 R=[A"g+[3, I

Jock R. Benjamin & Associates,Inc.

Consulting Engineers D

l CY DATE PROJECT PAgr g y-CHKD. CY nATE CUDJECT JE NO.

0R 4

- /

9'

/\\

4

\\-

/g d,,

b yj f /v

.. a c-Nf y

N

,-6 V=.

y f-s a

g C-.

l

.l i t

^

\\/ p./.

rc i

a i

-(

k b b%

su b tyt1

+

Sh

e. usm 4+ m.%g v

$h Nf4 %)f' r

i I

'P

!.L f. I, + _T

~7

.E.,

M}-

sgj

,c

]

s i

F _,

i a

2.

Mh 4>

dbd i

I Jock R.Benlomin & Associates,Inc.

Consulting Engineers 9

l i

l 3 1^

"#^

JO3 N 2. -

5'0N b

~

CHK CY MATE CUOJECT Nf$ d b h/ hM'

'F v.

dN A

W iw O

^

O y

R O

ls A j"'

Sj

, Atr l' 4p Md bYk bg/qd

}

k '5 5 v o ( $ g m ) 6 8 L W f"tddo Yeh W

lymf t. =

p; OA f d bM Adr hk$ + OM feo 4-hVd ) " o M

pA2@

~

Ade=

Av.b A4r _ Pys +m Pe 64r Jg)

D k/2 a s., GJ hs,kd F S

%s=k4,3 u

j 4

c (lirfgy,g}*

.A :

f hQ b.;

Nh N5 O 2a i

9 as.

g.4

/

Jack R.Bergamin & AssoclM, nc.

E Consulting Engineers 9

b CY "JATE P23 JECT PA 2 8' QF CHK3. GY nATE CU" JECT J33 NO.

l hook s~

1+

+

i u/

ae t

2.

hovs3

~ d

~

i

.V M

t 7

i I

'n

~

x<s!.bL.% 6 h,= kNT.... kM L

g m S<wL)

Os Y Vtw

~

3'#

(A%.LQ LJ h4 o

b

hA z

h ht).F5 SS t

o.4 6 _

L

'N 7_.

Yi d-i bs.)

2 d p.g 7tT [ SSJL, Q) N

.4%L A

m y

8 Jack R. Beniamin a Associates,Inc.

Consultire Engineers

~

l i

CY CATE PR OJECT PACE CF l

l CHKD. CY DATE SU3 JECT JO3 ND.

l I

I j

i i

i 1

4 4

O

)

l N

W 1

l l

Jack i Benlomin & Associates,Inc.

E Consulting Engineers D

l l

i

1 I

C:\\WINMCAD\\ FILES \\A8 COMPS 2.MCD 12/17/93 kip := 1000 lbf kai := 1000 psi fExed := 10 Hz Fixed-base frequency ofcabinet f =7.5856 Hz f,= 9.848 Hz peta := 0.05 Frequency and damping of the entire cabinet as e

0.375290~

~2.5 '

O.60 5

SA :=

g F :=

Hz Groundinput response spectrum

,0.723530, 25,

I I

f' g h0.05 SSa(f,p):=exp linterp(inp i(Hzj,In

,In

-(HziA 4 (f

and damping values f

f f

fSA I

Values of Sa at different frequencies i

i (g4 Wt := 10Lkip Weightofcabinet l

Spectralacceleration Spectralacceleration Spectralacceleration inn /s duection inc/wdirection in verticaldtrection j

Sa, := SSa(fas, peta) ew := SSa(few, peta)

Sa Savert :=

0.40 g

n Sa =0.65%1 g

\\Sa,=0.69991 g}

Savert =0.26667 g as e

l h := 109 in Height' ground to top ofcabinet Ins := 91 in Base lever arm for north / south direction axial force at support 1,g := 96in Base lever arm for east / west duection axial force at support at first bolt e

Iew2 := 19.2 in Base lever arm for east / west duect on axial force at support at second bolt i

Sa Se, Wt hi S* vert fi t

-Wt-l h ns e.

-Wt t

i 8

(21 8

(24 g

P p

P as -

ew -

vert -

g 4 7

p 4

ns ew2 1,,g + I 2-(

ewtj P = 1.97519 kip P,, = 1.91031' kip Pvert=0.66667 kip ns Yg:= Pas + 0.4-(P,+ Pvert)

Y2 T3 8

Y g =3.00598 kip Y2 = 2.5 kip Y3 = 1.64901 kip Y3 c g := Y o g =0.54858 1

Y *T3 2

/

c 2 :=

c 2 = 1.37144 kip Yg

STOPl L

ELASTIC POINT ANALYSIS This program determnes the allowable capacity of the plate bolt anchora8e system for the 4160V Swithchgear A8 at the i

Pilgrim Station l

i Plate andsteelproperties 8

E :n29-10 psi Modulus ofelasticity for steel E

0:=

Modulus ofrigidityfor steel 2-(1 + 0.3) p.:= 0.25 in 1hickness hold down plate t

} w I= 2.25 in }

Widthhold down plate p

d :=0.625 in Diameter bolt holein plate h

i p := 2.75 in IW, late from end to bolt (maxunum value) a bp := 1.375 in Iength plate from boltto support yp := 44 ksi Medianyield cap 4' plate o

Boltproperties 5

Eb := 2910 psi Modulus ofelasticityforbolt Ab := 0.196 in' Areabolt(Oross area for stretchmg) gp := 2.5 in Effectivelength bolt for stretchmg Channel properties C6x8.2 (AISC 9th Ed.)

d := 6.00 in Height of aannel b := 1.92 in Width channelflange f

t :=0.343 in Thickness channelflange f

t,:= 0.200 in Thickness channel web o := 0.599 in Distance outside edge channel web to shear center e

d K := 0.08 in Torsion constant l

l Cw := 4.72 in' Warping constant I := 13.1.in' Moment ofinertia about x-axis x

d Iy := 0.692 in Mament ofinertia about y-axis L := 38.375 in Channel span between assumed supports a :=b Distance of anchorage to support along channel 2

q := 0.25 in Distance from edge of flange where point of contact between plate and channel flange occurs

O Oeneralproperties c g and c2 are used for the case where the shear force c g =0.54858 on the bolt is proportional to the the Fem the c 2 = 1.37144 ka,p V=c g F, + c 2 pp := t + 1-(l4a== Meg l)

Offset in plate which causes manw=' in plate from t

f sheer force on the bolt (equal to thickness of channel flange plus deflection upward) g t

Y := 1 - c g. PP Y =0.85807 Variable reduction factor on the force Fe due to shear b

mktt p

t PP.C

( = 0.35482 kip ranarant reduction factor on the force Fe due to shear

(:= b 2

p onthe bolt t'W I:= P P

I =0.00293 in Mnmant ofinertia plate d

12 2

h:, (wp-d)t h P 8

g h = 0.02539 in Plastic section modulus plate at hole Z

4 M = 1.11719 kip in Plastic mamant capacity plate at hole M

Z yp h

h h

CASE 1 - MOMENT BETWEEN PLATE AND CIRNNEL FLANGE IS ZERO This calculation finds the value of a such that the slope at the end of the plate is zero and the mamant at the p

metachment to the channel (i.e., at x = a + b )is also zero. This is the limiting case where the channal offers no p

p resistence to rotation See derivation for definition of tenns and theoretical basis (Reed 11/22/93)

Mh Q be(* p) :=

p Allowable prying force corWie to plastic i

mnment capacityplate at hole J

2 EIgP k,(a ) :=

p 2

b A E '"p b b p

k(a ) := 2b + 2, p

b 3 (bpj p

I

~

fa I

r (a ):= gobe( p)+c e p l

i I

g 4

Constrain A to equal zem, wluch wi@ to zero slope at the end of the plate l

Find ap Q be(a ) a,.(a,+ 2.b g- (Fgag.Y-(). P.

b

]

2 9

,a,

a, ;= root f

b I' b

b l

F,(a)- Y-k(a)- 1 + Y-

- k(a ).Y-

- ( (k(a ) + k,(a ))-

p e p p

p p

l Pl.

P.

1

+

l 2

2 i

b p a =0.75511 in p

Findlimitingvalueof 6 I

b I' b

b F(:p)- Y-k,(a)- 1 + Y-

-k(a)Y-

-( 1 - (k(a ) + k,(a )).

e p

p p

p 2EI b,=

0, =-0.014004148 Tension force on anchorage at edge of channel flange F,(a ) = 1.3604 kip

~

p Prying action force on plate Q be(* p) = 1.4795 kip F (* p) :=F,(a ) + Q be(*p)

F (*p) =2.8399 kip Bolth w b

p b

bolt (a ) :=c g F,(a ) + c 2 Ybolt(a ) =2.11773 kip Shear force V

p p

p Plot deflected shape, slope, shear and nwnent of the plate 2'

Qbe(a )-a -(ap + 2 b )- (F,(a ) Y-() b P

P A(a ):=

+EI0 p

p p

e p

Y(X *p) 2*-* }-Q e(*p) x + A(a ) x+ (F (a ) Y-(+ Q be(a ))-(X- * )8 I

3 I

1 P) b p

e p p

6

-Q be(* p)'-+ ^(*p))[ + (F,(a ) Y-(+ Q e(*p))~(x-a )2 f

2 x

P p

b

_ (x>, p) r(x,a):=

p 2

V(X.*p) *-Q e(*p)+ (F,(a )+ Qbe(*p))'(#8p) b p

M(x,a ) :=-Q be(* p) x+ (F,(a ).Y-() + Q be(*p) -(x-a )-(x>a )

p p

bolt (a ):= (a,a ) A E

-y p p b b Pbolt(*p) =2.8399 kip I

P p

8P

5 Ax :=. P + b a

P Ax = 0.0213 *in n := 100 n

i := 1.. n + 1 x, :=(i-1).tx def,:=y(x;,a )

slope,:=r(x;,a )

shear,:= V(x,,a ) mom,:=M(x,a )

p p

p i p y(0 in,a ) =0 in y(a,a ) =-0.00125 in y(ap + b,a ) =-0.01636 in p

p p p p 8

i i

in U

h -e.01-L o

0.5 1

1.5 2

2.s 3

in r(0 in,a ) =7.06720925910" r(a,a ) =-0.004%3919 r(ap + b,a ) =-0.014004148 p

p p p p 0 01 D

in o

-e.01 -

o c.s 1

1.5 2

2.s "A5 V(0 in,a ) =-1.4795 kip Pg(a ) =2.8399 kip V(ap + b,a ) = 1.3604 kip p

p p p 2

i la i

_kir 0 L

I f

I I

o c.5 1

1.5 2

2.5 b

in M(0 in,a ) =0 kip in M(ap,a ) =-1.11719 kip in M(ap + b,a ) =0* kip in p

p p p o

b k

.muung-93

_5 l

o_

-t -

1

-1J o

c.5 1

1.5 2

2.5 sg E.

L

CASE 2 - ROTATION AT CONNEC110N BETWEEN PLATE AND CHANNEL FLANGE IS SET TO: 0, For thIS case a value of 6, between zero and the imutmg value 0, = 4.014 is h and the plate responseis calculated. Again, the slope at the end of the plate is set equal to zero.

1 0, := 0assumedet Assumedvalucof 6, 2 E I0*

obe(a )-(k( p)+=e(a ))+

+c p

p 2

,P F (a,) :=

e e-k(a) c p 2'

Qbe(a).,p.(ap + 2 b )- (F,(a ) T-(), p

+ E I 0, b

p A(a,) :=

p p

1 y

I (x-a )3 I

p y(x,a ) :=

-Q be(a ) -+ A(a ) x + (F,(a ) Y-(+ Q bc(*p))*

. [p, p

p p

p 6

- o be(a ) 7 + ^(a ); + (F (a ) v-c + o be(a ))-(x-a ")2 1

f

/

g r(x.a ) :=g-

. 1

(* P) i p

p p

e p p

2 V(x,a ) :=-Q be("p) + (F,(a ) + Q be(* p))*(#8p) p p

M(x,a ) :=-Q be(*p) x+((F,(a ) Y-() + Q be(*p)]-(x-a )-(x>a )

p p

Vbolt(a ) := c i F,(a ) + c 2 p

p Given Constram slope to equal zero at end of plate r(0 in,a )=0 p

p := find (a )

p a =0.87422 in a

p 1

i F,(a ) =2.53015 kip p

Q bc(a ) = 1.27793 kip p

F (a ) := F,(a ) + Q bc(* p)

F ("p) =3.80808 kip b

b p p

bolt (a ):= (a,a ) A E y p p b b Pbott(a ) =3.80808 kip P

p p

Ip ybott(a ) =2.75942 kip p

1

1 p+b P a

n.= 100 Ax:=

Ax = 0.02249 cin n

i := 1.. n + 1 x; := (i-1) Ax def;:=y(x;,a )

slope;:=r(x;,a )

shear;::V(x,a ) mom,::M(x;,a )

p p

g p p

y(0 in,a ) =0*in y(a,a ) =-0.0016749 *in y(ap + b,a ) =-0.01275 *in p

p p p p 8

i i

b in "I

[-0.01 8

o 0.5 1

1.5 2

2.5 "i

E r(0 in,a ) =0 r(a,a ) =-0.005747713 r(ap + b,a ) =-0.00362 p

p p p p 8

i i

i U

in Mi-c.cos-o

~~~'

i o

c.5 1

1.5 2

2.5 s

in V(0 in,a ) =-1.27793

kip Pbolt(a ) = 3.80808

kip V(ap + b,a ) = 2.53015

kip p

p p p 4

i i

U i.

almr; 2 8

e, o

1 I

f f

0 c.s 1

1.5 2

2.s M(0 in,a ) = 0

kip in M(a,a ) =-1.11719

kip in M(ap + b,a ) = 1.38013

kip in p

p p p p 2

i b

l h

"'"i Win 0

-1 i

i f

I o

c.5 1.5 2

2.5

Calcut:te rotation of channel due applied shear and moment and compare to assumed rotation. Also, calculate the l

[

total deflection of anchorage system and channel.

Rotation and defledon of Channel Flange..

1 Assume moment ofinertis is equal to sum of:

1. Plate and channel flange composite for plate width wp w-(tp + t )3 p

f d

Ifp:=

Ifp = 0.0391 *in

2. Channel flange width equal to twice distance from edge of flange to ins'de web sinface 3

2(bg-tw) tf 4

I :=

I =0.01157 *in f

f 12 Distance from ground to center of area of plate channel flange composite l

A :=2-((b - tw) t ]

fp :=w -(f.p + t )

A p

f f

fp p+t +Ar f t

f t

A c :=

c = 0.23784 *in Afp + Af Totalmotsent ofinertia 32 f

tf32 f

tp+tf IIfp + I + Afp'8-

+Af c-I = 0.06045 *in, t

f t

2 1 Calculate rotation and vertical deflection of flange / plate composite at edge of flange

,,,3 F,(ap)-(b g-tw-q)*

1 EI 2

t 1

t Y

-(b - tw-q)

M(a + b,a )+ V bolt (a ) t +

+

p f

-c f

p p p 6 g =-0.00325 F,(a )-(b - tw-q)*

p f

,3 ivm i := E I 3

t I

t i.

-(b - tw-q)*

M(ap + b,a )+ Vbolt(a )- t+

-c f

f p p p

Avert 3 =-0.00277 *in i

l Vbolt(* p) = 2.75942

kip

A Rot:. tion of cha.aici due to twistmg and deflection ct edge of flange.

Torsion moment at shear center T o :=[M(ap + b,a ) + F (a )-(b - q + c )]- V bolt (a ).

- tf-p p

,e p

f n

p T o = 0,13417 kip in At distance "a" from the exterior channel intersection to anchorage a = 19.1875 in Rotation of channel The following formulation is taken from Roark 6th Ed (as given in Mathcad Roark's 1 Handbook) l This file corresponds to Table 21, Case 1, and Table 22, Cases le-Ig,in Roark's Formulasfor Stre t andStrein.

Concentratedintermediat torque ag 4-8 T,

To

{\\

(

T, Case f Left end free to warp but not twist, right end fixed (no twist or warp) g/

/

f0 NOTE THAT THE 'LEFT' SUPPORT l

CORRESPONDS TO EXTERIOR CONNECTION AND THE 'RIOHT" SUPPORT CORRESPONDS TO AN INTERIORCONNECTION ONTHE At Io SWTTCHOEAR CHANNEL BASE FRA!

1

' K'O I*

p-0.080n.l_

p:=

C g E in

(

j F g(x) := cosh (p x)

F (x) := sinh (p x) 2 F (x) := cosh (p x)- 1 j

3 j

F (x) := sinh (p x)- p x 4

F,g(x) :=(x>a) cosh (p-(x-a))

l l

Fa3(x) :=(x>a) (cosh (p-(x-a))- 1)

F 4(x) :=(x>a) sinh (p-(x-s))- (x>a)-(x-a) p i

C 1 :"c 8hCD'L) l C 2 := sinh (p L)

C 3 := cosh (p.L)- 1 C 4 := sinh (p.L)- p L i

i C 3 := cosh (p-(L-a))- 1 l

C,4 := sinh (p-(L-t))- p.(L-a)

Vector ofend constraints j

0 forthis case. Recall:

T0 fCyC g4-C C,3)

A 4

l CyE5 4 CgC 4-C C3j deg

'0

)

2' 2

A l

I" 0

8 0.02022 A

I=

0 f C g C,,4-C C,31 g' A 2

-Tg-

-48.22618 2

lbfin

~

TA (CgC 4-C C 3j 1

I b.$'8 I-3 2 g

g

' bf m.

To

-F (x) + ' -

C yE p, F (x)+

jF,4(x) 0(1,x) := Igdeg +

F (X)+

2 3

4 2

p CgEp Rotation at point of anchorage at'eh-at to rhnnel 02 :=.0(I,a) 02 =-3.5109910 ' rad

~

j

i 11 Range of x-values:

x := 0 ft.1.. L 100 Torsion Torsion Pinned end Fixed end 1

Angle of twist,0 0.02 ikx.1 o,,,

0(1,0 A) 00 rad

.fs

~4 0

0(I,a) = 3.51099 10 rad

-0.01 o

1 2

3 0(I,L) =0 tad k

Venical deflection of channel at contact point between plate and flange due to channel rotation (b + e )-(

Amt2 =-7.96644*10 ' in

~

Avert 2 := 02 f

o Vertical deflection of channel between cross channels - assume fixed / pinned span as effective boundary conditions 3

-7 F,(a ) L p

Avert 3 :=

Amt3 =-0.00343 in 768 E I x Horizontal deflection orchannel between cross channels - assume fixed / pinned span as effective boundary conditions 3

-7 Vbolt(a ).L p

Ahoriz :=

Aboriz =-0.07083 in 768 E l y Total rotation from channel suppon system and deflection from concrete to switchgear 0Tel:= 0 g+02 0Tel =-0.0036003 rad j

Compare to calculated rotation at end r(a p + b,a ) =-0.00362 rad j

p p of plate (connection to channel flange)

Totaldisplacement Abortz =-0.07083 *in AvertT := Avert ; + Avert 2 + Avert 3 AvertT =-0.01974 in

+y(ap+b,a )

p p Cantilever flange Aven 3 =-0.00277 in Channel twistmg Avert 2 =-7.9664410~4 in Channelbendmg Avert 3 =-0.00343 in Plate bendmg y(ap + b,a ) =-0.01275 in Ael g,g, := y(ap+b,a) p p p

p p 1

Force at ends ofplate Q be(a ) = 1.27793 kip

,(a ) =2.53015 kip p

p Forcein bolt Pbolt(* p) = 3.80808 kip Abar,g := y(ap+ b,a )

Vbolt(a ) =2.75942 kip p p p

t

STOP2 l3 Calculate scale factor F (a ) + T2

\\

e E

\\ FS = 1.67338 \\

l FS :=

(

T-1 l

l Calculate frequency of cabinet in East / West direction l

l SSa(few, peta) FS f0.4Py) AvertT l

h A g:=

+

Act =0.1181 tn e

(1,wg - 1 ew2) ( T 1 j (2sfQ*

2 l

SpAcc g := SSa(f,,, peta) FS SpAcc g = 1.17121 g e

e S Acc g P

1 f_newew := 2 s-f_new,, = 9.84806 Hz Calculated frequency A,g 3

f

= 9.848 Hz Assuned frequency ew Calculate frequency of cabioet in North / South direction h

I SS*(Ins, peta) FS

\\

IPu+0.4Pvert h+ j AvertT +

Aei = 0.18759 in \\-

A,g :=

l T 1

)

(2sfQ2 as(

SpAcc g := SSa(f33, peta).FS SpAcc,g = 1.10377 g e

S Acc,g P

1 f_newns;* 2 z~

f new

=7.58572 Hz Calculated frequency Ag ns e

3 f =7.5856 Hz Aaanned frequency as l

l t

I l

1 i

13 STOP3 SECANT POINT ANALYSIS l

Spectral acceleration Spectralacceleration Spectralacceleration in n/s direction in e/w direction in verticalducction ns*SS*(Ins, peta)

Sa,, := SSa(f,,,, peta)

Savert :=

0.40 g

Sa Sa

=0.62852 g Sa,, = 0.67716 g Savert = 0.26667 g ns l

l Sa Sa, Wt. h ft Sa

' i

-Wt-l h

vert ns e -

1

-Wt 8

124 g

(23 g

p p **..

p vert -

'~

ns -

f 2

4 as Iew2) l 1

2 2 l,g + I e

(

ewij l

P,3 = 1.88211 kip P,, = 1.84823 kip Pvert =0.66667 kip Wt Wt Sa ns T i := Pns + 0.4-(P,, + Pvert)

T T

2*4 3*4 g

T 3 =2.88807 kip T2 = 2.5 kip Y3 = 1.5713

kip T3 c i := Y c g =0.54407 l

T 'T3 2

c 2 :=

c 2 = 1.36017 kip T 1 i

l j

M Generalproperties c g and c2 are used for the case h the shear force c g =0.54407 on the bolt is proportional to the the F, m the c 2 = 1.36017 kip g

l Vac g F, + c 2 l

l pp:=t + 20(lAbar gl)

Offset in plate which causes moment in plate from t

f e

shear force on the bolt (equal to thickness of channel fl8n8e P us deGection upward) l Calculated properties t PP Y =0.76341 Variable reduction factor on the force Fe due to shear Y := 1 - c g = b t

l p

on bolt t

t

(:= BEE c 2

(=0.59147 kip thdant reduction factor on the force Fe due to shear j

p on the bolt 3

1:= P ' P I = 0.00293 *in Moment ofinertia plate d

12 2

Z :, (wP-d)t h P h

g h =0.02539 in' Plastic section modulus plate at hole 4

l M " l 11719

kip in Plastic moment capacity plate at hole I

Mh :: Z yp h

h CASE 1 - MOMENT BETWEEN PIATE AND CHANNEL FLANGE IS ZERO

" Ibis 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. ' Ibis is the limitinerase where the channel offers no p

p i

ra=*ance to rotation See derivation for definition of terms and thentical basis (Reed 11/22/93)

Mh Q be(a ):=-

Allowable prymg force s.-, v Eg to plastic p

p moment capacity plate at hole 2EIg P k,(a ) =

p b'AEa l

p b b p j

k(a ) :=2-

+ -

p b

3 (bpj p

fa i

F,(a):=I

-Q be(a ) +(

p p

4

Constrain A to equal zero, which w.syacds to zero slope et the end of the plate Find ap b

Q be(a )-a -(ap+ 2 b )- (F (a ) Y-(). P p

p := root a

p p

e p

,, p f

bf b

b T-k(a). 1+Y p- -k(a)T

- ( (k(a ) + k (a ))

F(a)-

j e p e p p

p e p A

Pl.

P.

+

2 2

b p l

a =0.80506 in p

Findlimitingvalueof 0 e I

bY b

b F(a)- T-k(a)- 1 + T-

- k(a ) T-

- ( (k(a ) + k (a ))-

e p e p p

p e p

(

N.

P.

Ge :=

2EI P

0 =-0.014331831 e

F,(a ) = 1.83908 + kip Tension force on anchorage at edge of channel flange p

Prying action force onplate Q be(a ) = 1.3877

kip p

F (a ):=F (a ) + Q be(a )

F (a ) = 3.22678 + kip Bolt force b p e p p

b p bolt (* p) ; 01 F,(a )+ c 2 bolt (a ) = 2.36075

kip Shear force V

V p

p Plot deGected shape, slope, shear and moment of the plate Q be("p) a a + 2 b )- (F (a ) T-() P

+ E I 0, A(a ):=

p e p p

y(x,a ) :=

-Q be(a ) -+ A(a ) x + (F (a ) T-(+ Q be(a ))-(x-a )3^(

P) f y

1 1

g p

p p

e p p

6 f

2

)

(x-a )2 P

r(x,a):=

-Qbe(a )

+ A(a ) + (F,(a ) T-G+ Q be(* p))'

.[x>e) p p

p p

2 p

V(x,a ) :=-Q be(a ) + (F (a ) + Q bc(8 p))-(x>a )

p p

e p p

M(x,a ) :=-Q be(* p) x+((F (a ) T-() + Q be(a )]-(x-a )-(x>a )

p e p p

p p

bolt (a ):= (a,a ) A E y p p b b Pbolt(a ) = 3.22678 kip P

p p

8 P

\\v P+bP a

n.= 100 Ax:=

Ax = 0.0218 +in n

i :: 1.. n + 1 x, :=(i-1) ax shear,:=V(x;,a ) mom;:= M(x;,a )

def,:= y(x;,a )

slope, := r(x;,a )

p p

y(0 in,a ) =0 *in y(ap,a ) =-0.00142 *in y(ap + b

,a ) =-0.01698 *in p

p p p 0 01 i

i i

h in dd; o

T_

o_, -o.01 o

c.5 1

1.5 2

2.5 3

in r(0 in,a ) = 1.465303015*10' r(a,a ) =-0.005291601 r(ap + b,a ) =-0.014331831 p

p p p p c.01 i

i b

in o

W_I

!. -o.o1 o

0.5 1

1.5 2

2.5

i E

V(0 in,a ) =-1.3877 kip Pbolt(a ) =3.22678 kip V(ap + b

,a ) = 1.83908 kip p

p p p 2

b io i

kip o

-2 o

0.5 1

1.5 2

2.5 81 E

M(0 in,a ) =0* kip in M(a p,a ) =-l.11719

kip in M(a p + b,a ) = 0 kip in p

p p p 08 i

i i

i D

o e

"'""i g in -0.5'~

S._

_g _

-1.5 o

c.5 1

1.5 2

2.5

'i

CASE 2 - ROTATION AT CONNECTION BETWEEN PLATE AND CHANNEL FLANGE IS SET TO: 0 e For thlS case a value cf 9, between zero and the limiting value 0, = -0.01433 is assumed and the plate response is calculated. Again, the slope at the end of the plate is set equal to zero.

O, := Gassumed Assumedvalueof 9, p

2.E.I.0

l o be(a )-(k(a ) + k (a ))+

+c p

p c p

, p, F,(a ) :=

p t-k,(a) p A(a,) = 0 *(* 9) a,.(a, + 2.b,)- (F (a,).v-c).(9

+ E.10, c

-Q c(a ) x + A(a ) x+ (F,(a ) T-(+ Qbe(a )). - a )3' ' p)

)

(x f

3 3

y(x,a ) :=

b p

p p

p 6

p (x-a )2 I

2 i

1 x

p r(x.a ) := g

. - Q be(* p) 7 + ^(* p)j + (F,(a ) T- (+ Q be(* p))~.(x>, p) p p

2 v(x, a,) :=- q u(a,) + (F,(a,) + o w(a,)). (x,a,)

u(x.a ):=-ow(a ).x+ :(F (a,).v-c)+obe(a );.(x-a ).(x>a )

p p

e p

p p

bolt (* p) := c g.F,(a ) + c 2 V

p Given Constrain slope to equal zero at end of plate c(0.in,a )=0 p

p := find (a )

a =0.92366 in a

p p

F,(a ) =3.04146 kip p

1 Q bc(a ) = 1.20953 kip p

F (* p) := F,(a ) + Q e(* p)

F (* p) = 4.25099 kip b

b p

b 7(a,a ).A Eb p p b

bolt ( p)

P a y = 4.25099

kip P

8p y,g (a ) = 3.01492

kip f

p

N P+bP a

n := 100 h :=

h = 0.02299 cin n

i : = 1.. n + 1 x;;=(i-1) Ax def;:=y(x;,a )

slope;:=r(x;,a )

shear;:=V(x;,a ) mg :=M(5,a )

p p

y(0 in,a ) =0*in y(a,a ) =-0.0018697 in y(ap + b

,a ) =-0.01383 in p

p p p p 0

4 %

i i

i in M i

_Y -9.01 0.,,,,

0 0.5 1

1.5 2

2.5 "i

5 r(0 in,a ) =0 r(ap,a ) =-0.006072769 r(ap + b,a ) =-0.0049 p

p p p 0

g k

-0.005 -

-0.01 Y

~ ' ' ' ' '

0 0.5 i

t.5 2

2.5 "i

5 v(0.in,a,) =-l.20953 kip Pbolt(a ) =4.25099 kip V(ap + b,a ) = 3.04146 kip p

p p 4

I i

w shear; 2

--N 0_

0 I

I I

i 0

0.5 1

1.5 2

2.5 "i5 M(0 in,a ) =0 kip in M(ap,a ) =-1.11719 kip in M(a p + b,a ) = 1.26212 kip in p

p p p 2

g i

w i

""'i

_kipin 0

.i -

i i

0 0.5 1

1.5 2

2.5 "i

5 I

Calculate rotation cf channel due applied shear and moment and compare to assumed rotation. Also, calcul:te the total deflection (f anchorage system and channel.

Rotation and deDection of Channel Flange...

Assume moment ofinertia is equal to sum of:

1. Plate and channel flange composite for plate width wp w -(tp + t )3 p

f d

Ifp :=

Ifp =0.0391 in

2. Channel flange width equal to twice datance from edge of flange to inside web rrface 3

2-(b - tw) tf f

I = 0.01157 *in, I :=

f f

12 Distance from ground to center of area of plate channel flange composite l

A :=2-((b - t,) t ]

fp:=w-(tp+t )

A f

f f

p f

t f

tf fp p+t A

+Af c :=

c = 0.23784 *in I

Afp + Af Totalmoment ofinertia p+t/2 Arc- /2 f

f t

t d

I =0.MS in I ;"Ifp + I + Afp' t

f 8-t 2 j Calculet: rotation and vertical deflection of flange / plate composite at edge of flange F (a )-(b - tw-q)*

e p f

.g 0

1 E1 2

~

1 I

t Y

-(b - tw-q) p+b,a)+Vbolt(a )

t +

M(a

+

p p p

f

-c f

0 g =-0.00351 1

l F (a )-(b - tw-q)3 e p f

.I Avert } :=

EI 3

t f

t I'

-(b - t,- q)*

M(ap + b,a ) + Vbolt(*p) t +

p p f

-c f

+.

Avert g =-0.00304 in bolt (* p) = 3.01492 kip V

2.0 Rotation of channel due to twisting and deflection at edg of CanGe.

Torsion moment at shear center T o :=(M(a p + b,a ) + F (a )-(b g-q + c ))- Vbolt(a )

- t f-p p e p o

p T o = 0.5294

kip in At distance "a" from the exterior channel intersection to anchorage a = 19.1875 in Rotation ofchannel The following formulation is taken from Roark 6th Ed (as given in Mathcad Roark's 1 Handbook)

This file corresponds to Table 21, Case 1, and Table 22, Cases le-1g,in Roark's Formulasfor Stress andStrain.

Concentrated latermediate torque x

a h f.

8 s Ti To fN i

e I

I L

s i

Case f Left end free to warp but not twist, l

right end fixed (no twist or warp) 1 1

{

i N

fa l

NOTE THAT THE "LEFT" SUPPORT CORRESPONDS TO EXTERIOR CONNECTION AND THE "RIGHT" SUPPORT CORRESPONDS TO AN INTERIOR CONNECTION ON THE A8 Io SWITCHOEAR CHANNEL BASE FRAMI

2.1

.1

' K'O I '

p -0.080741 p := jC yE; in F (x):= cosh (p x) g F (x) := sinh (p x) 2 F (x):= cosh (p x)- 1 3

F (x) := sinh (p x)- p x 4

F,g(x) :=(x>a) cosh (p-(x-a))

F 3(x):=(x>a)-(cosh (p-(x-a))- 1)

F g(x) ::= (x>a) sinh (p-(x-a))- (x>a)-(x-a) p C g := cosh (p L)

C 2 := sinh (p L)

C 3 := cosh (p L)- 1 C 4 := sinh (p L)- p L C,3 := cosh (p-(L-a))- 1 C,4 := sinh (p-(L-a))- p-(L-a)

Vector of end constraints O

for this case. Recall:

f C C,4 - C C,3) g TO 3

4 CqEp CgC4-C C 3j deg BA 0

2 2

I;*

0.07977 g

is o

fC g C g - C C,31 0" A 2

3

,.g.

-190.28083 lbfin TA 4 (CgC 4-C C 3j 2

If

-F (x)+ gIbfin To I

F (x) + C yE p, F,4(x) 0(1,x) := I, deg +

F (X) +

3 g

4 2

2 p

p CgEp Rotation at point of anchorage attachment to channel 02 := 6(1,a) 62 =-0.00139 rad

Range of x values:

x := 0 ft,1. L 100 Torsion Torsion Pinned end Fixedend c.1 Angle of twist,0 l

o.os 4I,n) 0(I,0 A) =0 rad i

-des 0

0(I,a) =0.00139 rad I

I I

-0.05 o

1 2

3 0(I.L) =0 red E

l A

1 l

yertical deflection of channel at contact point between plate and flange due'to channel rotation 2:=0 -(((b +e )-it))

Avert 2 =-0.00314 in Avert 2

f o

l Vertical deGection of channat between cross channele -===nne fixed / pinned span as effective boundary conditions

-7 F,(a ) I,'

l p

Avnt3 :=

Amt3 =-0.00412 in 768 E I x Horuontal deflection of channel between cross channata - me==ne fixed / pinned span as effective boundary conditions bolt (a ).I.3 l

-7 V p

Ahonz =-0.07738 in i

Aboriz :=

7686I y

Total rotation from channel support system and deflection from concrete to switchgear 0Tg := 0 g + 02 0Tp =-0.0049 red l

Compare to calculated rotation at end r(ap + b,a ) =-0.0049 rad 1

p p of plate (connection to channel flange)

Totaldisplacement Ahonz =-0.07738 *in AvertT := Avert i + Avert 2 + Avert 3 AvertT =-0.26523 *in

+(t-tpp) f Cantilever Gange Avert ; =-0.00304 in Channel twistmg Avert 2 =-0.00314 in l

Channelbendmg Avert 3 =-0.00412 in t-t

=4.25493 in Plate bending f

pp

[

Force at ends ofplate Q be(a ) = 1.20953 kip F,(a ) = 3.04146 kip p

p bolt (* p) = 4.25099 kip Forcein bolt P

Vbolt(a ) =3.01492 kip p

13 STOP4 Calculate scale factor c( P) + T2 S = 1.91874 FSS :=

Yg Calculate frequency of cabinet in East / West direction T

SSa(fewg peta) FSS f0.4P I avert h

ew

+

A =0.17525 m, Ap :=

(1ewg-1ew2)(

(2sfQ*

2 Y l j SpAccp := SSa(fewp, peta).FSS SpAcc =1.2993 g p

1 SpAcc E f_newp,, = 8.51512 Hz Calculated frequency f_newpew := 2 s $

A P

few, = 8.51538 4h Assumed frequency Calculate frequency of cabinet in North / South direction h

I SS*(Ins, peta) FSS 1

fPas + 0.4 Pvert Ahonz+ i

{A

=0.31351 m, \\

AvertT +

Ap :=

p

~

(2sfQ2 I

T l j

m(

SpAcc p:= SSa(fup, peta) FSS SpAcc =1.20597 g p

SpAcc f_newp, := g f_newp, = 6.13345 Hz Calculated frequency 1

E 2s A

3 p

f

= 6.13377 4h Assumed frequency nsk 0g 0 in SPACC := SpAcc l AA := Ag e

e SpAcc A

p p,

i := 1.,3 Force / Deflection Diagram 1.5 g

i i

f0

)

f0

)

1 SPACC =

1.10377 g

AA =

0.18759 in

{l.20597j (0.31351) o.5 t

i t

o o

c.t o.2 0.3 c.4 in Calculate ductility factor and slope SpAcc - SpAcc g p

e A-Ag mu :=1 mu = 1.67125

=

s = 0.13794 p

e SpAcc g) i Aet e

i Age j

2.%

STOP6 Estimation of the Median Ductility Scale Factor, Fmu I

The median ductility scale factor, Fmu, is obtamed by several methods (See Reference 1):

The following variables are used in the analysis:

f = 7.5856 *Hz The A8 Bus frequency ns b := peta b = 0.05 Elastic damping fk := 2.5 I Response spec' a knuckle frequency see s = 0.13794 Strain hardenmg ration mu = 1.67125 Ductdityfactor Riddle-Newmark method for Calculating FuRn 9

The details of the methodology are given in Reference 1 i

L:ts first calculate Fmu at the peak ground acceleration level (zpa)

The zpa valueis:

zpa := 0.4 g and the spectral acceleration at the fundamental frequency, f, and elastic damping, b, is:

Sa := SSa(fus,b)

Sa = 0.65%1 *g Fmuis estunated as:

Fu4 := 8' mu"'"

Fu4 = 1.74485 zpa i

Then, in the acceleration range of the spectrum Fmu is an'imat~i as:

Fu3 := (2.67 mu-1.673)**II Fu3 = 1.52439 1

Finally, in the velocity range of the spectrum. Fu is estunated as.

Cf - I k \\ f ik <l.0

+

ik i

f

\\

f 21.0 Cf = 0.32957 (Insj (Ins j

(I ns j

Fu2 := (2.24 mu-1.24)' Cf Fu2 = 0.5774

The median ductility scale factor, Fu.RN, using the Riddle-Newmark methed, is obtained as:

Fu t := Fu3-(Fu3 <Fu4) + Fu4-(Fu3 2Fu4)

Ful = 1.52439 Using the ratio of ultimate static capacity to yield static capacity, R, defined as:

R := 1 + s-(mu-1)

R = 1.09259

. (Ful-(Fu2<Ful)+ Fu2-(Fu22Ful))

7 FuRN = 1.3952 Modified Riddle-Newmark method for Calculating FuMRN 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. 'Iberefore, the modified ductility ratio is:

mu ; := 0.5 + (""- )'I +

)+

mu ; = 1.50718 2R At the peak zpa level, Fu is:

Fum4 :=b mu 3

33 Fum4 = 1.72513 zpa In the acceleration range of the spectmm, Fmu is:

Fum3 := (2.67 mu ; - 1.673)"'I3 Fum3 = 1.42101 In the velocity range of the spectrum, Fmu is:

Fum2 := (2.24 mu g - 1.24)"33Cf Fum2 = 0.52402 where Cf has been defined previously.

The median ductility scale factor, Fu.MRN, using the modified Riddle-Newmark method is:

Fum1 = Fum3-(Fum3 <Fum4) + Fum4 (Fum3 2Fum4)

Fumi = 1.42101 Fu MRN := Fuml-(Fum2<Fum!) + Fum2 (Fum22Fuml)

Fu MRN = 1.42101

Effective Riddle-Newmark method fog Calculating FuERN Because the modified Riddle-Newmark method only accounts for second slope of the force-deformation cunte, the last correction to pc: form is to account for ground motion duration.

The factor to account for earthquake duration is:

CD:=1.0 and the median ductility scale factor, FuERN, using the effective Riddle-Newmark method is:

Fu ERN := 1 + CD-(Fug-1)

Fu ERN = 1.42101 Effective Spectral method for Calculating FuSA In this method Fmu is calculated using the effective firquency and damping ratio of the structure First, the ratio of secant stiffness to elastic stiffness is reime~i as:

Ks g := U + D'"

Ks g = 0.65376 mu Next, the ratio of the secant frequency to clastic frequency is estimated to be:

fs g:=]Ksg fs f = 0.80855 Compare to:

fs ;= fs rf fs =6.13335 Hz f

=6.13377 Hz as ns The ratio of the effective frequency to clastic 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-( A150.85) + 0.85-( Al>0.85) fe f := (1 - A) + A-(fs f) fe f = 0.93036 The effective damping ratio, be, is calculated as follows:

Cn := 0.15 Coefficient to account for short duration motion bh.= Cn-(1 - fs g)

Hysteretic energy dissipation damping bh = 0.02872 be:=

g)2 ffs

-(b + bh) be = 0.05945 (f* fj

'Ibe median ductility scale factor, Fu.SA, is estimated as follows:

' effective frequency, fe,%

fe := fe rf fe = 7.05734 4h ns l

Sa, := SSa(fe,be)

Sa, = 0.59505 g Sa.e is the spectral acceleration at the effective frequem, t,

e l

and the effective damping, be ffe f32 a Fu SA = 1.46763 Fu 34 :=

fs fj Sa

(

e l

l Final Estimate of the median ductility scale factor 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 average of the ductility scale factors found using the Effective Ridd!c-Newmark method, Fu.ERM, and the Efrective Spectral Method, Fu.SA.

FuERN + Fu SA Fmumedian :=

Fmumedian = 1.44432 2

l l

l

fmalfactors Assumed values Calculated values Elastic point fas=7.5856 Hz f new

=7.58572 Hz ns j

FS = 1.67338 f,=9.848 Hz f_new

=9.84806 Hz ew Fmumedian = 1.44432 els 0.01274 in ael g =-0.01275 in p

Gassumedels- 0.00362 GTel =-0.0036 FS Fmu Man = 2.416897743 f

s6.13377 Hz f_,newg, = 6.13345 Hz.

asp f

a8.51538 Hz f_newny =8.51512 Hz g

An===ned s 0.00490 0Tp = -0.0049 p

IFS Fmumedim

-81 1

BETA:=g-In BETA = 2.57973 10 1

4 2.416897743 j i

)

l l

l'

9

'Ibe variability due to random scatter of time history -aNM Fmu versus predicted Fmu values using.yy.w.... ate medwie (for example, the spectral averagmg method) is:

br Fmu := 0.4-[0.06 + 0.03-(Fug R-1)]

br Fmu = 0.03124 h uncertainty due to story driA aWdM with failure is calculated,====nmg l

that the 10' elastic deflection is at 1.0 seandant deviation fhxn the mean, to l

be 1.25673 using the same equation as above. Therefore:

f mua,,n!

l F

In

( l.25673 4 bu Fmu :=

bu Fmu = 0.13913 1.0

'Ihc uncertainty due to inelastic energy absorption model is:

bu Fmum :=0.1-(Fmua,,n - 1) bu Fmum = 0.04443 The combined variabilityis:

b p,.nu :=

brFmu + bu Fmu + bu Fmum bFmu = 0.14935 i

I l

l

uv.

wout5 taa xL55tr gyy33 g

CHKD. QY

" ATE SU: JECT JC3 NO.

l i

l w&cd 6 AA 1

l l

Jock R. Benlomin & Associates,Inc.

O l

Consulting Engineers 9

c CY MATE PR2 JECT PAGE CF CH K3. CY DATE CU" JECT J33 N2, b

& NY. xs &

a kA v

g u - a g. u q t,.

et)A A AA gHL 4w,& aat" 4 % % ' W E 4 =," b,L. a b 0

,i.e.,

u aLw w.a au&

cW.

% 0~A bA 4

Wd W.

I b$

U Jock R. Benjamin & Associates,Inc.

'E)

Consulting Engineers 9

S.ca::v Arc

~

souac,/M 6vs thae mu i

1

.l O. $

5 o

~~'

O O

c5 O O

O O

O O f o f

M 4

rd M-7 C'

f h,.

y r>

en M K hU

~

R 2

7 m

f.

~

ps 6

aa

& as d

b 3

6' t

Nw L E

b 1

Y Z

C J M O

p;.>. C W

Y #

C 2-n t

v.

d r

~

4 a

a a a 4

-J 4

l l

t4 C

W

^

w u

s 7 l

C C

d E. >#

w 5

L U

R t

e w

E t

M

- ~

~

a

~

=

~ ~

~

a a

)

k 2

s e

ym A E z

u-e e

u p

,g

")

i n

G a

r q

g 8

3 8

4 8

9

.n C

C f

o o

d 8

o o

o d

f, p

e l

r s

A-2 4

e t

r n

t l

s 2

G To

~

v s a+

a aa

+-za e =.

6 f

a hh S C O

r b

h ke D

^

(

?*

n, am

= c p

om 4-e-a a e a a w e v

e e e

2.O i

.I r

V W~

r

(

A C

A 9

m

=

K 6*g N

c C

e A

+

3 ih t

(

2 5

3 4

8 3

4 8

Y Q

o 6,

4 o

q 4,

q o

-6 o~ g 8

i a

i e

e j

k" $&

4 0 S i

E p

h c%.

_A.

e e

g o

3 td

.f. -}

r-s 3

c!

. e.

8I M sf o

o o

g Q

3 9

t t

f.

9 o

s, o.

?-

6, g

9-o.

O.

.o w

A i

g g

^

t

.t

.g y Ci f..$ g 9$

.N V

1 92 7,.

e t.. #

4 c4 s. 9 ;K. 8i.

_. F..a ~~M..

+ a:

e a-e a-e a-v-

y 1

~

1 Ul

,M e..

I G..a

,.. p 3..

._ }__.

C s

G

__g, 3

.., _.... gy w

\\

rw g.

. c.

g.

s:g 4

e d

w r

r: s r:

r-n: e e

r<

w e r s-

't e

..Q w, l

~

D

,J.,

'9 1

P p

4-

}

,t y

+ 6 8k m

e-. e m v

q 9

-5

v d., 7 W

i> /~s e

t-to f

~~

N.

d v; l

p.

.M _,2,.,-

.g g d

v v i.

3 k

M.

'fl j

b

'~k

s. o e:4,4 M e[

d a.. [ 8 %. "fg ;i s

}U

$5L 8

.c % 7o

. n g.,

^l

?

?.

),, [.ch +L l' ':

}

e a v$ v5 e 6 5 s:.,

.' t" n

M a.-

.a....

a.

c v

u,

,,d

[.,'. *

),

,,i g

hhhh h

~

c...

- - r :.. :.,.-:. :.=. r..

y.

i-;tiivw'.: C':-..

.........c t -

t

CY DATE PROJECT PAcr OF CHK3. DY DATE SUOJECT JCD ND.

L un/

D l

Jack R. Benjamin & Associates,Inc.

I)

Consulting Engineers

1 0.091 3Mkd N *.

2.5 0.375 5

0.6

~

\\

x :=

10 y :=

0.702 25 0.724 33 0.4 l

,50 0.4 i

l := 1. 7 1

N a g. o.s62 vi 0.1 0.01 1

10 100

1

r

}* I 1 := 1 7 l

l 1

' 0.0s1 l

2.5 0.375 i

5 0.6 x :=

10 y:=

0.702 j

25 0.724 ss 02 Y

zg := i 50 0.4 l

I 1

i i

bg

, e 2.6 N5 0.1 l

Zg i

Y' O.01 0.001 1

10 100

i I

- - - - - - - - - - - - - - - _ _, _ _ _ _ _ _

ML20247D364

Text

8.16 bFmu :

==

Navigation menu

Search