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Ing.Csar Arans
Anlisis devigas ylosas continuasarmadasenunsentido
1
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Column
A - -""TT-- A11 11 II.............,.--r+- - It -rl- ....._____.
11 :I Floor beams1t II Slab spans in ths d irectlorII l ------11 I I
11~ ~ - - . . . . . ~ . . . - -- - IT--ILl_ _ _ _ _ ,
(a) f loor plan th one i termediatefloor beam.
(b),Section A-A.
Girder
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(c) Floor plan Mi two i te ediateocr beams.
Co ur nSlab spans In t is directJon
F oor beams
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th Fl[ l 11 - t 111
Floor beams I\w X
N
lht L. ._
I
kP jr n 2 tt
z y
Girders ------+ 24 tt
I28ft + 4ft + 8 tt
I
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S andr I oedge be m
Consttuct tonjo t Gir er
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Tribut eas, Patten! Loadings, and Li e LoadReductiotls)
12
28 2 28
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II
R tt r11L--.. l i lt'- 1 ft.
rl~~~ /~~~~~~~~28 tt + 4ft + 8
Lrip andne-wa .,
24ft
24ft
I
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n .11 _ d fin d i n ~ p tt n t.m11n m xtn1un1 ign n1 n1 nt f r ntinuoul. F tored dead I d n all pan with f red live I d n t\VO adjL ent pan and
n liv load n any th er pan .t red dead Ioad n all pan
gLve the m mom nt at th n1id_p4j n f thead d p n . the n1axin1un1 neaative tnotnent and n1a i n1un1 hear t the e terior up-nlent. whi h uld be neg tive. at the mid pan of the
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g 1 J - ~ 2~ ~ ~ ~ ~ ~ ~ ~ ~ j u ~ ~ w ~ 2ft
N
r m ~ u u ~ M 2 ' ~~ T ~ ~ ~ r " ' ~ ~ ~ ~ ~ t 24ftM3 24
: ;
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CI Moment and Shear Coefficients-
The requiren1ent for using th CI n1 m nt and hear oefficient are given in , CICod ection .3 .3 a :I . There are two or n1or ontinuou pan .
The pan are appro in1ately equaL with th longer of the rn adja e nt pan notmore that 1._ tim the length of the horter one
....
J . The 1 ad are u n i ~ rmJy di tributed .~ The unfa tored live load doe n t e eed thre tim th un a tored de d I "d.5. The n1 n1be are pri n1 ti .
~ - - - - - - - - ~ 1 - - - - - - - ~I I I II ~ _ n_1 ~ , I I - . t - - _ n_2 ~
2
n(avg) = ( n1 + tn2)12
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En d spanDiscon t inuous end
(a ) Termi ology.Inter io r face ofex erior support
Exterior tace otfirs t in terior support
In erior s pans
Cm - - 1/9 tf only 1wo spans. 1r o 1 three or
(o) \ l foment and shear coeffJc rents-Disco" ' t ouous end unrestrained.
(c) Morrent and shea r coefflcents-Discont .nuous end Integra l wl h supportwhe e support as a spandrel gi rder.
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Con ider the cont inuou floor beam A-B-C-D in Fig. 5-4. e the AO momentcoefficient to find the de ign moment at the critical ection for one e terior pan andthe interior pan. Then, repeat the e cal ulation for the floor beam E- F-G-H in Fig. -4.
ume the floor lab has a total thi kne of 6 in. and a ume the floor beam have a totaldepth of 24 in. and a \Veb width of )_ in. ume the co lunm aJ 1 in . by 1 in. Finally.as um the floor i to be de igned for a live load of 60 p f and a uperimpo d dead loadSOL of 0 p f.
S f t ~ 12 fi12ft
24ft
24 fi
28 24ft + 8ft
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1. nfinn th.., t th 1 111 111 nt ffi i nt n I u d. There are t\\' or 111 repan , the load are un iformly distribut d, and th n1 n1 w pri n1ati . Th r" ti of th1 nger pan t the h rter pan i .... /24 = 1.17. \\ I'hi h i le than I._. The floor 1 b i in.
thi k and thu \veigh 7 p f. h e r e ~ th unfa t red Live load d not e eed three timethe dead 1 ad.
.... et nuin liv I d r du ti n .a t E teri r pan - B: For the terior n gative n1 n1ent and the n1id pan po itiv rn 111 nt the tributary are i qual t the tributary \vidth tin1 the pan length.Thu .,
T = 1... ft X ... ft = 3 6 ft21 = K LL X T = k- X 3 = 67 ft2
+ ~ = Op { . +--= 0[0 .... - + . 79] = 49.7 p f > 0.- X 60 p f (o.k.)
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b1 Interior p n B- C: r the n1id p n p itive mon1 nt. th [! II \Ving applir = 1_ ft x 4ft =I = K LL X r = = ...7 f r
L, = L J= + 0.-_.5 p f > 0. X p f (o.k.)
g tiv n1 111 n t B: r the in terior upport. th n1bin d length f tht\v adj nt pan re u d to find th tri utary r a. Thu .
r = l _ ft x ( _ ft ,I = 1 .- X 60 p f (o.k.)
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3. ~ I t r d 1 &C d .1 J xteri r negative 111 n1ent and n1id pan po itive o1on1 nt of pan - B: Thedi tributed live load a n ~ on th n1 i.....
L = qL (redu d) X tri utary\vidth91 P f X 12 ft = r l /ft = 0. 96 kfft
The di tribut d dead I d fron1 the lab and uperi1npo d d d load ar6 in . 3 _ ...q ( lab) = . Jft X I 0 lb/ft - 7 p fL .. an
t( l b + SDL ) = (7 p f- ~ p f) X L.. ft = 11 Olb/ft = 1.1 k/ftThe de d I ad of the in u d e d ~ ut we need t \' id dou leunting the \\'eight of the Ia \ h re it p over the top of the 01 w b. The\\'eight f th run web i 1 ul t d fro n1 th h d d gi n h \Vn in tg. -- L ..1 ( afll W )= (
The t tal de d lo'"d iin . - 6 in . X l _ in. X ll+t in . lft2
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o w ~ the totru factored load iu = 1.2tv0 -r 1.6tvL = 1.- X 1.37 + 1.6 X 0. 96 = 2.60 klftor. wu = 1.41 D = l.4 X 1.37 = 1.9_ k/ft (doe not go em}
1b Mjdspan po itive moment for pan B-C:The di tributed live Joad acting on thebean1 il L = qL ( reduced ) X tributary width= 52.5 p f X 1_ ft = 630 lb/ft = 0.6 k/ft
o, the total factored load iWu = 1. .. D + 1.6wL = 1.2 X 1.37 + 1.6 X 0.63 = 2.6 k/ft
S in.
24 in.
I 12 in. I
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4.
c Negative moment at upport B: The di tributed live load iWL = qL ( reduced ) X tributary width
= 40.5 p f X 12ft = 4 6 lb/ft = 0. 86 klfto. the total fa tared load i
1 tt = 1.2wo + 1.6t L = 1. X 1.37 + 1.6 X 0.486 = 2.42 k/fttM Negative moment at fa e of upport A: From Fig. 5- lld. the c ffi ient inegative l/16.
1 in .n( - B) = 2 ft - 1 inlft = 26 .5 ftMu = - 1/16 X _.60 klft X (- 6.5 ft )2 = - 114 k-ft
b Po itive moment at mid pan of beam A- 8: Fron1 Fig. -- 11d. the coefficient ipo itive 1/14.
Mu = 1/14 X _.60 klft X (26.5 ft)1 = 130 k-ft(c Po itive moment at mid pan of beam 8- C: From Fig. 5- ll d the appropriatemom nt coefficient i po itive 1116. and from tep 3 the total di tributed load i2.6 - klft.
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l in.n(B- C) = -4 ft - I . Jc. = ft)J l Jt1M14 = 1116 X _,6 k/ft X { - ft )- = 83 . k-ft
td . egative n1on1ent at f e of upp rt 8: Be au th bean1 e ti n de ign \\'illn t change fro n1 on id f the c lumn to th th r. th final de ign at fupport Bwill need to be forth larger of the two n gative n1oment fron1 the int ri r
and e terior pan .Th aJ ul tion of th 111 111 nt \ ill u e the average lear pan .o the larger f the h.vo mon1 nt fficient wi ll govern. Fr 111 1g. - l ld it an beeen that the coefficient fr m the e terior pan (negative 1110 [! r 111 re than t\vopans) will govern. ing at tal di tributed load of _ 4._ kfft.
n(avg) = . ( ...6. 4- 2_.5) = _-t.S ftMu = - 1110 X _.42klft X ( 4.5ft)2 = - 1-tSk-fti"n n1 n1ent ~ r b n1E- F- - H. i\ quick revie\\' of ig . -l l
and - IId indi at that the nly hange in n1 m nt c effici nt o cu t the exterior end ofthe exteri r pan ffi ient change t negative l/ __4 . However. depending n th ize ofthe gird r u d in thi floor ten1. th lear pan a1 ~ c h a n ~ e un1ing the 'Nidthof the girder are L .. in.. the lear pan and the a v e r ~ g e clear pan are
n(E- F) = ...7ft . n(F- G) = __3ft. a n d ~ r thee terior and interior pann avg) = _5 ft
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M11 ( ) = - 1/24 X - klft X ~ 7 f t 2 = - 79.0 k-ftM (mid panE- F) = 111 X - 0 k/ft X ( 7ft - = 13 k-ftMu (n1id pan F- G) = 1/16 X _,6 ft X (- ., ft )- = 7.6 k-ftMu ( = - 11 X .... w k/ft X ( _. ft )2 = - 1 1k-ft
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For gravity I ing n th ntinu u t1 r am- B- C-D in tg. -l an .. ept ble analy i m el i ho n inrepn ent th olun1n I ngth o and be l ~ th fl r y ten1 ing anal rzed. \erti a1ller up rt h u]d be added at ither joint r j int D to p1 vent horizontal di pL en1 nt
t th f1 or e v l.
a
A 8 c D
28ft 24 28
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ig. 5-14Rigid end zone in frameeJem n .
Fi f!. 5- l..Final de ign hear andm ment at fa e f uppon.
ode i odejwlX X n I-
Face of support
M node(Vnode X
Vrace = VnodeMrace =Mnode - Vnocte X;
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Fig. 5-16Effe tive beam Lion forflexural tjffn anaJy i of an r beam arrying gravityloading.....
bflange{s iffness)
For the initial analy i ~ e ign y le. prelin1inary 111 1nber izeon p1ior experience \Vith in1ilar floor y ten1 . Total b an1 depth . h ~ are typi al ly in thrange f 11 t /1 _, \Vhere i th ent r-t - enter pan length of th bem11. In typicalU. . pra tice, bean1 depth are rounded to a whole in h unit and often to an even nun1b r fin he . Beam width, b, or web width, bu1 ., commonI are taken to be appro imately one-halfof the t tal b an1 depth and are r unded to a whole in h unit. r hi t tural h1nitation on
ible di n1 n ion and required learan e al o 111a affect the ele tion f pre liminary
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unt [! r fle uraJ r k i n { ! ~ th . iiTO 111... I; ;ment of in rti pi ally i redu d too tain a value t r th rn ked m 111 nt of inertia., Icr . ;nm1on pra ti e f r an1 tion i t un1 that l cr i appr imately equal t 0. Ig .
1 3I cr(T-beanl) - - ballI_.Thi pr dure elin1inate the need to define th eff ti ve flange width and th re ulting~ n1ent of inerti ~ r a tlani! d b am ti n.
... - - - - - - - - - - - - - - - - - - - - - - - - - - - - ~
E
t'hI 28F G
A fJ,+ 2 tt +ntinu u an1r n ~ a ' lab upp rt d bya
H
~28 I
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wa don in Exam pi -- I. e wil l fir -t on id r the ontinuou floor beam-B-C-DinFig .S-LTo det rmine th fa tored de ign m men at riti aJ I ation alongthi ntinuou b e r u n ~ w wll u th an I, i mod I in Fig . 5-11 wi llu II of tharne mem ber dim n ion , d ad Ia, d , nd r du d live load aJcuL. ted in Examp le 5-IForouranal i w \ ill u theappropriat patt rn live lo d tom, ximize the mmrnt. at
the criti all ation . fter we have fini hed the na ly i of floor b am -8-C-D. w \ illm,ke . imilar al ulati n. for noor b am E-F-G-H n ig. -4.To determin th fa toredmom nt at riti I I tion alo ng thi ontinuou n r b am, \ e wll u e th anaJ imodel inFig . -1 7.
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1. n ly . m d I fi r n r b an1 - B-C- D. The beam pan length are given inFig. 5-13. and we will um that the coiUJnn length above and below thi floor y tern are11ft . The gro se tion propertie for the olunm are
Ag = 18 X 18 = 324 in?I g = ( I ) l/12 = 750 in .
A di u ed previou ly. we wi ll as ume that the approximate cracked moment of inertiafo r th e b run an b taken a the gro moment of inertia for the extended beam web with
urn d di1nen ion of 12 in. b 24 in. Be au axial tiffne of the beaJn will
Fig. S-13Pe rmi ible analysis mode lfor con tinuous beamsubje Led to only gravityloading. I
8
28ft +
a
c D
24ft + 28ft I
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hav , lmo t noefTe ton the analy i re uh for n imum mom nt and hear . th amappro imation an b u dfor th beamarea.Thu .
(beam) ::: A(w b) = 12 X 4 = . ,ln .-rall of the bean . \Ve wi ll as m that there i a rigid end zone atea hend of the beam Fig.-14 equal to on -haJf ofth olunmdimen ion. i.e.;9 n h . . Lm1ing acon rete ompre -ive rength of p i, th el, ti modulu for th bean1and olumn tion wi II be taken
Ec ::: 57 ,ooov' p i =3.60 X 1 6 p i =3600 k iThi hould be aJI of the information required for input into an appropriat t1u tur t anal -i oftware progran1.~ - - - - - - - - - - - - - - - - - - - - - - - - - - - - ~
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... nul) i r ntaxitnum m 111 nt ._ t . nnd tnid pdl f m ntb r . -B. Theappropriate ive load pa ttern omaximize themome nt at and at h mid pan of th m mb r -B given in Fig . -I a. deten11ined in E amp le --Lhe di tri buted de dload forall p,n i 137 and th redu ed live lo,d for thi lo,ding p ~ t t m(, determined forpan -B) i 0.596 k/ft. ing th lac df tor of I. for dead lo dand 1.6for Jive load, thean ly i re ult for the model and lo,ding how n n Fig. 5-1 a are M = - IOw k-ft dld11(mid pan ) = 10 k-ft. Thee re ult are ompared to tho obtained u ing the Cl
Moment C ffi ient in Tabl -I. II of tho e re ult witt be di u ed in tep : of thiexample.
J. rtt ly i r ntaximunt 111 ntent nt rnid pun f ntem1er B- ..The appr priatlive load pattern to m imiz the mom nt at the mid pan of the m mberB-C given inFig. -I b.The di tr ibuted dead load i un hang d;and the redu ed liv load for thi loading pattern (a detennined in E ample -J for pan B-C) i 0.63 k/ft. U ing the load fa torof I. for dead load and 16 for li ve l o a d ~ the analy i re ult for th model and loadinghown in Fig. 5-1 b i M(mid pan) = -6.9 k-ft. g a i n ~ thi re ult i compared to thd
obtained u ing the Cl Mom nt Coeffi ient in Table -I.
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4. . n, i r m, ximum 111 m nt , t fu f upp rt B. The appropri te liveload pattern to maximiz th moment at th f e of upp rt Bi given in Fig. 5-1 . Thdi tributed d d load i un a n g e d ~ and the redu d liv lo,d for thi loading p ttemd termned in E , mple --I for pan -B nd B-C) i OA 6klft. U ingthe load fa tor ofI. for dead load and 1.6 r li\e Ia the an( 1 i re ult forth n1 del and 1ading hownin Fig. 5-1 i lvtn(e terior f, e) = - I ft-k ip and Mu( nterior fa e) = - 12 k-ft.TABLE 5-1 Comparison of Factored Design Moments for Continuous FloorBeam with Column Supports
Moment Face of Midspan of Faces of(k-ft) Support A 1Member A-8 Support 8Re uh u ing - I I4 1 0 - I-tACI MomentCo ffi ientRe uh from - 102 105 - 14
Midspan ofMember 8-C
6.9
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Th top te Iu dtore i t th n gative mom nt will be ontinuou through the olunmand thu will be de igned to e i t the larger of th two momnt . ony the larger momen t i ompared to hatobtain du ing he ClMom ntCoefti ient in Tabl -I.
:. n1p, l'i n f r ult 1fl r b nrn -B- -JJ. omp ri onb twe nth fa t red de ign moment obtained from tru tural an ly i in thi e ample and from
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' mpnri n f rr ult I' fl r Icatn E-J -11 .Th nalyi. Jnod I or th iontinu u fl or beam. whi h . upport d b gird r. i given in P.i; . -I . For all of hbeam we will a um that h r i a igid endzone at h ndof th be m (Fig. -I )equl to on-half of h width f he upport ing gird r I in.) nd , umd b 6 n.for thi.an l i .Th pc ttrm live I d and he rcdu dvalue D h liv I dwill ntialy h.ame ( h o u d n part 2 3 and4of thi .ample. marion etwe n he a toredd. an moment obtain dfrom tn1 tural naly i u ina themod 1 n Fig. -I and hoefrom ppli ti nof th Clmomnt fli i nt inEx mple -I i.given in T, bJ ..- f r hontinu u fl rb am (EFG-Hwth gi rd r be 111 up ort . . n t d n thepf lou. tcpforth 1nd pcUl po itive mom t of an interior sp IL th n: ult obtai ndfrom th CIm-
ment ffi i nt are normaly ignifi antly a r ~ e that ho e btained fr m n1 uralan I _. . The re ut at the in terior upport (fa e of upp rt F ommonl l r hgher fro n1 thetru tual analy i 111 thod but f ~ l a ly lo to tho e btaind fr mhe Cl m mnt
effi ient .
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Use of Structural Analysis to Find Design Moments inContinuous Floor Beams
L D
28 24 ~
LL
28
(a ive loa pa e to axm1ze egati e 0 ent a A apositive orne at idspan of e ber A- 8 .
a
Db
d
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24 28ive load pa e to . axmize positive moment aidspan of me be r 8- C.
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(c) ve load pa em o maxi ize negati e momen at B.
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Th f( ult for the xt rl r.pan \ ill a ted . gnifi anti, y h a 11111 dpinonn ti natthee terior upport r hi onlnuou n r m i . .. -I .Th al u-
lat dmmnt atthef of th t rior uppOJi i. 7 ro, but . not dpr i u I , h authorr omm nd that the momnt bt in du. ng th CImoment c ffi ient Cm, qu Ito- 11 4 hould b ued in q. L-.. ). Thi re ult i ho\ n n parenthe i in 1 bl .. -...B u of th zero-momnt r i. tan tth ndof th ext rior p n. th midpan po itive mmnt wH larcrer forth tru tu ral analyL omar d o the re.ult from the 4 Clmomnt fT1i nt. . . ated pr viou. y, thi . anal i. pr dure d e re ult in anoverd sign for le ural trenc: th in th xtcrior. pn.but it I o an 1v tim whn h king the tor ional trcncr th of the ndrel b(am. If during th tor ion, Jde ign it i foundthat th pandrel bam il ra kund r a tor dtor th CI ode \vould r quir ared i tribution of mment int the xt ri r pan of th floor be mHw r. if theanalyi pro eduredi u . dh f wa. ucd t detcnnine the fa tor d d ign mm nL in the tc rior pan of the floor an1no r di tribuion of mm nt i. r quir d. I
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TABLE 5-2 Comparison of Factored Design Moments for Continuous FloorBeam with Beam SupportsMoment Face of Midspan of Faces of Mdspan of(k-ft) SupportE Member E-F Support F Member F-G
R ult u na - 9 L5 - LICf Mmn.Cefi i ntR ul from (- 9) I 9 - ]
tru turaln ly.
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