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  • 1. iN MEMORY OF AMI ~.­ _.- - --- -- --- .. THOfVISON I _.- -.. .- _...--- - -•• -'-.. ~. ~­! co. <neu. Sl {_.""..." II'i6llMe,,,, "" '-"P.,.o 10 c..1101M''Ooil ~ '50" Mod"d ,
  • 2. Contents PART ONE INTROOUCTION TO STRUCTURAL ANALYSIS ANO lOAOS Introduction to Structural Analysis I' !l 1.2 R, u , U , 2 loads on Structures ...., , , " E " (:I< J I U , L, " ",. 'nJ I " ,., ,,"" I, ,,' ' ,.. (.1t(I1.'l""~~ I "V ll)dr, 1.'U, _,n,l "."ll',e ,me ",.. I hCITlhll ,m<1 Olh I IlL"·!> , ,.. 1,>.,,1 (,'mb'n,III<' I> "
  • 3. PART TWO ANALYSIS OF STATICAllY OETERMINATE STRUCTURES ~I 3 EqUIlibrium and Support ReactIons .... u u - .. Beams and Frames: Shear and Bending Moment Ib! " "., "".. " Plane and Space Trusses " " ., .. "..., ..u "'.1 I) IJ' " ",,11 >lW Sh(ar .md Ikndltl~ 'lomenl Ih~ <;h al ,,,d Ik",lln~ M"m;:nl Diagram, I(,~ " 6 7 Dellecllons of Beams' Geometnc MethOds s• "..u u u Dellecllons of Trusses, Beams. and Fra-'...... Wor1l-Energ, Methods " " ""u ""u Influence lines •..
  • 4. u ..u 9 AppUnbon of Unn " .. 10 Analysis 01 Synunetrk SIrUctUm ,a, m ..., ,.. '", • '.. 13 Meltlod of Consistent OeronaatiOfts-FOfC8 MIUtod •,. 14 Three-Moment Eqaation aad the MettwMI et least WOft '"", '" -- PART THREE AHALYSIS OF STATICALLY INDETERMINATE STRUCTURES '" 15 InnutBCe LinK tor Stamally Indetemllnatl! SlNl;turn '" '" IntrodueliOn to StatlCall, Indetemllnate structures 4 ; 111 ' nJ [} ," nla ' r ". 11 11 2 • llde1 ,,".,., SI, ',' 16 Siope-Oeflection Method ,..'"'u
  • 5. ---- _c__ • •
  • 6. Preface
  • 7. " ,I n, I (, .don II II I ( rk UIl l Goo, Illa,,,lt.,,,1 {",,,,,,, A ,,111,41 ,
  • 8. Part One Introduction to Structural Analysis and Loads .. -~ ~, ...",~ - .. ' I •
  • 9. Introduction to Structural Analysis 1 " "".. .- -- . I . .
  • 10. fI'I.1 h , , II (It -..........-..-.~ --- ...,. • 1 11' ./,~ , t:' ./" "" "• • -c;.- _ --- mputcr hut 1,,,,,( U} '''lei ,n'rLl1 0'." J no" he p",h,nl <"<.l ,n '"' the ,'''IClll "omrU1"r·,>ncm,,,j meh,,,j, "I ltllhutcd I" ,,,,,,'n~ "the" J H I~YIl
  • 12. • Compression Structures Compression structures devd P mainI :,~E~~;acllon of external loads Two common c fum and ar h Columns are compressive loads as sh "'II m Fig. I subJocled to lateral loads and or m called a oIumn An arch I a curved structure w th verted cable shOWDI.F,g I b s:::~:~~r~::~::: upport brid and long J AG. 1.5 Tacoma Narrows Bridge O4,;illating before Its Collapse in 1940 '~'!'Ilf ~ 'n~tIMLOtl P!l1O No n 187 AG. 1.4 SU',pen')lon Bndgc Ib' P, P " 7't'----={P, P, la, Tension Structures The memba of ten 1011 stTUl.:tun:s aTC subjected to p~re lension tht' .tclion of elI:mal IlMd . Becau~ the tensJle sin:...!'. IS thstributed 10nn" Ol'T the: cro ,·....:ctional area.. of members. the material of a ..(r~cture I' utili/cd in the most efficient manner. Ten~ion t lomJXhcd of 11r.:lbk 'lcd l..tblc... are fre~uentl} e~pl~)ed to U bndi!c and long· p.m roof Bt.'Cau~ 01 their flexlblhty, cables negligible bending liJlne' and can dce1op onl} h:nsion. Thus ctcmallo,Hb. ,I c.lble .ld0ph a hare that enable~ It to 'upport the b ten,ik forl:cS alone. In other ord':i. the shape of a cable c a~ the lo.ld, .Kung on it ch.lJlgl: A an elample the ,hape~ Ihal a cable rna) a"uml' under tO different loading: conditions are shown Fig. 1.3 Figure 1.4 ho~ a familiar t)-pe of cable strw.;ture Ihe WJ: hrid!jt In J "u"pcm.lon bridge. the roada} i., suspended from Iwo cable... b)- me"n, or crtu;al hangers. The main cables pass over a of lOH~r., Jnd are anchored mto solid rock or a concrete foundabOD their ends. Be-cau.,c su"penslOn bridge., and other cable structures .,tifTnc,., in latcral directions. the) are su~eptible to wind-induced lations (~c Fig I 5,. Bri.KlIlg or .,tiffening systems arc therefore pro to reduce such oclllatlon,. Bcside., cahle ...truclurc other examples of tension struclures lOci crtical rods used iPi hanger., (for e>.ample. to support balconies Or ta and membrane .,tructurc su(,;h as tents. • CHAPTER 1 Introduction to Structural Analysis he rl,,'<Ill1~J lh.tI till) tll or !l1t)re of the b.I"1l.: Slru~tural I)-pes descn II ' 1-.., c('l1lhlnl'J In .1 ~lI1gle struLturc., ~uch as a III Iht: 10 l'IO' 01,1) I~ • . C , I' n1 "I IC' ,truClUTe ... functional requirements. IIlg l)f t 1,r1U~l' l II II 11&.1.3
  • 13. .... -- ~lllILLLLW ~ ~ T ILL ""'I hl;UI)
  • 14. iii. 112 t .. MALmeAL MODELS ~, I" <!k.. ',,,'!I<6n>rM,, r • mox! IS."'"' 1 Ih< "IOSI ,mr<> ,. I .., . "'Gu,m <'pt"" "" nd '",,,,J.;,J • t " • '.1 lh"f(lu~h Ul"k'Ola1ld,n~ .,1111< b<h" b<:J ',,"Ilhl: lr1101U1"1 IO,ponso plO<I"I"t I",", 01" '4hd "ol, ", Ih, ",tenL thai ,It< mudd 'Cl're<enl lJ",ch'l'menl <>f {hc .n.I>li,al mood Jl'r ",,, n "t Iii<" HI"" ng f."" < line Diagram Th '"'
  • 15. reu r used , /I." ~nd 2 fir hi ,,,n temled a 1>1, the II Connections Two I ~ ! <"nil" lure' I J r lh'rd type, ,"~ 1< • , "' , " " . Ith, h ",
  • 16. ....ons 2 Loads on Structures """,. """,." SUMMARY - .,_ .. 'It "'1m, ,11"1 I'r "lh~, 1I III< 1 ,I ,.j ., ull 1'0: 01"-..1,,, ..L,nce on 1"".1 I" t,({"",k 1/ f) "
  • 17. .a.t 6 1 t 150 490 40 Malenal ExIlnPle 2.1 Alummum Bnl:k Com:rete mnfon::cd Structural teel Wood TULl 2.1 UIlfI WElIIl11 Of COIlITRUCTIOII 1IA1DIAU • The numbcr'l brae: ref1 I~r.:;~=r".:~~ t llcms 11s1cd 18 the bibtiop'apby II 1 BeD be purtlwcd lrom the American Drive Reston Vuguua 201914400 BllIldm~ m,d (Jllu, Smu tim (SEt ASCE 7-02) II· Manual L I'Ll l)tdmltml Srt llfIC dllom lor H,ghn,a Br II Ol f:-flflIII( (,mtl ....l and Jnlf "l(IlIon(l/ Blllldmt/ (mit )1 Sj Although the load requirements of most local buddtn. generalh based on thllSC of the nallonal codes listed herein 1 rna' co~tam addltll1nal pn.l IslOns arranted by such re dlu~ns as earthquakes tornadoes. hurncanes, heavy snow and Local bUlldmg codes are usualll legal documents enacted 10 pubhc "elfare and safcI} and the engmee~ must become famdiar "nh the bUIlding code for the area 10 whIch Ih struetlll be budt. The load descnbed on the codes are usually based on put enee and stud, and are the m",;mum for which the anoUi structures mwi be designed. HO"eer. the engineer must decidIi structure I to be subjected to an) loads in addition to those by the code and If so, must deSIgn the structure to reslSl the loads. Remember that the engineer is ultimately responsIble for design of the structure. The objective of thIS chapter is to describe the types of monly encountered in the design of structures and 10 introduce concepts of load esumation. We first describe dead loads and cuss Itve loads for buildings and bridges. We next consider the effect or the impact. of live loads. We describe environmental indudmg wind loads, snow loads, and earthquake loads. We IlIWl discussion of hydrostatic and soil pressures and thermal condude with a discussion about the combinations of load design purposes. The material presented herein is mainly based on the A.S dard Mimmum De>ign Loads for Buildings and Other Slruc ASCE 7-02), which is commonly referred to as the ASCE 7 and i perhaps th most widely used standard in practice SUIllII tent here IS 10 familiarize the reader with the general tOpiC of structures many of Ihe details have not been included. Need the complete provisions of the local building codes or the Slandard must be followed in designing structures. Dead loath are gravlly loads of constant magniludes and fbed :"::' permanently on the structure. Such load CORStSI of truetural sy lem Itself and of all other rnatenal and 2.1 DEAD LOADS 1. ClW'18I2 Lua on _rea
  • 18. .. 2,2 LIVE LOADS SlCTION Z2 ltf'I L..- " Uve loads 'Of BUIldings T nlu 2 l '" IjI/AlM FUIOIl lin ~OAIIS FOR IUft-OIflGS • H ,. ) , ,. , '" He "'1",1' n,nUl cd I,), b) 'h , , .., , , " An~ ,"
  • 19. .:2 C!lAPIUI 2 Lo.da GIl St,uclUl" live loads for Bridges I ...... 1M L ,, IlG 2.2 I 11, fir tn<! ..... , -"Ll----,-I-,-,I,,-,IJet,-----,t.....Jt----,-1=--..!.-....t LJ..--L..L-1.. • •
  • 20. " CllAPTER 2 lDlGllNl SII"'IIl... .. 2.3 IMPACT 2.4 WINO LOADS "Xlu !lsI''''' n In anJ l~ gwm~l" ."hl1h~ (,", If Ilh"u~h the procahLM d I Ih "n ou oJ. f>r 1M em Un" "f w nd I, ,-I ,~u.,lI, 3l)" <kt,,,1 Ill.;>!;l of Ih m Jrc b'l-ed <'n the "'me 1;, l. rr IB'" hll' 110:1 n II wmd pc,-J I JllJ Ihc d) IlJOB" pr" lr'" '''<.incoJ "n .1 tiJI I/'!.',- I 'mIJI to Ih~ mJ n"w "hL,h "In be "hl,n!lc<! b Jl'l'ljtn~ H"",,,,,lh' pntKlplc amJ ""pre ",-"j a
  • 21. D I II I I 1 II 11 Exposure Urban and suburban areas Ith closet paced obstruction) of the size of single mil h or larger This terram must pmallln the up'"'lod dlrec.1:lon for a distance of at t 2630 ft 800 m or 10 urnes the buddin height whichever I greater Appll s to all bwldangs 10 "'bK:h ••1""'..... B or 0 do not appl) Flat unobstrUL1ed areas and w out Ide humcane-prone regl must f" II In the up" nd di dlsumcc of at lea I 5 000 ft I ".. m ,.... til< bUlldmg bcigh. wlu< Souru' ·daplcd "ilh permi~ IOn from AS( E 7.t1~ #mlnl/lm J) from ASCE 7·(J2; for further tnlOrmdlllln the I;ompleh: le'),! html'l(}9911~O) OI.:l.:Up.IIlC) or u:-c BUlldmgs rcprc<;cntmg loy. hazard to hum.a"-':n~I~'l:'~,"':-;lh~':-::'=::;:--~~':".~-=-=-:-~~~---~~~-~~~- fallurl.:. uch.1'; agncuhur.ll and minOT h'rage lal:lhtu: 11 hUlldmgs oth~r than those 11 tcl.l In (utcgonc 1 111 and IV Building rcprc$CntlOg d ub-.tantlal ha/.ud t human hrc: In the -':<l I.: of failure. ~u..:h as those "here more th n )00 people congregate in one area: da~·care f.u':lhtle Ith I.:dpal,;lt gTe-,Hcr than 150: ",hooh "jth C<lpall!) greater than :!5fl c(111cgc" y,nh capal.:lt) ~rcalcr than SI.Ml hO<lpltcil without emergenc) tre.ttmcnt or ~urgcry fUl:lhue but oath patient capacit) greater than SO; Jail,. ro"er stations and utilitic.. not c....cntlal in an emcrgcnl1. and hUildlO~<; l.:ontaining haLardou!> and explo~IC' material E'i...cntial facilitic'i. indudmg. ho"pltal fire and pollet ,tatlon'i. national defen-.e facihtle'i and cmergenC) shelten commUOIcatlon center.... po"'cr ~tati(lns and utlhtll~S requircd in an emergen') TABLE 2.4 EXPOSURE CATEGORIES FOR BUILDINGS FOR WINO LONIB c 1ECTIOIl2A _ ~ " ,~ I "~ ~~ •~ ~ " .¥ = f. -::.". V" -::: .,... '" -;;:: .. .-"'t ... "..----- , 26 CHAmR 2 LOid. on Struclu"'.
  • 22. 2lI CIW'TElI 2 lDlCls 01 Slruc_ JllIllllilli~ = r= ~ IlIIIUIl1ll1 - = ~'lllIlllllllr =/1f!/ftIi/Jlt.'iJ "-- _--:0 -= + " , " ro .. '"' .., '""" "", " {furl M, nlrndl ,.j. P. "",II, lndoo.:.! II lor ~lllh h" ,., ,f"- 2.5 I , -IJlIlliIUTllllUlliIlIL' = E=- § 11IIl1llllllllJIIIlllfiE, u ( I II
  • 23. , 0, t ,1>1 ,~t r Ilh In. <lew 1" h'll1o -
  • 24. 2,5 SfKIW LOADS
  • 25. • 2.6 EARTHQUAKE LOADS , , r 1111 " • J~, ,.. I,I dl flllllTTTTl • , . EQ"'" 23
  • 26. • C....PTEA 2 lOldlon Slruct"". ,. '" 2.1 HVDROSTATIC AND SOil PRESSURES 2.8 THERMAL AND OTHER EffECTS , the bu ,,, 2.9 LOAD C(lMBrNATlONS d lr J Ihe 1''' Jed the I al Iale lhIH I Ing Ihe al del I>"ul lh , ht (,,<oJ , ..~)( I ~I""'"
  • 27. CHAPTU Z LIlIU on SlrUCIIJ," SUMMARY .. .. • fill. P2.3. P2.e ..... Ed ~ .. " ---- 12ft. 1ft . , PROBLEMS I'll; 1'2.1. P2.2. P2:5 ~b ~5 1+. A A "' .."" , " n-«I{' !o.:l.J< ~. U"'~ " r T~ J,' I ,,,,,<I-f( 'II • ~'r'" >J
  • 28. Part Two Analysis of Statically Determinate Structures Ih ",nd· , ' 10 .....,1'1 hi .. •--, n•• • 2.11 1)" Ih. <" 1.1" kl J ...,,1 """",,,,I "'all< "t ,I<. h",I,Io, II P, " fI'- P2.9, P213 • ~ I m * ~" .. d. a, m /II If "" I P h"d,l "i ' 011 I" hi d, t pd O,t,,,,, th If I ,de,l(, ,n I 2.11 'Ihe "nd I, U> .... II I' 2.1 the ,,~)f I Ih<, 1 ~ 1 "h ,I,' l I, 1 m nth.: t><" u n~ ~,. 1,1"4 ,. f'2 4, P2.1 40 MAPlEII.' l_1 on Slf!lClu,es • f ....._ "" P2.IO, P2.ll. P2.12 • prn<U" th< '110 I ''C d"...n f 1',9 rh ,lI. ("hl<~~' III "'I'" "'" B I h. >oJ J.n Ll" , "( I ....· h tu.d ,"" '110 q It.. Jlf nJ I" d ,n UmL Ill< nr<"Uf< t, I"" "
  • 29. 3 Equilibrium and Support Reactions "".., u "30 " 3.1 EQUILIBRIUM OF STRUCTURES " "" t ~Ir In 4uihh"uln
  • 30. .. CH'PT(R 3 EquU,boUl1llnd Support RelClKln$ $ECTIlIH 3 I lqulto"""", 01 ~ IS 3.1 h 'rue, r m ,h r lJl<; wm "aCllll~ on m hal anJ 1 , n I' an: ult.un I,',.. n r n: ,Il-,n Ollu,., 1''''1 .... 11. to'", ru/!. lh"fl" "r'''' thee t'ud.... , I I I" tlu" ··,!I""n lOn.,I) Ill' 01 t.>I" hll l I I I ,It I"" and I>c d'" '1c "lon.,l. tc nd ""u- "th", ,1ln~ JlI '11 [<>nidl ~I' 1 >I'rlo> lit ,·<lU· "~ r<:c "m, '" n r b" 'urn 110 'd, l<llh ale. Atlernatlve Forms of Equations of Equilibrium of Plane Structures ", :c lhe F " F " • pr,'"de IruelL r nl one Ih~ = Concurrent Force Systems
  • 31. .. CIlIoPTER, (q"llobllUllllrtlI S"PlHlfl Re.cIlOftI IftllOll14 ....... ~ ...Io .J .... ~ 3.3 TYPES OF SUPPORTS fOR PlAJlIE STRUCTURES 3.2 EXTERNAL AND INTERNAL FORCES External Forces .. Inlemal Forces /"'"lUI" rt lilt I, 'd ,Ie ~ertl:<l • t>er p"l lam (II 111 tr I , 11, Ihe rc "llhe tlLlo.(urc Ihe~ I"" d ,1"1' 3.4 STATIC DETERMINACY, INDETERMINACY, AND INSTABILITY
  • 32. .. CItAJ'T(R 3 Equlhbrlum and SuPl'ort,ons , .. fl5. 34 R " "!'"n ,," "" •m ,:":':'''''><' II I Ll uaJl) , "nl 10 rr .1 II h) n I 1'''1<1 R ndll II lll<k II 11<1/1. I. 'I. .~. fIG 3.3 T ", , " '"
  • 33. .. .. , SUtic Determinacy ollnlernally Stable Structures
  • 34. " " ~-- M.:ll0 ., '" Rli 3.11
  • 35. lIIIulIaIl Ii Bod) D ram See Fl. ) 19{b A A EF 0 I 2S 18 IS 0 2 A " &actfcms Del mllIUIC The frame 1 Internally _cally determinate ~12ft;-t6l6ft I S kIft 6ft In-t-6ft I BtB, 2Hilt .'aU I) (h) ... S.1. .......__ 1.. 11...
  • 36. ,= "",...U' 2 J 200 o o E 840k o o ISk C4lJltidlrinl the eqwllbrium of portion AC we wnte EF 0 A 0 0 A 0 ~EMA 0 125 ill 200 0 By 408k EF 0 A 80 408 225 0 A Ion
  • 37. (d) HIIl§C (oj Jt Hmr (e) Jf (b) (nnmatc If the structure 15 stahcally ind....... n 11 (hen detcrnllOC' the degree ofexternal (d) "" the 1ructun:5 hOll nat or talK II md JL b :: If .... • _ I .... 11I1......" - ' - ..-U I.I ....... U 1&Pl.1 _PU 1&P1.2
  • 38. • ft ft -t I I I I Hili ,j I I I I 15 A B .... R 10k 10k I ilift 2 klfl ",el. ~jp 1-1- - - - 10ft -----ll 25k_ I I I I I I I I I I I .... P3.11 15 It 5m 24 k ~n ... 4 al6 m =:!4 m 70 k. 8k 8k 24 k 24 k 6a12Qft=1201l ~-----4u20ft=Wft------~ FIG. P3.17 .... '118 3.1. tbroUlh 3.•1 Dctcrmml,: lhe TI:.tctton ,II the urpons lor the trm:lurc, ho""n b(n ft 1---- ,-----1 25m .... P3.15 , L"_ -'1 -". - FIG P3.10 I r-'5 ft -t'----50 ft ----I FIG P3.14 A 3.1' Th~ clghl or a ~·Ill·Ill1g Irolh:~. mo.. mg peed on a Otam hndge: 1$ modded it a m 1.11 tnbuted load a ho n In f Ig pl 15 Dc:l.rmil~ pre IOn.' ftlr the crlu.:al rcactl~ln al th u 01 the po IlIOn 01 the tr IIIe') .l,~ mea un:d b ,.md rIot the ¥raph hi mg the ....arlatlon nf,,""'•• a~ fUnl1l0n (ll m 10ft~o ft rtUUIIJJt' . B , t A ~ 1 - 1- - - 24 ft - - - - CHAPTER 3 Equlllbrium.nd Support Reactions I.. m A ~.' !41llU~ .... ~2 m-jI---5 m-+-J m~ ... '11 ... AG. P3.12 .... '113 .... '111
  • 39. • .......-_.• : I . . . . .' I I ft I ft • r--I~•.-~-I~ft ft L-_ _ AA ...PUI ...PUII Hill 25ft-+--2J
  • 40. ...........-
  • 41. .. F T
  • 42. · . ··'f.r " . " . _.:~.~ , I:~>' ....':' ~ ,:' f~ shown m Fill 4 12
  • 43. '" ' I '" + - @,B F ... -13 8 ".=11 J=7 ,-) m+ ,.2 19) StatlcaJly DetemunalC' E StatICal) IRdct rm - 4 ",=26 J= 15 ,=4 (e Staucall Determinate ",=16 1=10 ,=4 ".+,=2) t) Slabcally DetmniIWC ".-19 J=12 r=S m+r-2J ",_17 ) If) ' .. 2 mt,<,q hi Stabcally Delcmunale A~ SECTION U Stag 0 c etenninlCV, Indetermtnacy, 1M InItIbIItty of PIIne T.- E nr=1O j=7 ,=3 nr+r<2) (0 Un~table E (d) Un"lable m= 10 j=7 ,=4 m+ ,=2) (h) Statically DelmninalC' m=17 )=10 '=3 m+r=2J (a) Sldlleally Delemunalc m=16 J=IO ,=1 m +'< 2J FIG. 4.15 1 Ikcau..e m , 2 tallc Indeterminacy , that the lru S ;onlalPJ fI III dod,I' )no(b) hlT thl tru un table I II" ,,IIK llr',I.ll-lIill of tru~,e.... ,i' bOlhncees Thr.: hr ~ ..'C'nJIUtll1 . M ~l.:lh:1 If III 21 -- r. the '.'u., .. ,.. dclinncly ta dod ,uthucnl III thl.: SCI1~ • d r • th' rem llllllll! 1-0 l:on Ilion.... lor ,talle del ,,:all} un"t;.lhk IloCHf. t.: • _ - ' l . . • , ld mJc:ll..'rnlll1.lC (m ":'J J . •lre nCl:csary minJI..' m T' al - , . . I In other Hmh. thc...c (0 equation, ..imply nol ...ullKu:nt t(111l11hl n , .' - , . h I· n -mha and rc.lclums I' "uffi.uent lor stabiti w, that the mOil (r ' I l.: , d d - -oy mform.llIon regardmg their (lTnml/(m 'Iff The) (1 ntll pro!' I c.: .... _ h II " Ill numl:x'f 01 mcmhcr... and e'ternal reach Iru.... rna) Jloe.1 'U h. t.: , • hut ma~ 1111 be un...tabk du.' 11..1 Improper arr.lOgemcnt 01 members or e"lema! upp..lrt... _ ' . e cmph.l Ill." th.ll In ,,'rdcr for the Crltena lor ~1.l1il· ~e(emll and mdelcmlmaC ;l ~l(:n b~ Ell." .Ll and 4.4 to be ·ahd. the lrUII O1u..t be !>tabk an"d act a a. lOgIc rigid bod~ under a general system 'oplanar IllJJ hen atl'lchcd to the "upport-.. loternan} ~table tfUllel ~U"l lx= upportcd b~ at lea t three reactIOn... all of v. hlch must be ther parallel nl1r concurrent. If a tru,,~ l~ mlernall} unstable then It m be ..upporteJ h~ reactions cqu,l1 in numhcr to at least three plus nu01lxr of cll.uallom of comhtlon I.l + t, and all the rcacUons mu t neither par.llld nor concurrent. In addition. each joint. member portlon of the Iru..... O1u"t be con"tr.lined against all po!'>~ible rigid bod, mOemenb 10 the pl,tne of the tru"". either by the rest of the truss or external "urpart If a Iru" contam" a sufficient number of mem bUI the~ are not properl} arranged. the lru'" i'l said to have aiticalfi For ..orne Iru.."e~. it nhl) not be ohious from Ihe "'hether not their members arc arr,tnged properl). Howeer. if the member rangemcnt is improper. it will become eiden! during the analysis of tru..s. The anal)"is of ..Ul..:h unslahle Irusses will always lead to meOD tent. indetcnnmatc. or infinite results. --------":~CIa ..if} each of the pl,lnc tru..~....hlmn in fig. 4.15 a" un"tahlc statlcall lermlllat~ or statical!} ml.!ctcnmnate If the tru" i.. ,tatll.;all) indet~rm then d~krmine the degree of 'fati!." indctcnnin<ll'). (c) for thl lru m 21 J 10 and, lattt.:all) mdetemllnate with the degree or 2 4 It should be oblOt15 If( m fig 4 IIii C membe lhan reqwrtd tor tablht Solution (a) The tru..s ho.... n In hg 4 15 a, contains 17 members and 10 Jom I upponed h) J r al."l101i I hu m , 2, Sml.:C' the thn."C reactions ther parallel n r l rn.:ulTtn' d th •. .an e m('mlll.'rs 01 thl: truss are properl) am..... It I tdlll.;ally dclermlnate CHAPTER" P1,ne Ind Spice Trusses11M ~mplt4.2~ ~ _
  • 44. -...~ ,,,, ,,, ,,,, CI) (b) BecaUlll the JC component of FA• •1 zero the secoad tioD, EF 0 can be ..usl!ed only if FAc il also zero. The IOCOIId type of arrangement .1 shown 1D PIt _ of three members AS AC and AD COII_lili joiDt A Ole that two of the three members AS IUl4 9tI ...... from the liIure that 8UIIlC tbm IB DO IppIiId to the jOIDl to MI,_ the com~ JiIIIilIDD equaltoD EF. 0 can be satisfied only ifF.
  • 45. A 1ECT1OIl4.5 _ _ ........ T_" .. _ , , _ F Rea ',ons 8 usmg propom collmear they cannot be dermn ned ~F 0 A 24k lOk 12k 411 It-110ft • F G A -A, = 0 t B 48 04=36 24 lO 12 bFIG. 4,19 F( 0 Fr m the dunen510n f lopes of 4 rJl FO the truss we find that all IDchned wn m Ig 4 19 a Th f I shown on FIg 419 b A e n:e-body dIagram of not be lime r aJomt with two or fewer unknIC'''....- ... Mot be found Ie I J n1 Gha DI tO unk l;a uate the suppon reac nOWn force FF and F. H actmg on It Solution StallC Dt'lemu/UJ() The I h 11 b russ as - members and 8 Jomts and y 3 reactIOns Because m 2 d .tru r 'J an the reactions and the JDl:mlIla are properl} arranged It IS statically detenninate. Zeru-Fi r Ut!mber It ca be members CG FG and G n seen from Fig 4.19{a that at and (G I not SHare connected of y,.hich FG and GH zcr r. rce mem~nce no external load IS apphed at jomt G maD Detennine the force in each member of the Warren truss shown m P by the method ofJomts. .._I, th II'T,lIn of lhe selected joint. ShOWID 6 8 {)r.I.1 Ircl.:·llVU. • r h' ... . pullinl ' ,t.I from t l.: JOint and com hlfl,.',"" in arH)" . ~, . .' I'. . pu..hlll11 1010 lhe JOIOt. I IS usually lorce' h .lrnl" e . I"' llnklH)11 member forces to be tensile l<l 3..,UI11C 111.: I . n.. . Ih' unklH'I" n [lm:cS b app ymg the two b. LA'ICrmIl1t..: t.: r-: 0 A .. . m MU Ilion.. ..... F 0 Jnd L f ::::. posItive a nu .,' - be"a member force mean.. that the mcm . r I,S In tension as d ,h 're'" a neoatiH anscr indicates that the as...ume t:.. e IS 10 ..omru·c..,ion. - ' If at lea... one of the unkno n torces actmg at the Joint I" in the horizontal or ertical ~ir~tion. the uno be coOenicntl~ dctcnnincd b) satlsfymg the two eq cquaUons b) inspection of the joint on the free-body the truss, 7. If all the desired member forces and reactions have been mined then go to the next step. Othern ise. select another JOIDl no more than t"O unknoy,.m. and return to step 6. 8. If the reactions "ere detennined in step 5 by using the equa eqUilibrium and condition of the "hole truss, then appl maining joint equilibrium equations that have not been u far to check the calculations. If the reactions were com appl) iog the joint equilibrium equations, then use the eq equations of the entire truss to check the calculations. Ifthe has been performed correctly. then these extra equilibrium eq must be satisfied. 1" CIW'lIR 4 PIlAt .... $pICI TIUSSIS
  • 46. Ans 'IS o • " ')•• r4 .:& 1) ,,, • o l 'pT pion F SECTIOIl' S ""ysIs of ,.... T..- by'" _ of __ F R, (',,, B • All 4.19 RlC'mbcr (G at JOlnl (j oF. ;1::~~~:";::: The lru ha l' member-; .tnJ 8Jntnh and is upporc r 2} Bnd the rcal:tlOn'i .md th m.:mher a.,anlll'd. It I statIcally deh:nnlOatl: th u "no"n fl..lrl.'C actmg at the tal or ertl.,;al dlrecUllO the un~no" med b~ sal! I. mg the w,o w n.pCC1lion of the Jomt on the Ill.::" body dlit D:teMnlnC the In e ch member of the ,lrn:n Illh'" hm-o In hg ~ 1 the m hod ofJ 101 7. mber fllrl'C and n.: ctlon~ rune bc:cn the tiP Omen' 1~ sek.OCl anothlo:f J I U known.. and return to h:p t. .. d t mnned an !h:P 5 hv usmg th cqu and nd lion of the" holt: tru~ then ~prl) th: nuum lOt cquihbnum equall".ln~ th.ll h.t," not he..:n UII} f; r to ~heck the Iculallons It the fe.Kllon l'fe (omputaj appl mg the JOint cqulhbnum equallon~. thl.:11 u thl.: cl.jUlhbri equa ODS ofth entire truss to .·heLk the l.:JkUl.ltilllh It the alJ h been perf< nned correct)) then the~ l',lr.t cquihhnum cqu I mu I be 1I jed .... E....pltU
  • 47. 20 JF 1~51 1 1 4 F",F 4.. 4.. (b)
  • 48. II 81.. 4.1 « 7'.......,.._ .,1......'. . . . Example 4.lI 20k 10k 20k (b) 20k A tSOk 20k20k B 10k 10k 60ft - - _ 20 ft + 2011-1 (a) I 20k 2Ok. 20k 20k ~6al20ft= 120ft ------1 A '----x 0- - A A-~-Jf_--f--+-+-_+-~ I •"a. __....-
  • 49. __4.7 .......... - - EllIImpll 4.10 4.7 ANALYSIS OF COMPOUItD TRUSlEI o3 5 625 25 8) - FHa 12 - 0 FHJ 62 Sk 50 8) + 16.67(12) FH 62Sk C m the horizontal dim:tion we obtain 25 16o By IIIIIIJIlIIlI FHa -1667 k FHa 1667 kN (C) 1Ion hb U 109 FIg 424(6 we wnk From F 424{a we ,an obse'f'Ve that the hon onlal section ao p8SIIDt the three members of tnt rest FJ HJ and HK also cuts an additioDal Fl thereby rete 109 fi ur unknoos hl&:h cannot be determined equanOll of eqwbbrium Tro ueh as the onc hemg conSidered the member> arranged 10 ,he fonn of ,he I.".r K can he anal)'llOd by CUf'Cd around the rmddk Jomt Ilk section bb shown ID Fig 424(& 't the kuI bon of upport reaction III usc the upper portlOD IKN. InISS abo .e<Uon hb for analYSI The free-body dlagram of !hi shown 10 FIg 424 6 II can be .... 'ha' allhough section hb hu merobe FI1J JK and HK fo= 10 members F1 and HK can be by summmg moment about pomt K and I. respectIvely bccaUle tile action of three of the four unknown pass through these pomts We fore first compute F A b) consu1enng section bb and then use sectiOn tertnme F and F S. 1Ion .. The f.....body diagram of 'h. portion IKNL of the sectJ.on aa IS shown In Fig. 424(c). To determine FHJ we sum about F whICh IS the pomt of intersection of the lines of action of F, Thus M N _aa _bb L F F • _.&. _.....- ...... td
  • 50. ...NA (d) (b) RG. P4.3 (I) • /ll,PU If the truss 15 statically indetenninate degree of static indctenmnacy. the plane _ shown as statioall indeterminate ·.,..--.....-The degree of stallC mdetermmacy IS given by (III ,) 2} The foregomg conditions for staUC detenmnacy and IDIdc1ernllinl necessary but not sufficient conditIOns. In order for these en vahd Ihe fUSS must he table and act as a single nllld bod general s stem of coplanar loads"hen It is attached to the u To analyze lallcally delerminate plane trusses, we can method of JOInts whIch e senllally conSIsts of selectmg a jom! more than two unknown force aclmg on it and applymg the hbnurn equallons [0 delermme Ihe unknown forces. We repeat eedure unlll we oblam all deSIred forces. This melhod IS mOlt wben forces 10 all or most of the memhers of a truss aR: desired. The melhod of secllOns usually proves 10 be more con forces 10 only a few specIfic members of Ihe truss are method essenllally mvolves culling the lruss inlo two pomona 109 an unagmary sectIon through Ihe members whose fon:es 111'1 and delermmmg the desired forces by applying Ihe three eqwhbnurn to Ihe free body of one of the IWO portions of the The analySIS ofcompound lrusses can usually be expedited a combmallon of Ihe melhod of joints and the method of A procedure for the delerminatlon of reactions and member space trusses IS also presented. •1l1li._ ..... 40' .., ....... U
  • 51. PI. - LKJ 6a1Sm=30m ----- 4.32 1IInuP UI Det nmne Identified by x of th , s«tions FIG. P4.31 I B 40kN B C I I 7 I 7 r-- ~5 ft--t- ft +ft - - 25 fl -- 2at8m=16m---- I 15 IN/m Jm L 1 Jm _I A~~===~~===~1'fRoof Purlin -. 12 It 1, IHt ~=~~~~;:{ FIG. P4.3D AG. P4.29 111 th(; Illl 1 hu lhl.: uniformly dl Inhut d I dm on lh t , l.1I mlth:d hv Ih purim <I onl,; ntrated I lad 1nll) . Ih~ uu Illint 15 kN/m Aoorbeam Deck 4.21 MIl 4.3D Oetennme the force 10 each ",,:naI1Il' root lru 'OJ hOl'n. The roof IS ~Imply supported whICh IR tum are attached to the jOints of the 'm 'm lF G H I 2"+2m-I-2...1-2....2....2m+2m+2 FIG. P4.28 4.28 Determine the force in each member of the "urporling a floor deck as shown in Fig. P4.28. The 1 ~lmp1) ~upportcd on floor beams which. in turn a neeted 10 the jomts of the lru"s. Thus. the urnform! tnbUlcd loadmg on the deck i, transmitted by the hearn'S as l.:onl.:cntratcd load.. 10 the lOp joints of the CHAPTER. Pllne .nd space Trusses 30' lOk JOk JOk 1----- 6 aI 1611=96 II -----1 A~kf~~± B C D IJO k IJO k 1 - - - - - 4a13O t= II I 1~ l lOl,~~~~~'f=7lr '12 fIG. P4.25
  • 52. ,.Pi, 15k 15k - - - - b II 20ft = 120ft - - - - I I. FIG. P4.39 bm 6aI4m=24m 4at4m=16m FIG. P4.40 FIG. P4.31 FIG. P4.36 III It HI II G 11111 1m 1m 60kN J 1m 3m 2m 3m 3m G III ft 25 k 4 at 20 ft =xo II .....1I--40kN --"'E m 11 k 10 II III ft CllAPlBl 4 PIa.. ami Space Trusses "'OlN 2ULN J b~====~~====~;"=~::==iF====;~r~(H ~~=~b~~~==~==dlE ~5 k 1M '1'1(.:=====~~~:=====~~~=======l'[) .... 1'4.33 .... P4.34 .... P4.3li
  • 53. "9' I A ..... 4.7 .............. D<te (he tru h en 100 kN ..:;.r)=Z.~ r-4 m-j SOkNIM SOkN SOkN_ N---r 4m L ---.l. I 4m :{'p.;1'5-l1t -t D 4m 100 kN .::.t.~ti..../l F -+4m A ~d~"",,"~c 1 CHAPTER 4 PI,ne Ind Space Trussesl!i1i All PUI 9f1 9ft 9 It AG. P4.44 :'0 k _~O II: J K 9 fl 20 ft t 20 ft 9ft 40 k )Ok 4 at 15 ft = 60ft 9ft AG. P4.46 FIG PU2 Yll 9tt FIG. P4.43 9 f1 9 f1 , 8 C D E -Sm+sm-!-6m+6m+Sm+sm FIG P4.47 20ft All. N.4lI
  • 54. • A c. tAy et C. t.,. I-IOft-1!Al-6ft1 -... I-IOft-~6ft ... y I , 60kN ID momber or sm-/ _~4_"'''''''_ ,3m )IUdlil~~~~2lJtN-+H 3m ~~~~~2lJtNt Sm ~;..._a:.~~ -, -, k ~20 ft 10 ft J__ c C If J e,2t~0 .HAI A 5ft ~tA , ~t.IOftrlOft II,
  • 55. 111 Unlike trusselol, considered in the preceding lhapter "hose mem~ always subjected 10 onl) aXial rorcc~ the member llf nglo fmID beams rna)' be subjected to ~hear fOTl:C-. .!nd bending moment a as ax.ial forces under the action of etemal load The detemun I n these internal forces and moments stre.,s re ultant IDe r) f r design of such structures. The obJecti'c of Lhi chapt r I t prese anal}sis of internal forces and moments that rna) ded p to be: the members of plane fnlmC) under the ac.:tlon of c pi r f ('xlemal fon:es and coupl~. We begin b) defining Ihe Ibree I)'pe of sue force!'l shear forces and bendmg moment that rna sections of beams and the members of plane fi ames n t di;"w" conslrul:llOn of the shear and bendmg momffit d m b the m,'th,ld of sections We also consider quahtatle deft(Cted h J'C'$ f Ihe between loads he rs and bendmg momenls. lion we dCelop th PfOL--edures for Lon tnk;Un the he r moment dJagnims using these relau n hi f II 'At P sificauon of plane fn,mes as tauuU del nm t loo."erm,nale, un table and the anal SJ of lallLB ly ~I:errnonate Beams and Frames: Shear and Bending Moment 5.1 Axial Force, Shear, an~ Ben~lng Moment 5.2 Shear and Ben~lng Moment Dagrams 5.3 Duailtat,ve Defleet~ Shapes 5.4 Relationships between Loa~s Shears an~ Ben~ ng Moments 5.5 StatiC Determinacy In~eterm nacy an~ lnollabllity of Plane Fra es 5.6 AnalysIs of Plane Frames Summary Problems 5 ,., 5t C jj.' r Sf! Sf! Sf! Plan c ...~~~:=~f=5~~=- c. --r- tC 10 fI , / ------,l-- - 10 (t / CHAPTER 4 Pl,ne and Spice Trusses Elc=allon • • t.-l'lift _-l'ift-, 180 36 ft A . . . . ' IA --" : B uLu_IA, IB x , ) __ Ie; 11----15 ft--l fIG P4.55
  • 56. III IfCT1ON5.1 _ _. _......... _ The mtcmal aXial force Q at an) section of but oppo lie in dUe1:t1on to the algebra!; urn re ta in the din.'Cllon parallel to the aIS of the beam of a th support reaction.. acting on clther side of the II n und r Using similar reasoning e can define the hear and bendmg mcnt as folloy.s: The shear S at an) 'ieClion of ~ beam I eqUdI In m,alll""Ld<butt :~:~:o: direction 10 the algebrall.: sum resultant the Io;:onlpooc,"s II perpcnd",;ular 10 the aIS of the beam of al reaction admg on clther SIde of the SlXII The bendlOg moment I at an 5((11 but oPPO lie In dlft:tlon to the algeb cenlrold of the ..TO !tCC1.Ion t the bca of all the ('1 mal I ds and uppon a §«tlon Sign Convention The SIgn commUon commonl) used ~ r the ax bendlDg moments' dep.C1Cd ID F,g' ,pana,,, C""''t'ftl'ioft. .gn convenllon ..h h. flen rd rred t Ihal II ),elds the me posit" or no p tu=D=oQIC"'('ltB S P B (d) Ibl I') C Ie) " i! !l.LL.UBA'--!l=,d=d=~'{a I (. P B, A Internal forces crc (kilned In Section 12 as the forces and couples ened on a ponion of the '!Iructure by the rest of the structure. Consider for esample. the Simply supported heam shown 10 Fig. 5.1 (aj. The ~ body diagram of the enure beam i~ depicted in Fig. 5.I(b), which sh the eternal loads. it'! well as the reactions A and AI, and B at ports A and B. respcctic1) i'i di'!cu'l'icd in Chapter 3. the support actions can be computed b) applying the equations of equilibrium t tree body of the entire beam, In order to determine the internal fo acting on the crOS:l ...tclion l)f the be.lln at a point C, c pass an I oaf) section {{ through ( . thereby. cutting the beam into two parts and CR. as shown m hgs S.lIe, .tnd 5.I(d). The free-body dIagram the ponion -Ie Ihg. 5 Ifc sho:l. in addition to the external loads uppon reaclion acting on the portion AC. the internal force Q and ./ exerted upon portion -IC at C by. the remOed portlon tructure ote that nhout Ihe'Se internal forces. portion AC I equJlibrium ~Iso under a general coplanar "y stem of external and reaclions three intemallorces (10 perpendicular force com and a couple are nCf.:e ary at it section to maintain a portlon beam m equihhnum. The tO internal fnrce components are onented in the dirCt:llon of and perpendicular to. the centroidal of the beam at the secllon under l:On ideration. as shon m FI The lntemal force QIn the dirt.'l:lion of the l:entroidal axis ofthc CHAPTER 5 Beams and ff'lmn: Shelr and Bending Moment112 '" 5.1 5.1 AXIAL FORCE, SHEAR. AND BENDING MOMENT
  • 57. - s _ ..-.wAaialForoo. - ...-...-_.- . . of!be __ c:oasilImd for c:omputiDl the mlllnlll pcIIilM dinlcliODl of tbc mtcmal fon:es acting on !be _ .... OIl ..a.Iide of tbe IllCtiOD are shown m FJI. 5 From a computalional vteWJlOUIl however .t IS IIIU~ _ t to expreIII this IIlID convention m terms of the IIIId 1ll&dI0DI Il:liDll on tbe beam or f1'llllHO member u to 52 d indicated m FIS. S2(b) the In,"'"t!tRIIIdIrfd ID poIll wMn tho ex,mrat[orca lit: ",.dIM:e IIIIUimt or hmJe 1M Ierumu:y 10 plitt 1M _ .
  • 58. E til (d) Bendmg Moment Diagram (t·ft SECTION 5.2 - .... 1IondInt __ .... E FIG. 5.5 ,0 -]4 Ora"" 55. (e) Shear Diagram (k) 8 L.....,.,,......;C,,- -14 46 A • b d 60k 180 k·ft •I ~ A _ 46 A kl E } A LJ b LIO ft -----!---IO ft -J-10 ft-L IO ft-l I , I (a) b Example 5.3 I SOO ,(14)(,1 ]40 kN m / .140 kN m The reader rna) r.:hcr.:k the re..uh.. hy ..umming forces and momen tlon AB 01 the beam after computmg the reaction... at support A '-__ 6m A Shear and bending moment diagram"! depict the anattoDl quantities along the length of the member. Such diagram trueted b> u~ing the method of '>eetion~ described in the tion. Proceeding from one end of the member to the other from left to right section arc passed. after each succesu loadong along the length of the member to detennine !be eq pre ing the hear and bending moment in terms of the dis seclion from a fixed ongm The alues of shear and beodiDI determined from these e~uations 3re then plotted as o,rdiDIlllll the po ilIOn Yo IIh respect to a member end as abscIssa to hear and bendmg momenl diagram This procedure the folloong e ample '). ':!O ... S HO k . ote that the 500 k~ m coup!..: dtlC" not any effect on shear Buuli"y .[omUIi Con..tdcring (he counlt:rc1ockisc moment yoc Yo rite Solution . ~ To ,nollJ computmg reacuons, we .')"u 1"" hh ' '..' ~-lg :-, ( hICh I" 10 th..: nght of the !>Cetion bb nalh un"upport..'J pomen B PUtl~f' the mh:mal force Sh(ar Con lJ..:nng the cxlt:miil forcl.: acting nte CHAPTER 5 Bums Ind Frames: Shear and Bending Moment EIlII.... S2 FIG. SA 5.2 SHEAR AND BENDING MOMENT DIAGRAMS
  • 59. B III:lIa.I.2 _ .......- . . ....._ M M from which 13 25 ft; thaI from end E or 40 13 25 26 FIB- SS d Solution Rea I. nJ See Fig 5 6 b I Example 5.4 10for 0 54 1410 IBe M 46x 60 M M F'OI ._",1 DC The abrupt change m shear from 46 k at an mfiniteslmal distance to the to ~ 14 k at an mfimteslmal distance to the right of B is shown on the agram Fig. 5 5 r.; by a ertlcal hne from +46 to 14. A honzontal IS lhen dran from B to C to mdlcate that the shear remains constaDt alue throughout this segment. To detennme the equations for shear to the right half of the convenient to use another coordinate "(I dmx:ted to the left from the the beam as shoon in Fig. 5.5(b). The equations for shear ID 8egmaa DC are obtamed by consldenng the free bodies to the right of secti ce respectively. Thus. S 46 60 -14 k for 10 ft < , s 20 ft S 2" 54 for lOft x, S20ft These equatIOns mdlcate that the shear IDcreases linearly from 20 II; at an mfimte Imal distance (0 the fight of D; It then dropI 34 It at an mfiDlleSunal distance to the left of D; and from there lmearly to 14 k at C This mfonnatlon is plotted on the shear shown m Fig 5 5 c Ikndrng M fMnl Dra(Jram Usmg the same secbon and """... p1oyod pn:v1OusI for compulong shear we detenmne lhe foil_ bending moment In the four segment of the beam For segment A M 46 forO 10ft Sht r D ram To detc:nnlm: the equatlon for ~hear 10 segment beam e pa a sa:hon aa ,It a dl~ta",:e : from sUpJXlrt A as Ii 5 b Con ld nn~ the free tx"ld to the len of this section W obtam s .u; k lor 0 10 ft As this equation andlcah."S the !l;hcar IS ~'Onstant at 46 k from an dl lance to the n~ht 01 pomt -I to an mfinlteslmal distance to pomt B At pomt A the shear mcrca~s abruptly from 0 to 46 k, hne IS drolwn from 0 to 46 on the shear dIagram Fig. 5 5 c at A to ehange TIus IS (011000 b) a honzontal line from A to B to Indicate hear remams con tant an this segmenl exi b) u mg $CCtlon b hg 5.5 b m segment BC as S 2"(1 for 0 S "(I 10 ft aod 171 Cle .&1 _ ... ~ SMar and BendInD Moment
  • 60. mSECTIOII U Q ...... DIn, ....... 5.3 QUALITATIVE DEFLECTED SHAPES o -W.75 (' .;) 60.75 , -.......:~~-;;8+-....L.T,..L....I.-JI.-..!.--l a h ...-'- B .- 60.75 l.'l , I s -1725 (bl h , (e) Shear Diagram (kN) , 75.5 Id) Bending Moment Diagram (kN rn) from hleh 636 m The alues 01 S computed from IheoC equalion~ arc plolted to obtam diagram hon In Fig 56 c The pomt D at hich the shear I zero from the equation fIG. 5.6 '" m-~-- hm CHAmR 5 Bums Ind Frames: Shear Ind Bending Moment
  • 61. 5.4 RELATIONSHIPS 2k1ft F (I) Beam ' 180k ft c) Qualillbve Deflected Shape > _ ~ 1IOftIIIII 60k
  • 62. .'AU ..dS d, m "hich dS d., represenlS the slope of the shear dUl&J1111 5 2 can be expressed a which can also be wotlen as dM dx S - _•• the slope of tbc bbelldillil ted as ApplYIng the moment eqwlibnum equation to tho beam element shown m FIg 5 8 b we wnte EM 0 M wdxdx2-S tIS By ncgIecting the terms containmg second-order dl8'" dM Sdx ope of shear diagram ta poUlt Dl1ding by d, ".. " nte Eq 5 I) as c:IwJae m&be r between AandB To determme the change m hear between points A &XIS of the member see fig. 5.8(a)), we mtegrate Eq to obtam fdS S8 - S. =J:w dx in which (S8 S.) represents the change in shear and B and J~'" dx represenlS the area under the diltri gram between points A and B. Thus, Eq. (5.4) can be ..... .......-_..............-.fled me" no hon onlal force are automallcalllYsatt~~' qU~'bnum equallOn E F. 0 we'"menl pp ymg EF. 0 S .. d, S dS) 0 dS .. dx
  • 63. 3. SEClION 5.4 IIetItJoaoMPI_ ~ - . . .......... • magnitude of the moment of the cou change in hending moment al the pomt of application of a couple hKh can be ...,.!ted a~ Procedure lor Analysis The follo"mg step-by-'tep procedure can be used for cons <hear and bendmg moment diagrams for beams by applylll' going relaliomhip, bet"een Ihe load" the shears, and the moments. t ~ L 1/. - 0 .1 7 ( I + ,1.1/) - 0 d./ .1/ dong Moment BHmslnd Frames: Shear and Ben I COU les or concentrated Moments . . P 'Ix'lc:c:n the 10,luo,; dod shears ltlwugh ,he rd.ltIl1ll,1~'~"', I" III. ,Inl! .. 5, I~)) arc alid at the- I I'h .<'>, .' l~lr h.p•. :'.1 I mll ~ .... 'lIlo.:ntr.ltcd 1110menlS, the relatl f .. 'lil,ie... III Lt. b E arrlll:.t1ilHl l' U Ixm.ling JnlHllCIHS as gicn.) qs S lH'c:n thl.' ...hc:.If' ,lIlL! h Plin'" ., Illustrated In Example ltd It ,ut: (.. .. :i.ln .II''!.: not .1 ' . .. pit' the bcnlllllg mom"nt chan . I' tlit)1l (11 .1 lllU. . pt.1ml 01 .trr It. th' 111.1l:!l1itudc 01 lhc moment of the h) all .tnWlIlll ('qu.d w ,~ ('llll'" ida the equilibrium of a ddli dcrit' Ih" rd.ltllln.~ll~~)m the lx.tlll of Fig. 5.8(a) b} passing ment th.tI I l(llah: I.' 1<lIKe, to the left and to the right I mfillltClffi.1 ul' .. -,ec:t1on .1 .. ,_ 1/ fhl' frce-bod} diagram ofI 1 n01 the coup c (If ,tpr Kalu I~ 5 s J prhmg the moment equilibrium cq I" "htl n m fig. .< • J. Calculate the ~upport reactions. 2. ConMrul.:t the shear diagram as follO. s: a. Detenome the ,hear at the left end of the beam centrated load 1 applied at this point the shear pomt. go to step 21 b). Otheru i,e. the ordinate of t gram 8tthi, pomt change, abruptly from zero to the of the concentrated force. Recall that an upward the hear tu mcrca'ie. I. hcreas a downward fo shear (0 dttfcase. b. Proceedmg from Ihe poiOl al "hieh the shear w the prelOU, top to"ard the right along the length Idonllfy the next pomt at "hich the numerical rile CHAPTER 5178
  • 64. and nunlDlum value of bend,ng moment 0CClII' the shear IS zero At a pOInt of zero shear ifthe from posIDVC to the len to negative to the rip bending moment d,agram WIll change from of the pOInt to negative to the nght of It; that mmnent WIll he max,mum at th's pomt. Con ofzero shear where the shear changes from to posItlVC to the nght the hendmg moment will For most common loadmg condition such loads and umfonoly and linearly clistnbuted I zero shear can he located by COIISIdenng the shear diagram. Howevcr for some cases of loads, as well as for nonlinearly dlstnbuted bel ery to locate the pomts of zero shear by JftSSIODS for shear as Illustrated in Example 54- Co Determine the ordinate of the bending moment pOIOt selected m step 3(b) (or just to the left acts at the POlDt by adding algebraically the shear diagram between the prevtous pamt utd rcntly under conSIderation to the bending VJOUS pOIOt orjust to the right ofIt, ifa couple .. Determine the shape of the bendmg moment the prcvtous POlDt and the point currendy by applYJng Eq S.8) which states that the moment diagram at a poInt is equal to the Ibt8t e. If no couple IS acting at the point under proceecI to tep 3(f). Otherwise, determtne the bending moment dIagram just to the nght of mg ....braically the magnitude of the 1DOlMi! to the bending moment just to the left of dw bencti... moment diagram at !hi pomt challlt amount equal to the magmtude of the JDOIIIIlIi£ t Ifthe pamt under COnSIderation .. not mal' of the lam, then return to tep 3(b - diagram has been completcd. If dw ClIIried out cornctIy then the value of the rig1n of the ri&btlllCl of the the """"'4enon. Tbe JlI'OC*Iure can he IIIlld for ....- ...... bypra-'U11 cr- cad, ..c oed,... the JlflIClIlCbue be
  • 65. 240 40 IJ IJ P Inl( Pin, D Potnt 8 k A P In' A Draw the shear and bendlDg m shape for the beam b wn ID F g. S I lECTlOIIu .......Oi........ _ ~........ - . ••_ _• I Example 5.6 24 k 18 k IJ (I (10)lt1 ll1 24+0 -24k ~ r n '::,r (I I~ 12 to n () IJ Purnt D SOlution R ttl" s. I 18 +0 - 18 k Bl'Caus( a ncgalic (dol1ard) com,;cntTatcd load of 12-k magmtude point B. the ..hear .IU'I to the nght of H ., S.. 18 12 6 k Po;", C S( I Sa H. .. area under the load diagram betl,ccn just to the nght jn'll to the lell of C 5,,_60_6k .~( It - (, - J() ~4 k S/l It 24 +- ~4 - 0 The numencal alue of hear computed at points A. B. C and D to construct lhe shear diagram a hon in Fig. 5.9 c The shape or beleen these ordinates has been C'Stabh hed b) applying Eq S 3 lhal the slope of the hear dlagrdm al a poinl IS equal 10 the load thai polDl Because no load IS apphed 10 lhe beam between thac s1 pc of the shear dlagnuTI IS zero beleen Ihese points.. and the n is of a tien of homontal hn a ho n in Ihe figure ole diagram cl Ie return 10 zero JU I to Ihe nght of the nght beam lO<bcatmg that the anal 1 ha been earned oul colnCtl 1 • up" <lrd I.:om:cntr,tted fOTl.:C of 18-kP Int ., SIO~c:' a po 1 11; . h h aT diagram Im;TC.I..e.. abruptl) from () to 18 Itad at pomt .J I C I; pomt P ml B The ,h~ar Ju..t 10 the kft tlf POint B i~ gien b) _ ,"", • _ an:a under the load bety,ecn just to the right 0 SB L I jU'ot tn the kft of B h· h h h ·"pl'" /" and' R" arc llsc.d to denote "just to the Im",)"tc!>u~, - ' " •. h· hi" rcvt'Cllch A.. no load " apphed to this segment)U"1 to t e ng. t·~ • heam. CIMPTER 5 IeIms .nd Frames; Shear and Bending Moment112
  • 66. .00 -200 • M M ~ m/8 Me P, Inl C M a..ding Mommt D , p, "" ( ~h r D m P, nt A p, tnl 8 ..IJDII IICTIONU ft •• a I•••• ~ . IEllImple 5.7 70 kN I (d) Bending moment diagram (kN m) m (ej Qualnallve Deflected Shape A -620 om 4m (a) 70 k H4:::.620~ m I ~ B - ::; 70k ' tb) Zero slope 70 I A B (e) Shear diagram (kN) A CL: M, 0 M 06 200 0 MA 620 k B ding MomentCHAPISt 15 Bel.... and Frames: She.r end en114 ... 5.10
  • 67. 10IlN/rJI
  • 68. A_ _ .1.. IICT10IlU nl" ,,, 1111_ -....1 . Ind the q.uoH'-II! EF 0 EM 0 I 3 6 2 02 B SO k See Fill 5 12 b O'i 1to; 1'168 472 47 kN m P ml D Ill 47247 472 68 0.21 ~ 0 The hendlng moment diagram IS sholl m Fig. 5.11 (d 'The diagram between the Mdmates Just computed has been based on thai th lope of the bending moment dlag-mm al. any ~mt I equal at that plint Just 10 the nghl of A the shear IS positive and of the bending moment diagram at thiS poml. As we move to the the shear decrea linear!) but remains positive) untd It beco... Therefore the lope of the bendmg moment diagram gradual) becomes Ie steep but remams posl1lC as c move to the ri&ht It becomes zero at E ote that the shear diagram in segment AE the bendmg moment diagram In thl segment is parabohc or a ,un because the bending moment diagram IS obtained by IRtegra diagram Eq 5 II Therefore the bending moment curve will degree higher than the correspondmg shear curve. We can see from Fig 5.II(d that the bending moment maxunum al pomt E where Ihe shear changes from paslbYe to the live 10 the nght As we move to Ihe nght from E. the shear bc<:oa~ and It decreases hnearly between E and B. Accordingly the slope or moment diagram becomes negauve to the nght of E, and It uously becomes more steep downward to the right) between E aDd left of B. A pOSitive clockwise couple acts at B, so the bendina crea abruptly at thiS pomt by an amount equal to the magmtudl: ment of the couple. The largest value (global maximum) of the over the eonre length of the beam occurs at JUst to the nght of .. no abrupt change or dlscontmuity, occurs in the shear diagram • FInally as the shear In segments Be and CD is constant and bendmg moment diagram m these segments consists of straight bDII bve sJopes. Quolilotwc /kfler"d Shope See Fig. 5.II(e). Draw the shear and bendIng moment dIagram for the beom shown In FIg. 5 12 I _ air 'M' _1IId _ _ 1... 1I...lng Momon'
  • 69. AIls til B1931 fI249 ft and D A B Po M Draw Ihe shear and bendlDg moment diagram and th qlualitalli.. della;;led shape for the beam shown in Fig 5 13 a Solution Rl!actlOn~ (See Fig. 5 13 b 0: II: 0 20 10 5 C 10 - lOll 15 0 C LF A 20 10 250 .00 0 A t Qual"., lk ,n! Shup< A Qlua'lltativc ddlornd hown m Fig 5 11 e The bendm@ momm beam I bent c nca e upward In tb n ment I negative m segment AF and GD In these segments 1/ from 1lihKh or hathelr pomt F. T0 l~at Ih pamt f mflectl hon fot bcndmg m ment 10 upport pomt B hg 5 12 b RClION U 1Itl-.,._.-.-............_ ... IExample 5.9 16.. k II c , 0 II 1 12 IR k L 2 "II Ii IR 50."' )~.7 k , ." ~ 20 21 3 kL '- S • 2" .' + ~td 9 k I ~t 6 0SD 9 2 f' nl ., .h r l>ull}r. un Plml R Poml ( Paml f) Pomt A 1/, II Point B II. 0 72 n k·ft Poml £ 1/, 72· 17822 - 10622 t-f. Pomt ( 1/, In6 21 11412 18 k-ti Pomt [) fJ) -18 , 18 () The ,hear ,h.tgram I' ,hon In f-i£ 5.12(c) The shape of the t"fen the ordinall:s just l,;omputed is obtained by appl) ing the condi ~Iopc of the ,hear diagram at an~ point i'i cqual to the load In P0ln!. F'or eample a'o the load intell'ilt~ at A is lCro. so is the slope diagram at t. Beteen ~ and B. the load mtemilty is negative and. I Imearl} lrom ll:ro at I to - ~ k· ft at B. Thus. the slope of the shear negatle in thi.. ..egml:nt. and It dccreaCs (becomes more steep from A to jU'ot to the left of B. The rest of thc shear diagrdm is COD using ,imllar rea..oning. The POlllt of lem ,hear. E. i, located by using the similar tnan the 'ihl:ar diagram bcl"l'Cn Band C To facilitate the con..trucllon of the bendmg moment diagram the ariou'i ~gmen... of the 'ihear diagram hae been computed aDd m parenthe-.e~ on the ,hear dlal:!faffi (Fig. 5.12(c)). It should be n areas of the par,tbolit.: pandrel.., A8 and ('D can be obtained by mula for the area of thi, lJ.hape gicn In AppendIx A. II The wpe of the bendmg moment diagram beteen theIe tuned b) ~ng: the condlhon thaI the !llope of the bendmg m an) pomt I equal to the r.hear at that point The bending momeat con lfUeled I shown In "II!! 5 12 d It can be seen from Ih fiIS gure that the maximum De ment OCCUR at POlOt B h erea the maximum poslUve bendinl &ndlll!J fun/tnt 1>/lJIJfum CHAPTER 5 Bumllnd Frlmu: Shear Ind Bending Moment190
  • 70. B E BA -500 S L SO 20(10 ISO kN S R ISO 250 100 kN SD L 100 0 100 kN 100kN I S R 100 100 0 The diagram II shown In FIB 5 13 c &oding M 'DID ram p, ,j M SOOk m Po B SOO SOO 0 P, IE 0 62 62 kN m SOO m Po D 0 20 1m fIG. 5.13 P, urI C P, urI D lOOkN 20kNIm I j , , , 'l'"="c----:D ',-iC =250kN (b) ., m----+----lom~5m~ IIIU •• _ .. _ _, .... 1IIndInlI ~ 1# -
  • 71. 'IN CHAPTER 5 lei.... lnet Frames: Shu, Ind Bending Moment ... 1../11 '-""'-""'-'-'--r., IECTIOflU _111,,_ ~ - . .........".._ P 'nl n .4 24 ~1I1t "It (.Il ~ l.J1l f!n TTl=rpPTI, H C , =:!..J I.. R ="7~ I.. ,I» '" J" (() Shear Diagrilffi (k) ~IXO '9D D =24k -24 rExample 5.'1 Th ar dl m Bend" II ,n P nl A It p, ,F. It '44 44 P .,8 Pin' F It PI "'0 II '44 44 The bend n m ment d (/wI I iN Sha f, Draw the hear and bending moment dJ shape for the tatlcall mdetenntnate be m al;Uon dctcomne<! b) usmg the proccd r lennmale beams presented In Pan Three (th J 1lIfI 8lkft(IA I f I I I I I l:;t) kJt8 1m 154k I k 1----18" I I ft • Id l Bendmg Moment Diagram (k.ft) ( ) ()uahlall e Detlc''tcd Shape fI6. 5.15
  • 72. - ,. D b ,. B.:::~B/~ B:%~~:c ~ ~ IECTION 1.1 ..... Da.. ' u:..'111.101101.........,... P $ 2S, ......"- fIG. 5.18 (d) QualJtatle Deneclcd Shape Solution Regardless of "hether a beam is staucally determinate or iJulletelllli. the support reactions ha...e been detemuned, the procedure for shear and bendmg moment diagrams remains the same. The shear moment diagrdms for the glen statically mdeterminate beam are 5 IS band (c. respectively. and a qualitative deflected shape of shown;n Fig. 5.15(d). As defined in SectIOn I 3 rigid frames, usually referred to frames are composed of straight members connected either moment-resistmg connections or by hinged connectlons to configurations The members of frames are usually COl_*'! j01D1S although hinged connections are somellmes used prevents relative translatIOns and rotatJons of the member nected to It so the jOlDt IS capable of transmllllng two components and a couple between the connected memben. aetlon of external loads the members of a frame rna ubjected 10 bendmg moment shear and axIal tension or A frame, cons,dered to be slalkally delermlNll mome", shear a"d a 101 forus '" all 1/ member. IU lernal rearl/OII ra" be del ""''''ed by "''''9 lbe eqI-m-1i"1On and rondlllon Smce the method of anal)'Sl p.......led In the foll'.....l...' be UIed onl I anal u tallcally determlDate frames, tt the studenl to be able to recognIZe tallcally detcl'l1liDale proc .. l,ng WIth the anal , ... 5.15 old 5.5 STATIC DETERMINACY. INDETERMINACY. AND INSTABILITY OF PLANE FRAMES
  • 73. m J member end forces but in oppoSlle dnectlons in accord Ion s thIrd law Th analySl of the frame mvol~es the Ihe magmtude of the 18 member end forces SIX per JIIelq three uppon reactions At A, and Dr· Therefore the unknown quantllleS to be determmed IS 21. Because th entire frame IS m eqUllibnum each of and JOlnlS mu t also be m eqUIlibrium s shown m F... member and each jomt are subjected to a general ""PI8uii forces and coupi must saIl fy the three equatillQl num EFx -0 EFr -0. and EM O. Sinoe the three members and four JOint mcludmg the two jom uppons , the lotal number of equations available IS 3 3 1bese 21 equdlbnum equallons can be solved to calcu1a known The member end forces thus obtained can then .. tennme 8X181 forces, shears, and bending moments at along the length of members. The frame of Fig. 5.I6(a stallcaUy detenninate Three equations of eqUIlibrium of the entire frame aa could be wntten and solved for the three unknown ralClilil and Dy . However these equitibnum equations are II8t from the member and joint equilibrium equations and do; any addillonal information. Based on the foregomg dIscussion, we can develop the static detennmacy indeterminacy, and instabiltty of frames COntalOmg m members and j joints and supporfllf ber of) external reacttons. For the analysis, we need to member forces and r externsJ reactions; that is we need tota1 of 6m r unknown quanlllles. Sinoe there are lit JOInts and we can wnte three equatIon of equiltbrium ber and each JOint, the number of equilibrium equa m J Furthermore Ifa frame conlalDS mternal hQ.na1 roI1ers, thae mtcrnal condillons provide additioDal can be IIICd m COIIIunCllon WIth the equilibrium eqlq:::::: the unkoowna Thill, ifthere are e equations of Cl the tota1 number of equallons eqwllbrium equatiOla CODdition vaiJable IS 3 m J e For a frame, if equal to the number of equ&tiOllS, bat 6m r _mil ..... _ ............. _1
  • 74. M ......_ ..c B_-Ho-Q s ......- .....c D AA hi"""JOml lbe number o(equaUODl _ ..., ... tbc number of mcmben rncelll1ll al tile ....lIIIPle, conJider tbc binpI Joml H o( the (rame biqe C8IlII I traDJIIUI momenl the m......"'. three memben EH H and HI meetIng at tile M 0 M H 0 and Mill 0 HI__ III ale IlOt IIldepcndcnl m the that ifany ale tisfied aIonI wllh the momcnllt~IlI~== H tbc mnauul1ll llIuaUon will aub tile binpI JOIDt H pnmdca two iDdcpendcnt eq similar IIJIII, .1 can be shown that all pn:Mdiel tbc llIuabODI of condiuon whose numblr _ben meetJl1ll at theJomt I ...... AppI'lIICb altalDlli"" approach that can be UJed (or delID!lJtlli indcIcrminacy of a (rame IS to cut enoUBh by nc nn"IIl'ry 1leCU0DS and or to remove IaIdcr the _ Illltic:a1Iy delInnmate The IOlI1 and tenIII I'CIIrI1D thus removccl equals tho an example consider the (rame sbowtl ....-................
  • 75. ClW i8i I . . . II1II F.-: _ Ind IIndlng m=6 J=6 , 3m.,> (e) Statically Ilnde.... j = 3 (number of giJden) (I) .... ... m=4 ;=4 r= 3 e, =0 3m.r>3)+t' (b) Staheally Indetemunate (. =3) .. • .. ,110, I frd.mes shown In Fig. 5.20 IS sta Vmfy thai each of thCdP =or~tatlc mdeterminacy. nate and dctennme Its g. • = 3 (number of girders) = 3(4) = 12 (e) SlIIulIon See FIg S 20(0 through f The foUowlDg step-by-step procedure can be used for member end fon:es as well as Ih. shean bending DIClIIIIiIi fon:es mmemben of plane Iatically determinate f1nm-.. '" I. Check r. r IaIU: detennmacy Usmg the proeed ptIle:eding lIIlClion detennm. whether or DOt the II&. 5.20 Hinge ".-5 =6 r=8 ~ =0 3m.r> lJ+~ Slalicall loclelenninare •=s) I.n m= 10 Jz9 r=9 e =S 3m+T>3J+t d Staticall Indelermi.... • = 7) - U _IL'...OF PUlE FRAMES
  • 76. ................ ..
  • 77. 2lII ClIAPTER 5 ....... end FnIm11: $hI1f Ind Bending Moment IIECIIGII U - ... _ .... -y T T"~- ~-_xI 15 fI 10ft x --'-- 18 42 x IB 42 JOh- ---1 360 C (a' Y Y-~t -- D lB~ 18 t8 C L 42 18 (d) D tA,- A Dr Ay (bl C 42 ~r 8~~~~'BC CBe 8 C 360 8 r Be 18 B -sf c:'4,tN:-AO 8 C x MiD'P Ii M: B" Miff. • B:' 8rc C:c D C D A A 8AO Cr (e) Shear Diagnms (t) r (I) IIeadiDI Momenllliql-. (l rt+8:' MAO 8 C • B 8 ArA:- 42 -18 AAOr I! r:o D A A~ (J) A>ial Fon:e IliIIn-It) IB ... 5.21 ...121 (e)
  • 78. AIls AIls qualitativeDd Boca lenD A 'ui 11, Dill ranu From the ~ S 21 d we observe that the axial ~ comprewve WIth a ~ tant magmtlde r gram for lhl member I a stought hne pa a hOWD 10 f II S 21 Similarl 11 n be forces In members Be and CD are al 42 k l'espectlcly The uUlI fora: dl3 ben are shown 1ft FIg. S21 g Quo/"G" ~ ltd ~ From the bendin ":::::.::=~:'memben of the frame Fig S 21 f we observe thai the bend concave to the left and conca"c upw rei b menl dnelops ID member CD It don not bend but terniU bve deflcetcd shape of the frame obtained by In t the three memben allhe jOlnb i shovin In Fig S "I h lhe deftecllon of lbe frame al support A 1 Due to"~hc~~=:,:~~~::.: JOint S deflects to the nght to B Smcc the axial defol neglected and bending defonnabOns are umed t be only 10 the honzontal dlJu:tlon and JOint C deflects joml B: lhat I SS CC ole thai lhe cUfUlu of SlSlent With their bendmg momml diagrams and thai the onlPna lween memben at the ng.:d jomt B and C ha c been rna mcd Draw the shear bending moment and ulal force diagram deftecled shape for the frame shown In Fig S 22 a EF 0 A 2 Solullon SIal' IN, rmmQ)- m 2) 3 r 3 and 3) ~ and lbe frame I geometricaUy stable III tall II IExample 5.14 f:C = -360 k-ft 42 42 0 18 2(30) +cf' == 0 213011(5). 42(JO) +.ftC ~ 0 ~'omldcnng the equilibrium of member BC we wnle C:c 0 C:c 42 k M!C 0 o 360 B, lemhI' BC . elll + - [F, - 0 . [h 0 ·C[,f, () Sumlarh b .Jppl)mg ~ f! o f' obtam .f " I k d of .f' and A:' no,", kno":D mernbIr femht'r .fB llh the ~f~u f'd foil uhkh can be detennmed by .fB ba Ihm unkno",ns Br r an ~ Th I 'F (J'"F, OandE(,t-· us.p)mg ~ ~ s(, 18k Bt " -18k fi B ::..360k-rt, J IJfI 8 Proceedmg De! 10 Jom! Band com.idenng its equllibnum obtam Joillt C Appl)lng the three equilibrium equations. we obtain (CD 0 cflJ -42 k .f,(D -= 0 Iemhi'r CD Appl)lng. L Fl 0 and L F) - 0 in order, we obtam Sinct' all unknol,n lor",~ and moments hac been detennined we chcc:k computatIOns b) u mg the third equlhbnurn equallons for member CD Slwar Dwqra11lJ The x coordinate 5)' tems sel«ted for the three melD of the frame arc shov.n In Ftg S 21 d and the shear d18gram for the ""lIIII~:< co truckd bUSIng Ih pr~urc descnbed In Section 5 4 are depictedS21 c BnuJing I nt D (lranu The bendmg mOment diagrams for mcmbenofthcframeare OVtlltnflg 521 r : ~ //) - 0 Jo"" D Cbeclung computation dl- Momenland FnImes: ShNr "net Ben ... ...... bod d agram, of .all the memben Bnd If tnd f r, Th~ ln~' ,', I L..'<t1O the: computatIOn of Inm "i~l, f'l.anl'<.e of the fr,tme ••rc hown In "Ig - D both (,("hleh hale anI} three unko t 0101 ~ or al 1"101 . (OfL'C§ elther,j, J ~ from ih free-bod) diagram h mol -4 u"t.:an J In' -l lk'pnnl11g. t }l " ',t lthe n"hl llith J. magnitude of I- O.fmu,tall e In order to satl f L 'I of 1~ k to the left. Thu.. to balam:e the honzontal rtlU.. Uon .f:' IS k
  • 79. 111 Shear Di,acn....ttJ D A 8 c IEl:TIIII U - , , , , - _ (J) AJ<iaI Fa... DO...... (Ik 25 B ... 5.22 I 6 klfl I I 0 c (d) c x R V180 11m D t .no B~nJin~ Mument Diagram (kAt) (el ( 1.6 kilt sf· -E1-1o'lc....LI_ITJ .e IS .11. 8:( 2H A, • rh uta ,- ..25 _ A Ax 'fJ 410 24 x c y (" 1Sf' IJd./ll 8 aIAPJEfI 5 ...... _ FnmIS: Shelr ,ReI Bending Moment :no
  • 80. (i) Qao~" DoOocIId 48 1ECTJOI5.I .....,." _ "'- 2lS 338 C H~,'(P D 60 60 It 60 A (8) Bendinl Moment Diagratll$ 0lN E m C -96 B D -384 314 fit 5.2:3 Contd x ((, Shear Diagrams (kN) 611 I~ B __- 12 CHAPTER 5 ...... and Frames: Shear and Bending Moment "'5.23
  • 81. Ans •• Member AS Q Member 1K' Q Member 'D Q Member DE Q •-, m JOUII D Appl 101 the three eqwl bnum eq Do 12k 0 48k UfOnm" DE [F 0 [F 0 E 48k C[ I( -0 60 12 5 0 JOint £ [F 0 12 Checks [F -0 48 48 0 Checks Shrar and lkrulmg Moment DIQR1anJJ See Fl. AXIal Fore DIagrams The equauons Ii r axial fI frame are 'r -4RkN r .wiB - -60 kN· m4RkNB"I12kN 12 8 ~ 0 . ~1) n ... I: • E II ,,' 1I '- ' .f /Ii 12.$ 2 () 1: F, {I 12 £, (J £1 -12 L.~ B", R dd' Iemher AS Cono;.idenng Ihe equIlibrium of member AB. we obtain Bft -12kN B" 48 kN ':( - 60 kN· mI ftmha BC [F, -0 ef< :.. -12 kN [1', 0 48 1241 eBC 0 C·< 0I I • [ liB 0 60 124 2) ... 12 , 0 Ch8cIcI Vnnht-,CD [f, 0 0'" -12 kx [1' 0 SUMMARY124 0' 0 o p 48kr [ 110 0 /0 0 / p 60kN m loint B Appl)mg the three equilibrium equations. we obtain JOllfl C Conssdering the equlhbnum of Jomt ( 'ol.e detenrune (D 12 k cctJ 0, ftmhn End JQrH IS« hg. 5.:_~lc Joint A 8) arrl)ing the I,."qudliom. of equilibrium L F, -=- 0 and L Fy 0- c obtain 218
  • 82. 111 Jm J r iJ e Ihe fr..m I tahcall u tab) lm r 3 ~ the fram , tatoe II del nn 3m r 3J e the frame I tatlcaU md t on The degree of ta tiC mdetennmacy I glen by 3m r l bend~ procedure for the detemunall n of m mber end r. n~ moments and aXial force In the members detennmate frames is presented in Section 56 P ""I.("If (hc mcmt,,:r 01 .ill ,Ill: l Il'rn.11 Ill,ll.1o- .lOd re.lcllono,; .Icung ,,In either , I h II' "" "I,'f II W ~ Pll«IIIC. hen the c'tema.' forr-..IUl' 0 I l' Xlllln. l "'"' . ' _ IlnJ to rnxhlt:c 1",'11 j(ln, The ,IKlr .H.IIl) ,ct:t!on oj a m~mbcr IS equal In magnlludl but llPPll Ill' '" Jlredllln. W the al~cbr.':llc o,;um of the ",'omJl'.lnlnl in thl' dirl':lil11l p.:rjX'l1l! to the 3:10,; 01 th~ member of .111 Ihe e:~;t""mal fl'a",·ltlll1 al"ling on either Side 01 the lieC1ion l' l.·on'IJa it hl he rt1 ItI e hen th" e:lernal force:-. tend to push the J'l'rtJon tlf the nll'mbcr lln the kft of Ihe upard IIh respect to Ihe portion on Ihe nghl of the ,edlo!1. The bendmg moment at any sec.. u(ln llf J mcmtx'r I" equal 111 magllltudc. but l1rpO..lte III direction, to the algebrah-' urn of the mllmcnh abl1uI the o,;txtioll of .ill the eXlcmalloads and reactlt' on eilher Idc of Ihe ~cl1on, Vc consider it to be p0,iuc hcn [he ch:m.lllon..:!,;' and couple"l lcnd to bend the member Wned c UP .lrd. l' C(lmpre ..ion m the upper fibers and tension In the locr tibe~ at the '-Cl:UI.m Shear bending moment. and .Iial force diagrams depici the vana- tion.. of Ihe..e quantitic' alonf! the length of the member Such diagrams can be l.'On"ltrul;led h) determining and plotting the equations expressing these 'treo,;, re"lultant... III lenn'l of the d"tancc of the :section from an end of the member. Thc con"trul;t10n of ..hear and bending moment diagrams ("an be. con'lderabl) epcdlted by apply mg the followmg relationships thal eXltit betcen the loads. "lhearti, .tnll bending moments: 21' CHAPTER 5 Beaml and Frames: Shear and Bending Moment ..lope of 'hear diagram at a pomt intell"llty of distributed load at that point (5.3 l'hange in tihear bcteen points A and B change in shear at the point of application of a concentrall'd load slope of bending moment dIagram at a point change In bending moment beleen POints -4 and B area under the distributed load dIagram between point ., and B magnitude of the load hear at that point area undcr the "lhear diagram betv.cen points A and B (5.5 (5 12 58 510 change In bending moment at the point 01 applicauun _ magnitude or the moment of II couple of the (,;ouple 514 A fio:lme IS l;on Idered to be bendIDg moment Bnd aial lore tatl(,;al.l) determinate if the external reactIOn can be d e ID all Us members as well a all clennmtd b h num and condUlon If a plane f ) I e equations of equil I uPPOned b) r react rame contams m members and J JOID Ion and h . as t' equallons of condition theJIlf
  • 83. d Fro...., SbN, .nd Bending MomentCHAPTER 5 ...mI.n . I~k..: m IIIIIII ill 1111 (1A Hnge B 4m-1--4m-i--4m-1--4m~ p ===tI-----L.---I fI6. P5.18 fW r L ! A r== I L - _ B 10m ... P5.9 FIG. PS.12 B 20kNim 20ft 5 m----4-------- I m fIG. P5.25 :f--rrrrrr-r-rr-c ~ I 30ft I-15ft fIG. P5.2II fIG. P5.27 3m 6ft 10 f• 3m+ 12 k 1m Sf{ 6ft 6ft b===!B:="""'----....C., 20 kN 20kN air;=~!~-""":!!---" ./.- ISk 100 k-ft I ~-C--!--_IDA t"'ii'" c fI6. P5.19 fIG. PS.2O FIG. PS.21 IV A~ --L---I • Aprrrq B I L I A 1i====;)M I--L_----.,I FIG. PS.13 RG. PS.14 fIG. PS.15 fIG. P5.16 24ft 40k 1 "'-PS.l0 ... P5.11 .....u ....... U1 qualitaliw 10kNlm B ---+----7m---l7mfIG. P5.17 fi r af the resWu g ldiqram
  • 84. -• 1--11 11'-+'':'+ ':'-1--3511--1 .... P1L41 ...... "'1'U7 .. 40k II ~ t 10 kNlm 30 ft ----+---10 ft 2 kJft 10k 2.5 klft 3Sk ~ A~C D ~S ft.J.-- I51,--1-'0ft-l '50.N 60kN A~B C D I-6 m--l-6 m-I--sm 15 kill 4.',;-,,,---,--,':'::',::u:I:9: -I- I-f_ _ JOft _H_'...:...._+- AF~b6~;.J..u..u..u...r..;~ B I 12m--1-6 A 10 ft A SJ< I RG. P5.43 75kN 3 kill J klft 24 II -_. --.+-_ 2S kNIm '50kN I 600kN m 1CE B V :b. D 1--5 m--I--5 m--I--5 m--l B 6 m ---+---6 m,·- --1-- 20. I 7Sk-ft A 11======8 V D 1--10 ft--+--6 ft----l-4 ft-J fIG. P5.37 fIC. P5.36 9 ft --i- ----_ 24 ft -+_ "'PUI "PI.AO A m 3m 1 m---- • A _I_....__....--.- 20L M1k • 1-_ _1...._ _1B C 60 ;.--!l-----J...-..;,t.. flC.P5.36 ----~------ 10 ft , • 20. 20. !-I-L~J!-~Jl---,.4 8ft 8ft 8ft 8fl 1----3 '0--+--J m 3 m 50kN 'OOkN 50kN IJI-!:--!!--L-~ ...- .. ... P5.30 ,.,.1.31 "'PU2 ...PUI ..PI.M
  • 85. - 6 ft4-6 ft--l JOk I 1S m+---s m----l ~4m 12kN1m Pi 12ft-+--1 ft ... P5.8I ~I Sm "'P5.58 ....... 10ft 1 m c 90kN IS ft S m---<1-- 25 k 8 25 kN/m 15 ft soctI""U J7 ...... 5.71 Draw the shear bendmS moment and I I ron:c dIagrams and the qualitatIVe deflected ha~ Ii .1]<1 I' .... r the: frame !>ho"n. fIG. P5.57 (b) (b) Hinge (d) Hinge (d) J. (a) o (e) (a' (e) ... . AG,I'5.55 C 4ft 150kN I AJJ B .1.4 1-9m-+-9m-+--a-l 1--20ft--+---20ft_ _ 4ft+4fI~ ISl!ft I A BmmC D ~~~ OS klft 60k A 1===:Hi:~~=~~~-====91rI: 1II.P5.53
  • 86. D ISkNIm c 1----20II ---.I A B 1---10m---+--SllllilJ I ft I ft ....... ..... 9m --_.......-'
  • 87. 221 J 8 Deflections of Beams: Geometric Methods 6.' D,Nerenlrat Equallon 'or Beam Deflectron 6.2 Drectlnlegrallon Meltlod 6.3 Supelllos~JOn Meltlod 6.4 Moment·Area Method 6.5 Bending Momenl Dragrams by Parts 6.6 Conjugate Beam Method Summary Probtems Strut:lUfC like ,III lthcr ph} 11.:,11 bodic!'>. defonn and change shape h"o suh)ltCtcd to fon.:c_ Olher common causes of deformations of 'JtrUl.:lUn:, mdude Icmpa,llurc changes and support settlements. If the deltHJndllons t.I1.lrrcar .md the 'itructure regains its original shape when the action... l.:.1U mg the deformations are removed, the deformations are h:nncd deWit ,/(/omwflofh. The permanent deformations of structures are referred h) ..... mt/mlll or ,,!mli£ cJelormutions. In this text e WIll Ilx:us our alh.:nUon on l/nwr t/tlllic tIt1uT/1wtimlJ, Such deformations 011") Imearly "lth dpphcd loads rfor instance. if the magnitudes of the load altlflg on the lrul.:ture ar.: dl)Ubled. its deformations are also doubled. and so forth Rel:alllr)m Scl,;lion 3.6 that in order for a struc. lure to rc.: pond Imearh 10 Jpphed load'.. It musl be composed of lInear cia til. matenal ilnd II mu t undergo ~mall deformations. The pnoclple f UJlt.iTk.'SHlon I ahd lor ut:h trudures. h r rno I trueture e;l.:." (ve deformations are undesirable a they ma~ lmp.ur the tructurc s abiht~ 10 '!oCl"e its intended purpose f or example a high n bUlldmg rna) bt= perll.':tl) safe in the sen that the II )v-able Ir re d ft a n I,",eec ed )CI useless unlX"Cupied If It .... 51 Ie: I I) due (0 md l.:au 109 uack in the alls and window ruclUre re u uallv de d h d Ign 0 I at their deflections under DOrmal m;~ n III n 111 not e l.:eed the aJloable alues specified tn buiJd. SECTION 1.1 DI""..... ' ............. Dsn • 6.1 DIFFERENTIAL EQUATION FOR BEAM DEFLECTION •
  • 88. 6.4 MOMENT-AREA METHOO MDiapm - _ _-'--..J EJ lECTJONu --- m M, P, P, wlx) A Beam I--x I/Ix I T.....'"B in wblcb fl. and fI, are Ibe slopes of tbe ela tiC curve al pomt A and B respectIvely. wltb respect 10 tbe axis of tbe beam m tb und ~ rmOO (borizontal stale fllU denotes tbe angle beteen tbe t ng n t I elasllc curve at A and Band f: I £1 d< repre n th undo M EI diagram between polDIS A and B Equallon 6.12 represents Ibe malbern IIcal momtnl-a"o Ih "m wblcb can be stated a ~ II The chan In 51 pc beteen !be la polDl I equalt tbe a... under ,he I EI ~:::,: proVided that the elastIC curve I Un LU fIG. 6.4 J 'M El dxo, or - -- - -------- The momt'n;~~re.1 method for computing ...Iopes and def1ectio~sof beams 3" dedopcd 1:1) Charlc... E Greene m 187J, The me,thod IS based 011 tO thcorem.... called the mOn/tIlt-area ,I"ore",.'. reiatlOg the geometry of the ela...lic cune of a ocam to 1" f EI diagram. hich is constructed b) di iding Ihe ordmate... of the bendmg moment diagram by the ftexu- ral rigidII} £1. The method uuhle~ graphical mterpretations of integra) imohed in Ihe DIUlion of the deflection dllTcrentJal equation (Eq. 69 in term... of the ilreas and the moments of arcas of the .1/El diagram. Therefore. II is more comcnient to use for beams with loading dlscoo tinuitle... and the ariahle EI as compared to the direct integratlOD method de-.cribcd preiou..l) To derive the moment-area theorems, consider a beam subjected to an arbitrar} loading as sho",n in Fig. 6.4. The clastic curve and the .1 £1 diagram for the beam arc also shown in the figure. FOCUSing our attention on a differential clement d of the beam. e recall from the preiou... ~ctlon .IEq. (6.10)) that dO, which represents the change m slope of the e1a!lllc curve mer the differential length dx. is given by If dl/ EI dl 611 Note thai the lerm 1 £1 dx repreents an infinitesimal area under M Ef dIagram. as honin Fig. 6.4. To determine the change ID slope beteen (0 arbitrary pomts A and 8 on the beam we integrate 6.11 from A to B 10 obtam ' rdll r~d< DetIectionI of Beams: Geometric Methods . lIlJl idu.dl~ on the beam. The slope and to ""h.h '1 Ihe I(lad .ll.:tdm g · d .1 kllJ C,1O Ix l,:oll1pulcd by using elt"- . d tel -il HI 1 I u . · _ ICT dl,;lkxllI.m Ut' ll.l h' d' 'Tlocd prC10U'iJ) or one of the 0'''-- I In mel llU l.... ~ Ihe Jlftx:t mh:gr.t It b' 'Ill ~dlon' Aho. man) struclural tnDl h d d · J III ..u -.t:qUt; - ' eo Illet i..' I"l.-U 'l II nlUlI ofS(I.'/ ('Of1trlU fUm p.ubhshed by .L_ . - h ndh:ll1k o! ,1 _ 'UlI: nC:nng a" ~ I (QmITUl 1/ n wllta," ddkClIon formulas for ""til TIl all Inwfllt: ,: ..'1 ~,f lo.ld...and Upport condltion. ," hich can be t'leam.. fN .trI(lU ., rxs . -h fonnula... for ,Io~' and deflections of ....._-- d tor Ihl'" rurpo"C: .. m: . . .-ms u . 11 Illd .tOd ,upport conditio"" are glen IDslCle for ,.lmt' l,.'omnll1n t~ pt: . l • , .~ . _ '·,h. bOl1k for clmt'Olcnt rekren,;t' the tronl l:l" er 0 .. CHAPTEIII
  • 89. -SECT10IIu _ ProeedUJlI for Analysis 61 6 Sub-.tllution of dO (~) ,",U dl .11 8 t' [he ckment lh "here J the dl 1.101,.·..: In1m l ft.!1 mIl) Eq "I ~ Idd:lo r:d f'1 Hit ,U "r fitAs< ,d, ,E/ in hich ~'H fl:prc..enh the IIl1/e/l'II.';O/ dlrwtioll of Bfrom the tan at 4 hlch ... the deflection of rOlll! B In the dlrl:~t1on perpendl to the undcfonncd ,11'" l,f the beam from the tangcnt at pomt A r:,1 f1jrdr Tepre..."nh lhe moment of the Jrea undcr the M £1 gram bctwt.-cn poinh 4 itnd B about pomt B Equation (6.15) rcpre~nh the mathematical expression of Ih ond m01lUnr·/lHU Ihl oft m, hil.:h can he stated as folloYos. Tht' lang~nllal dClalion In Ih~ dtf('(tlln pcrlll:ndil.:ular to the undefi alll of Ihe tlcam tIt a POint on the (I,...u!.: ("une fwm the laD Dt cIa tiC tune Jt .molher pt1lnll" equal {{l lhl: moment or the area 11:/ diagram hel.een (hI; 10 pl1mt.. Jboul Ihe pomt al ",hi h tlon I d m:J pr lu,k"11lhal the cia Ih,; I.un I l;llntmUtlU between pomts It is Imponant 10 nole the ord~r of Ihe subscripts used for 6 15 The fir,t !lu~npt denutes the pt.)!nt "here the devlaUoD temuned and ahout "hl..:h the muments are caluated where ond ubs4;npt denote the poml I, here the tangent 10 the elastIC dral,n Also In the dlstanl.:e an Eq. 6.1<; i alway taken tle the IgD or ~B I Ihe same as thaI of Ihe area of the M gram belecn A and B If ,h area "f 'he If Ef dlagram beltwe_ and B 1 positne then ~B IS also po Ilne and pomt B I the po5ll1 direct h IOn t e tang ot to the cia tl4; cune at polo ICe 'rna . on the nl!ht~h.:md side of Eq. 6.14 represent ote Lhal Ihl.: h;~ • 1m 11 .trC.1 u"rf: pendmg to til: llbout B n1C'mcnt E cJ t~el~n ~~~lcen an... ((l arhllraf) point -f and B on granng q.•' . bc:J.m e: obtain DeflectionS of Bums: Geometric Methods _ J I t) thl,.' unddomll.:d aXIs of the I ........qxllllU.lfl ek'l11l n1 d, llO I 1Jll ,. lrllm a roml 81' gll;ll t'l~ CHAPTtR 8
  • 90. IECT10ll u .....--. DID........ " ..... ,,....---(r- Ft6. 6.10 - c ., GeOmetric M...odaDeRictiN.ClIAl"IER • 1Ilk ~ klft A B Ie Hill c::r:=Q!~C .;i I c + A .... I t Jk Jk J I t 4k 16k30 k. 3D k JOft I ID ft 225 169 8 C + CII A II h6k 46kt120 I JOft I lOft 120 fa (b, (e) 225 d) 120 Ill. ..9 6.-5-8E-N-D-IN-G-M-O-:-M=E:::N::T-=D~IA:'G:;R:'A~M;;;S;B;Y;-PIA~TS. . IpplKallon of Ihe moment J n' '(tll'O.• ... 10 Ihl: prc.."'(c I)~ of the ;In:.'" and mome~ts of areas h _.1 nH11L'''' l,.'llmrut.lll~ m It ill ~ ..hown In the fol~ 01('( lOU I . I f I dlagr.1 . ,. d fl nolrthlO' 1.." the - lh ld for detcnmnmg e eClionaanou" I"~ • be 1m me l - . h '-__ -I n that the (cmJugah:- < - I" thc:...e quanuues. W en a ~ -.ex 10 mputatu1n 0 . . f d ..... 11"'0 n:ljum:' .'t 1 . J m.:h as a combmatlon 0 lStI' • . 01 loa ... ... . ..ubJccted to Jltfl.'renl t~ rc~ dc{ermin.Hllm of the properties of the ut·d and COl1l.:Cnlralcd loa h' 'ombmcd effect of all the loads, t: - due to I I: r..: oded b ..ultant I 1::/ tlJagrJnl. Th- difficult can be 301 Ycon tl'UCt ba:ome d fomlldabk t.l"~ ,... .
  • 91. III p s/ lECT10Iu ....... _ _ 1---.---1 I------ L _ p L~l sl p c Ic b '"r1UaI ..A LB lniliaJ c _ equdibrium P position sl A~LJ p c fc Ii AG.7.2 see FIg 72 c The vlnuaJ work. don Virtual rotation (J can be expressed as W. P a By subsUlutmg Eq 78 through total virtual work. done as If EI' t Because tbe beam 10 I' EM tberefOl< Eq JI', P Ii - LI' tA II Pli ( li II Pli II AliII and W k Energy Methods CHAPTER 7 Deflections of Truues, Beams, Ind FrlmtS: or-210 Principle of Virtual Displacements for Rigid Bodies 1 he pnm.:lpk of mu••)!l1t'nh for rigid bodies can be stated 1011('1 :.: If a ngld boJy I' In ~"IUlhhnum under a ~~ ,tern of force.... and If II I ~"i,:t~d w .m> sm.llI ,rlu.11 n~ld-ood} dl,pla~'-'IT'lcnt. th~ lrtual work doae by thl" c1t:mal fon.:!':' I' I~ro rhc term I irl/1al 'Impl) means imaginary, not real. Consider the he.tffi ..hon in hg. 7 1<1' The free-bod) diagram of the beam i shown in fig. 7.2(h; m hich P, and P, represent the components of the ternal htd P in the." and y dlrCl:tions. rcspectic1}. 0 SUppO'iC that the beam is gicn an arbitrary small virtual ngid·bod~ di..placemcnt from it-. initial equilibrium position ABC to another ro..ition I'B'C' as 'ihm...n m fig. 7.2(c). As shown in this figure the tOlal 'irtual rigid·hody dhplaccment of the beam can be decom po!Cd into translations !i. and 6., in the." and y directions respec- tiel). and a rotation (J ahout point A. Note that the subscript L IS used hL're to identify the displaccment'i a.. Irtual quantities. As the beam undergoe.. the "irtuitl displal:ement from position ABC to poilU 4'B'C'. the f(,lrce.. actmg on it perfonn work. which is called uork Tht: total irtual ork. H . pcrfonned by the e"ternal forces mg on the beam can be c~prc"'M.:d a'S the 'Sum of the irtual work. Iv. and It donI:' during tram1<ttion.. in the " and y directions respectl and thL' irtual ork It' . done during the rotation: that IS. Dunng the Inual translation~A and < of the beam the work done b) the forl.:cs I~ gicn by 7.2 PRINCIPLE OF VIRTUAL WORK I I 111/ rli. Im.h .1' lI1trnduccd b) John Bernoulh The flllt If IINl/!1 ' l . , rl,,1 11111 11l.11 hllli lor mal1 prol.)ems of t I~I"" pro'lu J pt.l!; " . . . I h Xll~ln C ~tud) 10 lonnul.ltl(,lOS of Ihl Prm. tural nlIX h.mll. n I IS 'l.: I •• n Ipl nll/Twal ~/lpltl((nlUlt lor riC/ill 'kid, l,.lplc nam ) Ull prl I ( I "II r 'lor dt (tlfl/whlt- hod/( The latter fonnula th.. p'It/' 'I' al (IT Ii.. ' , h' r,ll,' m" ,,-'1.:11111 10 dc, clop the lilt thud of'"uon l U'CU 10 I ; II. eo , •• '11 IJ,·n.:J III ~ one: of the mo,' general methodsII ,,.. IIK.. 1"1 l,.( u ... dctaffiuung delkl:twn (If ,truelun:
  • 92. which is the mathematical tatement of the pnnclple f lnual Ii tee Ii r deformable bodies. It should be realized that the pnOClple of Inual ~ here IS applicable regardless of the cause of real deli mat deformations due to loads tempe:ntlure chan or n t be determIned by the apphcalton of Ihe pnn. pi How,.,·e,. mallon must be mall enough so that the Iftual Ii an magnitude and dlrecllon while pcrfi rmlO t In though the apphc tlon of thl pnnclple lD thl t t structures th pnJlClple IS ltd re rdl la lie or not The method f Irtual w fon:es ~ , deformable bodlCS rewntten IIEC11OIIl.2 1'ItnoiIIIo.. _ _ _ w w. In .... hleh the quanllt on the left-hand ide rep ....ork Iv. done by the Inual eMernal fo real external dl placement A Also real n th the & cos IJ are equal to the real Internal dl p1acernen members AC and Be respectively we an C fK:11Kk that the nighl-hand Side of Eq 7.15 represents the "Inual mternal.... rk JJ d "irtual internal forces actmg through the real mternal d p.Ia,:em,en". lhat is ~ ~11 F, 11 11 or W ~ A JOdlcated by Eq IJ stde of Eq Thus Eq ~11 F &e05O 7 I 8, 8, n 'qulhlmum under a _nual S) ..tem of Iibl trul,;tuT I , I; It • J t(lnna ll, :l,.fcd l(l an' Tn.11l fl.'al dc:lonnation (:ODII d pc ndlllfl'iO)U'" - to 1,;( U J unull lt1nJlllll11 (11 the: ,trudure then the llih the UpplS .•n ..on . d b lh- frlU.11 c'h:mal I"Tee.. Ian coupl Ie- I ori. done I; . . . I JI pJ.Kl.:m.:nh and rol,lIll)O" '''' I,.'quallo the thTl u h Ih. n.: I h;m.. . d , J n· bl Ihl. 'lrtual mtcm.l! 100.:':" an couples wI mll~rnal 1[.. 1..' I; . I ' I'm I dl'.plal'cmc:nt .md rotatIOn, thh ugh I I,," n: III l. I th' teml rirluaJ ,.. a......ociated with the forenIn thl ..IJh.'mCll c mdll...lle Ihe: force ...~ h:nl l' arhHrJr) and doe.. not depend on action .JU 109 ,hI,.' n:.t1 Jdoml.lU()O. _ . _ . r d tr I, Ihe: ahdit 01 thl" consider the two-.1 cmlln, ,It.: _ ., , ". I h""n ,n I·i.' i 1.1 The tru"s I~ m equlhbnum unOOtbemc:m~r ru.. v e." • J~llon llf a inu.1I t:(l:rnal fon':l: P a~ ,hlm n, !he fr~.lxxl~ diagram 10101 ( of th~ tru , i, ,hoJl In hg. 7.J(b). Sl~ce JOIO.t ( IS In equilib- rium. th~ irllial ch:n1al .1IlJ mternal lorce, acung on 11 must satisfy the 10110 lllg (l) I:lulltbrium equation, ~f 0 P F'H co,lh F;B( cos()~ - 0 - ' L F. () F, U 'Ill O[ + F,B( sin (h 0 m hlch F l( and F H( rcpre,cnt the VIrtual internal forces in members .J( and 8C n:'~l.:tlc1y and fh and fh denote, respectively the ang1 of mdin<llion of Ihc~ member, with respect to the horizontal FIJ. 7 J(a)). No. lei u, a"i"iume that joint C of the truss is given a small real Ji,plal.:cml:lll .: 10 Ihe right from its equilibrium position, as shown Fig. 7 ~I.t 011: that the deformation is consistent with the support (bl E.,u I ClW'TER 7 fIG. 7.3 m
  • 93. - n 10 Eq H When the defonnauons are caused by external load Eq 19 n be subslltuted 1010 Eq (7.22) to obtain W. LF . By equaung the Virtual external W rko Eq 2 t mlemal -ork Eq 721 in accordance '4uh the pnoclp forces for deformable bodies e obtam the r. II WIn e p I method of Irtual work for truss deflecuon 1 A Because the destred deflection A IS tbe onl UollO valu can be determmed by solvmg thl equauon Temperature Chang" and Fabrlcalion EmIlI The e pressl n or the Vlrtual w rk method qUit sen I In the sense thaI II n be ftecIJ os d 10 temperat= chan ~ b,rica.ioD elfeet r. r which !he member 1de can be luated be haud The I deformalion chan 10 temperat= Member) F. Member} (; _ FL M D o p B D B P, (a) Real Syslem (b) Virtual System FIG. 7.4 ~( lrlual mternal force - n.',.1 mlcmal dlsplacem ( lrlUal ('xh:m.a1I~IC~ ) '" ftal (; h.-mal 1.11 r1al'(ment ----~ h h h I r r " Illd d/pJaftll/tllf l . are used in a general In 11: I I: cnns " . and mduJe mlmlcnh .mJ rl)t;jl1l~n, r~'pccllcl), ~ote that because I r . nd 'no'oJ..:nt 01 the ,lctlons t.:au".mg the real defonna'trtU.1 on,:cS.lrI.: I l,:1~ . d n 'nsllnt durin" the rc.1.1 deformation, the c,,"premotlon an fl.-mal 1. • . eo ""I the ('''ll.:mal.mJ Internal lfto.11 ork In I::.q. { 18 do not contam f,H,:tor I .:! . . ['I. ~ Il IIldll:atc. the method 01 Irtual -ork empl 1"0 ...crant1c ,)sICm ,I Inual forc~ 'i),h:m and the real ~}s_tem ofl or other dk':h that cau...e the dc/ann,tUon (0 be determmed To terminI.' the dcRL"(llllll ,or slope) at an) point of a structure a Vln fon:c ~> lem i, '>Ch..":lcd '0 thai the dC''ilrcd ddlcl:tion ~r rotation WID be the lml unknon in Eq. 71M The exphcH expressions of the Virtual "ark ~cthod to Ix used for computing deflections of trusses be and frames are dr:c1opcd in the three sections. In hu:h 1 A and 1. dcnote re rectI' el) Ihe length c:........oec1ia1'*' area and modulu 01 c1asllut} of member J. T() determllle the Vertical ddla::uon d. at joint B of the Ittl a Irtual y tern' .consl IlIlg uf a unit load actmg at the ,..N'·_m the dlrecuon 01 the d d d fl .eire e ethon as sho-n m Fig fL At: To dedop the cxpn:ssion of the virtual ,,'ork method that can be uaecI to determine the delb::tums of trusses. consider an arbitrary statically determlllate tru !oJ, a!i !iho"" In Fig. 7.4(a). Let us assume that we 10 determinr: the ertlcal dcfle<.:tion. 6. at Joint B of the truss due gien extcrnalload~ PI and p, 1he truss is statically detenmnate axial forces in Ib member... ('an be determined from the method of described prc iou...l) III Chapter 4 Ir F represents the axial force I arbitrary member J e.~, member CD in hg. 7.4(a)) of the truss from n/( han/( of n/lltl'rllll~ the axial dcfonnation c) of this I'IICIIlber glen b) rk Energy Methods Deflections 0'Trusses. Beams, and Fllmes: Wo - II/will fl'{(m ClW'TEII7 7.3 DEFLECTIONS OF TRUSSES BY THE VIRTUAL WORK METHOD
  • 94. ....,.a ..........'-......-,_..... ,n n ,n 84m H r lIlt I I n I (J T1,.1 lal:lhtatc the compulallon of the defl lion the real and 1rtual membc'r lon.'CS are tabulated alonl Jcngths L and the cro~"-Sl.'(tlonal area~ -t of the members a hown III 71 The modulus of ('la5111,;11 E I!'> the same for all the members 50 II..""'.' ROC mcluded In the: table ote that the same sign convention I used Ii reo..) and Irtual S) tems that IS In both the fourth and the fifth colUlllDl "'''''..-.. table ten lie fon.-es are entered a" posit" (' numbers. and compreSSIve €I negatl( numbers Then lor eal:h member the quantlt) F, (FL A II COIIIIP. and It!> alue IS entered In the sixth column of the table. The algebraIC 111II the entnes In th '501 Ih fo:olumn L F. Fl A is then detenmned and I recorded al the bonorn of the sixth column. as shown. FinaJly. the deIirId flectlon.1 IS detennmed b) appl)mg the ,irtual work expression Eq shown In Table 7 2 ote that the positi e answer for .1G indicates that deftects to the nght In the dlrecuon of the unit load. TABLE 7.2 Member L In A (in ') F (k) AB 192 4 60 I CD 192 3 0 0 EG 192 3 20 0 AC 144 4 60 I 5 CE 144 4 0 ~7.7 0 BD 144 4 15 DG 075 144 4 15 075 BC 240 3 -75 -125 CG 240 3 25 125 _ 7 _ .... III - . ........... _ : W....-En.rgy M....od. -
  • 95. • Ana 4 I .n I In 1ECTIOI7A a.IUlI••• II. . .'" lie 11I1III'"~ MembeT CF E:F Tlill 7.1 7.4 DEFLECTIONS OF BEAMS BY THE VIRTUAL WORK METHOD f .l al ~O h - 60 h -----< E -O.-lm. F 4h, IC) Inual S)stem - F Forces 'a) r20 ft ~~~~Dl It'll Real Sytem - 8 . 'Ii. II ddl. Iu,," I t01l11 J) 11 the Iru h.'"n In hg Lkt t tbe l;n he f f '0 fl.-I III 100 shon i: "', {f ''0 (l In (I(' I""f! . nJ m 11l r mt:thx! f tu.!I" ri.. I k In Solution kat S m The re-dl S} lern ..onw'ls of Ihe I.:hanges In the leo members (f and E.f rth tru a ho"n In FIg 7.9 b llfun/S 1m The I lUI di Inua )' tem Call! I Is. of a J-k. load applied rtl.:tI nat J Int 0 a hov.n 10 "Ig 7 9 The F In mcmbe (F d . I.: necessa '-;;;'JjI n an Ef cotn be ed Ily computed by uSlOg the . Wortl...Energ~ Methods CHAPTER 7 DefttctiOItI o,Trusses. Beams, end Fremes· Ell lillie 75 IIG. 7.9
  • 96. .".._,....- Af HIli , 3011 EI E I ....... :::"'~I.;;"A--------'-_·'" ........ E••",pl. 7.9 2,193.75 200(300) ~ 0.0366 m r(;)75, d, ~f(:)(75, dT +~f(:) 75,+~)dT+ J:(H 75 2 193 75 k 2 m' £/ 2 19375 kN m' EI .1. 366 mm 1 1 kN .1. Therefore SoIIlIOII The I nd Irtual t 111~ rc luw.n til Fig 7 B{b) and am be n from Ilg I' a Ihat the Ill.: ural rlgldlt)' £1 of the abruptl) at pOint 8 and f) 1Sl,', Fig B b) and (c mdlcates and ,rtualloa.dlO~an: dl ,"IIOUO"", at p'-,mb ( and D respecti quenll th an tlon (If thl: quanllt) ( " ' f.I 1" be dllCOlltin Bend D Thu the beam musl ~ dl Ided mto four scgmen .... and DE. m I,:h gmcnl the quanllt) ' I £1 will be COIIti ~fore can be mlcgrated ~ coordmal sclCIW for deternllnmg the bending m~ are hon In hg I' band 4.: ole lhat In any particular ...__ beam the me coordmatc must be u!oed to ",nle both equations equation for the real hendlng moment 1 and the equauon (or bendJn, moment " Th equation for 1 and I for the four the beam dc:lenmned b' u 109 the method of sections are tabulated 76 The drft lion at D can no" be computed b) applymg the vartlW pression 11en by Eq '0 1 ,V V 1 ~ EI d'f TABU 7.' Sqmmt AS A o 3 -';IEI Be A 3 6 2EI D A 6-9 2EI ED E 0-3 EI 75 • _ .& 7 Da""_" nu-. ..... and Frcmu: WOIk-EnIl1Y Methods
  • 97. 6S01lkft £1 019410 6 SOIl k £1 ApplicabOD of the virtual work metbod 10 determine tbe 8octions of frames IS Slnular to thaI for beams. To determiIIi bOD or rollbOD 8 al a poInt of a frame a virtuaJ UDit couple IS applied al that pornl When lbe virtual sys1eID the defonnabOns of the frame due to real loads, tbe work performed by the unll load or tbe urnl couple II W. I porbOD of the frame may undergo axial In addibOD 10 the bendiDa defonnallons, tbe total virtual done on the frame IS equal to lbe urn of tbe mternal virtIW. bead!QI atId that due to axtal defonnabons. As diacuallll1J1 ma lOCbon, wbell the real and vlnual loodin.. and tba Elan COIItilllIOua over a seament oflbe frame the virtuaJ due to beadml for that seament can be obtained by .....,my JI, MEI over the lenatb of the seament Tbo due to beadial for the ent,,,, frame can then be the work for the iDdividual &eamenll, that 11, W. EIMMdx £1 ~ §~if~the axial fI F UJ4 F. dose to tbe niland the uiaI rilidity A£ tile _al tile u discuaood m !bat &epICItt due to axial . . . ',<.FL/AEl, Tbua, the mtanal work clue lie JIIae 1IF fU- BY THE VIR1UAL WORK METHOD ......_ , , - . _ .... _ .....__ MoIIHIdI 7
  • 98. • __,......_.II/__....__-E_~ Ikft G------::o!D I B A I, I I cr)(rc~ ~I I lli I EII_7.11 C 1 • -l B I • AJ ...n. I (c) VUlUl1 S)'IIem - At D III. '.11
  • 99. • W rk-Energy MethodSawrrEII 7 DeflL1IoftI of TruIIII, and Frames. 0 B ( TAIlE 7.1 - y ( ~ '-Y 4 ~--t!.- TIl<'t', k;H 4 ISO kN m £1 m m • -:.~ rh ~ 10 mm AnH ' 0 H 4 -t> M IS. 711 d ElIlllJlle 7,~•.11.'2: --:-_-:-_-:::-:-:::-i
  • 100. III 1. - D (<)VutualSy .... .1 - A B lIl:naIIu Dollica". Of " - " tis _ B 'MUfTO FIG 7.17 (l;ontJ) £:29000 I 1000 ... A = 35 .. 116/l I27S n IOU- b) k ISm AI F 125 t- I() ft -Jo,. IHI A 167_ II 5 115 107 tJ5 CHAPTER 7 DrttIIcttons of Trusses. Beam•. and Frames: Work-Energy Methods : lift 1.!!-l.,,1=,_~T,,=~r:=!191 c B 312 M717
  • 101. 4 To develop the expreSSIon for IDternal w rk or tra n truss let us focus our attention on an arbitrary member CD tn Fig 718 of tbe truss If F represent tbe member due to lbe external load P tben d. twed ax,al defonnallon ofIbis member I 8' en by FL Internal work or stram energy tored lD member t w w Strain Energy 01 TI1IIIIs ConSIder .he arbitrary truss b n tn FI a load P wblcb IDCr<a graduall fr lhe structure 10 deform as shown tn I fi enng linearly elasllC tructures, tbe deflect of applicallon of P tncreases linearly tb I cussed tn Section 7 I see Fig 7 I c tb tern dunng tbe defonnabon 4 can be expressed The tram energy of !be enlJJe SlDlpI stram..,...... ofall fils membeTs tUKI be ...~...... or beca each FLtllJU>l. ... Member} 6=t.l:. AI: D p AG.7.18 , I ~ 11 67"1: o - IVw II ('I'" ~o J""I,- ( , . u )~ 01 !..-It 9. r'5 I.At 1 If EI 510S 9.rSII:! +- - - . '5 29000 29.000 1.000, OI)()()()5 0.04655 o,n466 ft 0559 In 1 [ I h7 d f/ 2 1 If " I~ I I k ~ Before we t.:an dcclop the next method for computing deflectio ~lructurcs. it is nccc..sary to under~tand the concepts of conservabOD energy and ""tram energy. The (ntryr of a ,lrUl:tun: can be 'iimply defined as its CQpacr doiny 1I0,J,; The lcnn ,train tnat/y ., attributed to the energ t (ru("(ure hm h((uu{' 01 ;H t!t'/ormutiun. The relationship between ork and train c:ncqn of a strUl.:turc i~ ba~d on the prrnc,plt -,_<. {'rtatum of tturql v.hit.:h can bt:' 'tated as follos: The otl peth rmed on an ela tiL Mrul,;lurc in equJlibnum by I1OJdIIi~i~ gradually applu.-t.I external lot(~'i is equal 10 the ork. done by t tC(' or th tr.tln energy hm.:d 10 the Iructure. ThiS pnnciple Can be mathc:maIlCall} expre!'osed as Sote the magnitude of the il'lal defonnation tenn i!'o negligibly smaD compared to that of the lxndmg dC'limnation tenn . W k-Energy Methods Deftections of TrullleS. Beams, and Frames. or t I t lI11e l,ln tx- ddl?fllum'd b~ .lprl~lOg the lrtual fle.,:lll,m It Illtn' ( (l 11 r. o:prellllllhnhly 10li ClIAPTEl1 7 7.6 CONSERVATION OF ENERGY AND STRAIN ENERGY
  • 102. • •~.!..............- ... - " " - " " ' " ,
  • 103. II' r Eq rep • nd Eq IJ. ~PIJ. 2 "lotredmth { P, Ial enl C:"';Iiano's dl or By subslllulIDg Eq 7 51 IDIO Eq 7 56 we oblaln U dPIJ. l and by <qUlllng Eqs 7 54 and 7 5 we wn t P, l P 8ECTIOII77 -"_'. _ _ l l dl l I - dP, dlJ.. 2 • which lite madtema Since dP2 rematns constant during the additional deft u n . poinl of apphcallon, the lenn dP Ill.. on the ngllt-hand de 7.55) does not contam the factor I 2 The term 1 ~ dP d6. sents a small quanut} of second order so It ~an be neglected (7.55 can be "ritten as and the total tram energ) amount dP th n lhe me 111 ram apphcatlOn or dP can be en r Hlllen a II il t,ft • 'r <llhHhl.:f cnl'rg~ mCllllxl for dctenninlDl I Ihl SC... tl111 C 1,.011 luI: . n 1111 mdhoJ. Im:h l.:an he .lpphed only to 11 IIOIlS ,11 trudun: .. IL, - , I Imll.dh rrc,cntcd 0) l>crto (astlgha .:Ifl ell III,. slru"llln: . - . I L , ' I • 1 1 n ' (t.lllylul1JO t'{ mil 111l1lr( m C I -, !OJ 1 ... OlllilWIl 1..1 l • - . " , I hi -11 l" In be to c",I,lb1l,h equatIOns of ghaOl fir I t H,'MCm l ' _ _ ' - . , I n 11 Cllll...1JacO In tim, tc'L C.htl[!hano hhnum 01 Irultun:s ... rhcorcm Lan ~ tatcJ.I 1...,11(1 in y.hid 1 ~ and .Ire the deflections of the beam at the pOlO applil.:allOn of PI p, and P, rcspcctivcl~. as shown in the figure 7 51 i indil.:atcs. the ,tralll cncrg) L IS a function of the external) and l.:,m be Cpn.:sscd as 10 hid l ,tr.un cneqn. ~ defkl.:tion of the point of apphca of thl,; fl'fl,:~ P In the dlrcction of P:; and 01 -= rotatlon of the pomt apphl,;.ltion of thl' couplcl in the direction of Xi,. To proC thl'> theorem. col1,idcr the beam shoy. n in Fig. 7 20 beam i, subJcl.:lt:d to cternal loads PI P~. and P,. which increase uall) from .fcrn to thell' final alues. causmg the beam to deflect ,ho n in the figurc The strain energy (L stored in the beam due to external ,Ork It ) pcrfonncd b) these forces is given by J PI P~, P,t OY. a ume that the dcf1cClIon tt.: (If the beam at the polDt plication of P i~ to be determined. If P: IS increased b~ an IDfinil......~i' I I 1 leu turl: thl.' partial dcri<lIi1.' of the tram llr 1m r l: JS II; ... h ~I" n 'rrhcd fllrcl.' ,)T couple i... cqu.d to the di p1acemen"It (I; 1__ l . ' f h c'"C or ,'ourlc aillm! 11 hnc of actwnrolaU, n ,1 Ie Il..... ~ In mat!ll.:mal1cal hlml. thi thcorcm c.m be" ';Ialed 3!'o: W k Energy MethodS CHAPTER 7 Deflections of Trusses. Beams, and Fr.mes: or- 7.7 CAST/GUANO'S SECONO THEOREM 3'8 P, P P, .... I I J~~Ill. 7.2l1 'I I_i__
  • 104. and 0.. Ihe flJ.rtl31 dl'n<J;t1e tF 'p ~,. tF tP. the c:pression of llghanl)0 ~'':ond thn)ft~m fM tru "e Cdn be nllen a!> Application to Trusses To dc.:H"k'p tht: l"prt: ](In (,I Ca'tlgli~no· .... ".:cond theorem whM;:h he used fo Jl'lt:nmnl" the dd1cctll'll' 01 trw·....e" I.e ....uh!>titute Eq for the lram cllcrg) l I of tru"se mIll tht: gt:ncral cpression of ~hano'" ..n'onJ thl'tXCI11 for Iktkdil1n" as gicn b) Eq (7.50 to 0 320 CHAPTER 7 Deflections of Trusses. Slims, and Frames: Work-Energy Methods ,,(,!) FL L.- ,p U .. 8 A PrtlCldure for Analrsls A stated prevIously the procedure ~ r ture by Casugliano's second theorem I suml work method The procedure essentiaD In I t, When the effect of aJUal deformatIons of the membe g1e<:ted In the ana1l"ls Eq 7.62 and 763 red t and 1 L (, M) M,I El d< 8 .- JI ,fz ("'i I' I' OJ) ")L'ltl" and {/ - --= -dx I (/ _L t'.1 0 2EI The- foregoing epre ".on i!> ".milar In fonn to the e"presslon method of lflual ork for tru!>~, Eq. P~3))0 As illustrated by ....ohed eample... at the end of Ihl' ~ction. Ihe procedure for COlD defla:l1on.... b) (.tstigliano'" -.ccond theorem is abo similar to that lrtual ork method Application to Beams Bi substituting Eq, (7.44) for the stram energy (L) of beams In general exp~e'ts.ons of Castigliano's second theorem (Eq. (7 SO obtam the lolJomg expre........ions for the deflections and rotab spcctlel). of beam...: or and
  • 105. AB .. TUII712 1IC1IOII77 ew.......'......,...... •1 Sk/l, AI ~ I kill D c D t8 El - 4Ok_ E 29.oooUi B I 2JOOm' A A40- 30ft t(aj 6S-!! 6.S " 480ck'i• -l 40- B I • 4O_ U '-6.S-1 (l ... 7.23 ==~"= ~ 00129 rod 6487 Ik·n El o 00129 r.ld PL .lEI Solution This frame .... as prciousl} analylcd by the virtual work 7.10. No external wuplc is acting at joint C. ,,"here the rotation IS desired, apply a fictitious couple .'01 ( 0) at C. as shown in Fig. 723 b) The dmates used for dctenninmg the bending moment equations for tbe tbnI ments of the frame are also sho"n in fig. 7.23(b) and the equauoDl Ii temlS of (j and i I i I obtained for the three segments are tabulated 712 The rotation of Joint C of the frame can no be detenmned by I 0 to the equations for f and I f I q and by applYing the ex Casughano' second theorem as gicn b) Eq. 7.65. fl, LIC~)~~d' The Jelk IWO al R l II nll be ~lht,llncJ b~ appl~ing the ex,on"';'. 'lCn h~ Fq. 7 (10 .t foIlO5.C t1ghano's ::ond thlort'1tl J. ,1/ ,r Delenmne the Hllation of joint C of the frame shown in Fig. 7.23(a ghano's -.econd theorcm Solution ""(h C nlc the equal10n l 1111 the llWld,natc IWl1 III Il~ Nnllmg R111m lit In the I1t' 1ll.1 1/ r I "llh rc pcd III r I~ glt"11 b The p:!0111 JI;n.IIl ('I k Energy MethodS J24 CHAPTER 7 Dettectionl of TruUlI, 8.8ms, 8nd Fr8mes: Wor -
  • 106. ., w. For a Imearl C be tbe ron:< and coup a<tmg through the de,romlatio ron:< and coupla I equal the through the deformation due to lhe P or Max II aI MaxweU m 1864 pia aD unponan mdetennina.e stnJctum 10 be 00t1lido:red MaxweU law WID be derived here Bm /a which was praeuted stated as foU 1ICnotI7...... • lIw·......lIw......· ... a.......'WIlD? p.P, I (a) P Syslrm (hi QSYMeID P, I l---+-' ,'---+--..--- 'L---+- 7.8 BETTI'S LAW AND MAXWELL'S LAW OF IIECIPROcAL IIEFI.ECnOU (b) 10+ P, ,4 (-15+Pl+0.43P~) -15+043P .~5 P.- B 1PI I =0) P,t = 84) D .l5- Solution Thl~ Iru..., .1 pn:Iou I) an.l1)zcd b) (he Irlual ork method in Example FIG. 7.24 (a) D ----------r Dt:: n'fl~t.llli I £=ZOOGPa 4m -t - I.ZOO mm· CHAPTER 7 Deflectionl of Trussel, Beams, and Frames: Work-Energ~ Methods L 4m -....- EumpIe 716 TAItl 7.13 ,F ,F For PI - 0 and P2 I F {"'PI t'P, (i"'F/t'PI )FL Memhc'r m IkN) rkN kN) rkN kN) (kN m)._--- 4B 4 15 P, OAW, I OA] 84.48 B( ] 15 04]P 0 04] 0 4/1 566 ~"'2k- 061P. 0 -061 0 BD 4 P 0 1 0 CD 5 15 - (J 71P, 0 071 0 Ee)FL 8448 ~. ~AEC;)fl k m 7Y708 £A £A kN m 0 5m 7Y708 2UI) II) o(K1I2 flf,MH32 m Ans. ~.. 332 mm Ans.
  • 107. I i .lfQ Ifp d< o EI ~ L n'Q - L Qt1') I I Equation 76 represents the mathematical statement of Maxelrs 13" of reciprocal deflections states that elastic structure lh~ dejfeclwn 01 a POInt I due to a un" lotIII pomlJ IS equal to the deflectIOn 01 j dut' 10 a unll load at I. In thi tatement the terms defierI/on and load ""' UIed eral sense to Include rotalion and couple. respectively prevIOusly MaxweJrs law can be considered as a special law. To prove Maxwell's law consider the beam shown ID The beam I separately subjected to the P and Q systems, the umt loads at points i and j. respectively, as shown m and b As the figure indicates. f, represents the deftcc:tiOJl the unit load at J. whereas fj, denotes the deflection at ) d.. load at i. These deflecuons per unit load are referred to a e/fieumlS. By applying Belli's law (Eq. (7.68)), we obtatn 1(/0) - I(fj,) which tS the mathemaueal statement of Maxwell's law The reCIprocal relallonshlp rematns valid between caused by two un,t couples as well as between tbe~;:::~ rotauon caused by a urnt couple and a unit force n f, fj or oung thaI the nght.hand SIde of Eqs. (7.66 and (767 we equal the left·hand Sides to obtam In Ibis chapter we have Ieamed that the work doIIe by :=:~~OPla=oent.1 or rotaUon OJ ofI ofI hoe of acbon IS gIVen by W rPd.1 _ CIA lSi 7 _ _ ... - . ....... Inti FrI_: Wort<-EnlllI' MelIodl ext "e assume (h,H the beam "ilh the Q forces aetia& 25 b IS ubJf1:led III the deflccuons causc.-d by the P 25 a B equaling (he IrtUOI) external work to the VirtuIl "ork "' obtam A + I I J: kd- J.l. IL (b QS~ ...,.8
  • 108. F 4m ... P7.3, PlA7 .;........,._.......~........,......... _~::::=~~::~d theon:m ror IlDearI
  • 109. •ft. 1le4000m.4 E- k ...~A_ _!~_~: _ 15 It fIG. P7.29, P7.57 A ~,,/-1Iiqe-~1~_...:1: 1---16II 8II ---1~8 ... P7.311, Pl.. ....7.. 7.J1_ 7.11 1 5m IOOkN 21 f =con tant 250 GPt I = 600f IW', mm4 10m l()()kN ,m r- / ~~.m (f;;~====I=OO:!:t_=-.....-.-~ C A 8 i---6m-!--6m--I 1--__ L-12 m ----I E./= _ _ E. - 200GPa rr===~8=======:::!C~--- D 30k 30k All 1 ~ tf 1 8ft I 8f1-t-8f1=1 L=24f1 - £1 = constant E. = 10.000U. 2 kilt '~c ! 15f, 6f.--I £1 =constant £ =29.000 ksl 1=3.500m 4 AG. P7.2S, P7.56 fIG. P7.26 FIG P7.24, P7.SS fIG P7.21 1.28 tllrough 7.28 Detemune the smallest moment or lOer. ti,~ 1 reqUired for the beam 'iohon. so thai It maxunum deflcl.:tion doe~ not exceed the hmil of I 360 of lhe pan length (I.e. 6ma~:5 L 360) Use the method of "1r1ual work. 2/ E =conSlal11 ~m 1-:1 = .:on:-.tanl E =70GPa 1 =1M (lot') mm4 HKlkN 2. Ult t.1 =loO"lanl 1 =21.).{)OO I.. I I = 3.fX)() in:~ p ==::::::8,===....:1C L L 2 T I A , JF========-....:~ $KIlon 7.4 7.2O.M 1.21 l~' thl lrlUal orl... mcthod to de lhl IClfll' .I1lJ ddk..:(i~'" at r~'lllt HoI thc beam h RG P7.20, P7.S1 fIG. P7.23, P7.54 AG. P7.2I, P7.S2 RG. P7.22, P7.53 7.22 through 7.25 Usc lhe irtual y.ork method to mine the deflcl.:tion at roll11 C orthe beam shown 12 II H a=6~(10 If f G H a b'l 10 ,.. the rucal deflection at Jomt (j 01 the tru A l============t------P 16 II member Be and CG are 0 5 m. too 8 method f rtual Yo lTk h m 3 m ~ 1 f =con!>tant = 70 GPa I = SOU ( Iff') mm4 I", ~Fi 8ft 'IF==~Di8 ft ~ CMAPTBI 7 Deflections of Trusses, Beams, .Ad Fr.mes: Work-Energy Methods ---- 4 11'11-4~fl tal dd1ectt n .11 JOint II 01 the n F P- 1 due ((.. a temperalure lOlrca l f c beTs BD DF nd Flf l se the mdhClJ fli 'lr fII. P7.16, P7.18 RG. P7.17, P7.I'
  • 110. Hlft '-- A D, ~- Pc I 18 ft --. I I I .-fF.dbd~=zJ.,~,- B c ~ P7.31, P7.l12 3m 200 E / 20kNim c D 8 fIG P7.38 and P7.61 7.38 LC the virtual work method to detennmc lhe rotabOD ofJoint D of the frame shon. ~I 5 m+1 5 m-l--- Sm __--I AG. P7.36, P7.37 E1= COn!itanl £ =200GPiI 1= S()()( lOb) nun" Il sc Ih lfltl,ll HlTk method t d 1.3 I H(II the If,lIne hon III 1,'111 1l Ihe Iflllal "'l~rk method I de nnl 1. 1 lilm at Jomt B,ll the fram h n In FI PJ 1J..t ..,- Sm 1/ 'm c ,m £1 =con,lant £ = 711GPa I =1.03011((ilmm"' -Wk- 2/ 8 AG. P7.33, P7.60 7.34l'-.c the Irtual ork method to detennine the of JOint D of the frame sho" n. 7.35 L;,c the lrtual ork method to detenmne the zont••1 deflection at jomt £ of the frame shown ID P7 3~ II&. P7.34, P7.35 'O~ I 1 kif. '~~~~~~ET c- D 2/ I ilt tIl_ Lift--1---16 ft -----I E/ = (on'laOl E =lO.nuO bi I =l'I.Ib() in." 15 ft ~o fl c 15 (I £1=1.: n lanl F =290111J I.. t I _ ~OCIOm." 8 ft mual rk method to dc:temune the hon- Int C f the framC' sho.... n Work-Energy Methods CKAPTER 7 Deftectionl of TnllIeS. Beams, and Frames' D 336 ~ P7.31, P7.59 ~P7.3Z
  • 111. _,.., ...11I.__11I'--_ .... _ _ -E"""'--- 15kN1m fIG, P7A3, P7.44 ..... 7.7 ,AI ....... ,AI Use Casngliano I ":;~ lermme the honzontal and vertical 0 ftectlon at Joint Bof the trusseIi shown m ,.II Usc Casbghano's second tbeoJaa honzontal deflecuon at JOint E of thI P78 ,.It .... ,.u Use Casbgll8DO IOCOIIll mme the slope and deflection at point B in Figs P7 20 and P7 21. ,.13 ....... , .. Use Casbg!iano' detennme the deflcctlon at POlDt C of Figs P7 22 P7 25. ,.I, ... , .. Use Casbgl.ano mme the slope and deftcction at poiDt ID Figs P7 29 and P7 30 , .. U C'SbgllanD S secood tbo_lIl~ mtlcal defIect,OO at JOIDt C or tbo P71 , . U C bgllano second ~=~ borizoolal dcftection at Jom! C ft P7 3 ,.It U Castigliano second tboc~a ~ talioo of)DIm D of tbe frame 1 C u D U 3m +65_ kN 20ft 3m 1: ·A -310+- 510 20ft c c D D --+--IOft A 1----,1011----1 EI - ' .ooobi • " ..."A1
  • 112. o~'" a L L Influence Une for Bending Moment at B When ,he unllioad IS located to the left of point B (FIg 8 pressJon for the bending moment at B can be convemend uSlOg the free body of the ponlon BC of the beam to the ConSldenng the counterclockwise moments of the external reactions actmg on the portion BC as positive m accordaDcl 'altl Ign com: nllon (Section 5.1). we determine the beDd , . at Bas The mfluencc Ime for 5. fig 2 e shows that the zero "hen the umlload IS located at the left suppon A oftt. the umlload mo,e from A to B the shear at B decreases It becomes L "hen the umt load reaches just to the left As the unll load eros pomt B the shear at B mcreuea I L It th n decrca hnearly as the umt load mo until It becomes zero hen the umt load reaches the ngllt ob med b mph sUMIIlutmg Eq . Ms C (L - aJ 0 S x S a When the umt load is located to the right of point B we body of the ponlon AB to lhe left of B to detennine M. the clockWIse moments of the external forces and reaction portion AS as posItive we detennine the bending moment Ms A (al as < S L Thus the equa',ons of the IOftuenee hne for M. can be M s {C (L a 0 S x S a A a aSX L Equat,on 8 5 IOdtca... ,hat 'he segment of the iD1I_ betwoen potnts A and BOx a can be obtained by onIi..tes of the gment of the mftuenee hoe for C by L a Also according '0thIS equation the segmeal Itoe for M. between pomts Band C a x S L can mulbplYlO8 the oro,..... of the segment of the iDII_ :;::"Sand Cby a The iDlluenee hne for M. hili ......, lines for A and C IS shown m Fig. 8 2 of this iDII......, line 'n tenns of the JIOSIbOn of the IIDiI obtaiDed by ubstitubng Eq 8 I and 8 2 mto Eq ... _..._-
  • 113. SECltON 8.1 lnftuenct Unn tor Beams and Fmnes by EqulIlbrtum MetbOd r L· 'f ( Solullon r( l ( JIUIlCLI l~, " ( ... L3 [_-7.---=-==-- " £IIIIlPII '.1b. f' _ unol II the desired IOllucot;: d I 1m td .' nl .. e ~nt I l.: ~n tOld th~ ulllucncc hn r. r momc:: t b) u m th mllu n hn II r upper. bel pr mg'Althlh.. coru tnl..lllnof ntnn r bending m mcnt Jt a J"llOf l n lhe tflJl.:tun.: mflucn.. hnt; fll( IIlhe reJl.{1 ns lm III (the Idl I h rumt undl;rt,;on Idcr.tuon -trt.: Jdllahle OthCN d '0 the qu ero nnuence lines IN n;,h:1l0ns ,,) 109 the pr du de nix'! In the prclOUS h:p 0 tnllul'n~e 1101: f(l" Ih she, r bcndlO£ IOl1meni .11 a ["llOt ~ln lh..: ..trudure 1;.10 hl: con tnll:tcd s follll.... a. PI.1cc the UOIt load l)O the ...trudur..: ,II .1 arl.lhk IX"'llon :t f the t /r flf Ih.. POIllI under nlO...IIJerathl n.•lUd dell:nnin~ the ( pr.. II'n hlf the "he.lf 'M bcodlllg IlHlO1..:nl If th..: mfiue[l~c linl"'S l,)r 11 th' fe.I,:lIoth arc knmn. then It i~ u...u.t1I) (ome· 01.. 0111' u the pllrtillil of Ih,: ~truelure tll the rillhl nr Ihe l"'Jlnl hlr JctcrnulIlntt the cprc......ioo for ~hcar (or 'lending mome01 hl....h "1111,;llOI.II0 terms im 01 iog: nnl) r..:.l:tilln~, The ..hear C' lxnJmg moment IS ,,:olhukred to Ix po..iti,: llC ne~ahe III ¥i. l.urd.tOC llh Ihe hal'" S III j om. tlt,o" t:...tuhh hlo:ll In Sl.'1.:tll .:;;1 Ig~2 b. .. t pi cc the untt 10<:ld to thl,; rt /11 Ill' the pOIOI uOOa I. n uh:ralll 0 and ddennlOt: th.... t: pre ...,on fllC the !>hcar or bendtn~ m ment If the IOllucn..:c hol.~ for .111 the rc.KllOns are k "" then II IS usuall~ com ..ntenl to u thl' ponion the (TU ture to th I of the f')int for dO::h.:nTImtng tb d", m:d e P on h h 111 conl,nn teml lmolvm ~ (lolv rca..:UQns c. (f e pre n for Ihe shear or bcndlOg moml:nl l:on rm n; htng ooh reactIons then It IS ener.1l1~ Impkt l.: n truC1. the mfluence hne f('lf shear or ~ndm moment b'l comb ng the gm Db of the C.:.II.1.IOO IDflurnc lin 10 tlC r et h these e pion Oth"f ubslltul 1 cxp 10 r. f the aC' on mto the c presslOns f r the heal be dl m m nt nd plut the (I; uluog e p 100 hKb be 10 terms ooh 01 (0 ohw.m (hI: mRuenl hnr.: d. R pc t tel' until all the deslced mtlucnl.... Itn for shc.a bend n m cot.. h.. heen detennmcJ CIW'1tR 8 Inftuence Unes
  • 114. CIW'IEIII ""'...... Una , { ( II ~ , 12 ft I~ ft :!t1 rt __ L' ...... .............., , . rh~ nRuen hnc CIr • hcm n III I Ie J' I II L r U br I": th..: umt load at a po Ilion f B od dd mlln' lh~ bent.hng mt ment ,II H b:- u mg the lree bod II n II th beam to tht': nght 01 8 II ( o :::. I:! ft ~xl the uml I ad I It Il,.'d 10 the nghll'f B. and 1: use the free body pom n of th beam to the left of B to dclcmllne fB: " I:!~ 12ft-sxs:!Oft Thu the equ.lllon of the nflucm:c hnl' Illf fa arc L: ~ 1I~oc:;;I2f1 II, h I~ , 12fts:s;20ft The ml'luenl"C hne lor /" .....ho""" in !-ig. IS 3(1' Ora the mttuen.:c lines f(n the crlical reaction and the reaction ",upport A and the ..heM and bending' moment at point B of the cant shon in rif!. X4'aJ Solulion Infillllll l' lor A, FIG. 8.4 The IDftllellOO tine ht}Iunt Line M. M 12:.,1' 0 A -1_0 A The mllue-oLe hne for'" 15 bo" n In fig S,4(c Injfulnlt'Lm /or 1 f >L·II,O /~ If-O Examp118.3 The inft line 1/, I , x The mfluence hne for I "hKh I obtained b) plouing thl equa m FIg: 4 d A 11 (he ordmate of lhe Innuence line are nepb (hat the n of 1 ~ for all th ~Ihon of the unn load on the be... II j unterd kin, ad of dockwi a IOnia11) a umcd den mg th equatl n ( f th mOuen!,; hne
  • 115. Ans -EG m m IECTIaIII.2 -air I.............. " • . -............. 8.2 MULLER-BRESLAU'S PRINCIPLE AND QUAUTATIVE INFWENCE UNES o F , 10 - 05 -0.5 ,)10 o D (0 Influence Line for SE(kNIkN) (ellnfluence Line for A. and B~ (kNJkN) {dJ Influence Line for B (kN/kN 05 OHJ3 c,-----:""""o:"""'=--+__-!:-....:..!D c 05 Lf () A H I) B • 3 ., I III ", :(I 5 I~)• J • 10 < 6 " III m The inftut=nl.:e line lor B I hm.. n in hg. l:(7(dl d Inffutn ( I III Iflr ~ e '" III U'C the cqualion of condition L M hctennme the e:..prc Ion 1M ~ hr~t c platt the UOIt load I the mge 1. that I on Ih fl!,!ld part ('1 01 the frame to obtam "L 1' I /I Influent I' I II/(' for B, AG.8.7 ,. Hon D E F G (el Influence Line for A (lNlk~J ,+-----,,'/4 -- 8 m S m ----t- 5 m --+-- "m IkN I D I E F G c _I, , 81_8A I IA B b> 354 CHAPltll. , _ UIlOI
  • 116. -IEt11oIII.Z m _&-. 25 o c Released structure for MD Influence line for Mn(k·ftIk) o (I) A I k A 1_ _~_-L'_-r;.E_ c D fMD=2.5k-ft £y=OSk F i~o F C £'"""'""""J -05 1.0 Solution It,tiUI ntl Llllt jor..f To detcrmine the general shape of the Inft for A . "e remoc lhc n:slraml (:orre!iponding to A, by replacmg lhe ftIIld pon at A b) a rolll;r guide that preents the horizontal displacement tatlOn at A but not the crtil:al displacement. Next pomt A of lhe structure IS gien a mall displaccmenl 6. and a deflected shape of lbe dran as hon m fig 810 b Note that the deflected shape IS co the upporl and contlnully I;onditlon of the released structure. The end beam hlch I altal;hed to the roller gUide cannot rotate so Ihe poI1lil1lU" must remam honLonlal in the displaced configurdtion. Also pOInt E i........ to the rolkr upport; therefore It cannot displace in the 'erucal diRlCllOlL th portion (f rotates about E as hon m the figure. The 10 rigid AC and (F of the beam remam straight m the displaced configurabOD late relatne to calh other at the mternal hmge at C which permlt1.~:.~~; lion The hape 01 the Inftuence hne IS the same as the deftected shape leased truclure a ho n m ~ Ig 8 lOb B) recogOlzlng that A 1 k. .... hen a I-k load IS placed al A we ..alue 01 I k k for the mftuence-hne ordmate at A The ordlna and f are then detenmned from the geometT) of the Inftuence bne TIle l;I~li Ime (or A thu obtained IS hon m fig 8 10 b Influence line for SB(kIk) I k when I k is at just to lhe right of B (el AG.8.10 (eonld.) t£ =0 o when 1k is at just to [he len of B I 0 0"----,1A B 5,= "A I ICli'---_----::i4:i:---F r----'B C CHAPTER 8 Influence U....
  • 117. - In the prcYIOUS section w con den:d t were ubjected to a moving wnt I In mo.1 bndges and bwldmgs there are nol ubjected to lave loads dlRCll bu t tnmsnulled vIa ftoor franung tern T bndges and bwldings were de!aibcd re'p«:tivc' Another cumplc f the fnlnon. hown tn FIg 8 12 The deck f the brid " which are suppofled by by the ,. r Thus an li gardlcss f where they Ioca """""nlnlled or distnbuled I as ootICCtIlnlled I applied I tIoor beams T and A .........I 8.3 INFLUENCE LINES FOR GIRDERS WITH FlOOR SYSTEMS 4m f o fllnge J) J) " ' 10m ( Rd~·J t"d Beam tor A Inl1u nee lme for ( IIt ILN) (" -04 Rdl'a ed Beam fur C Innuell(t" Lillt" lor A, (kNIkNl (h) 4m . 4 II. hn ~ I III " B o 4m .--~L--------:i:J)i:---<5;'-- F '-1-'~r; JlOO" I 0 ~-:'-o_-::-- -;;------;:--~O, B C D £ F A CHAPTER' InOuene. lin.. Ex..... 8 8 A&. 1.11
  • 118. o lot .F IfI:TllIII u IIIIIaonco Uooo.. _ _ ... __ In d .....13 (e) lnnuencc Line for F x~',!_x ~ Br=r(2US) - • F F 8 =-- C riD E :.. Q Aoorbeama .J / Deck ; StrmgerGmkr G der r b Slnnge L b F rbt m<; floor ~am - " / --"::;-Ir::L ~ W- I I / I I -1--- .J r----- - ----,---- I I ---- I -....... Girder -~ "-- " Q L CHAPTER 8 Inftuence Linn
  • 119. an L oa, L L F L-aM IIECl1OIlI.3 1nII_ Uooo lor II-. When the UOlt load IS located to the right of the pan I POlOt C bendmg moment at G 15 gIVen by Influence Un. far Bending Moment at G The mfluence bne for the bendmg m ment t pol t G in the panel BC FIg, 8 IJ a can he nstructcd by procedure When .he unit load I located t .he len f t the bendmg moment at G can be expressed Thu .h quail .h 10O F I F 4 L A L LL L O$x~ 5I F,SHe I L I I l 0 A• 1/ 0 L,~ F ,1 F L 0 . L f ~ 0 L Influence Lines lor Reactions 1 '1",,- for Ihe ,.crtlcal rcaclton AI Ill' 1Il1 UI,:IlLl "- The C"-IUatlon ('I 1,. •.• h rrl)lI1g tht' eqUilibrium equations f lao hi:: ddcmum."U V.I l' • '111' In,tt:.ld. the "UinpCf!l. and thed I.' S - • , 'llh..:r , Jt'pldl' III J Ih II th..:.r lOP l...dgc, .Ire een With II 1 IIIIllll' '1 • I h N lin.. ,Ill U II I Pi I " It tht.: ....lIne 1cC ;,IS t e top ""II'''~. I 1'1:1"11.1111 ' . olhd .llld In: ('It Kf I , the: glrd..:rs see I l~ 8 1_ Influence Line for Shear in Panel Be 'C!. "uppose that "c "i...h to construct the influence lines for shean pomls (J and JI which.ire located III the panel BC as shown m R. n(a). When the Unit load j, located 10 the left of the ~nel po the ,hear at an) point wllhin the panel BC (c.g.. the pomts G and can be exprc~scd as 1 bt 'ned b rlQuing these equations are shThe mfluence lIle" (I <II # - " d . I , d (,,' ole the"" IIlfluence lme~ are I entlca to Flg:.. I.t h .10 d beam to '" hleh the urut for the n:::u.:1I1111 01.1 "'ll1lpl~ ..,upporte applil'd din..'dl) CHAPTER 8 Influence Linesm L 2L 5 M When the umt load I loca.ed wIthin the pane 8 13 d the mOmeDt of the fan:< F. exerted n beam at B about G must be IIICluded 10 the "I_we mOmeDt at G 4 1+- L ") - I. - I. A Similar!). 1,1, hen the unit load IS located to the right of the panel the ..hcar at any poilll ithin the panel BC is gicn by When the Unit load IS located 1,1, nhin the panel Be. as shown 8 13 d th fOfl.:e f B exertcd on the girder by the floor beam at B he mcludcd In the e'pre Ion for shcaf In panel Be
  • 120. M ............ L 0 L 2L «I 5 0 4 0 (I DO When the un,t load I located to the letl of C rll- moment at C I gIVen by 3 Sx When the unit load IS located to the right of C MAcn (I ~)2~ ~(L nUl the equanon of the Influence hne for M. M l~C~) ~x A c~) ~ L x The iD8l1C11CC line obtained by plollUlg these 1 that this iD8ucncc hne IS ideQIk:II IIIDIIlCIlt c:orl<ljlOillljDg beam WIthout ........ Iar. . .
  • 121. Plan (deck not shown) Ix E60 0 A60160xOEM 0 connected at their ends to the JOints on the bottom chorda longlludmallrusses Thus any lie loads e.g. the wetght 01 regardless of »here they are located on the deek and wbe!lla concentrated or dlstnbuted loads are always tranSlD.ltted to concentrated loads apphed at the Jomts. Live loads are tl1lll_111 rooftrus In a Imllar manner. As in the case ofthe girder the tnogen of the floor systems of trusses are assumed supported at the" ends on Ihe adjacent ftoor beam . ThUi hoes for trusses also contain straight·line segments between To Illustrate th construction of Influence lines for uu..... the Prall bridge truss shown in Fig. 8.18(a). A uml (I.k from left 10 right on the stringers of a ftoor system attached tom chord AG of the truss. The effect of the unit load .. the truss at Joints (or panel points) A through G where tU are connected to the truss Suppose that we wish to dill lines for the venical reaclIons at supports A and E and force In members CI CD DI IJ and FL of the truss IntIU111C8 UIIII for R••ctlons The equallon of the Inftuenee hnes for the verti<:aI ....._ can be delermined by applYlJlg the equilibrium equa _ .. i....1.lul ..... .....17
  • 122. +<LMo 0 FIJ 20 E 15 0 FII 075£ <LMo 0 A 45 F, 20 0 Influence Unl for Force In Dllgonl' Mlmber DI 1br e pressJons for Fn1 can be determined by ~ Fill 18 h and by apply,"g the equlhbnum equaliOD of the two ponlOn of the truss When the wut load II ofJOint C pphcalJon of the equdlbnum equauon I:F. porn n DG of the cruss yields 4 LF 0 SFm E 0 FD/ 12S£ When the I-k load IS located to lhe right ofJOIOt D ..... 4 LF. 0 A SFD/-O FD/ I.2SA The sqments of the mOuence hne for FOI bctwcea A tween D and G thus ronstructed from the mllueace IiDoI n:speclI...t are shown 10 FIg 8.1g I . The ordiDa... then c:onnccted by a straIght hne to romple.., the in1I_ as shown 10 the figure Innuenee Une for Force In Top Chord MIIlIbIr IJ By c:onlldenng section bb FIg 8.18(h)) and by pllCllll tint to the left and then to the nght ofjomt D we oblItia exPreallons for FIJ Fu 225..1 The iD1I_ line for F thus obtained I - .., -
  • 123. .. o o£ 16 J SFH 8 FHL. 1661£ 12 F IIlCT10IIu -... ~Ior_ II 0 F £ 0 F 66£ ted to II1e ngln D LM F 12 A 4 0 F -2A 24m m A 16 F. 12 When the l-kN load I to the ngbt of D we blaIR " 0.208 C ......... r ~7 D C ol.:U -0417 (e) S~clion bb (g) Seclion aa (f) Influence Line for FUti (kNIkN) Ih, Influence Lme for FHL (kN1kN o FHL = -I 667 £) - I 2S Fl'tI • -0 833 E o A A, A 7----"!----"~ B C .II • a L CttAPTER. Influence Lin.. 1-----4 paMl I m ~ 3~ m - - - - - j (b) A 1 kN I-- ,----.j c Influence Lme for A IkNIkN, 1(I ~oA D f. 10 ~~O!5_JA C E d Influence Un< I E kNlkN) R&.1I.2O 310
  • 124. - B AI'F"=~======"D fIG. PI2, PUI ...1'&3 ;m PROBLEMS FIG. Pl.l, Pl.59 SoclI_ 8.1 _lid 8.2 '.1 llnugh ... Dralo the influence line for vertlCal reac- tIOn.. at support!l A and C and the shear and bending mo- mem at POint B of the beams shown 10 FI P8 I throusb Pl<4 SUMMARY B (b) Real Beam (d) Conjugate Beam (e) Influence LIRe for Vertical DeflectIOn at B (M) M ( k-ftlk)(c) Ei Diagram D B 1«1------ £1 ... constant J:, .. 2~.OOOhi 1 500 in.4 (3) A [ o )5 £J 15 £1 Ik X IAj--~-----iB x~ CHAPTER 8 Inftuence Unes fIG. 8.22
  • 125. CHAPTER 8 Influence Unn FIG 1'8.4, 1'8.61 - au B f U at mom nl at E/)CH '.15 Dr,l'· the ..,t1u"n:c hne for the f nd bendm TTlll Dll"l1t .11 pomt /) 01 the ho.... n In hi P8 I ,.16 Dr.I Ih, influcnl,,'e hnc fur the n I r lIrfltlrt I .lOd F .md the: hear and hendm roml 1> of the Irame hO"11 10 118 P8 16 ,1. Llr.I' thc mllu..:n,.;c hnc for th he f ilnd be dm . ,,·ut ,11 roml Hoi the hcam hol." In h P8 1f11111 Oft R I~ It FIG PS.9, PS.58 I.' 1)1,1 1h... 1I111ul·IKe [1I1C Ill!" thl" Crll:.a1 reaction l 1l,:(lvll l1l)lllt.:1l1 ,II UppMI I ,nul Ihe ')hcar and lllonleni .11 ,~'1Il1 Hill tll.' :.Ollk,er hcam ..hown ID r " hnl.: r l,:IUlal n:.J~lJon') nd bcndm morncnl 011 rs "and I' b n 1,,:=====R~=====;i;===-P 1.10 L)r.lll Ihl' mftucnct' hnc.. for Ihe hear and momt.:nl at pomt ( and the ..he.IL jU')1 to the left and Ihc n1!ht Ill' UpJ"'.lrt D of the beam ho" n in hg P8 I FIG. P6.5 R 4m n H c D E 'm AI. P8.1. R c n f 'm 6m 401 -:m-4 .A Hi. 6 IIiDae B 6m 6111 IS ft-+--15 ft FII. ..... PUl. P&22 fIG. 1'8.19 ... Draw the mtlue upports Ii 0 nd F 3mSm 12 ft Sm 11 ft ·A FIG P8.17 FIG. PS.16 12 ft • 3 m '.17 Draw the mfluence line!> for the "ertleal reactions al 'lupports A and B and the shear and bendIng moment al point D of the frame sho"n in Fig. P8 17 B A--.~.----~ "C===";D;,,,==_-iEi- F- - r 1.11 Ora" the mfluence hnes for the [C"dlun moment al upport A and n},lment al point C of the frame 1M ft D 4m R Hinge ~m 4m B C FIG. PS.I3, 1'8.14, 1'8.15 A:c: LP ~4ft 6ft 20ft AG. PS.l0, PS.ll '.13 Drd" the mfluem:e line.. for the vertical ,..rillll!l upport -I and E and the reaction moment at u the beam htl.... n lD fig, P8.I3. FIG. 1'8.12 '.11 Dr,I the mfluem:c line.. for the !l.hear and mom,nt at POIllI £ of the beam ..hown in Fig. P8 10 '.12 Draw lhe mflucnce line.. for the ..hear and bcndIDr' moment at pOInt B and the ..hear.. ju..t to the left and Ihe right of .,uppart C of the beam sho"n in Fig PBI 'm 5m Xmm m B A.Li" . :b..( 1--- ~111------t-I 0 ft ~ 'm FIG. P8.6 1.7 Ora" lhe mtlut.:n:· hnc f()T thc crtll.:al rc:.u.:tion at uPI' rt -I and ( thc hcar at JUI to the right (If ., and the bendm ' rnomc:nt at POint B of the ho" n 10 hg: P FIG. 1'8.7 FIG. PI.8 U Dr Ih I ftuenlX lin f r the hc:ar and bendmg 010- otlle er beam hoo In hg pg
  • 126. • G Hop F 8 10 (t 4m-f-6m 6m 10 (t --4m j1;:=XC:::::1:f =i=l D E ·A II&. P8.3lI ....... ... Ora 10ft upports A B ( poInl E (the I10 ft 81 I , B 5m--1-- 5m 6m E /) Hmge A D 5m---I--5m T8~==;iC~=...;;D..........;E;'-'_"!iF::.. Hinge IOJ FIG. P8.37 L 1.31 Dra- the mfluencc hnes Co r the upport.-4 the ..ertical reactions t upports A '.37 Draw the influence hne~ for the reaction momenl at support A. the vertical reactions at support A and F and the ,hear and bending moment at point E of the frame ~hown In Fig. P8.17. . FIG. PB.36 F16 PB.35 C F A. .... IOft--l-IOft 20 It 1.30 Dr;! th... IOtluence hne, for the ,hea~ and mC'mt.'ntc; .tt JX'JnIS ( and F of the t'leam s.hown p, -0) 10 (t A 8 1.31 DI m the mfluem:e line· for the reacllon mo.... .Illd Ihe u'rtKal r...action' at '!upporh -I and F of the h,,n III } l~ PS JI AG. PB.31, PB.32 AG PB.29, PB.30 8.32 Ora" the IOfluence hne... for the shears and moment'! at poinh Band E of the beam shown PR.)I fIG. P8.33, P8.M 10M Dray, the mfluence hne for the shean aud moment 011 POints ( and E of the beam sit PO) 8.33 Dravo' the influence line'! for the vertical reacti ,upporh A B. G. and /I of the beam shon m FtB PI ........ Draw the mfluence hnes for and -erllcal reat.1ltlns Oil uppons It and B ~o",n In hp Pll 3S and P8 36 .,(j Hinge , .J E F JIL h G --+1-II---i15ftl5fl InftUt."T1 hnt.' or lh~ ertKal re l:. nJ (, • f lhe: t>.:.lln~ ho n In 'Oft --IOift lHI- Hinge A 11====,==,;;C==iD~="".~E==*Fr==:l;G 3m JJO 4ml.mjJ _ CHAPTER 8 Influence Unea I.ZI Ura" the ml1uenl,;t: lines tor the ~hear and bending mom t t pomt !> (It the ~am ..ho"n in FI~ P8.1' La Ora" the ml1uence hne.. for the ..hear and bendmg moment at pOint! of the beam ..hown in Fig. P&,:!4 LZ7 Draw the mtluenle lines for the reaction moment at upport of and the crtll,;al rCdl:tlons at l>Upport~ A D. and F 1t Ih bcdm htl"n m hg, Pl17 II&. PI.270 P8.28
  • 127. D-- .. L 4m-+4m-+-4'D-j4m E F" G 8 J K ;------ 6 pane 114m 24 m -I AG. P8.54 AG. P8.55 "'-1'151 T~A I 3 panel at 4m 1 m -----I 10 fl 10 ft G ~A " R C D - - - - 3 panel!lo at 4 m - 12 m _ f 10 fl 10 ft T ~G PS.49 - - 6 panel!. at 30 ft = t80 ft ------I ~. PS.50 J ~~ 4fr H " " " L 8ft ~~12ft ~G. PS.51 1----- 6 panel, at 16ft =96ft----_ L M N 0 P ~A~8 C 0 E F G "'" PS.52 1.53 ..... 1.57 Dm- the m8uence Ii Ii r the In the member... Identified by an of the 01. figs. P '" PR, <;7 LIe loads a transDll ted I chord, of the IruSoses o 5m ,m ;.1'p==F ,m ~panel,al~m=9m J) f - ~ panel, at 20 f1 = 80 (I ---t F G" H ~±8 C D ! - - - - ~ panels al 16 fI = 64 ft ----"' AG. PS.45 "'- PS.48 AG. PS.47 AG. PS.46 I 2. F t i FD ll'ift-90ft---- - l,: 'I' Ihe he,ll In panel £F 8nJ rh :trde' Ith the lloor ,. ll,:m I r the h~ar In pand B( nd f he rdt:r "lth tlle 11.)0)1 ) tern Ihe I ftuencc Ian Ii r Ihe for In n f the tru ses hon m I Om! uaJ t the boll m ( n i 1 Hmgt 4pane1 811511 =60h (, ~,;::~::~ n I I hIm r n I ( n ,Jnd n 1 ( Ilh gtrlkr Hh th~ fl{'l, I (elll 8 i A i CHAPTER 8 Innuence Lines ,---~--- '; panel III h m - '0 m ----_--1 8 R6 '8.41. '8.42 .... Ora" Ih~' mflucnU' Ime Itlr lhe hear in panel cn and the bendmg m ment 1/) 01 th~ gIrder "llh the floor ,,) ,1em llmh'P844 "'- P8A3 "'- P8A4 _1.4 IAI"",,- a.u 1>, cd l , .....
  • 128. 402 CHAPTER 8 Influence lines el 116ft -ISU A6 P8.57 , Section 8.5 8.58 Dr,1 111l' lnl1u~nl: hn..: I(lr thl.' (TIKal deft pOint BIll Ihl.· 1..1ll1Ik"..:r tx',lm ()J Pr('hll:m ~llnst.tIll ll. 'Ig px 9. 8.59 .Itd '.60 Dr,1 thl.' lllllul:lH:c hnl.' lor Ihe fk twn t r0ll11 n)1 th~ lInpl~ 'UPporll:d beam I ms and 8 2 U Clllht.1ll1. ~c h P8 I and 8.&1 PI'''' thl.' mtlul'nl.'l: hnl' for the l:rtKdl d pc nt J) ,If 'hl: !'learn :'ff P(ohkm SA I:J 1,;0 h' P .. 8 Application of Influence lines9.1 RespOnse at a Pam Load eular LocaIJon Due to a SIOgie Mov ng Concentrated 9.2 ~=nse at a Pameular Local on Due to a Unltormly Dstnbuted Lye 9.3 ~~~.:~aL=eular Locabon Due to a Sones 01 MovlOg 9.4 Absolute Maximum RespOl1se Summary PrOblems ~ Hlq/nml Bridqe SlIhjc'cll'l/to lIariny Loads s~ofthe nos oepartmentof Tr tl· In the preceding chapter y,e learned ho. to I,; n trul,;t mfluence lin for various response functions of structure~ In thh chapter y.c con Idt'T the application of influence lines In detcnninln~ the malmUm ",lu of response functions at particular local1on~ In Ir1l;ture due to i n- able loads. We also discuss the proctdurcs for caluatm the a lut maximum ialue of a response function that may occur an here m a structure. 9.1 RESPONSE AT A PARTICULAR LOCATION DUE TO A SINGLE MOVING CONCENTRATED LOAD As discussed ID the precedlDg chapler ea h rdin a i,n,8",,,.:e gives the value of the response funcllOD due ( a n c load of urnl magnitude placed on the uuelure al the I ardmate Thus. we can state the f; II wing I. The value of a ...ponse funcl,on d I -1l'" load can be obtained by mulnplymg the mOl the ordinate ofthe response funcl n 1D8uel1C the load ..
  • 129. - H n 'c Hmg OkS ~,g;--j~--:i£i::-- -<!4';;D_--';5--~ b Inti 4m 4 14 IIooIIonIo "'".......... LocatIon D"'10 I ~ DIIlIIt_LM!oId H S£CTlDN U FIG. 9.2 p I D , C D c 1/ P r t 'lJmum B B n of Lo.1d P fl r MaXimum a eMs CHAPTER 9 Application of Influence Lines 1' lh:llllllllll: thl..' 11M l1llUll1 pO...itlH' aluc (11,1 rcspon fi 2. JUl.' h' ,I 'S1Il~1t. 111 Ill!,! I,.l,.lnl.:I,.'1l1r.lh:d 10,IIJ the load must be .11 thl' II,. .llhlll of the 1l1.1llllUI11 pI'...lllC ordmatc of the r:::==-funClIlll1 11111UCIh.:1' hnl hCIC.1 tll dl·tl'nnl11l~the max.lmum n dlm: of thl rC"1 '11 l lunditln the: IlMd n~mll be placed at the tlt'll (lIth..' 11l.l ,"Will !ll'g.ltlC ordllldh..' 01 the lIlflucnce hne (('l1Slder fl'r l' ampk .1 1x:.1111 uby.:ctcd 10 a moing concea h.lad of lll.lgmtudl P .1 ShlH n 111 J If! 9.Ia1. Suppose: that we Jdammc the bt:ndll1£ mOlllent .It B "hen the load P IS located dl,Un(1; : Inlm the left uPfX)fI I The IIlfluencc hne for t. &1 Fig i). I .1 h.1 .m llnhn.Hc at t~c PO"I~lon of the load P ;" :::~ that a umt 10;10 rl.hXd .11 thc po Ilion 01 Pcau~'S a bendmg f ,BI.:C.lU'l the prnh:lplc of ~urerp0'oilion I~ alid the I mac.nitudl: P mu~t I.:au~ a ocndmg moment at B. "hich IS P largl: a~ that l.:.lU cd h~ thl: IO<ld of lImt magnitude. Thus the belllll.' nll)ml:nt at 8 dUI: tl.:) the lo.ld P 10;; .18 Py I: t UPP{hC that our ohJl:di c:: i... to detcrmine the maximum. ni1: and the m.limum negatle bending moments at B due 10.tO P From the mflucl1l.:C:: lll1c for .18 (Fig. 9.lla)) "e observe maimum po ilne and the maximum negatie influence-line ard OCl.:ur at rx)inl'o B .1I1d D. re'opt:ctie1} Therefore. to obtam the mum POSltlC bending moment at B, "c place the load Pat POlot B shown in Fig. 9.1(0) and compute the magnitude of the maximum itic bcnoll1g moment <to;; .18 PYH. whcre YH is the inftuence- ordinate at B (hg. l).I(a}). Similarly. to obtain the maximum ne bending moment at B. we place the load P at point D, as shown m 9.I(c). and compute the magnitude of the maximum negative bendidI. moment as .1s Pro. A FIG. 9.1 9.2 RESPONSE AT A PARTICUlAR LOCATION DUE TO AUNIFORMLY DISTIIIIUTED LIVE LOAD 70 k70 k0 ) 4 Solution Influt'nCl! I InC 1he InflUcnLC hne for the ertical reaction at u thiS beam 'las prevlOU Iy contruf.:ted in Example Its and IS showD 9:! b Rt:all that ( .;1 a umed to Ix fklSllIC in the upward di :On trUl:tton of lhl mflucm:c hnc fa lnlwn (pit rJ R tJl tllm at ( To Oblain the maximum poll ( due (0 the 50·" conl.:entratcd Ilc I()ad l,e place the load at B Yther the rnalumum po IIle ordmate 1.4 k kS of the mflucocc B) muJllpl IR~ the mdgmtude 01 the load h) the alue of thl t nmne the maximum up'l<lfd rcaL110n al ( as For the beam ~ho.n In hg. 9.2(a determine the maltlmUm upward ...1di"j!J;i s.uppon C due 10 a 5()-l. conf.:cntrated Ile load. Eumple9.1
  • 130. H H oc 15kNm Hinge c Hn bl lnflue B B 14 4m 4m SECflON '.2 .....DItIi It • hi $ "lie I.ICItIen Ow" ............, 011*7 " ...... LIIJ ., ~,.mp.::le~9~.2=-- ;:;:;:;::-;;:~~~~;==========j FIG. 9.4 0.125" .lDL ·18 I area under the influence line beteen C and D 1I(~)(J:!5L In Ba~ on the foregomg discus~ion. e can state the fall I. The alue of are ponse function due to a umfonnly di tn apphed oer a ponlon of 'he structure can be obtalDed pl}ID@. the load inten It) h} the nel area under the com. porh n of Ihe re ponse fundlCn mftuence hne 2. T detemllne Ihe maximum po!'OlIle or negall re pon funl,;hon due 10 a uniforml) dlstnbuted Ii I ad mu , be placed 0 er Ihose ponlOnl'i of the tructure rdmate (f (he re ponse function mftuence hnc ne lIe .18 II (area under the influence line beteen A and C) ",G)(O.7,t)( 181 037,,, 1'8 L III hllh th...· IIlh:gral J 1.1 repn;-.('nt the ,Ire.' under the 5<I:molltli thl.: mlluc,:oll'lme hl("h o:c,lrn;'r~'nJ to the loaded portion of the Thl MC' I "hoo J .1 haded MC'.1 ~1Jl the influent.'C line for M 91. fqu.ltl(ln l).2 .11 (1 mdK.ItC" th.It the bending moment at B Ix maAlmum po Itil,~ If the uniforml) dl·.tributcd load is placed th(1'1,;' p',rtwn" c,lf the ~,tnl here the mfluencc-Iinc ordinates tilt.' and Kc.: CT.•1. hom fig. 9.J(.1 e Gill sec that the ordiiD8I11;~' the mtluenlC' line for lB are po,it1c bcteen the points A and negatle bc.:1I..'en ( ,lIld n. fhereforc. to obtain the maximum po~tr(. bending moment at B. e pl.ICl' the unirormJ~ distributed load A to C a... ,ho11 in hg. 9.J(b). and compute the magnitude mi.limum bt:l1I.hng mom..:nt <1' Is I1 c/ II JId Similarl) to obtain the:: maximum negatie bending moment at place the load from C to D. it......ho n in hg. 9.3(c). and com magnitude (~f the maximum negative bendmg moment as ( ell ""A 0 1---<-021 CHAPTER 9 Application of Influence lines hlIT I I the mllu":l1li:' 11l1l.: MJIIl.Itl· ,It hit:h i.. the POlOt ... u n I r.h SI1l1 n III Ihl' llgure 10. ckr...-mllne the tOlal mt."lm..:nt at R Jue t,l thl' JI Inhut...·J 1~1.IJ ll,~lm II to h glJt~ • i.I. 9 1 tx't1,;l n 1111.: l 11I11It III ~1htalll B ~'n _ ("J A An1111111111 ffik .tJ. B A lJ 1 - - - 5/ 1 0'<1-< CD ''iV,.ngemcnt fUn formly o tnOOted LI e Load I for M urn Pili M
  • 131. ClW'TER' AppI_ of Inti..... Un" h • l:e 11Il('.lre £X' llnc. hcrcas the 20-k m the (rdmah: (,1 t ,",,11&:1" b .• d d I I pI l' d (" ..r (he ..nlln: kngth ol the beam TheJI,tn utL-u ..3 II Il' • JX»111C bcmhn~ ml1m of I ( I' ~IH'n h~ ta lmum po IIlH" II 9(1 2 ~oC) 9 2 "' :0:G) J -2) w} kN· m l mum t atl BlnJlIJ!/ fun! II' or ( The loading ~lIlII'" obt".l.In the maximum ncg.ttlC bcndmg mllment at CIS shon 10 FIg. 9 maximum Dc-gam 1(. glH~n b) Ma'imum Degall.:' I, - 9U -2 t- 40(~) 3 2 20[G) J) -2 + G) 9 2 -180 k m 9.3 RESPONSE AT A PARTICULAR LOCATION DUE TO A SERIES OF MOVING CONCENTU LOADS As discussed in Section 2.2. live loads due to vehicular traffic way and railway bridges are represented by a series of m centrated loads ",ith specified spacing between the loads (see and 2.3). Influence lines provide a convenient means of anal tures subjected to such moving loads. In this section we dlSCUM IOfluence line for a response function can be used to detemune alue of the response function for a given position of a sertII centrated loads and (2) the maximum value of the response fi to a senes of moving concentrated loads. ConSIder for example. the bridge beam shown in Fig 9 that we wish to detennine the shear at point B of the beam wheel loads of an HS20-44 truck when the front axle of the tated at a dIStance of 16 ft from the left support A as showa UTe The Influence hne for the shear at B is also shown 1D the distances between the three loads as well as the locahon of are known so the locatIOns of the other two loads can be hshed Although the mfluence-hne ordmates correspond.ina: can be obtamed by usmg the properties of the SImilar lrillllll'- by the IDfluenee hne It IS usually convement to evaluate nate by mull1plymg tbe lope of the segment of the inIIl...._'. the load I located by th di tance of the load from the FIG. 9.6 SECTION 9.3 .......... II ........... ~ .... II ........ -..00_..1..__ &.- 8 41'
  • 132. k S. 8 20 (3~) 18 Sk ext the entire senes of loads IS moved to the left b 4 nt pta the second load ofthe senes the IO-k load. at the locallon of the maximum positive ordmate of the mfluence hne as bawn m FIB 9 d The he r at B for thiS loadmg position IS gIVen by S. 8 6 (3~) 10 20 (3~) IS 17 (~) 12 IS 567 k The n of load I then mo ed funher I third load of the senes, the 15·k load us The shear a B I now gIVen by SECTIONU 3ft (.j b) Influence tIDe ...;r,.I_ { c c 12 tt '011--- 20 (t B .ft IOk15k 5k 8 k u,. (lk 5k -0 1 ..r R" 101.151.: 51.: I C ~~~---151,---1 c '~~+----~ hear at POint B due to the senes ot four 4.:oncentrated loads the figure 1 he mDucoee line It,lT ~8 is !ibo",n in Fig 9 (b that the load n moves from nght 10 lefl on the beam we from these figure thai as the series mme from the end C t "aed pomt B. the he-dT at B Increases continuously as the CHAPTER 9 Application of Influence linesel2
  • 133. Ans "7 T 44H TMaxunwn F lECT10IIu _ Mo:d *4t::::===dr_·__....... :..;.SI. 15ftl--l_ --4---- 40 A 8k H A~t:::==:::~'"---=----'-_~ ~2Oft--I-1- - - - flOft ------I r I...o.ain Pus 4 For loading position 4 Fig 9 Thus far we have COIISIdered the maJUJJlUDl l<Spon.. a parI 10 tJOn 10 a truetun:: 10 sect ,"'...." delCmUue the I max ... a1ue. ~f~,:110~~ " oa:ur al an IocaUOD through uf a • 'a:::~ r.:=upponcd beams CXJDSidered m this herein be used to develop ~~~::.~ IDWIII1IIIl 1Jl' SIngII CaiWiib.1Id lIId CoaIida:-~:'~::'beam:-,~ • 9.4 ABSOLUTE MAXIMUM RESPONSE Ali 9.8 contd J::! ~11 (1_) :!80 k T '0 00') OJ B 'I Iladmg f't Itll)n I Ig: t} s d ,. 16111(:) 1::!6t1 I ,.T I ng fklMllOn 1 I t!f ~ ~ • • 'I--------T-------~J , 1 I I , I I I I 16l 12k • In : 32k ~( cbb;,J) i ill E B I I A ~20fl__+_lurl_+_l'i n-l-- 20 [1----+-15 f1--1 ------T--- ----~ , I I , I , I "16k 2l 8l 32k : ~t::=W=4i , 1-10 I 10ft 15~ I I,,-l--- 20 I. --1---25 II-----l d I loading PositIOn 2 (:1 Loading Pu Ilion J 16k J2k 8l E ih::::iD cbB C D 1 2 J -- •< at~ft= oft ---I 1-,01,+-15 ft I 2011 .16 CHAPTER 9 Application of Influence lines
  • 134. 5 b maxunum DOOiitive Uniformly DlltrlbutId IOId exl let us deenDi.. the aboolu.. IDlIXImUID menl ID the 5IDlpi supported beam of Fig 9 di5lnbuted live load of mellSl • ~::~:Emaxunum pclOIbve or negab shear I the by pbu:iDc the load over the porb • f the~beam~.:.: the shear iDftuoaoe line Fi 9 b ,.. multiplying the load 1IICII5I the Ioodod porbOD ofthe beam maximum bendmg moment pa(1 ~) The envelope of maxImum bending moments constructed by pIonlng Eq (9.5 i••hOWD ID FIg 99(e It can be .... that the aboolu.. maxI mum bendlDg momenl ocam at mIdspan of the beam aDd has magm- lude PL 4 IEChOIU ~ .. 7 Similarly the lllaXunum negab e rna unum negah bear Pc L 4 L , rr(L- ar () () L , (g) Envelope of Maximum BendlOg - Unifonnly Dlstnbuted Load (e) Endopc of Ma.lI:imum B ndUlI Mc....:: - Single Concentrated load (l J Envelope of Maimum Shears Di"tributed Lood A~( I - JL--a r L-Q- ~ Influence I me tor Shea• ..It SeLl..lon aa FIG. 9.9 -f p" L a ,.;J;."'==~~.,*=~ -0--< I . and c, ft:' J'C':1Ie1~. RCl.:all that thee mfluence lines were 1m ,e1op<d In S<xllon 8 I fig. X.2Ie) and (fll. Suppose that lot:' lol",h to ut:'termme the absolute maximum the beam due to a ~mgle moing concentraled load of magDltude discussed in Sel.:l1on 9. I. the maximum posithe shear at the "':OOD' IS glen by the product of the load magnitude P and the poSltle ordmate J u L of the influence line for hear a a fig 99 b . Tbus maximum ~1t1C shear p(I ~) .k_-=:::::::::::::"',':-L"() 0(1 .)L A~Ct---a L-a-j (I.:' Influence Lme for Bl'ndmg toment at SectIOn (I a fd IEmelope of Maimum Sheaf'; -- Smgle Concentrated LUdd .18 CHAPTER 9 Applicltion of Influence Lines
  • 135. P M A (~ ) P I bemWmum, Thus the bending momenl under tbe load P I SfC1IOIl U _ IIocl 11...._ Lit II can be sh .L_ L_ n Utilt tall; absolute rna Imum N"020 section located J II .L_ I f ._~ t1cd L_ 0 U~ e I of Ihe nghl uppon C po UIl;am SlOce the location f the ab!.ol :,,~~~~;~~knov..n the...--a u m d y·~uure f r compuung maximum ue to a senes f concentrated load de loped ID emplo ed 10 delemulI< the magn tude of the a luI Because the Influence hne for hear Just, Ide the left caito the IOftuence hoe for reactJon at the left uppon t convementl used ~ r detemumng the magmtude f the a mum hear To detcmune the locauon of the absolut ma m momenl conSIder tbe slDlply upponed beam ubjected I sencs of mO109 concentrated loads P P and P h 9 10 The resultanl of lhe load P P and P don ted P I located al the dIStance from the load P hown m tbe fi bendmg momenl dIagram of the beam con I Is of tral t I ",gmalls between lbe load pomts regardless of the postuon of lbe I absolute maJumum bending moment aceu under De of the 1 sumlng that the absolute maximum bendlOg moment aceu under t load Pour objectlie is to detenmne Its position fr m the nudspan the beam a shown ID the figure. By applymg lbe eqwlibnum cq I n ~ M. 0 and usmg the resultant p• •lISlead of the md Idualloads I the eqwhbrium equation e detemune lhe ventcal reach nAt be +~~ If. 0 AL P.(~ C ) 0 A -",---+--"."--"1 ,.-l"---! I AG 9.10 Series of Concentrated Loads The ffid,,pt' of maXlmum bendmg moments due to a umforml tribult:d Inc load constructed b) ploulng Eq. 9~. I hown 9.9 g It :an be seen Irom thl cmdopt: that the- absolute ~::~ hendmg moml.:·nt OCt.:uTS pi mid pan of the beam and has m II L' The ab~oluh" m.llmum .lIuc of .1 rc pon..e lundilln m any 1b1lCl"'~ ~uh.tcl,;tcd to a serics of IlltlVlIlg cOllccnlr.ltcd IOtuh ur any loadm!! condltioll he ddcrmlilcd from the em-elope of ~~~!~ aluc~ of the re'pon~ lunuioll. SUl'h an cmc!opc l.:an be: co by ealualmg the 1ll.1lmUm ".t1uc~ of the: re:~ronsc fundlon at n of poinh along thc Icn~th 01 the ~trUl"lure: by u~mg the pr de~crihcd III SCI,:tion.. tJ.1 thTllugh 9J ,md b) plotting the m alues BCl:iJu..c of th ct'lhlde:r,thlc amount t'll:tlmputatlOnal ohed. Xl:tpt for ~OI111: o"nnpk trul:lurcs th~ anal} IS of absol irnum rl: ponse IS u u,tll) flI:rlormcd u'lIlg l'omputer In th f~=~ Sf't:uon: -f dN'U Ihl.: dlrl.: t method Ih.1I an.. (t1mmonly emn d~{(:nnme thl; "bs(llute m.tXlmum ~hc,tr and hcndmg momeD pI) upported uhJeded [0 .l sc TIes 01 111m mg l:oncent As In th 1.:a5e (If In 'k t.:onl,; ntr.lled md umforml dilllriblil I.,ad the absolute ll1a mlUm he'H In ,I ~Imply upported beam senro: 01 mmmg' nl,;C'nlraled load al-.l tll.:lUrs at section Ih I .e uppon rum the mfluent' line lor hear at an arbl au 01 d Impl upporled heaIn htl- n In • Ig 9 9 b w n order to de lop the maximum p 1 IU I: he u at th u place a man load of lhe TI d po thle on the portl n fur y"hlch (he Inllucnw Ime I po Itle and a fe load po~1IIi th ponlon h n: lh mflul:nc hne I n g3tlC MOTeO U U I hJlil-d ttlard the Iclt uppon 01 the L .•, the max.lmum h "",om po 11 e ar ll1l ntmuouslv mcrea beca and the rna Irnum ordlnale I the po ltl~ roruon of the inllIul_ In ria Voh rc Ih I thl: neg lI..e ponlon dec ut rna Imum po tl h II ted ar Vol I tX:4.: ur hen the ",,:licle oc JU It Ih n hi f Ih I ft uppon 4 l Ing a CHAPTER 9 Application of lnftuence lines Th~ :Odopc 01 m Imum hl.:ar due w a unift'mlly dl!~tnbuted Jwd t.:on tru..ted '" p](1wng Eqs. I} 6 and 9. l. hon In FI n.-'-''" II t.:an Ix n that Ih .tbsolulC rna lffium dcdops al secll""'IJI.~ In Idc tht: SUppOI dnd ha magmtude" L -: To ddcmllnt: the I.: pre Ion for the malmum bending m xXLL n a'a (' mu!llpl) thl.: 10.11.1 mtcnslt))I b~ the area oftbe mg moment mthll.:n"c hnl.: Fig I} <} t.: t() obtain " u rna Imurn hendmg O1OO1l.:nt -,- I L a
  • 136. • 12k 12k k W cb!:;;;;;;;;;;;;;;;;;;~ 120ft---+--- ... Pl.15, PI.21 -tOO ~I-l5h ... Pl.14, Pl.ll, Pl.18, PI.23 ... Pl.13, Pl.17, Pl.ll, PI.22 18lN 72kN ~ib;;;;;;;;;;;;;;;~m. !--- 43 m---l 1.13 For the beam 01 Problem 8 I ddefmine .... OII_'~ po ItIC hear and bendmg momeol I poIDt 8 FIG PI.1Z, PUll SIclIo.9.3 9.12 For the beam of Problem 8 2 deternunc the mwumum JXhllie ~hear and bending momenl at POint B due to Ihc ~hecl loads of the moing H20-44 lruck shown In FI M.l:! 6 I <lr the I'>t:.lln III I~Hlhkm 10 de renm t ,. , and neg"dtlC hcar tAd th m tmu p>SIU uC bendmg ml)rncnt at pomt (d I "" d of 1)0 k a umlorml d l"bul ~H fIl and a uOllonnly dl trabuled d d I 1 ror lhe hl'dnl of Problem 8 2J de mil t. IUC and ncgdtle he nand th ma Imum r:Jue bendmg: mOffit:n 011 POint D d a '0''''''01''''«1 load )1 )0 k .1 umfonnl) dl tnbuted It J ilC ,"J.l uOllonnl) di tnbuled dead load of I It t' 1or the beam of Problcm 8 29 delellillne lhe ma,im,um ~hllle ami neg4tle hear!> and the ma mum posI IlC!.lIle ocndmg moments at pomt F due to a h~ IO.ld of .to k.•1 umfonnl) dlStnbuted h load dod a UOlfonnl> dlstnbuted dead load of I k fl t,9 f'or the tru...s of Problem K47 determine the rruwmum ten,lle and compre......le a tal force In member eN due concentrated IiC load of 30 k a umforml dl Inbuted li e load of ~ k fl. and a umfonnly dlstnbuted dead load of I l. ft. 9.10 For the tru.... of Problem R.50 determme the rnUlRlum comprc...sie axial force in member AS and the maximum tensile a"ial force in member EF due to a concentrated live load of I~O kN a umforml) distnbuled hve load of 40 kN ,'roo and a unifonnl) distribuled dead load of 20 kN m 9.11 For Ihe tru..s of Problem 8.51 detennme thc maximum (cn...ilt: and comprcs..i'e aJOial forces In member JJ due to a concentrated live load of 40 It a umfonnly dl tnbuted live load of .. k 1ft. and a unifonnl) dtstnbuted dead load of 2 k ft. SA for lht: beam of Problem HS determme the POSltlC and ncg-alle hear and the maximum poll negatlC bending moments al point C due to a co.......~ hc load of IOU l. a umforml) d...tnhuted hve l m dnd.1 umlormJy distributed dead load of • .5 f-or the t:dntllet:r beam of Problem 89 dcla1""miUlmum ul'"ard ertl",al reaction and the .ounterdock"'lse reJr.:tlon moment al support ,"Onctntr4h:d Ile load of :!5 Ii;. a umforml dilllri~. load of ! k rt and a umforml) dlstnbulCd oSk It h II lhl: Ill,ld To Ode:rmllle: Ihe ma'tmumhilI.. al I I.. ptl..llh1ll I . .1 1 ft n: Iloll,e Illlll..'lltlll uue to.1 lOgle movi or 1 ~L1II( .1 UI: 1 • h I • I I lh' Illd n10 t tx' pl.ll:l..'d .11 I e rx:.ltlon of thcl..I:nU.llr.:u l'1 I.: I. _ . , ,rdUllle of lhe: n:'pl.:lJl~e: lum:t1(lIl lOflucnce JX'SIIII.:' or negltl1.: t ' . ,. I rh I I a , n.11l~ luncllOIl due to a ullIi(lnn. ) dlstnbuI.: .1 u...1 t.: , .• I I r" In 01 Ihl' ,lrudun: elll be l,ht.lIned by multi4lpp I l ,n cr 8 ['l.' , IhC' 1000Id Inkn Il~ b> {hI.: Ill..'t .m:;.! under the .corre...pondmg POrtion I "'n mlhllncc hne To ,",."Iennllle the malmum posjn:pnl:un.. 1l . I . 01 a re rt1nX lunctllln due to a u.mfonnly dincgatl(· 'I Ul.: 11 c I,'ao the load mu t he pl.ICl:O o~..:r Iho-.e. portion of the h..'n; thr.: ordmale III thl.:' n:'JXln c tunctlon mfluen",-e hne are PCl.~ I.)r ne:1!.tll e The: m.tXlmum ,alue (If a re,pon..c function at a panicular I-·de"", in a ..tructuTt" due to a xne... of TTIO lIlg concentrated loads can tt'mlined h uc..:e 1 d) pbl"mg ca..:h load of the "Cne.. on the at the loca"tilln of Ihe rna Imum ordin.uc of the re:"!ponse fUOCli flut'nt:e line b l"omputlng the alue of the re..ponse funcllon for po ilion of the" "l'rtt:... Ihfl1ugh algebraic~lI) the ~roduetl lo.ld magnitude... and the respcctlC mf1uence·lmc ordmates aad comparing th~ alue... of the respon!'>e function thu.. obtained to mine the ma"imum alue of the respon!'>e function In Imp!) supported beams (a) the absolute maximum shear op:"! at ...ecuons JU'lt II1sidc the supports. (bJ the absolute m bendmg moment dut: 10 a smgle concentrated, or a unifonnl tributed. Iie load occurs at the beam midspan. and (c) the a maximum bendll1g moment due to a series of moving concen loads occurs under one of the loads near lht:: resultant of the loads the midspan of the be..lm i!oo located halfway between the load and resuhant m f Pr blem 4 delCnmne Ihe maximum I poInl 8 due 10 a l·k it umf< rml dl d CHAPTER 9 Applicahon of Influence lines PROBLEMS Sn" 11.1 •• 1.2 1.1 I- r the beam f Problem 8 4 detennme the maximum e bendlDg moment .1t poml B due to a 15-k ..on. trate<! hid be4m f Problem 84 deternul1C the maximum n I upport,of due to a l-k ft umlonnl I d
  • 137. ..... 'IN' • "" L-. -::::~:":-.momeot m• boom due Ibe wheel Ioods Cll~Ile .~.. It»... ., F... P9 12 :"'~~5~:~;::~JDUIIDI"D beDdinI moment Indue to Ibe wheel Ioods ., F'I P9IS &II Determine the absolute rnaxiD...HiI • IS-m lon8 SImply upponed boom three movlftg concentrated loads sbDwD. t.D D<ltlllUll" the .bsolult 111lWIIl_ .60-1-10.8 SImply upponed beam due mOVIng concentra.... load shown
  • 138. •1E1:YD11o.t Ii t ... I • I'II.lll.2 I)MI 11IIC S1IUCtIIII al AduaJ 51 c A lh R n II n hout A I lJ( I lh'7;-'==-'1i Reflection The d OIU n 42lI ClW'TER 10 AnI'ysI. 01 S",",,",", Structures 10 1 SYMMETRIC STRUCTURES ... 18.1
  • 139. o B I ft~1 E A .COI""" Ro_ I N M L K J G F 0 2511 W E Ret1ectioa If) • 1KTIOII10 1 Ii 'iii" I w lAl W E-COIUtant Frame A B 1011-+-1011-1EIt _ _ "'- L M N H J K II I E F G II BCD - '" 25 II - L:: 25 II ., 0" I I I ...AI 15 II -t- ItI 15 -t- lA 15 -'- FIG. 10.3 (contd.) 28 II 21 , £. A = constant ReflectIOn E. A ;: conslanl Reflecnon E.l.A Ellt Reflection I I CE====fiB=__ L L (al ( b) •- 30 ft ------I ~~E~D~====",F F r.Hmge 8f! 16 fl Be;tm , E. - =- con..tant Beam E. A =conMant Tru~ ~~====ll~Bi=====':'fL ~--;o-~L..,----+---;"'IT-i--I£ I, - E I. ~ •..-:m It ----t .4 IHJn~el Q) CltAPTEll10 Analpl. of Symmetric Structures (e) • •ilO ft---j ilO II---jC 0 o '-, I 21 21 12 h I 1 A B B ~ 20 h-------l 0---2011 E.A cun l&nl E.It= __ ""10.3 E fS rom I.OC Frame ReflccbOD ld,
  • 140. Q2 CHAPTER 10 AnI'p'o.1 Symme1rtc Structures E. A =constant (b) s A 8 C D ~25 It..l-25 It~25ft..).. trUl:1UfJ.1 )mml'tr lor lilt: pUrp(l l' or.tIl .tnal) ,is, II is 0'..,..... l'lln IdeI' the: Il1l1ll'tr~ l,r olll~ thl''''C ,trlll:tur,11 properlles that dic-d Oil I't:sult... llf Ih.1t p.trll( t~p.: of .tnal) In other IflKlme ~ l:llll ldl'rt:d 10 !x ~Illmelnc for the purpose of I'" If Its ...Irul'lural rnlllt'rttl' h.le .In el'tect on the resul anah "'1'" drl' mmdm: (:on hjt:r for l' J.mrle the ...t.tticall) dctemltnate lruss su ertlc.i1loalh.•l'" h n in hg. lOA. e C,tn '>Cc from the figure ~cl-'ml'tl) llf the tru... Il.e the dime:n...ion... of the truss and the Ill.:nt of tru ... I1ll'm~r and Ih materi.ll and cro...s-s«tional E .tnd 4 an..: )mmelrl( llh re"'pcct to the . a,is bUI the 101all' ...~ mllle:tr~ lx-c.w,< Ihe 11Ing.:d ",uppon at A can exen both IonIa I and t.'rlI(JI re.Klion . herca... the roUer suppon al C can. onl~ a t.'rtJcal re,lclllm. HOeH'f. the tru s can be coosl<iered ...~ mmetrll: hen ...ubJl'Clcd 10 t:rtil:al load onl) because under load. the hOrizonI,ll reaction al the hinged suppon will be ., 01, therefore. It ill not hac any effect on the response member <1'1.11 force:... and ddlectiom) of the truss. This truss con"'ldercd 10 be ,) mmClrlC hen subjected to an) horizontal hOeHT. Solution e c,m see from 1-1.. I(J 5 b L_ th c tndl the dlmen Ion the arrangemeat e matenal and l:TO ~tlUnal propt'rtlc £ and A and the lIII...rlI glen trus are all mmetn h the m mber ((I of the tr l: U ropecl to the ertlcal aXI aXI U Thu Ih~ Iru IS s)-rnrnetnc wllh The tru.........ho",n in bg IO.5(a) i~ to be ,malyzcd to detennine its member fon.:C and deflections due to a general s) ~tem of loads acting at lhe JOID the tru,... he wO'ldcrcd to he ...) mmclric for ~uch an analysis~ lao F G H ~~It ~l A B e D E l--4 panels '11 25 II _ 100 ft--' E , - ,on 'ant .. - Iet---L L- £. 4 - con unt £ample 10.1 ... 10.4 fIG. lD.5
  • 141. a--+--o---I cgob o,freflecti... M 1---'1 '81----+--- Rcnecbon (d) s Rcflccli... 0) b Rcn 1--0-_- SECTION 10.2 "'IISble MIl Ai,,,, Ie _C 4' .... ' & • w cD:r:r--e oo::::c:cuA B, I--o--l--a --lW C ...... f~A I--a--+---a--l Rcllection £ .. 10.11 (......./ b .l B ,. 1---0----0-- M M I--- '1 ---I M ~ ~-+-------'t! W Loadmg Loading Loading Loadmg Loading i "fl:r:r--.x-.. --+--o~,. , I " [1j';-'--'-""'I~fPTIJc _,,----11-- a--l .< I " 'u:r== :;==rtJc "1--,,-I-a--1 .< M I---~---l ~ - i I-----a-----I --1---0 1-;------1 D s I--~-----l ~.I' - : - - - - a - - _ - - I Rel1ection RefleCl10D ., P B -",----- a-----of~ Reflection Id I Ie, II) R n dlnn I ) FI5. 10.10 E mpl I A nil )- rnm ttll: loadmg t - - - -----< pi B;-I------,t~ ----a_ --I Loadmg Loading Antisymmetric Loadings A loading I con Iden:d 10 lx' J.ntl)mmctnc v.ith respect to an plane if the ncgatll' 01 the n:flcl.:llOn of the loading about the lllal to the loadmg it -II Some e'i:imples 01 antis) mmetrit: loadings are shown in Fig 1 each loading ca~ the reflection and the negative of refteeuon AG.10.9 leontd." Loading -Q---<--Q- , J----~_l ~ ''',--,-- , " , ' I •-'I---i ~~It~ -,-,-,~ f----a-----;I n ClIAPTER 10 "'"I,... 01 Symmetric Structuri. r--_-----"I:
  • 142. Solullon S mm In Loud", C ", The I:~:;~~~:~~~~and tbe half loadl.B are bowo m FI I I b loadlDg about the UI 19 drawn In Fig 10 I The lbe Slyen loadlDB I detennined by addlDB the b If I reOeotl•• FIB 10 13 • as showo ID FIB 10 I d IECTl0II1D.2 ~''''''.U:''.'''~101c .... t'.1101••_10 C'-IIOO.1oIII blaC. $I kif' 50 Example 10.6 FlG.l0.13 (eonld.1 2st ft Q I b, Half Loading , I X.lll~ , I lJJItlDtm Q- 20 It--+-10 ft-t-IO ft-4-- 20 ft---l (a) G,..rn loadmg ~"(I HITm (ric 1.(J(1('I11~ ( mpOllUI! The anti~}mmetric componan loadmg I ,lbtained h~ ..ubtractmg the ~} mmclric loading component 10.1211.1 from the IOlalloadmg hg, 10.12(<11) and is shown in Fig 101 ~40 -10 -In 0, O. 0.' (til Half Loadmg FIG. 10.12 NOI~ that the ..urn of the ..ymmclric and antisymmetric componen to the gncn loadmg ~-lO ~O 80 ~O ~{) 0.' oS 0, O. Ie : beam is ..ubjel.:ted 10 Ihe loadmg ..hown m Fig. 1O.13(a). DetenDlRe metn( and antlsymmclrl( components of the loading lAith respect to the ,,> mmctry or the 1x,Im ld IS) mmetnc Loading Component CHAPnR 10 Anllysis of Symmetric Structures LZ''<:2''C''''X:'~ ---T "m ~"'l~~~-'- j j j 81 0 0 000 6nctm41m- aGnnl mg Eumple 10.5 IS. 10.13
  • 143. 411 I H 18kJOk Due S)'I1IIII<ll'lC ~ Coa_.. G 18 k F64 G64H ~A~~= ~ IE I 24k JOk I k ~k k b _ Forces Due IJlk c MembcrF Jlk 6k 6k}k (g) _ ForcesDue"AaIi y.-ux I.-liDcC:"'.....,- 1ECT1OII11.4 ProcIdIn f8r ....... at " ••"'" ...... ISk , I F ,G H ~6k 6k (cl Anli~)'mmetric loading Componenl , f G H ~ft I 2H JOk 12k I - .t panch al 20 fl _ 80 fl ----J (a) Gi'cn Tru~s and LoadIftB v~18 k JOk 18k (b) Symmetric Loading Componenl I3 k (d) Sub'lrUcturc With Syrnmetne BoundarY CoDdi1JODl ...,... ~"",Ple 10:.:,9:--- -:==--_-:- ==========JOclCmunc lhe ~ 10 • CftAPTER 10 Ana'ysls of Symmetric Structures . I h "U ., LI" III til' thl.' "tfUl.:Wn:', on either side of the2. Sc l'('1 Ll U" l ' . " "lll"I"', JilL' l.:W"-'CCIIOI1.t1 i1r~i1... and morn-"Il1IT1L'lf) l I • ' - h "-q . . '·,h """,.",- (1' tIll.: 'llh'lrw.:tufC. teh are locatedinertt,. (1 l l 1'",- _ h• . ,. ","'1,'(n ...lhlUIJ hl.~ r('JUl.:c:d b) h.llt. hereasfuU [ l .11 ( "'- . . II I be h I,',..,·" hllUIJ h.: u'l'.d hlf.1 O( leT. mcm rs.l,r t e..c pn.... l D ,h ' 1'1 ell IlllJlIlg 11110 ..., mmctrll,; and aOII;.Y1",.,....3. l'l.-("llll pI".: l c ' . ' ... , ""h ,,-,""'1.:1 to the .t'I' 01 'i,mmctrv of the stnlrho_((lmf' ' 111.'11 .. .. I'~ , _. • - - ...... U'!OI:! the pnx'cJUTl' dt'~nht.-J 111 Section 10.2. -t. [)c,;mllfl tilL' fC pl'n'L' of the ,trlll.:lUrc due to the 5}mmetnc: Ill!! l.onlptlncnl.1 fllI!O . " . a. 1 c.II.:h Jllllll ilnd end 01 Ihe ,ub,tructure. "hleh IS loc:a the <1' of )mmctr: .Iprl) rc,lramls to prcent rolaltoa dcfl~(twn JX'rJX'ndil:ular 1O the a I~ of ~~ mmetl). If there hmgL'.1t uch a Jomt or cnd. then 0~1). the deflection but rot.ltlon h"uld lx'<lIlled at that Jomt or end. b. .rpl) Ih~ ...) mmetril; I:omponent of loading on the sUbs,tnll:1"'~ ilh the mac.l1llUde... of the concentrated loads at the smmetf) n:duced b) half. c. _~nal)le the ,ub,truclUrc to determine its response. d. Obtam the ...)mmelrll' respon-.e of the complete structure fleeting the re...pon...e of Ihe substructure to the other Side nI ....~· als of~) mmetr). 5. Determme the rc...pon-.c of the structure due to the antJsiYlTlIIIIlldl/!. loading component as 1'0110 ...: 8. AI each Joint and end of the substructure located at the uil~!f" ...ymmctry. apply a rc:-Maint to prevent deflection in the am""". tion of the axis of symmctry. In the case of trusses, the forces III members located along the axis of symmetry will zero. Rcmmc such members from the substructure. b. Apply the antisymmclric component of loading on the structure Ith thc magnitudes of the loads and couples alPPlill~ at the axi~ of ~ymmetry, reduced by half. c. Anal)lc the substructure to determine its response. d. Obtain the antisymmctnc response of the complete strut rcflectlng the negallc of thc response of the substructure other side of thc axis of s}mmetl) 6. Delermme the total reSp()n~ of the structure due to the gIVen ing b~ sUJXrimpo~ing the s)mmetric and antisymmetnc rei obtained in "tep, 4 and 5. rc'pc:t;tle1i. The foregomg procedure t.:an be: applied to statically dcterminatt l.ell as mdctennmate ~)mme(rit.: struc(ure~. It "ill become o,bvi~,j sulhequent chapters that the utihJ'ation of structural symmetry erabl~ redu~, the I;omputatlonal etfon required in the anal cally mdetermmate Mructurcs
  • 144. A_IO ~1~ qc 0t 1 U 15 $10 I~ ~1~ 1010 10-qC 0t- -'~D::-,.;j 15 15 I'15 ~IO 1~ 15 Member _ Due tD,Aatioym_1rI ...- ':---::-
  • 145. - k k - •.------,_+. . , -'- _L -'- -'- A,Rb.ymllllClnc looohni Cempno... (d) SublrtIUc~ for Anal of Symmemc Response S ",m" and An"u'''....,., C'nnlj_'" and c Solution SUMMARY FIG. 10.23 (eonld.) II I Sft -+Sft -+Sft 1 I I I I I k It I I I -25 ft --t---25 It -----t-25 f1---l E, I. A =c()n~tant (3) Gien Structure and Loading t--l75It_ I b, S}mmctnl Luadmg Component I IbTdl I·I I I I 2. klh I I I I I I II I I I I - -'- _L - '- ,2 k ft I I I I I I I I l ,2 II. ft I I I I I I I I -'- -'- -'- -L.... ,.-.-.-r-hlk ..:.:ft~~_ 3. _ 11 " I I I 'I• -3k 12"- 12"- I I U I hI.: uhslrlldulC'> fI', the an.llpi~ llf the ~ymmet fllmdnl,. n: J'Il1nsc .In. hO11 In I-Ig 10 ~~ d .lI1d ('1 re peell I Solution "L I 11. I Thl N.IIllI~ mllldrK Ith II.: 1''1.":111 the crtlcal 1I1 11,1): 10~'" b I h kf h.lI! "j th... 0..:.1111 1 ckded lor ,malYlil Ddanunc: thl' ut'! trudUI 1M the .lI1:l.l~ 1 of the ~~ mmelm: and antill1lll"iii n: ron~ "r thl.: w.tJlall~ indeterminate frame; ~h(l n lI1 Fig 10.23 a ExlmpJe 10.12 456 CHAPTER 10 AnalySiS of Symmeh1c Siruciures fIG. 10.23
  • 146. - .. "" DHinpE F A .1I--1-l 3m-+-Sm 'm---+ III E./A-_ A 1-1'11 Eo/A 24k- C 2H E fII. PI.... Pl0.23 fII. Pl0.7, Pl0.22 fll.PllIJI,PlO.21 E. A=con tant 60 (,() f>() 30 JO k'l kN kN kN kN A 18 Ie ID IE IF G ~m H / J K r---t-'-r---- 6 panels at 4 m =24 m =----1 E. A =constant lUI ......... Il1.D Determan< the member eud the fr.lmc.. lohon an Figs PIO 6 PIO U tural )mmt:'try. FIG PlO.5, PlO.20 FIG. PIO.4, Pl0.19 F1G PIO.3, PlO.18 soco- 10.3 .nsI1DA 10.1& ttwougfIl0.ZO Detennme the force 10 each member 01 the tru'..c.....hon in figs. PIO I PIO S by ubhzJDI ,trudural ) mmc{T)i. 4ft Ht I:! ft FIG. PlO.2 InsI Plo.17 PIO, I PIO.IS with rC'lpect to the of symm ..trUI..{urc. . IO,ldinc: I~ n'll IdcTl;d It' Ix: ,,~mmclric: uh respect to an It rlane If th~ n:tJC:dll'l1 l,f the IllJdm!,! about the axi is Identteal loading It~lr !o.ldlOf! l~ Cl' l1 H.h:rc:~ to Ix anu,,}mmetnc WIth tlhln .1'.'" In 11 plane If the- nc:g.ItIC olthc.n:llcl'1Jon of the 1oad1lll the 31 I Idenlll,...i1 to the: )O,lomg Ihdl An) general uns.YttUllllll loading c.m be dc(,:ompll cd into ..) mmctm: and antis} mmetn nenb lth n: p...xt wan 31 hL:n J mmc:ITll: qrudure I" ..ubjected to a loading that mctnc: "ith rc:'l""Cl to the: ,tfm:tun:', ai~ of s)mmctl) the res the ,trw.;turc I...ll'l) ') mmetTic Thu.. c can obtain the response entire 'lructure b) .ma1) 7mg a half of the ..truc-ture. on either side a ,.. of ,)mmetr) llh ..)mmetric bound'H) ,,:onditions; and by mg the computed T6pon,c .ll;out the axis of ~~ mmetr). hen a }mmdril..' ...tructurc- i......ub.ll.'Cted to a loading that ..)mmetric Jth re'lpcct to thl..' ...tructun:~"" aXI~ of s)mmetry. the of the ~tructure i... al 0 anti,) mmetric. Thu~. the response of the ~tructure can be obtained b) anal)7mg a half of the structure on ...Ide of the aXl~ of ...) mmctr). ith antisymmetric boundary COI>dilii and b) reflecting the ncgaUc of the computed response about of symmetr) The re"'pon:-.e of a symmetric structure due to a general unsoy""",. loading I.:an be obtained by determining the responses of the due to the s}mmetril.: and antisymmctric components of the metric loading. and b) superimposing the tO responses. 5m 80 f. A - ron tanl 'm 5m .. CltAPTER 10 An.lysli of Symmetric Structure. ........ 10.1 _10.2 10.1 ......... 10.15 [)l:temllnc the :,mmClnc and anti- rometnl.. rnponem f the loadi~gs ..hoy, n in rig.. PROBLEMS fl&. PlO.l, PlU8
  • 147. o B •11-+-2011 ........ _1'18I11 -..........__... I . 35 I 10 m ........... 15m E.IA=~ ... PlG.13 ... Plo.2l B C 2t1ft 20ft 10k J K L"TI ............... Hlft ,t10k G H 3k1ft :Mil 10k 0 E F E A B C m 1---3Oft I I I 30ft .. PlG.14, Plo.2l
  • 148. SF/ncr Harhour, Australia v glIal Vis 11 Introduction to Statically Indeterminate Structures 11.1 Advan1aQes and Dtsadvantages 11.2 AnalysIS of Incle1ermlnate Struc:. lnlleterm nate Structures Summary es In Part Two of this tex.t. we considered the anal) 1 of tatlcall) deter minale structures. In this part Chapters 11 through 18 we focus our allention on the analysis of staticall) mdetennmate tTUCtUres As discussed preiousl) the support rea,lIon and mtern I f, n: statical!) determinate structures can be detennmed from th equat of equilibrium includmg equations of condluon If an H er indelerminale slructures hae more support reacli n nd r mem lhan reqUired for static stabdlt) the eqUlhbnum equall n I oe a 0 t sufficient for detennining the reaction and mtem I r. f h strw..'tures.. and must be supplemented b add II n I I n nilu on Ihe geometry of dcformallon of tr1l.'1ure These additional ",Iallonships "htch a t nned .he ,·..,polihiliIY ("(}lid", ns c:murc that the connnult) of the dl plllt.'tITlC'n throughout the structure and that the trueturc v gether For example at a ngid Jomt the deftecti the members m«tlDg at the Jomt must Ix the an mdetenmnate trUCture ID 01: III ddi arrangement of members of the uuelurc nal propertlCS ueh as cross-sectIonal a of .1as1lCl' .'e which to tum <!<pend
  • 149. •• -'......... £/ L ... 11.1 nol ncc:essanly collapse and the loads WIll be redistribu~ocI lacenl ponlons of the ,Iructun: Consider for detenmnate and indetemunate beams shown m FI 1 specIIvely Suppose thallhe beam are upponmg brid way and that the nuddle pteT B I destroyed when a ...11.1 1EC'hON111 ...... e'''DIIld, $ e 1I.,:h1l: A••• In_. if AAgj B S_s1ly~Bam Inoaasl"'" ifA~r The adantJ.ges of staticall) indeterminate structures over det structures include the following. 1. Smaller Stren'e~' The maximum stresses in statically I nate structures are generally lower than those in comparable nate structures. Consider. for example, the statically detennmate indeterminate beams shown in Fig, 11.1 (a) and (b), respecti bending moment diagrams for the beams due to a uniformly dis load. Ii. are also shoy. n in the figure. (The procedures for indeterminate beams are considered in subsequent chapters. It seen from the figure that the maximum bending moment and quently the maximum bending stress in the indeterminate beam nificanlly loer than in the determinate beam. 2. Gr~at~r Stiffn~ssef Statically indeterminate structure hae higher stiffnesses (i.e. smaller deformations), than thoac parable determinate structures. From Fig. 11.1. we observe maximum deftectlon of the indeterminate beam is only one-fiftIl the determinate be-dm. 3. R~J"""'d~f Statically indeterminate structures If pnlpl~ SIgned have the capacity for redistributing loads when certatD ponlOns become overslresscd or collapse in cases of ovcrloadl eanhquakes tornadoes. Impact e.g. gas explOSIOns or vchidc and other ueh events Indelerminate structures have more and or suppon reactions than reqUired for static stability or member or suppon of uch a struclure fails the enbre sU1111l1l11i CHAPTER 11 Introduction to Statically Indetennlnlte Structures 1 h ' "I 11 III structure is therellstrw.:tun.: C Ul I!! ( • . ) t 'nl"nnCI dh..'rd~, the (rclatle SizeS ofthclHll 111 an IIl:r.1 11.: .. - h ." , 1',11, Is,Unll..:d ,lI1d to anal}lC I e structureml,.·ml~1 .In.: IIll I. . • . h I " . Ihu.. lhl IIIlt'd .m.' u..ed lO fe I"C t e memberIIlh:rna lm..e l. . .. , ., Il.... In." 11llt dose to those Inltlally asswnedTe') ~u menh>t:f ... '"'" , . " Ilal, " ..1 U 101' the I,tlcsl member sizes. The Itcraslrw.:IUfC I' Il,l .... U e- Ith 'nl·"" ,iln t'l.l cd on the re~ults of an analtmue... unll c me.: '; to Ihl)~ .•"uIned lor that an.i1~"I., ... Despite the !i.....,n:g(lll1g lhtlkult) m. dcslg~mg mdetermma lUre". a great maJorll) of ,trul:lure, bemg ~Ullt today are sta determmate: I.)f C'.impk. most modem remfon:ed concrete afC 'itaticall) imktermin.tlc. In this chapter. .."c dlsc~ss some or portant .IJ anlage .lnJ di ad antages of ~ndetemunate strUCIIUri lX)mpared III Jeternunalc ..tructures and mtroduce the fun concept-. of the anal) .." of indetermmate structures. 11.1 ADVANTAGES AND DISADVANTAGES OF INDETERMINATE STRUCTURES
  • 150. B B - F a T B a a.l.T L E A I (a Slab all Detennirwe B IlCTION 11.2 AIIoIroIo aI h,d".i_'" '11 • A 4 F al~T)AE~ 11'"========== B lbl SIJtically Indelemunate Beam Internal hln~(, a) Stath.alh Determinate Beam CHAPTER 11 Introduction to Statically Indeterminate Structures AG.ll.3 11.2 ANALYSIS OF INDETERMINATE STRUCTURES r mdeterm- f Iall E. A. I lb, SlalocaJl Indctmni.... Beam from deforming axIally by lhe fixed uppo temperature change AT a c rnpressl c a I ~ F • dT AE develops In Ihe beam as hown In Ihe figu", fabncatioD erron are SImilar to those of temperature ha rnmate and mdetenninatc structure Fundamental Relationships Regardless of whether a structure IS statically determmate nate its complete analysis reqUires the use of three pe shIps. • Equllibnum equations • Compatibility conditions • Member force-defonnauon relauon The eqUlhbnum equauons ",late Ihe r. Its part eDSunDg that the entire structure w In equillbnum. Ihe compoubihty condiuon ",I I Ihe strueture SO thaI Its ,anoDS parts fill defonnaUon ",lau os. which In' I e Ihe material proper1JCS E I and A of Ihe members..~~~:: Iween Ihe fon:es and displaeemen Ihe~ In tbc anal of III detenm of equilibrium a firs. used I blainilhe~1§P~§~§~~fo f Ihe trueture then Ihe memftberi.~i 10 Ihe companbili ndi .,. em di pJaeemen F r eumple AG 11.4I B,. 'u~ Ihe: "l.ltlcalh detemlinate beam is sUpportedram IOto I. l .. ", ~... iu...t tht: ...utfillel1t nUll1ha of reactll)n~ reqUired lor ~tat1c stabil ·remo.llllf urrl)ft 8 ill l·au..e the entire ~tructure lO: collapse aslllll,. in hI!. II.:!'J . Hoca. the indeten1l1nate beam (Fig. 11.2 b has etra~reih:tillO In the crticdl direction: therefore. the slructure wiD nc-ce :.arih collar".' .tIld rna) remain ~lable, cen after the support B failed. ~umllll! that the he.lm h.h been de...igned to support dead onh m C.l"C of ;ueh an .tceidel1l. the bridge will be closed to traffic pie; B i" reraired and then ill be reopened. The mam di~Jdanlage~ of ...tatlcally mdetcnninate structures delenninatc ...truclurc'l. an: the folloing. I. Strene Due W SUppOI·t Settlements Support settlements cia cause an' strc:-.'lC'l in determmate structures: thcy may, however ..,ignillcal;t in indeterminate structures. which should be mto account hell de...iglllng indeterminate structures. ConS1der the tennmate and mdeterminate beam shO.-'1l in Fig. 11.3. It can be from Fig. 11.3(a) that "hen the upport B of the detenninate undergoe... il ...mall settlement tJ.H• the portions AB and Be beam. . hkh arc connected together by an internal hinge at B me"". rigid bodies .ithoul bending that I .... they remain straight ~Iresse... dedop m the determinate beam. Howeer, when the ous mdetennm,ltc be.lm of Fig. II 3(bJ IS ...ubjected to a Sll1ular ~lliement it bend. as ho. n in Ihe figure; therefore, bendID8 dcelop in the beam. 2. Streur, Due to TrmprrQture Clrllngt's lI11d F"bri~ like support s.:lllcmcnts the'<C effeels do not cause stresses m natl' tructures but may mduce ignlhcam stresses in indetemu·'-'...... (on Iller the dctennmate and mdetcrminah: beams shown m F It r....n Ix s«n from hg. 11.4 ill that "hen the detenninate beam Jected to a umll..'mn temperature mcrease tJ.T it simply elloopl""'~ the aXial defonnatl)n gien b~ () 1 tJ.T)1. Eq. 724 dedop In the d tenninate hearn. smce it is free to elon eer hen Ihe indetenmnatc: lx-dm of Fig. 11.4 b whICh
  • 151. !OUt M".1 If 'SInD.·
  • 152. m 10 48F.F. , F. FA, The member aXial defannatlons can now be compu ~.~-<~f~:':~~= these values of member aXial forces Into the member f( relallon. Eq. 11 10 Ihrough 11 12 I obtai <lA' <l 0136 ft 1629 In and Fmally by uhslltuung the values Ibe oompallbilil oondlllOO5 Eq dlSplacemenl ofJOlOl A a <10112 or 1K1IOtI1t.2 ", . Substilutlon of Eq c f. - = (on lant - 20.000 k. A DeI{lrmcJ hapc (e) 500k (d, '" 5ll0k i)lB J-rc 1 ~jn (J - 0.8.- i)w " 12 II Ib' 500 k 16 (I erl1lo,;allh~plal,;eml:nt ' 01 Joint • Th d" I d·IS h . ~ .1 e J~p acement lagraDI 00 In 19 ,11.6 CI As~ummg the dlsplal..-ement A to be Yo nle the l;ompallbllity l,;ontJition a, CHAmR 11 Introduction to Statically Indeterminate Structurese71l FIG. 11.6
  • 153. en CttAPTtR 11 SUMMARY Introduction to Statically Indeterminate Structures 1 h' INl:1.: IllclhoJ.... , hu.:h an~ .prc'Cnted In ~), 'rmng l,"yu.llllm", : " I F( l 'r,11 ((l1l,"I1Cl1t lor .Ina )/lIlg 'mall st ler... I ~ and I~ ,IfI," I!cl1l;· . . 1 . ' ~I I' kc:r 1.'0......... mcmber... alllLor reactlo Ith I k (l'uUllu,tll'" .I; ' . , I I hl 1ht·...C' mcthod" .ln~ also u't:d to den requm.'d It'lf "t<.HIL .1'1 • .1 . , .1 'j' nlillon fI... I.I!ltlll' llL'l'dl.'d to ul,"t~lop the displ member loru:-ut: l'f ' " " 11' 'I,plal:....ment method... ,m: 10 Chant- mc:tht1u.... 11.: u ...-. I Th ' m ·tht)lh Me 111('r" ,),tcmi.ltlc can be easil throul:!h, c.; I.: t: . " 1 ~ -~ .'1mputcr ,Ind .m~. thr.:n:lore prdared lor the anal r emenll,"u ('n u large Jnd hlghl~ rl'dundant tfw.:lun:. In thl I.-hapla c It:amed th.1t the ad anta,ges of statical tenninate ,lrw,;turl,"... O'er dl,"h:rmmate ...trUl.;ture_~ Include smaller mwn "tn~ t:". grt:atcr "ulrnc~"c.:... and redundancies. Suppon sett! temperature ch~lngc"'. and fabricauon. errors may mdu~ 51 ,e" in mdt:tcrminatc ...trUl.:ture". whIch ...hould be taken IOto hen de igning ...uch ...tructurc . The anah"I" of ~trUl.:tun= imohes the usc of three fun rdatlon...hip.,: ~ equilIbrium equation". compatibility condittolll, member force-deformation relation,,_ In the analysis of inde ~tructure.,. the equilibrium equation... must be supplemented by the ratibilit~ condition... based on the geometry of the deformation "tructure. The link bc.:t-een the equilibrium equations and the patibilit) conditions is c.:...tablishcd by means of the member fl deformation relation", of the structure. The methods for the analysis of indeterminate structures classified into two categories, namely, the /iH("(' f/Ii'xihility) meth the dill'lac(,lJ1l'''' srilll/f_'.' I /IIethods. Sr. Louis Gate1my Arch and Olel Courthouse Jeff :lI1 Na'" ,beans 01' Mel'lOl N PtServe 12 Approximate Analysis of Rectangular Building Frames 12.1 AssumPtIOns for ApproXImate Analys,s 12.2 Analysos for Vertlcal Loads 12.3 Analys,s tor Lateral Loads-Portal Method 12.4 AnalYSIs for Lateral loads-CantIlever Method Summary PrOblems The analysis of statically indetenninate strudures uslDg the force and displacement methods introduced an the preceding chapter l;3.n be con· sidered as Hact in the sense that the compatibilIty and equlhbnum conditions of the structure are e,,-aetly satisfied in uch an anal Hocer. the results of such an exact anal}sis represent the ac 081 structural response only to the extent that the anal 1Ical model of the structure represents the actual structure. Expenmental result ha e demonstrated that the response of most common type of ruct under senic.-e loads can be reliably pmlicted by the fnrce and d p menl methods provided an accurate anal}t1cal model fthe troet used in the analysis Exact anal) i of indetermmatc truetUfiS IR I of del1ecuons and solution of simultaneous equations time conswmng Moreover. such an anal depends SIZeS cross--secuonal area and or moment of merna f of tbe trueture Because of these ddlicullleS at<d " analysIS the prehmonary design nf mdetermm Ie t based on the results ofappronnwi ana In "hi h In are nmat<d by maJul18 ""rtain a wnpn os about the ~=:~: and or tbe dlStnbunon of forces bet n the membe 01 thereby avo,d,ng tbe neeesstty of computmg dellecti
  • 154. CHAPTER 12 Approxllnlte Analysis of Rectangular Building Frames h ." rn)(" hl ht: lJlIIte l"omcnu:nt to use 'rrn'l:lm.ltl.: 111.1. I ' h f rro1 xl hen ('aal a lern.tlle deSigns plannmg r .1 t: ~l J.: . Th II ., 111I Ih.'J for rd.lIl~ c~.·l)nOm). e results ...trudurc ar~ lI"lJ.l 1.:. • I In 11,(' he u cd 10 (',ornate the Sizes of PH) mMIl: an,. l'i (,:. • I I .;.. 'f "l'('Jed 10 Iill[.lh: lhe. C:iK.'t ana }SIS. Thetnll.:tur.l mlnh...: .. . I ' J I' "l·nl..·r~ In: then fe I,,"-,d I.tcr.tllc}. usmg then.ln (' u:m II .. t.: l~ . , . - . - " an ,I,..c hl tfmC at their fin,ll deSigns. FunlIcec ..1 C e .Il .... n "1" IIlth 1 I (meumc... to roughl) check theJrrr,,I...... '. . . of (,.'.KI .m.ll) r hll.:h dUt..' to 11 l"ompICH) can be prone to hoalh. III n."Ccnt ) car... there 11,1'" ~cn an Increased tendency ren(l~tinl! and fClfohtung {,Ida ..trUl.:tun:.., Man} such strueturea ,lfUdcd prillI' to 19(i(). inl.:luding man) high.ri~ buildings. wa. hmed okh on thl.' b.I, of Jprro'-Im<lt~ analySIS. so a knowlceIIe u~dl'L (amh'ng of .lrrrlnmhlie method, used b) the original des.i u~uall) helpful in a rem.Hallon undertaking. . l nlikt.· thl.' l',-,Kt I11cthod~. hll:h are general In the sense that can be applied (0 arious I) pc~ of ~tructures subjected to vanoUl in!:! condJtion~. a pecilic method is usually required for the appro an:lys.i, of .t p.trtlcu!ar t) pe of struclUre for a particular loadiDJ. example. a different approximate method must be employed ti anal)sls of a fecti.lngu!ar frame under ertical (gravity) load than the anal)si~ of the samc frame subjected to lateral loads. N methods hae been deeloped lor approximate analysis ofindete structures. Some of (he more common approximate methods pe to rectangular fmmes arc presented in this chapter. These methoda be expected to yield re..ults within 2()'~/o of the exact solutions. The objectives of this chapter arc to consider the appro anal)sis of rectangular building frames as well as to gain an standing of the technIques used in the approximate analysis of tures in general. We prescnt a general discussion of the sun assumptions nccessar) for approximate analysis and then con ider approximate analj-"is of rectangular frames under vertical loads. Finall). e present the 10 common methods used for proximate anal~s.i... of rectangular frames subjected to lateralloadL 12.1 ASSUMPTIONS FOR APPROXIMATE ANALYSIS As discu sed In (haplers "' through 5, stati<.:allv indetemunate hae .more uP.JX>rt realtions and or member~ than rcqwred tablht): therdore. all the realtions and internal forces lOci ~omem ~f uch truetures cannot he determined from the .equlhbnum. The eXl;e s reactions and internal forces of an mlTlale IrUl.:ture ar' f I dd . c e errc to as rt dUlUlant, and the Dum undants I e the dltfercn(,.'C beteen the lotal number of un"""".the number 01 eqUlhbnum equations is termed the degr IECnoN 12.1 _...1119...... tor .................. Assumptions about the location of Points of InftectIon
  • 155. ~C~) I - - I 1lCTIOIl12.2 ....,...ror_ ~ CIt c F 8 w E H w (b) TypIcal Girder L - - - - fa) BUilding Frame (c) Simply Supportod Girder fll.1U w ~~ --- - I A D G ~_--L - - - - - L ----1 CHAPTER 12 Approximate Analysis of Rectangular Building Frames I .J 1"1(ICI.'Jurl." Il'" .1ppI0'1I11atl' anaI)SIS of n""'.....,. l'(lll1l1hlll ~ l"t.: . . ) . . J , . ~lIbll'l'lCJ Il) l'rtH.:.lI t!!r.lll) lo.ld~ Involve ,,'"-.lal hull log 1.11111.: . , . , I.. lut (hI." heh.1 1M of c.lch girder of the framthrn.•1 ,ulllplh1n" Pl _ _ . • '."J III Ul1lfllrlnl) lh,tnbulcd loath", as shILI!..'r .1 fnll1H;' lIl'I!';L l _ • I' , Il,' '·'I:l:-h!.ll.! JI.lgram tlf .1 t) pH:al glTlh:r DE of the Ii .-1' - - a 1 . : . J I . h I I "" ...... hlml lilt: JdlCl:I.1." · 01 I. e girder '1Ql1d1lt.I hO1l In "It! __ v _, In thl.' figun: e (lb~nc: th.!t tl~ mtkl:tlon POints Clsl near both oj the ~lrJcr Jht.' I: mllL'cULlIl poml dcdo!, becau-.;e !he columna thl. IJ'.I(('111 ~lrJcr (tll1nl,.~h:J to the end... 0'. gIrder DE offer partial 'trJIIl! or rc l'I.IIKe .Ig.tin t rt1t.tlIon b) c'crtln.g ncgatl(' moments and tUl at thl.' !!Ird~r end.. D .Ind E re~pcl.:tle1). Although the IlX:3tllln of the mflccti,m ~)in... dc:pend' on th.e rclallc stlffnesses of fr.lme m~mbl,.'f' .md I."W be d~temllnC'd onl) from an exact anaIIYltL •• ~an e,tabh h the regll..ln...dong the girder In which these points III: I.att.:d 0 namll1l11,!! Ih~ lll extreme lo·ondillons of rotallonal restraint the i!ird"er cnd 110 n In I Ig. I~.::!!c) and (d J. If the girder ends ~ to r~lt.It~.•1" m the ca..e loll' a ..impl) ..urrorted girder (Fig. 122 c l'~ro h<.-ndlll1! m()l11cnh and thu.. the mtlection poinh would at the elld.._~On Ihe other eXlrcme. if the girder ends ere com fixed <l!!Jm..t rotation C C,1Il by the exact analysis presented "ub-.eq~enl ch,lpt~r.. that the mtlection points would occur at a eli of O.::!IIL from each end of the girder. as illustrated in Fig. 12 Therefore, ,hen the gIrder ends are only partially restrained agaunst tation (fig, 12.2(b) the mtlection points must occur somehere WI a di,tanl.:e of 0.211 L from each end. For the purpose of appro anah ..i,. it i.. ('Ommon practice to assume that the inflection pomts loca;ed about hall"'ay between the two extremes that is, at a d. of 0.1 L from each end of the girder, Estimating the location of two flection point-. illol'c~ making Iwo as~umptions about the behaVIOr the girder. The third aumption IS based on the experience gained the exact anal) ..c~ of rectangular frame~ subjected to erticalloads I, hich mdicate that the axial forccs 111 girders of such frames are ef)- ..mall. Thu. in an apprm;imate analysi~. it is reasonable to a that the girder axial force are Jero. To ,ummarizc Ihe foregoing di""us..ion. in the approximate ~~:~.~ uf a rectangular frame subjected to ertical loads the following a lion.. are made for each girder of the frame' I. The inflection ~)lI1t' arc 10l.:a1Cd .it one-tenth of the span from end of the girder 2. The l,!lrder axial force IS lcro. The effect 01 IhelM: ~impJiI) II1g aumptions is (hat the middle tenth of the p.m 0.8/" I 01 each girder can be considered to be uppoftcd on the tv.o end ponion!'l of the girder each of wbich the Icn,gth equal to nne-tenlh of the gIrder span 0.1 L). as shown 12,2 e ole that th girder arc nuv. stalicall} detemllnate end force and moments l.:<ln he dctemllned from statics a sboWIl 478
  • 156. hmlenor column 1ICIJOII123 ....,.......... ..--_ _ I. z. --.-! ---+•2 -+• -+•2 -L. (a) BUilding Frame (b) Simplified Frame , --., , ,, , ,, , ,, I • • •• , •, , , '~ , , ,, , -l, , ,, , , ,, , , ,, • • • i• , , I, , , ,, _L , .'- .L • L -.. " Intlection Point _ L _-1--- L --1--- L -----j 1'- (c) Equivalent Series. of Portal Frames s 1',- P,- P,- "-nnn "-nnn- -- -- - statical!) delermmale because II IS oblained by inserting onl 14 hm@:e I.e one hinge 10 eal.:h of lhe 14 members IOto die frame hlCh IS IOdelenninate 10 Ihe 18th degree. Thus the mdelermmaq of the imphfied frame of Fig. 125 b I 1 CHAPTER 12 Approximate Anllysis o' Rectangular Buildino Frames IS 12.5
  • 157. -1EC1'IDI1U -.... ....... 'DIIiI ........ b. a. ( )" I h t'!C.rlldlr 1/ Q --Cl----<c__ '+ II CHAPTER 12 Approl mite AnlfysiS of Rectangullf Bulldmg Frames RG 12.6
  • 158. - 12 1 4 Q 12 60 M 1EC1JoII1U ...... LIWII' II. ............. EF 0 F 0 EM 0 - CHAI'TEII 12 Approximate Analysis of Rectangular Building Frames ~ t-r::jZ ~-C---<; 1T~ ~ r- I:':. ~ , ~ - , - " f- b-' ~ 0 "~ § ~ ~ :.-ff"_ '" - -g (: ~ ", " ~ f0~ -c a 0 ~ ." ""tiJ '" '" 0 - ~ "'c v l .. E "v :; ~~ " ~ '" z~ S ~'-V:. ~ .. ~ ,*p~, v. v - :f: 0 ~~~ Z ~ ~
  • 159. RIG"" _ 60 IICI1OII12A AooIroIoIor'-nl' ••, : , , , / / / / I Centroidal axis ,~' Tension ew... I I , I I / f /0I r; 0: I ,, - - 12.4 ANALYSIS FOR LATERAL lOAlIS-CMTuvER MEnIOD FIG. 12.9 Column A /UI Fon e We begm the computation of column the upper left Joint (i from the free·bod) diagram of Joml G 128 e l,e absent: that the aXial force m (olumn DG must be eq site to the shear 10 girder GH Thus the aXial force at the upper cod DG I Q n I k By applymg L: F, 0 to the free: body of obtam the aXial force at the 1000er cnd of the column to be Q the column DG I subjected to an a:tlaltenslle force of 1 k ADa.I Temammg second tory columns £H and FI are detenmned ldenng the equlhbnum of Joml 1/ and I respectively ~::: ~ rces fi r the first tory column AD BE and CF are c equlbbnum con lderatlon of Jomt D £ and F respecbveJ thus obtamed are hown m Fig 12 8 r / H I , The I lal JOTle h'.'.lT .lntl Ilhlmcnl at the nghl end H can now be arpl~ mg the three cl.jUlhhnum CqU.I!lll I1S to the fr~ bod) of girder '2 e 'ppl}lng L: FlO. C oblam QNe.• = 7 5 k . From EP obtain 5 I k .md to l'ompule fH&. l,C appl} the eqwhbrium .. L/u II 15 IH ,-+130=O In otc thallhc guder end mlmcnh. hll and JtHC , are equal 10 hale the same dm:("tion. Next the end a"lion" for girder HI are computed. The equili tlon" ') Ft ~ 0 and L I 0 arc fir"t applied to the free body ofJ 11.8(f} to obtain the a'iial force QHf ~ 2.5 k - and the moma 15 k-fl ) at the left end If of the girder. The shear SHf - 1.5 k IS tbaa b} dlidmg the moment tlll by half the girder length. and the three equatlon~ are applied to the free body of the girder to obtam QIH Sill 1.5 k 1 and 1fH 15 k·ft .:> at the right end f of the guder 12.8([ ). All the moments and horizontal forces acting at the upper nghtJ no'. knol.n so I.e c<tn check the calculations performed thus far by L Fx - 0 and L .f ~ 0 to the free body of this joint. From the Ii gram of joint I shol. n in fig. 12.R(f). it is obvious that these eqwb tlons are indet:d satisfit:d. The end a"tions for the fir~t.story girders DE and EF arc similar manner. b} starting at the left joint D and workmg acrou The girder end a"tlons thus obtained arc shoy.n m Fig. 128 f CHAPTER 12 Approxl_1e Analysis of Rectangular Building frames f I'Sll t,.' .:.m .lbll ~c that 10 order tJll'r IIll )1 "'lllt (, Ig ~, . ~ . " (II thl 1,'lrdcr ,.'nd moment M H m 1 lll1enl ll.JluhhnUill Th If '" I I '·~-II ll']umn l."nd llHl1l1l."nl, us H I kmJ IN'(NI I,) I 1<,; I k . d . I 11'11 IlIl Illtnt (, bUI .1 r.; )( "ISC lrectlon at thel(luntenk,I..ISC t In'( I glrJlf (,/I - J h 10 Jt:tenmne the ',fda hC,lf , II (' (,:oml cr. t e moment equi the kfl half III ~lr&r (,1/ I H'lll the frcc-hod)" diagram of girder ,h·'t the he IT !tlrn' " 11 mu~t act downward WI"8 "lIOX'" k - / / . . t .... 11411 It oan dl.'l·]llp ,I "Xluntcrcloc WI mtude (II , - "'. h I • / •• 'th Internal hlllgc to h,llam.'C tee ockWimtuuc' I aouu " I Thus -
  • 160. A -IQM) QBI ~ 0.125 QAD (d) Section bb Q H - 0 I.:!'i Q(Xj Ie) SCl,:llon ad (e i,, L __- x y I- 15 It -1- 15 It-l ~ M'H~18.9 HV-- 6,R~ ' : Q Q(iH - G M ~189 HG 11(; . 5~1f = 1.26 SHG =1.26 Q r CcnlfPlJ 1'1 .tllumns CcnlrC11d of Ctllumm r- 1bb711-l Gi, __~_--,--;Hr-~_-;l ]OL-r- T12 ft ,0 L _ ~D!!.- -'-1rE'-~_--j F+8ft ..L - 1. fl ( ~ '0 ft H H 8 FIG. 12.10 I" f h) Stmrhlll:d Frame: CHAPTER 12 Approximate Analpis of RectanGular Building Frames • 502
  • 161. - 66 k The columa axaal forces arc hown m FI diagrams of aU the members and JOlD G"der Sh ar and Momml KnOWIng hunn can now be computed by OO1I5Idmng eqwbbnum m JomlS Starttng at the upper ltfl: JOint G we L Fr - 0 10 1he free body of Ih,. jOint F S" I 26 k .ttb< Idl end of 8mler GH TIl< detennined by muluplyilll 1he bear by ba the .",I.r1en~ih: It Q B I1U1lmi,aI m"""mls M. 0 Subs lUual Eq and 4 b1a a 60 II 875Q 50 0 QIX. I 26 k Q£" ~ :q 26,67 QDG 0.125Q1>(' "'-HQff ;~ ~~ Qo () 875Qo" 60. () 12~(h 1(1 .unp d F elm Th~' IInphfi..:J lraml: obtained b) insertlng bln~ at midpOint I all th~' mt.'m~f' (If th..: gi cn frame. I" shown I ~ I l'l (. IU11111 4 jJll~), To l-ompulc J.ial forces in the column oftbe _.,. tory of the tram~' "t' p.l .in lrn<lgm.ll'} -.ectlOn aa through the IDtcmal at tho: mlJht'l~hb 01 tolumn f)Ci. Elf and 11 a.. sho"" in Fig 12 I free-body Jj:tgram llllh(' pt1rtilln llf tho: frame aboe thi.. ~t1on is shown I~.IO r.:: Bel.aux the CUI!< the- r.::olumn~ al the Internal hinge&., 0111 nal hear:'> and Jlal fllfl:r.:: but no Internal moment.. act on the free body point "he~ the l'l,lumn.. hat' been cut. " ....uming that the cross~sectlODII of thc wlumns arc cljual. e Jetl"mlinC !he kx:ation of the cenlrold of tile columm from the left column OG h) the relatiomihip I{O) + 4' ]01 t 1(50) - - - - 26.67 fl ~A ~ 15 11;;- 095 (8' Support ReactJon.. ':- ---lH:.-__-;' 10,- ~,,-kD------+.E,...----j F The later.t1 load.. art' ,u.:tmg on the frame to the right. !lo the axial force umn D(j "hir.:h is to the left of the centroid. must be tensile. whereaa force.. In the columm 1:'11 and f1 located 10 the right of the centroid a.. ho" n in f-lg, 12.1 U(c). AI..o. since the axial forces m the are a....umed to be line.lfl) proportional 10 their di!ttances from the cell rdat(ln~hlp) bet'Al"cn them C.lO be c'!ltabh'ihed bv means of the Similar shl)"n m Fig l~_I()(v. thai i... • By swnmmg moment!< about the Ielt Internal hinge J. lAC lAnte L If) lJ III 6 Qw]O Q" 50 0 SUMlJtutlD~ fq I and ~ lAC blam . . CHAPTER 12 Approxlmlte An.'ysis of Rectangular Building frames FIG 12.10
  • 162. 117'ra'•• 15 ft 111.,,2.2 4m 6m6m AG. P12.1 IIII1 III11 f) E F 30 kNfm Section 12.2 12.1 through 12.5 Draw the approximate shear and bend· ing moment diagram.. for the girders of the frames hown m Iigs. P12.1 through PI2.5. PROBLEMS ( ""/11 1'11/1. 1/f lid ./1 af I ilh the f:mkr llll,.lnienh now kn I I on bI: dd...·nllllll·d l'l nll1'ldenrlf: mument .....ull,......1 limn nlt1lll... n , ... ......~ .. lOin!) HI,; 'mom' JI lh... s.......·,md 'hlr~ and .Irpl~ tIlg ~ 1 II to (he' fi ( '. I' III C 1 (ltol.un 'h...· Imllileol .It the uprer end of columnJ0m',rle-~ . tx: I I~ 'I) l·lt· fhe heir al tht.: uppt..·r ...·nd III culumn DG I puted b l!' I ~) h lit the t.:11lumn hel~ht thai 1 115 k 0"" llul "I, mu I. d h1 tht: nl!hl 0 it can tlec1op a c10ckwl to ~,jl.m ...-c me l'lmllt<;r...kll:kl--e end ml'ment "up, The ..hear and moment 10" ....T...·nJ narc lh...n ddermw...d h) .Irpl 109 the equlhbnum equations Oand""'" / UtothelreePlxiyoft:t1lumnDG -.eeFig.12.IOe Q moml;;;' nd !oheJr'o lor column '-"If .tnd fI an,' I,:omrmed In a similar thaeaft...·r [ht.: rnxedure IS rcpe.lted to detemllne the momenl and th...' fir 1- I ,t} ...·0]umn.. -l n Bf. ,jnd C1' '>t:e Fig I~ IOlIl , Gmf. T A 101 / OTt I e bcgm the computation of girder aJ.laI ft the upper lefllomt (, prJ) 109 L F, -- 0 to Ihe free-body diagram of hlmn 10 bg I~.JU e "e hnd the al.lI force in girder liH to be 68S k pre. Ion. The i.Iial force for girder III i... delemlined ...imilarl) b) consideriq eqUllibnum of J0mt /I i.Ift....r hlCh the eqUllibnum equallon L FK plied to the free b<xi~ of the right Wint I I(l check the calculation The force' for the IIr 1-..IOT) glrd....r.. DE .md EF are then computed from librium COiNder,ttl0n ofJOinI f) and F. In order. The axial forces thus 0 are ,hon 111 FIg. J:!.IO.f). + LE, 0 10 20 95 15 5.54 0.04 ~ 0 'LF, 0 7.5R + 0 95 + 6.63 0 ''0 L II 0 W(2ISJ 20116) 75.1) 7 SRI50 • 1202 - 0951201 +44 3 RW<f/flf/ The force' amJ monU,'nb at the loer ends of the first umn,.fD, BE .1I1d CF repre'ent Ihe rcat:tion.. at the fixed supports A B re'~diel)'. ,1'0 ..ho"n 1I1 I ig, J2.10(g). Chc'(klnfl CmJll'urlllio!l' To check our computations, we apply the equilihrium equatlom (0 the free body of the entire frame (Fig. J2,10 g In Ihis chapler. c hale learned that In the approximate sLaIIl;all} mdetermjnah~ lruclun:... t'-o I} pes of simplifymg 8IlUIapi are commonly emph}}ed: I assumplions about the locatlOD tlon p()ml~ and 12 a ~umptjon~ about the distribution of fo 508 CHAPTER 12 Approximate An.lysls of Rectangular Building Frames SUMMARY
  • 163. - D......... I " 2 G H F G D't--~""_'-'E f E 12 II .i. '-A T12 II 6._ A":'- 81" c,,~ 0< 1--8 m I 6m,-+-,--'DR, mm----l ... Pl2.12, Pl2.2lI 2OtH--;/ ..,Jr-_~K __~ .,.. 40tH -t.E~--:F~-~G~--~H4 ~ A 8 1--3011 --+ ... Pl2.11, 'l2.21 ... 'lUI, 'lUI I 15k-i---_I--OO+---..J K 8m c8 30 kN _.r.G~=~Hii-_ ....1 Ii====::'ll==--~FI- 40k T 4m 50 kN _II,.=_~_-"":!'I +Oli E F 4. A - L- S- ~ C~ • .1. 1-6m-+-6m-l X()_~D=======iEr_---~F,N Ir FIG P12.8, P12.16 16 It l~,- ~ ""~ I-24ft --}-24II--i FIG. P12.9, Pl2.17 FIG. P12.10. P12.18 4() l.~.-, _-rr£=====__.;;F 20 I. 5eclIon 12.3 12.1 ","",,1112.13 Do:h:mllno: the arpro~imale sl'le rs lind m(lmCnl for all the memben of lhe Shll"O mIlt; PI26 thn.)ugh PI2.13 by uslRg the nu:thlx! B -;:cF===--~ D D snk-if=====_4E;.. _ FIG. 'l2.7, 'l2.15 Xm I() It B :W ft f---8m-+-sm~ A fIG. P12.3 : k..:tl I I I I I I I I(J H .BAt 10ft n I I I I I I I [ f ~ _ CHAPTER 12 Approximate Analysis of Rectangular Building Frames 8m 8m nj;=!=='==!,dd=!::1s=g,Jd=;/F T I 12m ~d:;I:;r=i~IO:lt":/ml::r:dbr;:J(, I I I T I I I H /-,- J:I::I:I:i;:I:Jc'"O:lIN:"mi:rjI I I I I fJ I £ f --+- FI&. P12.4 N.'l2.5
  • 164. 111 A 0 IIfCT10II 11 1 -...- _ ....... __ II •••,......._ .." -...-...- - I» To determlDe the m1undant C by ng :::::~;~ we rem e the roller support C from t ill 10 ,:00'_ It IOto the detenDlDate cantlie' r bea I rmlnote beam I reti ned I the P~~~::;'1 Iben pplicd as 0 unl<o I n gIVen Iemalload p 2 k ccanhedctcmUned US( unl<n load C mg the prinwy same I of the C the upport I 13.1 STRUCTURES WITH ASINGLE DEGREE DF INDETERMINACYStructures wIth a Single Degree of Indeterminacy Internal Forces and Moments as Redundants Structures with Multiple Degrees of Indeterminacy Support Settlements. Temperature Changes. and Fabncatlon Enora Summary problems 13 Method of Consistent Deformations-Force Metho 13.1 13.2 13.3 13.4 In thl3 chapter c sludy a general formulation of the force (fleX! method called the method of ((111.';.1"1£'111 t/e!lmnClfio"" for the anal statically indeterminate !'>!ruclUrcs. The method, which was IOtrod b Ji.lmc'i C. Ma'(-cll In 1864. essentially involves removing enoup straints from the mdetemlinalC structure to render it statically nate. ThiS determinate struclUfe, hich must be statically stable rerred 10 as the prinwrr S!ru('/u/'t'. The excess restraints removed the gicn Indeterminate ...truclure to convert it mto the detenmna maT) struclure are called r('t/wuJwlf rl'.lraint!i. and the reacUons lernaI force~ a~~ocialed .-lIh lhcC rc~trainl~ are termed redundan redundanh are then applied a... unknQ n load~ on the primary and their alue!'l arc dctermmed b) ~ohing the compatibility cq ba-.ed on Ihe condition that the deformatIOns of the pnmary due to the combined dTcct or the rcdundanls and the gIVen loadmg mu!t be the same as the deforrnation~ of the original nate tructure Smce the independent ariable.. or unknm..ns in the method Mslenl dcformallon arc the' redundant forces (and or momen must ~ determmed before the other respon~ charactensh pla('cmenls can be caluated. the method is da~itied as afi In thi" chapter e fin,t ded()p the anal}~is of beams. Ii tru llh a lOgic degree of indetcrminaC) b) usmg the 510
  • 165. IIIIIfl:IIlIIII.1 _ a " ".._" of the )lnJJIary bea due and lhe redundanl m be t rnmate beam ppon C lhe IOdelenm..1e beam Prt1JllIry beam d tho lhe rodundant C m I be f tho pnmary beam due I t ~;~~~:;~::f.COD'eDlenli by pen ft<ct,on due to tho 1 VldUl1l on tho beam thaI I <1 10 which l1co and <1 repment '"::::~I end C of tho pnmary beam due t II dant C <a<h actmg alone n tbe be m :~~~:I~~used 10 denote the deflectIOns <1 nd <1 first ubscript C ,nd,cates the Iocau n f subscript 0 I used 10 IOdieale that <1 loading, whcr<as the second ubscnpI ( lhe redundanl C Both of Ih deftecll If Ihey occur '" the duectlon of lhe redundanl be upward a hown '" FIg 13.1 b Since the redundant C I unkn wn It I D 1 <1 by first evaluatlOg the deft<ct,on al C due I a~um~:~~te dundanl C as shown m FIg 13 1d and then III lion thus obtamed by Ihe unknown maglll f lhe <1 C (0 Bendmg Momenl Diaanm for Indetenninale Beam Ut.ft) '00 20 [---==::;IO~-,._ :it A B C Bending Moment Diagram for Pritury Beam Due to Unil Value of C (k ftIk) 120 Bendmg Moment Diagram for Primary Beam Due to External Loading (k·ft) + o + ________----~cc =OJ in. B Ie xC 1 k Ik dl Primary lkam Loaded ~ith RtdundAnt C p- J2" "*J~''''''''-O;_=_=__=_~!===-=-,5'''',-~~ II .. ------- t c I ft I 10 ft --; >----l. _ ~ ft --~~--: E- ~OOllk...... 1,=512,"~ I a lndet:ennU'lal~ B~am II&. 13.1 (r: IPrimlll) B~am Subjected to External Loadinl1 1.1 =21) k-ft 0-(. IA L p= l2 k I e ~ " .1 0 ,-.;;:- ---- "', '., I --- B ---------- A C I'b Prinwy Beam SubJ~cted10 ExltmalloadlRg and Redundant C A 32k A - -E- I,,-A---~IB!!.......- _ _4:oi M I kl'l I A 2k ( =10 k p= -'2 k , -E- f!lIA_-==~IE...B __---"C AI - '20 ".fll. --------------- :Feo= ~1 '" " =32" 512 CHAPTER 13 Method of Consistent Deformabons-Force Method
  • 166. kA Moment as the Redundant In Ihe roregomg analysis or the propped canul.. or FIg 13 1 a we arbltranly selected the vertical reacUon at roll r uppon C t redundant When anal):lng a truetuT b /h m !ornuJt,ons Itt' can choose onr suppor', a,1I n , ml Tiki menl as tht redundant prOf 'ded thai Ih , mo I Ih (':'::~::~ re. tratnl from the glt n rna terminal 1m iI sITU turt that sIan 0111 Jetem,mat and t CODSldenng agam the propped cantil which t redrawn on Fig 13 2 a we can Iha traml comspcmdmg 10 the honzontal labcall unstable Therefore A cannot be used H_ e.ther or the two other t ppon the m1tmdanl Let us COOSIder the anal or the beam menl M the m1tmdant The actual arbitran usumed I be coon lock obtaJD the pnmary beam. we rem the ~:::..:= end A by replaCIng the fixed suppon bi A An A C 0 A A 0 + A C 32 I 10 22 k ~ M. 1I.a + lI.c C 320 ole that the second subscript 0 i eXlernalloading only Fig 13 1 c wherea the nd u notes reactionS due 10 a unit value of the redundan C FI 13 I Slmdarly Ihe bendlOg moment dIagram ~ r the on term beam can be obtaoned by supenmposong the bendong 10 ment dl or Ihe pnmary heam due to external loadong onion the bendi ment diagram of Ihe pnmary beam due to a umt value rredundan multiplied by the ..Iue or C The bendong moment dIagram r r I detennmate beam thus constructed is shown 10 F18 13 1 r IECT1OIll3.1 .-...- ......... Degree .. In....' .,=, 111 LF 0 A 0 LF 0 A 32 10 0 ~ L II. 0 M. 32 10 1020 0 After Ihe redundant ( ha • _ oth u=> mputed Ih re~;;:::,,:~er response characten lJcs of the beam n be :PIOYJD.g uperpDsUlOn relatIOn hips ImJlar In3f< nn t perpo IUon relaUon hlp expressed 10 Eq 13 4 Th can alternatIVely be determlOed by uslOg the UperposUI see FIg 13 1 a c and d 5(.12)(20) , 48(3().OOOj(512)/I44 ~ -0.25 ft 5PL' 48£1 j" (see Fig. 111 (el) in "hleh a negatle sign has been assigned magnHude 01 6( () to IIldlCalc that the deflection occurs in the do direction that 1. in the direl:tion orpo~itc to that of the redu Simllarl} the fll:lbiht) coefficient !e ( is evaluated as L (21))' lEi J(lOO(){) (512, 1-14 0025 ft/k ~e hg, 13.Jld B) uh,tituting the c.prc...sions or the numerical oj .1({J and f« mto [4. 115, e dctenninc the redundant C C (5/'/)(lEi) ~ P _ 10 k 4Xt./ I' 16 The poiue ans", r I) (. d ... I,; (r m It.'<lte:> that our IOllIal assumpb the up",ard dlrcruon of ( "'.I~ corre("t "' Ilh (he rcacllon ( kno" n the three rC'maining reacUo be delcrmmed by r I h h ..hod a p) 109 tel rl:e e"lulhbnum equations } of the mdetermmate beam hg. 11.1 e) huJ r l"IU l" .1 Ot'ncdlon of A(() at end hll11 III ~'li! I .I("· SIJlI.:~ the: dd1c,:tlon <It C In the original minah.: ~an; I It:ll. the n:duno.IIH foro:C must be of h131:.!niludc: ((l pll h lhe: t,"no ( b.ll:)." 11110 Its onglllal position by In!:: ~dn urilrd ddlt'I,.'llt1 11 llf { I .It em.! C of the primary beaaL e,;IUJh.: rhe: dkd l.lf ( llll Ihe ocJrn. f' l:omputc the ftexibJln I.:ienc j; { hh.:h I the: ddkdllln at C due to a unit alue of the Janr fig IJ I J Sm(c ,Uperptl..llWn IS alid. deflection I prorx.rtlOnal {("I load: thaI 1'•. Ifa unll load cau~~ a deflection then a load lell HOlt: a" OlUl.:h 111 cau~.t deflecuon of IO/ce Tb up<lrd of magnitude C cau"c:-- an upard defledioa C II, at cnd C tlf tht: pnOlar) be.lm. Smce (he upard (r. cau't'd b~ lhc redundant C mu,,1 be equal 10 Ihe d dellL'dlon j, () duc 10 (hI." ctl."fnal load P C Hile C fcc .1co m hld both ddlcctlon... ftc and ((). arc a:>:>umcd 10 be JJOSI ""rd. 'ote that Eq 11.6" <quiaknt to Eq,. (134) and (13 5 prelou,l) Since the primar} be.lIn I... "taticaJl) determinate the dellleclliI .1l () and Icc can be computcd b) either u"mg the melhods p dC""-Tibed in Chapter" 6 and 7 or by the beam-deflection Ii gien m"ide the front cmer of the book. By using the beam-dcllalll fonnula.... C determine the deflection at end C of the primary beam to the etemalload P(12 k) to be CHAPTER 13 Method of Consistent Deformations-Force Method
  • 167. 117 120 k fI PracedUfI far Analysis Based on Ihe foregomg diSCUSSIon we by-step procedwe for the anal)'$l of tern U WIth a SIngle degree of IDdetennlnacy I. DetenntDe the degJee of indeternllllCY lhe degJee of indetemtIDlICY area lWe IS mternally indetenttJDllte then end The of mternall iodetemt "::~ WIth mulbple degrees f inde .. PL 16£1 L 3£1 from whICh ote lbal a nesal. e gn has been •••,,_... because thi rotatl n occun ID the cl to the counterelockWi ducetl n a 13 2 a By ubsulutlDg the numen compaltbihl equallon Eq 13 7 w nte o0075 00000625 M IEC1tOIlIU Duc1urIo _ ...... Dogroo" .._ ......._ _'1 The lopes and f an be delleclton formul wide tbe fr • B + .....-------...... ---,M ......... C " I k-fl t I A'A = 'Ri k (c) Primary Beam Loaded with Redundant MAo A~= 0 p= 32 k 'd>=0~ I .i.9140 8 ......' I ----- ---- t A",= 16k C.o= 16k (b) Primar)' Beam Subjected to External Loading p= J:! l- 8. = 0 I c "± I. --__ B _---- 1I. If. t' --------- tA, C rlOlt I 10,ft--=j r---:--L= 10 ft ---~ E = .10.000 k"', 1= 512 m.4 fa) Indetenmnate Beam Fig. 13.2(bJ. :'<Oote thaI the "mpl) ,upported beam thus obtained call) determinate and ~tablc. The redundant Jf~ is now treated kooy, n load on the primary beam. and ih magnitude caD be from the compatibilit) condl1ion that the slope at A due to bmed efftXt of the ex.ternal load P and the redundant M... must The primal) beam I!I subjected separately to the ex p -:. 32 k and a UOIt alue of the unknown redundant M,t Fig. 13.2 band c respectiel}. As sho"," in these figures !tenh the slope at end A due 10 the eXlemalload P whereas the flexlbilit} coefficienl thai i~. the slope at A due to a unit the redundant f.... rhu the ~lope at A due to .lA equals 9.. Because the algebraic sum of the slopes al end A due to 1b8 load P and the redundant .1~ mU!lo1 be zero. we can e prell paUbllll) equauon a CHAPTER 13 Method of Consistent Deformations-Force Method51. FIG. 13.2
  • 168. 518 CHAPTER 13 Method of Consistent Derormillons-Force Method 1 I - I,', the: rc:dund.1Il1 III imph that theIX JlIC' 111.ll!ilI Ull , - II • d " 1 l:Mrcd, hc:r....l ....1 neg,llle value of the113 .1 ume: • , nlluJt' III mO......lll: Ih.ll the dClu.11 '>('n~ 1 oflflO!lte to .1' um J lmuall~ R,'mo t:: Ihl re Ir.lIlll corrr.:,pnnding 10 the J. b h J!1.'n lIllkll'mllll.lll ,tfuetore 10 0 lam t c tructun: 8. Dr.J a dl;Jgr.lm of Ihe pnma~ ...tructurc ",ilh only the 4. nat huhn!! .Ipplled 10 it. Shh.:h a deflected hape of the lure. and h(m the ddkcllon tor slope. at the point ofa tlllO and In !he' din.'ction of the n..-<lunda"t b} an a ",mho!. b. .CL dr.l •• lh.lgram of the primar) structure WIth umt alm.' of the redundant applied to It. The uml fOlQl moment O1U,1 ~ applied in the positle direction of die dunJanL Sketch . deflected ,hapc of the structure. and <111 <tppropnat~ s)mbol the flcxibilit) coefficient represen ddk:lion (or ,lope) al the point of application and ID the rcxllon of the r~dundanL To indicate that the load as well rc'pol1se of the qructure is to be multiplied by the red hO the redundant pre(,;edcd b) a multiplication sign to the diagram of the tructure. The deflection (or slope IOl.:ation of the redundant due to the unknown redundant eq the flexibility coefficient mulliplied by the unknown rna of the redundant 5. Write the compatibility equation by setting the algebraic sum deftecllom (or slopes) of Ibe primary struclure at the locatton redundant due to the external loading and the redundant the glcn displacement (or rotation) of the redundant support al.:tual mdetcnnmate slructur~. Since e assume here that arc un) icldin!!. the algcbrail: sum of the deflections due to lernaI loadmg and the redundant can be simpl} sel equal to obtain Ihe compatibility equation. The case of support m is l.:on idered in a subsequent section.) 6. Compute Ihl' ddk·(:tion" of the pnmary structure at the I the redund.tnt due to the external loading and due to the of the redunddnt A detllXuon is con,idered to be posmYe the arne s.c:n'tC a, that a sumed fOf the redundanl The can be determmed h) usm~ an) of the methods diSCussed tcrs 6 and 7 for beams oIth l:onstant nexund rigldlly EI all} comement 10 determine (he~ quantities by uSing the formula ~!Icn In ide the front l:Oef of the book w~ flecllon vf trusse and fram~, can be comemently com u Ill@: (he method of lftual ork 7. Sub IJlute the alue of del1ection~ (or !lolopes II mto the !,;ompatihiht) l'quatlon and solC' redundant
  • 169. CHAPTER 13 MethOd of Consistent Deformations-Force Method AIls Ans AIls AIls /. L L EI 8o and I 1ECTlCJII13.1 ....... __ 11..............._" lIL 2EI + A 0 A --Ie~) )lI 3M A 2L A 2L lI. - lI- LC2~) M II M 2 2 The reactions are shown m Fig 13 J d The positive an wer for B indicate! that ur m ward direction of B was rrect R II The mIlalDmg mu:ttons f the indetmnl detenmned by superpoilbon of lhe reatbons f thc external moment" and the redundant B hown In FI pecUvely Shrar and Bendmg Moment Diagranu By wung the reactl ns the shear nd bendm8 moment diagrams for the mdelenDInale beam re nstructed The diagrams are shown in Fig 13.3 e Ana DetcmuDCthe reaenon and draw the sheara~~bc;::::~~::~::the bcom shown In Fill 134(0 by the method the ractlon moment at the fixed upport be the la mhuJ. 0 I IUdtmdant 8 UbsblubDg the ",...""" I.. mto the compaubilit equation Eq I determl ML 2EI ID which the IlC b e gn for 4 mdlcal that th I~:::':••::; downward dlrectl D that I oppooJltc t the Up" rd d redundant B IExample 13.2 M 8 Shear diagram ", '"., = ldl50pport ReactIOn for Indelenrunate Beam Bending momenl diagram (e) Shear and Bendmg Moment Diagnum lor Indetenninale Beam /.. _______....nM = o-f- I, I t 8 =~ Solution Dt'f/fl'( oj Incit (( rmin{/('y The beam is supported by four A, A f~. and B, It ig. D.J(a): that is. r 4. Since there are only librium equations the degree of mdclerminaC) of the beam IS equal Primun &um The crtkal rt:J.l,;IIOn B al the roller suppon B be the redundant. The sense of B i) avmmed 10 be upYtard as 11' a The pnmar: beam obtained b) remming the roller support • glcn mdctcnninatc beam i hmo,n in fig. 13 31h) Note that the lecr beam IS tallGtlly dctennlOalc and stable. ~ext the pnmary JC:ted separatel} 10 the external mumenl 1 and a unit value of redundant B as ho,n 10 fig 11Jrb and c) respective) hgurc. 6.0 denotes the defta:tlon at 8 due to the external moment ,•• denotes the tlell.lhJ1l1y l;octhl;lent repreloentlOg the dcfteeuon • umt value 01 the redundant 8 Thu the deflection at B due to redundant 8 equals JuS (ompatlhllJl Equal, n The del1l'Ctlon al !lUpport B of the mmate beam I zero SO the algehral... urn of the deftecuons of the at 8 due to the eternal momenl f and the redundant B m Thus the l:Ompattbl1lly equation l:dn he "ntten a 6" JaBS 0 lkfi llum 1 Prrmar Sturn 8y u 109 the beam-deftectiOD btam lhe ddkctl n 6s( and Iss 10 he 520 FIG.13.3 cOld.
  • 170. =----~- I IIICT1CIIll1.1 - . - .......... liliiii"'"011'111...._ 20ft EI t4H crt (b) Prinwy Frome Subje<1ed '" E>IaMI LoadiDB M_ (c) l'I-rFrome SSuldljj"".'IDlJai1_oIIPW....Ir .... A:,-·..,..._. = 30 ft (3) Indctcnninate Frame I I I I I I I , .A .~.:. B ----------------- C_1 + I ,,,,,, Ik_ A ~ BlFe=!==!:d....lU I I I I I A.-_ A FIG. 13.7
  • 171. - Bam ft I k 9~-----'1'-----41 gtK 5719 B B 5719 (b -1Ill MomeDl Diqnm I CoaIin A A_,- A _OJ-Iliapms for - . . AS and Be -"Ia.z _"'-__."'.'__ ~12~ A 714 r- 16 I B l) ~&d andSuppon _ _,for 1,........_ fiG, 13.8 (eonld.) ,!lIUl EI 12 k Tangent ju!>t to right of B B I kAt ~~~f,§BB~"~'::====::,~C~ f88R .... ----------- "!lQ ~ U EI (d, ConJugale Beam for EAtemal Loadmg (el Primary Beam Loaded .....Ith Redundant MB Tangen! JU{ to left of 8 '0l A,'F."J,I==;;,9=;~5f:E======.:'::ilC..........._..__ .!L.. ....... ,/ ,"0 " A11'F.==r==!f=; ....... __ pl!J- ... TangenlJust III Tangent JU' 10 left ot R right of B (bl Primal) Beam SuhJected 10 External Loading + 1:../ - l'lln,tant ,a) Inddl'mllOale B~am l- EI , i , Ji-C ' " ,fJ £1 i7 ii Ie) ConJugltlt Beam lor Unn Value of RedundantM. Method 0' Consistent Deformations-Force Method Tangeot at R 12 .. UH uH ~ I B .'__ j c IO-)~I- I~;:r ,_0_",,-,,-;:._._-_._:0_-_:_·.....-11 :0It _~o ft ClW'TfR 13 flI.I3.lI
  • 172. Intlmally Indeterminate Structures As Ihe foregomg dIscussion mdtc:ate lructures Wllh a ngle deg f Indetenninacy that are externally Indetennmate can be an I b lecttng either a reaction or an Internal force or m ment the redun dant However If a structure IS Internally mdetenmnate bu ternaU detennmate then only an Internal force or moment caD be used the redundant because lbe removal of an extern I reacI n uch .tructure WIll )'teld a .labcally unstable pnmary lruclure Con'lder for example the truss hown ,n FIg I conSl.ts of SIX members c:oonected logether by ~ ur JO ported by three reacbon c:omponcots 1b 4 4 the degree of mdetermlllacy f the I 6 3 2 4 I Because tbe three reacI the three equal100 of equiJibnum 0 the ternaJl indeermiDale 0 the finl de tba member ban required for internal stability. To lIDI1 the rUSS, we must memben 0 be the reduodaoL Su tba the dia member AD be the redlUDdanl. spood"'l F. ben mooved ~ 0 the member D no Ioopr d$ When the priJJlluy II deli and gap member AD 953,33 k-ft' £1 10 k-ft'fkoft £1 420 + 533.33 £1 ()HO rd IBBL 667k-ft k-ft VFI and IBBR c- hom Eq. 13.9. y,e ohtam 118 5719k-ft fBB f881 f88R 6,67 +10 16,67 k-ft'fkoft £1 £1 B~ Ub!>tllullng the aluc:s of II and f 'I th "'DII!'". BOrd BBrellOO e~ equation fq. I, I() J y,(' dctemlinc the magnitude of the f11 as ~53 33 fT or The nc,ihility codlkicnt ISBrr:.J can be computed similarly by conJugate beam for a unit aIUt~ of the redundant .la shown 1U(el, Thu, OBfl +- fss I la -" 0 hll.:h l,ln be l,hcd for the redundant bending moment M. after change.. of ,lope" (}so and f88 n::1 ha c been evaluated. SIOl:C tll:h IJf lhe "pan.. of the primary beam can be treated "Impl) .,uppMted beam. the ..lop..:') at the ends B of the left and the ~ran~ C.ln be c,I"'II) computed by using the conjugate-beam method. conjugate beam... for the eternal loading are ..hown in Fig. 13 8 calling that Ihe "lope at a point on a real beam is equal to the Ihal point on the corre~ponding conjugate beam. y,e determine the rJSI and OSR at ends B of the lert and the right ~pans, respectively 4CO ~-ft' , 7 , d 533,33 k-ft' (}SL t" "'/ an OHR_ £1 Thu" from Eq, (13,8), we ohtaln CHAPTER 13 Method 01 Consistenl DefOrmations-Force Method In hllh Iss .mJ BR dlilotc thl' . hlfX'~ .It the ~nds B ~f the the: nghl 'ri~tn'i (l[ thl' i:'l".IIll, fe pc.:dlc1) due to the umt value n..'dund.tnt fB . The: lomp.lllbllll ('I.jU.ltl(lll I' on the reqwremenl IlllX' of thl' d. Ul,; ,:Uf l ()f the JI:.tu.11 lIl~ctcm:malc beam IS COD. at H. lhat I then: ,... nl' l,:h.lIlgc of ..lope Irom Ju~t to the left of B III the nght (If B. Thc:n.:h1fC. the algcor.tlc ot the angles be11W1l. tanglnt at Ju..t (0 the: left and at JU"1 Wlhe fight of B due to the IO.ldlOg .100 the fedund.tnl .1B mU'..t ~ zero. Thuo,. 536
  • 173. + (b) Primary Truss Subjected to E"temal Loadmg -FoForces D A O..erlap =fAD A.D ~/ lD the actual mdetennmale truss we conclude that the mm' £. D must be of suffic,ent magnitude to bring the ends of Ihe lIOns of member AD back together to close the gap To feet of F'D m closmg the gap we subject the prmuuy value of F'D by applymg equal and opposite unIt .,,,alIOlIdI pomons of member AD as shown m F,g 13 9 c ote lba sense of the redundant fAD IS not yet known and II arbi to be ten de with the unit ..,al forces tending to elo.... of member AD a shown m the figure The umt value the pnmary truss and causes the ends of the two po AD to overlap by an amount I.D AD as hown In FIg I erlap m member AD due to the ....I force of ma,piitullllI f F. (c) Primary Truss SUbjected to Umt Value of Redundant FAD -"AD Forces fIG. 13.9 (a Indetemunatt Truss /l;C;===7.i D r p A.~==~~B ~ .~~nI-Fon:e Method_ 01 eonll._....,~,,-- C q- ClIAP1'EII t 3..
  • 174. 10 1.67 m ,... - JE(21) - ~ (~J) (~') • (21~~) ~ 1275 k and In 167 5 m EI EI EI 15 III l'i IfJ '.nm FI Ifill reI l'i 10 r..?" " m' '·UI U 60 III :'00 k, , I" Ifl ·m - - ~ - • ~~E ~I 1M 21 EI 10 ,lEi D IRBI CI~) 825 k B) .:On ad nng the tqulhbnum 01 JOint 810 the er11cal direction H H' 8" 975 1275 225k t. GJ 15 111 C1~~) '2.5 kN 8" (~) 15 1lI C25)-97 k10 5 Slmllarl for member Bn JIIIJllitullt (1/ th" Rllllllltlalll By !Iub!ltituting the values IBfirel Into the compalihililj equation (Eq. (I)), we obtain 1 125 • ( 5 ) /•. () U U -225kN·m fBI tSIlL IIIBR • Thus Rt'(HI/II1J!j The fon:cs at the cnd.. of the members AB and BD tlOUOU bt,:,tm can no.... be detcnnim:d by app1)ing the equations of eq to the free hodle, of the memocrs hon 10 Fig. 13.IO(d By co equilibrium of member .-18 c obtain The Ill-Il'llllt) ...oefhl..h:nt I fig 13 10 ~ t: "an 'o("C that Thu'i In hlCh CHAPTER 13 Method of Consistent Deformations-Force Method JfC Ih~ 1,pc .i1 thl; end 8 -.1' the left and the am rl; p..·dlcl due hI lht: ntt:m.ll hladmg 8 ,1.: '1:11 lin
  • 175. CHAPTER 13 Method of COnstlten1 Oeforma1ions-Force Me1hod t F T ~.. It A n1, 18 V 4H 01. In, . pand at IS ft _ 54U £- - Clln~lant allodelenmnale Tru .. E F 1EC1llIII1U _~ ... Mol I •••, ..... TMLllU AB BC CD EF BE CF AE SF CE DF .. I4(1 £.."i+---I/.__-:&.~=o::D 30 BI 2625 CI 2625 45k .10k I 35 lb) Primary Tru..~ Subjected to External Loads-- FoForce'i + E F F. Next the ftexlbdily cocffiClent exprallon see Tab) 13 3 k I computed n to to IA =40 (; PriIlW) Tru s Subjected 10 URn Ten lie Force 1Q Member CE. -"c Fone fIG. 13.11
  • 176. B ......__..-----0 _ ..... Loadcdwil~.R....Il~...te...... D, + d Priawy Bam Loadcd th - - + + (b) Primary Beam Subjcdcd to ExtmW Loodt A //-----F-h----F------t;.r-- :A tB C D--..::.':%l.... I « Prinwy Beam Loaded WIth ReduDdanI B FlG. 13.12 Th(' rot,thad 0t (")0 ..1 I('nt Jd('1rnl.ltion.. J~ eloped in .the ""eadil.. uon.. Itlf dn.II~ZJng ..tllIl'Wf(' llh iI. "mgle degfc~ of md~te'rmtiJu.., ('.1 I" lx' (' r..:nr.kd w tht.: I~ 01 ..tructure~ uh multiple Inde;('nmnal..·). C.'n ider. for n.lmpk. the four· pan c~ml1nUOUI Uhll':ted III .1 ulllforml~ dl..lnhuted Ill.ld ". iI ho~ In Fig I Th~ ~am I" "uPflI.'1rlcd b~ 1 "upp<.lrl reactIOn..: thus I1s degree tenmn.( ) equal to b _, 1 To .mal)7(' the beam. we mUll three Upptlft fe.ll:ll0n...1" rcJund,mh. Suppo~ thaI e select the rea("lion.. 8 ( .•lIld D ,11 the mtenor supports B, C and p.?Ctid~. to bI.' th(', The roller ...upport... at B. C and then reOlO ed ff(lm the gi en indetemllnatc beam to obtain the Ita detennmate and ,table pmn....) a......hon in Fig. 13 12 b three .tTl: no trc<lted as unknown loads on the beam. ,md their magnitude... can be detennined from the com condition.. th.1t the deflection.. of the primary beam at the I B. C and f) of Ihe redundant.. due to the combined effect of the e.ternal Io.ld I' and the unkllQ 11 rcdundanb B'. CI' and D m equal to lefO. ThiS i b~cau...e the deflections of the given inde beam at the roller supports B. C. and D are zero. To Clablish the compatibility equations. we subject the b<am ,cparately to the external load If (FIg. 13.12(b)) and a untt of each of the redundant> B,. c.. and D, (Fig. 13.12(c). (d). and speclid) A....hown in hg. J:~.12(b). the deflections of the beam at points B. C <lnd n due to the external load U' are dena t!1J(J, j,( (J. and j,f)(). respectld)' Note that the first subscnpt flectIon j, indilatt:-. the location of the deflection, "herea! the subscript. 0. 1.. u--ed to 11ldicate that the dcfleL:tion is due to the loadmg. The flclhility cocllkicnt.. representmg the deflections pflm.lI} beam du~ to unit aIUt:'i of the redundants are also double sUb..Cflph .I'i ho n In Fig. I~.12(c) through e ~ubscnpt of a Ilexihillt} coelhl:ient denote... the location of the and the s....-cond suh Tipt indl~atc'i the location of the umt load the detlcl:tion. lor e,ample, the fle;ubihl> coeffident j; B dt:flect!on at POJllt ( of the pnmary beam due to a unit load at Fig Ill2 c herca JB( d~note'i the deflection at B due load at C FJ~ 1".12 dl and so lln. Altcmalhely a ftexlbili Clent I ma~ al 0 ~ Interpreted .1') th~ deflection correspondiDI dundant i due to do umt ,alue 01 a redundant j; for example the ddk"Ction corrt.: pondin~ to the redundant C due to a UDJt the redundant B I-Ig 1' 12 c IBf denotes the deflection mg to the redundant B due 10 a unit alue of the redundant on. A deflt:lton or flexihlht} (;Oetlll:lent at the locatlon ofa con Idercd lO be po IIIe II II ha the Same sense as Ihat assUJDDCI redundant CHAPTER 13 Method o. COnsistent Deformations-Force Method 13.3 STRUCTURES WITH MULTIPLE DEGREES DF INDETERMINACY
  • 177. 1lCI1OII1U _ II 111 ." CHAPTtR 13 Method 01 COnsistenl Deformations-Force Metttod .It 'nllllll at ptlll1t B of the primar) beam wef lXU"'lIll:! llur. L: I I d I N'llli due. 10 the t:".tcrna 03 IS A the Jdki.'lll1n .It III ,-. : I", 12( ) I ~ ,., b thr.: d...·lkl'lllll1 dlll' to n, 1:-' ISb S (} Ig..-' c the . -. (.1' , It L~IJ .md the ddleclion due to D JUI,.' l) (I~/B{ ,1£ . d h 1 I , Ih the hl! II lklkX"tlon .It B. ue to t e combinedlu!I c u ' , (' - - II' and all (ll the rcdundanb 1'0 .lao -+ JBBB 01 me (".h:m,1 O,IU . . d . .. h d ·t1eclltll1 01 the: .lctu.11 In etermmate bcaaatspD ~ml.: let: , " ' f rt B t ",'rl' C' the ali!cbrau.: !'oum 0 theLU 2.1 at upro I · ~ . . h • 'am at B cqu.a1 to 7ero [0 obtam the com01 t e pnmLIr) ~ . • .I B t (.eC "DD, - O. :"exl. e fOCUl eqUJlIt1n. H() B I l , b . II .~~I __ Ie-nilon at p..ltnl C of the pnm,lr) beam; b) alge falea y ............ " t ("u' to the ctem,llload .md the redundants and byIIlxuon.1 U L: _ • • the equal W zao. e obtam lh~ second comJ>a:ubihty eq t (oB !tee ,,,,0. o. SImllarl). b) setllng equal the algebraIC '-om of the ddlecuon'l of the pflmary ~am at D the extemal load and the redundants. e obtam the third com.. equallon. DO t (o.B. IDC C t fooD, O. The three compa equalion~ thu~ ohtamed MC .IB01" (e.B t f" C leDD, = 0 .1'0 + fc.B. t .Icc C. + fcDD, = 0 :00 t hmO, • .lDCe!, +- JnnDy ~ 0 Since lhe number of compalibility equations is equal to the of unknown rcdundants. these equations can be solved for dundan!>. A, Eqs. (I) 15) through (13.17) indicate. the campa equations of struclUrc~ with multiple degrees of indeterminacy general ((Jupln/. m Ihe "iCnC that each equation may contatD more one unkno,", n redundant. The coupling occurs because the at the location of <t redundant may be caused not just by that tlcular redundant (and the external load) but also by some or the remainmg redundants. Bccau~ of such coupling. the com equations mu~t be sohed imuhaneously to determine the redundant". The primaf) beam IS SliltkaH) detenninate. SO its dcftectiom the external loading a... cll a, the flexibility coefficients can be b) u"ing the method, dl'l.:u..Cd pre 10llsl) in this text. The toW of deflections indudmg ftexlbiht) coefficients) involved m a compatlbllit) equationl> dt:pend'l on th~ degree of indeterm structure hom Eq 1315 through 13.17) we can see that beam under (ol13lderatlon. hlch is mdeterminate to the third the compatlbllit) equatluns cOnlam a total of 12 de6ections flection'S due to the eXlernalloadmg plus 9 flexibility C~oefficient:~:'= eer accordmg to I'HIU'1/ /em oj ru ,procul deflections JUI ISf JDB IBO and 1m 1<D. Thus. three of the effiCient can be ohtamc:d by the application of Ma well Procedure far Analysis Based on the foreg 109 diSC Ion Ilo by tep proeedu~ for the anal of truet 51 tent deformauon 1. 1. 3. 4. ..
  • 178. Example 13.8 550 CHAPTER 13 MethOd of COnsistent Deformations-force Method ~. nh.' 1.0Illp.IUhlhl) lqu.IIH1n for the 10c",lIon of each danl h ~ttlng. th.' .llgc:hr.IK 'lUll {If the dcllectlons or slopee tf' l ·{un.: due: III tht: ntanal loading and eachpnmaT) l dunJant l·~U.lI to (he "mmn dl pJ<Kcment. (or rotation at J n • Ix:alwn l'll lhl' .Idual mdctemllna.h: 'tructure n:Pl-ml~' numba of (omp.ttlbJlll) l'qUJUOIl thw. obtamt...d must be thl nurnlxr l)f n:dunJant . 6. <-ompute the ddll'I.:llllIlS and the tlc:ihhllil) cOhefficlents In (he;luhilil) CqU,IUl'lh b) u,mg. t c mel ods di lllU !) In thl' Ie t ,lOLl b) the, apphcatlon ~f. Maxwell la l:ipfl'l(.ll ddk,IIOIl ddk"l.:lIon (or f1clblht) ~~ClCD J(xation of ..l rcdund.ult I com,ldcred to be positive If It .tme 1.'0..('.1 lhal a.....umcd for the redundant. 7. Sub..Ututc the .IIUl:' of deflections computed in step the comr.ltlhilit) cquatiom. and sole them for the redund.lnh. 8. Once the redundant.. hac been determined. the other charal:teri,til:' (e.g.., reacuon, shear and bending moment gram... and or member forces) of the Indeterminate structure chtlu<tled either through equilibrium considerations or by pm.ition of the re"'polhcS of the primary structure due to the nal loadlllg and due to cal.:h of the rcdundants. lXtermme the fI:"J(:tlOn, and dray, the ,hear and bending moment the thr~e· pan contlnUOU beam ,hm..n in fig. 13.I3(a) using the con I tl.'nl deformations Solution D. n (' oj Imlt tl rmlllill I 2 Prl11ldrl Blilm The crtll.:al rcactlOm Band C at Ihe Intenor and C re pcdl"I) ar~ ~k"llcd as the redundant.:. The roller <l.nd ( arc thl'Tl remoed to obtam the pnman beam shown m F I the pnmary he.un I !>UbJCdcd !iCparate'" to the 2-k ft exterllll the UOit alues 01 lhe redundant Band (' a' ~hOn In Fig 13 t d r f'l"(tl I) (mlp,mhlhtl Iquau ns Slnt,; the detle1.110nS of lhe actual beam dt uprom B Jnd ( .ne ;ern e set equal to 7Cro the algebraiC del1t:tlOn at pomt" 8 and ( rc fIt.'Ctivel) of the primary beam d external I ad and t"dl,;h 01 the redundant to obtam the compabbili A, IlthB Jse C () 11 1/ liB he ( 0
  • 179. CftAPTfR 13 Method of consistent Deformations-Force Method IICTIlIIlIU ....... ""f1110••,1111...._. - Ans. 60 4444D D I k Ans II I Ie 16 B e 44k ~F 0 SJwar and Bntding Vommt Dio , The diagram of the beam are shown m Fig 13 1 The shapes ofthe hear andbendlD momen::::~~:::~:beams. m general are SImilar to th r. t th shown 10 Fig 13 13 f A shown lD thi figure neph erally develop at the mtenor supports ofconbn moment diagram I usually po$Itl e over the mIC1d1e bendmg moment al a tUnged uppon al an end of me beam m I generally Regatlve at a fixed end uppon Also the pe ment diagram lS parabolic for lhe pans ubjected to :=:~~~:~and It conslSts of hnear segments r. r pan bjected t actual values of the bending moments of depend n !he loading as weU .5 on !he lengths and lie> I gidi conbnuou beam Example 13.9 3111.J1IC 0 J 555 556C 0 1.111 III fl' EI 3.555.556 ft' f(1 - - - - E/ 3111111 fl' E/ 29'333333 k-ft' £/ foe 20 20 Bendmg momenl diagram (k·ft) 2911)1 133 • 15555568 291Hlll3 JIIIIIIB 8 F e fi tJ -20 Shear diagram Ik) (0 Shear and Bending Moment Diagram~ for Continuous Beam 16 By appl)int:! {dx,ell's la I"llnJtudl' if the Rt'dllIJdunl B ub'Stituting the values of the and Rexlblhty :oeth:lent of the pnm~ry beam just computed mto blbt) equation 'F:q 1 dod 2 e obtain D(1ft (lion IIJ rhl P"murl' Blum By the beam-deftectlon €I obtain As.13.13 wntd.
  • 180. CHAPTER 13 Method of eonststent Deformlbons-Force Metttod , tI EI H1Hm EJ 1) ~33 m U IEC1lGI IU ....._ .....-.........."IS p .----1...- b_--- A ft~I...A ~.~••_-i:o~,_-__- ~-_!!!"'-=-~Cl7- t ", I---L I El-_ .--.....• .._-~-----------~ o p I c Suhm' I &.I la lnll ,1 Imwtanl.:0U 1 tnr ( and 1. b 1ft R,,11 flOII I hl.: rern.llnlng f.. l~lhlll ~ III 11(1 he lktcnnmcd b three ..qualJlln ot yUJllh(lUlIl III Ih.. Irl.:C hl)dy nl Ih: md I mll lll~, . 1'1111( H 111( 1 111. f 82500 "'111 '-"XI ~fI I~ kN b Primary ..... Subjected to Exlcmal Lood + ~ ---------r;; 'i: ·-···-7"fA"t:=>c I~.Q; ... I I ~- I " 0 0 I~O I.' I~ 120 411714 0 < , ~ I " I 1'0 I. I. 10 I k m , nd& I 11 I. f ,. ,111110 ~ 1115 Primary ..... Loodod with RedundIlII'" + A~-------------"'~ r I Primary ..... Loodod _, t II
  • 181. - 10<0_-- (~ L I IO L E .. -..---_ ..-.._D$ IIIl:nalIIU ••__ 'lusw_...............,..,....., FlG. 13.16 (eontd. ---]______y-----tl111 D <too E I .. 0* E lie C ---_... - D ......... --- ... I I IC. D, Eo, + .'.'.-2L ......... ( h C :--=--:';--::.--:....--!-D~/DE!!.-_-d. II Ie) Prinwy Belm Looded with Red'mcI... L ( Ic II (d) Primary Beam Loaded with ReduodaDl D ( f-::-.l-:-:-:!:-t-!'D~-L-tE 2..L' IC ------ --l!O____ Am --2wL (e) Primary Beam Subjected to Extemal Lood (b) Substructure for Analysi £1 = constant + w (a) Indetennmate Beam ,,, " ' ApT~E ~L-l--L-+-L-I-L---l 510 CHAPTER 13 Method of Conilitent Deformations-Force Method ... ,3.,8
  • 182. IEl:I1aIIIU _ _ 117 + ,,0 I ,," flier " .- .-'-.- I0 B C Dr I JOtfl~A ~It ,....... .. r -FlG.13.18 d rn-y ID .. - c + ------ -, , c "'l I , II £J._I D -D. D I ~~"" I" I .0 ~ 60 1050 r60 Ib) Primary F..... SubJccted 10 Ex..maIl.oodui - Mo _ I I I B A t, I JOII--i II Indetcrmuwe F..... 'xD I • l~lk fD'l.DXI/ H "(e) Prinwy Fnme Subjocted 10 UOI' Value of, ....... D IOk- I.050k·1I The member rorcclhus obtained <Ire ,ho1l In Llble 13.4 and Fig ft mhcr 4'wI Fon l" The fon.:c In the n:maming members of the I mindIe lru...,> ':.111 he uClcmullcd h~ USlllg the upcrpoMtion relationship P,mar Irum The real.:lIon f) and D) at the hinged su kxted 4 the redundant I he: hmgce.J uppon () I then removed rnman kame h I n In I It; 13 I h Cll.t the pnmary frame Tald) h the h::mitllo.ujlng dod the UOit aloe of the redluotW..1 D hoo In Ilg I' l~ bland d re pectl Iy. (mpul I if/II I " ling that the honLOntal and I Ih L1 al mdeicrmmalc: tram I the hmged upport Dare rnpallbl II quail n Determine the rCdllums and dr,,,, the he.lf and bl:ndmg moment the h 0 In fig Il IS OJ by Ih~ mcthlJd of l:omistent defi"onDal'" Solution Dt r Ind( It milia/,. , :! & ...... ,3.,3
  • 183. C 8 o n I o ----i---............. D + + (c) Primary Beam Loaded With Redundant 8 A ._-------r;;----~-r---·---, 0 Jl' 8 Ct A I d Primary Beam Loaded willi JtedundaDI (b Primary Beam Subje<ted 10 E wnaI --------A,' Ji .&: I I I I I I I I I} A 8 ABO C A _---D~ -------- A A IEcnOll 13.4 "_1011111...__a" lemp11"11ur1 CII.g••, ..... FIG. 13.19 Thu, f.u "~ha1: um"dat.·d th~ .m.I1)"':-o of structures with unlYieIdliIit· rt d'......u -.('..1 111 Ch.lpll:r II. :-oupport moements due to SUpp"l ,. . . . - . fllundalwn and the IIkl: m.t) mducl: significant. stres~s In external detemlln.Hc ,truclure .md must he cllnslder~d In their deslgns:::~f ~ttlC'ments. hllH~H'r do not h.t~ .111) dftXt on the stress condi ~tructur~ th.1t an~ mtern.i1I) mdeh.~rminate but externally de Thi~ lad of etk,;:t 1:-0 dUI; III the fad that the settlements cause :-otrudure~ III displ.H.:e and llf rot.lIe as rigid bOOies without cIIIuqlilii their ~hapes The OlC'thod of consi...tent de.formati~ns, as developed the prel.'eding ~tions. can be ea:-oll) modified to mclude the effect ~upport settlements in the anal)sis. . Con..ider. for example. <t tv.o·"pan continuous beam subjected uniformI) di~tributed load I. as ..hon in fig. 13.19(a). Su~ the ..upports B .and C of the beam undergo small settlements A. and respectiel). a" sho"n in the figure. To analyze the beam, we ca.ii" the ertical reactions 8. and Co to be the redundants. The sulJPOJ'tl and Care remOed from the indetenninate beam to obtain the prj....;;' beam which is then subjectcd separately to the external load wand unit alues of the redundants OJ and C, as shown in Fig. 13 19(b and (d) respccliely. B} rcali/ing that the deflections of the actual dcterminate beam at supports 0 and C arc equal to the settlemen aod ,1c. rC5pectively. we obtain the compatibility equations v.hich can be sohed for the reduodants BI and C1 • Note that the hand sides of the compatibility equations (Eqs. (13.18) and 13 19 ~o longer e,qual to ~ero. as in the case of unyielding supports COIwll III the pre10US sections. but are equal to the prescribed values of ment ~t supporb 0 and C respectiely. Once the redundants ba detennmed by sol ing the compatibility equations. the other characle~stlcsof the beam can be e aluated either by eqwbbri superposition. Although suppon settlements are usually specified WIth the undef~nned ~ition of the indeterminate structure the 1IIl.... of uch dISplacements to be used in the compatibIlity equalJOlll be measured from the chord connectmg the defonned postti supports of the pnmaf) structure to the deformed postllOD dundant suppon A h. ny uc support d.splacement IS consiceIWd poslt1e If It has the same sense as that assumed for the red CHAPTER 13 Method of Consistent Defom1.tions-Force Method - SUPPORT SETTLEMENTS, TEMPERATU'!E CHANGES, AND FABRI_CA_T_ID_N_E_R_R_OR.:.:S:-_.J Support settlements 570 13.4
  • 184. .. lUi ....,,1_1II1II...".1II1II1.....1'_-,,._...-- •, ............ AM " ---_.... t t B., c CGmp1111IiIiIy eqo- 1..B.,+locC _4. laB +1cr: C -"at tho _bbility equabODS instead of tho • ..... Ac This IS bocause only tho c1ispl.. !MiI" cause .0 tho beam In olber words, if tho wooId have oett1cd ..Iber by equal amounll or by deformed poIIbODl of all of the supports would iii Ihon tho beam would mnam slraisht WIthout bon.... would cIooeIop m the beam DeImnine tbc _001and draw tbc ....... and bondiaI the three-tpon CODbDUOUI beam Ihown ID F... 13 1 dimibutod load and due 10 the support lOIlIanon~tI;.:;~::: and ~ ID .t D U.. the method of_ d
  • 185. ...._lyaMIyzod m Eump!t I ruction••'1 the roller 1UPJlCl'.:.:.-"".. IJ. JIl1IIWY beam .... .)l(Irjl.... ""....."" Tho ~ upport 1OlIII..... _ "canbe _ that the :....: ...~::: .-conned pClIIIIOII of the C aad E relalift 10 the c:bard _ _ thai 04m aad MM_ ..........
  • 186. FF The member fo~ thus btained SUMMARY At =O¥~_-::-:=__~ - r A 32.067 B o 0,0479 mm 111 nlIllllt vf lIlt Rlc/mu/mrt lJ) ub-.tituung the alucs of'&"DO'''''' Jr. mto the compatlblhty equation I::q I I.e ubtam I '12 lJ 1I479 flO 0 f. 40084 kN T 2·t.05 '41)0 192 rnm Ct the flexlblhty l:oethl.'lcnt Jto An 1 computed b) u~mg the virtual pre Uln see l 13.6 numenr.;al aluc'i lIt these quantillC an: tabulated in Table 13 6 .1 WI1 1;; deh:mllncd to Ix fiG. 13.23 o It'll Pnm~ Tru Subjected to Temperature Change.. 578 CHAPTER 13 Method a. Consistent Deformations-Force Method • Sf •( t"i lUI mm-) [) OX TlIIU lUi E ~O(l GPa M 1'(IO~l ( 06 AB" CO • A( 0- .. BO I t AD Bc 8 0 0 • m ((:1 Pnmal) Tru.... SUbjected to Ten de Force in Member AD a Inddemunah: Tru . Fon:e!'o (kN') C lOT D C 32.067
  • 187. 111 EJ-_ PI.'" £1 =consrant £/ A~ l 12m 6m / lJ £=<00",,", I k ~L+L----+-U2 8 Fa P13.17 ... Pl3.1• FaP13.16 ... P13.15 SOk 2Oft-----~ £1 =COOItanl 50 k 10 ft FIG P13.14 ~bId~l"""""" __--"CA B fiG. P13.13 I.S klft 25 kN/rn A~ I 10 1 m I 2~m £ =200 GPa / = SOOf (06J nun' AG. P13.12, P13.33, P13.51 13.13 through 13.25 Determine lhe reaction and draw the ~hear and bending moment diagrams for the structures shown in Figs. P13.13 Pll2S using the melhod of con Is.- lent dcfonnations. AG. P13.11, P13.32 -- £J =conSlallt SOk SOk I C I1l:'ri'==~B==Jb~=-Dr....-- 60k -' kltt l ~- - - - cO It 1--10ft £1 = ,on~(ant '12h 12ft 12ft t, =29.000 k!oi 1=1500in-'* FIG. P13.10, P13.31 RG. P13.9, P13.30, P13.50 FIG. PI3.4, P13.8 13.1 tIIroUlh 13.12 Determine the reactions and draw ...hear amI b<nding momenl diagrams for the beam m '·ig..., PLl9 Pl112 U'iing the method of con fl)nnatlon.... SclC:t the reaction at (he interior support the redundant c D )m ( 101) l I -+---6 m ----I £1 =- con tant ,, EI-l"omtant " 3m E ~UOGPa I_l ~"f1(lttlmm~ 8 60l I 'g "'---L--I m m i---bm 120 k.." ~==:=bl===B _13.1 13.1 ........ 13.. ~IC" m n the readlOn and dr.l the bendl m m~ t dlagr.lm for the Nam h. n PI' I PI'.1 U109 (he methl'ld 01 ...on I It:nt de· tl I ! h dl n ,lI the: roUe:r urp.:ln 1(-' he the red ndan( PROBLEMS FIG. P13.2, P13.6 ... P13.3, P13.7 n:"!,('1n"e'tcri...Ul.'" ,j lillo.' ".1lkICrmill;tlC ...tructure ca~ be ev l'llhcr thn'll~h CLjuihbnum «'Ill Idcr,ItIOn" or b~ ~UJlCrpoSUl0nor ~l'l('n~ ollhl' ,trudun.' due to thc ctcmal loading and l"3l.'h of the rcdunJ,lIlt U.5 tIInIIIIII 13" Dc:tennme the reactiom and dra the h r nd hendmg m lment I.hagram for the beam> hun In f I PI I PI '.1 h~ usmg lhe: method of conM"tent de- ~ rmat n lett the re ell II momenl at the he:d uppon t be the redundanl FIG P13.1, P13.5, P13.49 _ CHAPTER 13 Method of Consistent Deformations-Force Method
  • 188. - A 6 PI UI. • fl 29000 10k E 18 10k 'pane fA = tant fIC. P13.27, P13.52 20 fl /;./ = I,;on tant 1-_ _ 20 II ------j-- 15 fl--l FIG. P13.24 '0 It E/ =l'lln~tanl . FIG P13.22 I "Ill. _ CHAPTER 13 Method of Consistent Deformations-Force Method m r5m D f&P1129 ..... ,3.1 'UI....... 'U1 1«tJDg Ibe Ibe UPdaD FI6. P1U8 1- L~ ~A~===d;,,::===~( I£1 =constant A 30k-~.f: 60 40k 1----2 panel..! 8 fl 16 fl---1 EA constant j--151! - + - - 1 5 fl--I C 8. .D ,. 15 tl 2 kfft J~ FIG. P13.26 13.26 tbrough 13.21 Determine the reactions and the force in each member of the trusses shown In Figs PI3 26 PI129 u~mg the method of consistent deformanons F'G. P13.25 E D 90kN E =conslant c FIG. P13.23 3 kllt 8 A 1----20 rt--+--15fl 2J------j fIC. P13.21 E. =: CUll Ltnl AG. P13.19 2k1ft D C ~ 1011 ""l 8 Itilt • A IS h E/ = ctln~tant FIG P13.20
  • 189. 8ft B - k EJ A '11'.. • !Ok ft --.--- r-~B===iCr--OD;J 16ft 6ft-jt---- flI. Pl3A3 30k 50k 50k 2kift 2k/fl B I D I F~ A~CjfE ~30 fl-+15 ft+15 ft+15 ft+15 ft+-30ft-j £/ =constant 15 k FIG P13.40 FIG. P13.41 25 kNIm D I cJm EI +75kN B Jm EJ .J A 9m _'l3M RG P13.42 FIG P13.39, P13.54 21<1ft 10 I, E.I :;: c:onstant S{lI..N B IfIfl B :!5l~/m I 5m -~~ EA .-.....I'j$.. 5m B-1,I:h====a' FIG. PI3.36 - 13.3 and dIa45 Determine, the reactions 13.37 UIroUIII13. oment diagrams for the rJ Ix-ndlOg' m h ethodhear an p Pll.45 ll'.ing t e mhl.)"n 10 f 1£'_ PD.. . I~nl deformatlOn~. 8m E=70GPa RG. P13.37, P13.53 fIG. Pl3.38 10 ft I klft A I.:! It mm Methodstent Deformations-ForceM....od .f Con"ClW'TBI13 501. Am It- f1 E:- .. tXlOll 1001... 3m f..A = con lant Am 1J.M ..... U .. ll< , I h 1M flI. Pl3.34 flI. Pl3.35
  • 190. ,,_IIID........._-1~I11=......... • .PlUI B ~IrN 60tH a•• m - 21 ... PlUB 11M Sol Problem I 1 _ I of 30 mm al oupport D 11.11 Sol Problem 139 PI3 9 aad • IOItIemenI of I 1U1 SoI1e Problem I I PI I aad tho oupport lOlII_oi(l at" aad 40 mm at 1. . SoI1e Problem I PI aad tho 1UpJlOJ'I.aIo.....llaC aad I ID atD 1. . So... Problem 131'1'7"'_ PI aad tile IUpJlOJ'IIIIII.....irJ mmatC 11.14 Sol Problem 13 39 fI PI3 39 aad tho ouppon oettI_1IIM 3IDatHaad IolltO
  • 191. 14 Three-Moment Equatl the Method of Least W 14.1 DIIIvIIIalI of T111l11-Momenl Equation 14.2 AppIkIlloll of TIIn1t-Momenl Equalioll 14.1 IIlIl/Iad of LRSl Wurk SUmmIry PrubIems In this chapter we COI1Ilder two alternate ~ lIcxibility method of anaJy... of talically 1D'''1lii the ~I ~iDn and 2 the IMthod The thrcc-mOlllCDt equation wbich _ Clapeyron ID 1857 PfOVIwacon_ tool beama The thrcc-mOlllCllt equatlOD NJIi to, COIIIllBlibility c:oadilioa that the slope of the an _r IUpport of the COIItiDuoua -:=~tIIIIlc: momrnIs-the bcmdm, '~;iIX IDd at the two ~.~ :::.. ~at the lIIleri«~ .... arc tnaled ~a:the~md,,:,!""':':"~l1IaI JIIlIied at the III .......bhtllty cqua
  • 192. w w w ,. ,. . B _ (Ie_ _ ........IS_'_...,_ ..-..oIl.1at .....
  • 193. - 4 11C!UJ11141 .-..- 01 _ _ ....... Ulan 01 w oblam th ral ~ f I th, .",,,,,,nt 4 m :hlCh M bendmg moment at uppon heIDS COll5,dered If M bendIDg mo to the left and to the nght of pecJ E L L length of the pan 10 t fl and , livelli moments of menta f the pan t t of c r pcctlvely, P P concentrated load a I nght span rcspecJ el) k or k ra' f th d from the left or nght support to the pan len Ih distnbutcd loads applied 10 the left and the nght pan A settlement of the suppon under con uieratl settlements of the adjacent suppons to me left and t the n spectlVely As noted before the support bendmg m men ered 10 be positive In accordance with the beam (II 'ion-lIlllt when causmg compression 10 the upper fibers and ten 100 ID the I fibers of the beam. Funhennore the external I ads and uppert tlements are considered positIVe '" hen ID the downward dl I n .ho"n in fIg. 14 I a If the moments of inertIa of two adjacent pan ra Dtmu beam are equal (1 e I I, 1 then the thr~m ment equa pllfics to upport rquullUn c, - c L, I.. ') II' /.1 ..,. 24-£f II,~ c C. I (I , In IUl.;h the ..ummJIJOIl ~lgl1' h,n e been added to the first term nght 'Ide.. of the..!,; equ,t1I~ln . "tl th<1t multiple concentrated loads arr1it.-d hl e,tch 'p<.111 of a 'lIlgk concentrated load as In Fig... 14. JI J and b I'M .Implicit) ). ... contlnuou~ beams us,ualJy .. loaded uh unif('rml) dl tnbutcd, oer entire spans and __~. trated load... the ellect' of onJ~ the...e tO types of loadings generally con..idered In thc three-moment cquallon. Howeer. the effects of t) pt.' ,,"If load.. l.<tn bt: mcluded. 'ImpJ) b} adding the expresu ..It....pc.. due to the:-e IO<ld.. to the nght ..Ide, of Eqs. (14.6a and 14 The ...Iope.. 0 , and O,~ of the left and the nghl spans. res'''''Clht&; due to ..upport -.elllemen". can be obtained directl) from the deft po..itltm of the ...pam. depICted in Fig. 14.l(c). Since the sen1emen1l as umed to be mall. the l)lopes can be expre!sed as 1'1 A I ~ 6H The ...Iopes at cnd.. (" of the left and the right spans. due to red.._......· support bending momen". ,Hg. J4,l(d)), can be detennincd COJllII'c( nientl) b) the bcam-dcncction formulas. Thus, .[ L, .W, L OIl - + - - 6t:I, 3£1, PL'A 1- A', II I. L .- 6El- - t 24El C, c 1/ I If l_ + ~,!~ + M,L, _ 0 L, 61:.'1, 3EI 1£1, 61:.'1, B) simphf)lI1g the toregomg cqualton and rearranging It to terms contammg redundant moments from those involVlDJ 1, L, H,L, ---- +-- 31:.'1, 6£1, in which .f,. f and f, denote the bending moments at supporU and r. repectie1). As shown 111 Fig. 14.1 (d), these redundant moments arc considered to be poiuvc in accordance with the cOllum/ion that is, when causing compression in the upper fiberl tcmion in the lo-er fibers of the beam 81 sub (iluling Eq,. 14.6) through (14.8) into Eq. 145 the compatibility equation as CHAmR ,. Three-Moment Equation and the Method of Least Work
  • 194. cttAPTER ,. Three-Moment Equation and the Method of least Wont ~/'H I k Off" f" ... P I k f 4 I 41 I The three-mument equatulll . a 1!lCn b} Eqs. 149 thro y,ere utrled to atl Iy ,hI." I.:ompatiblilly condition of slope the IDlenor upport 01 contlnuuus beams These equaU The: flln: 'lllllg Ihn':C:-l1llllllc:nt c:qu.ltlon Me applicable to n ,. urMrt ( .100 r, 01.· oj l'llllllO.UOU' pro...(ln~':lI Ie: t· . Ihere .w: n,l OI~(lntlllUJlIe: . !i.m:h .1' mternal hmge.... In the beam the kit urrort .1l1d tht' n1!lll uprx1rt r The follo m~ 'ter·b~ -...tep prOl.:cdure can tinuou lx.iITI h~ Ihe three·moment cqu.illon. I. Sdecl the unkno n bendlllg moments at all interior supports bc:am .t... the: re:dundi.lIlh 2. B) tn.'.ltlllg cach Illlcnnr support successive!} as the In ate ",urr0rt ( ritc a thret>momcnt equation. When wntIDI equatllln', it ...hould be re,lil/ed that bending moments at the end support!'> .Ire known. For ...ueh a support with a cantil hang. the bending moment equals that due to the external actlllg on thc cillltileer portion about the end support. The number of three-moment c"Iuations thus obtained must be the number of redundant support bending moments. which m the onl' unkno m, III the,e equations. 3. Sohe thc ,),tcm of three-moment equations for the unm pon bending momenh 4. Compute the span end ,hears. "'or each span of the beam a fn:e-bod) diagram ,ho II1g the external loads and end and bJ appJ) the: ~quation ... of equilibrium to calculate fon.:es at the t:nd.. of tht: ~pan 5. Determine ,upport rcactlon~ b) con..idering the equlhbnwa uppon joints of the ~am 6. If so de Ired. dra !.hear and bending ~am by u..illg the hwm H"" nJllt tnliml. Fixed Supports 14.2 APPLICATION OF THREE-MOMENT EQUATION- --
  • 195. CHAPTER 14 Th....-Moment Equation and the Method of least Wont - ,,, , I I,=- '0 oJ B~B_ "'(}!l B =h:!.1 (ttl Span End Momem<' and Shears IoIUl1an -'1U ''fa .... II TIll. 7 ...... • -1 = tl A 10.4 (1 k D :!1H.. .Q..B B =62.2k Id Suppon Reactions 32.6 -9.6 2.5 klft III. 14.3 Shear diagram (k) F, B Bill •• _icoI.__..,...... Bendmg moment diagram (k.-ft (d) Shear and Bending Moment D'I;·,--
  • 196. C = ]7~.7 10m 1-7IX)(I(tlmnl'~ 10m II =' ,-on lanl l"ktl.:nmnl.: tht: rc'h.U n h,r lhl.: t,.t'Il11I1Ul'U' lx.lm hln In Fig 14 the umlOnnly Jl lflhutw )l' d .mJ du: tlllhc ~urrllrt :'>CttlemC'nLs of 10 '0 mOl at R ~n mOl ,il ( .wd·1O mm II n L'il; lhl.: thnx·momrnt eq (b) Span End Moment' and Shear, E=lOOGPa IlJm B, =~77.9 :OJ..N/m A =0 L ~') I I t I, = I]~ H 8 = ~JJ 9 'N C = nXHr< D = 104.9 kN RG.14.4 SoIulion R( dundalll The bendmg moml.: nt~ fBand f( at the 1D.t~""" ... and ( r..:' pet.:tlct .JTe the redunddnt'i lhw.'- '10m nt Equatton Of Jomt B 8) consldenng the lIII(IIIIln and (d {~md, re pectlely and ubstltutmg 1 10 m £ 2.00 10 k m I tUO III mOl" 700 10 m" M ~ ~l Iflmm OUI m - As "Omm 005m A 002mdndP p, (J Intol:q 1411 "('"Ole (:) SUpplll1. R..:'adl<m CHAPTER 14 Three-Moment Equation and the Method of Least work ED""" 14.2 )1"" 1~15 llbA II"" ~)1.2 195.1 ~51.2 rr:D:::i:D')- J-8~ ( o::iII::o') J-C~ (("....t-.;30L- kN .....,m..l....J:;; 'I 8 I 1152:ft:JI152 18 C 14512Ji:J451~ ( A - 1'H5 8'" = 1"1 5 I B,H' = 11".4 e"' = IX)" I C;IJ= 95.
  • 197. AIls AIls AIls 9 k 1ECl1CIII14.3 _ It ~ _ .. II II II Equa beam The bend.. '-..~B --_.... t B, 1t' , I--- FIG. 14.6 14.3 METHOD OF LEAST WORK 25.25 18 W90 I 90 C, = 43.25 - 10ft ... ).M "'.Itt ·n.2S" Thrf:' . ttln nt l;quutlon at Joint B SImilarly, upJXln A B and ( nle "I 20 2. .1,20 ,() /( 30 45 211 I 2 I FIG. 14.5 EI ::: con..t,ml Ij, InJ.Clcnmnak Hcam - IOf! - Inll (t:) Span End Mnmcnh and Shear, (} IOtt - IOU ~OI1 lOft (b) Equnalent B(am 10 he n.II)/eJ by Thr'C'-Moment EquJtion Cj--......l.._-4;.L.LLJ...Ju...J.....L.I....L.1.-1-ci: c i..1..J o I -I" k __....;'_-..J-l_,( I ! ! .- .,bB ',' I 2475 2K.75 J.;12 S 90 .. .. 142 5 I.H kilt 1--"'1'"9f",~T:.,.,rr-r;=TT""T-'-T:~ B, =53.5 Y7 ~ b.·ft 9 , 2025 CHAPTER 1. Three-Moment Equation and the Method of leasl Work600
  • 198. the Jl,,·tkl,,·u~'Il .tl the J1i.llllt of dprli~ation of tht.: rt.:dundanl B ilrrl) mg CI'.llgll.IIW "l,,':~lIld Ihr.:orcm. c can nle o tl." 0 tR1 tL t R~ 0 ,L ,R 0 ,L ,8 nu, R,. R,... . R,,){ v.hKh reprc~m d ystem of IT Imultancous equation and t:an be sohcd lor the redundants. 1 he procedure lor the analy~is of mdetenninate truetunII method or lea t ork IS diu trated by Ihe following examp III whil.:h II' rcprc..,cnh all the known loads and RI • R2..... R" denote redundant... ~~xt. thc prinCiple of least work is applied separa each redundant h} partially differentiating the strain energy ex (Eq. 14.1411 ith respect to each of the redundants and by set partIal dcriatlc equallo lero; that is It hlmld be: rc.lh/t.:J Ih.ul:q. I·U~I r~prcsents the compatibility lion In the dirl,,"l.'thln ll redundant B, and It I.:an be solved fl redundant . ['I 141. IIldil.:atc . the first partial deriatic of the ~;';:.;~; erg~ Ith re fll..'ct to the: rt.:dund.lnt mu~t be.e~ual to zero. Thi that fllr the: alul: 01 thc rcJund.lIlt th,ll saustie... the equation or hbnum and n1rnp,ltihillt) the tr.lin energ) of the structure IS mum ,)r rna IIllum, Smo: for a linearly elastic structure there m.n.lmum ,llue of "tram encrg~ becau"c it can be increased illtdeliaill4ii b' inl.:rc,1 ing the alue of the redund.tnt. we conclude that for till ~lue of the n..:dund.tnt the slrain energy mu"t be a minimum Tbia du IOn t kncmn as the: primiplt ol/tmt lI"ork The moqlliwdt'" oltht rUl/llldalll5 ora stutically indelermmQ tlln mInt ht /1( h tltat tltt' .Hraill Ult rqy .lwrt'd in th(' truclur~ mllm ,. t (Itt i"(t mal II ork dO1/{ i. fht' /t'mt The method of least ark .IS described here. can be easily ex ttl the an.ll},,!s of slructure" ith multiple degrees of indetemuD8CJ struclure i~ ind~tenmnate to the mh degree, then n redundants kxted. and the ...train cncrg) for the structure is expressed in tenns kno n c"ternalloading and lhc IT unknov.'n redundants as 102 CHAPTER 14 Three-Moment Equation and the Method of Least Work
  • 199. &04 CHAPTER U Three-Moment Equation and the Method 0' Least Work _1403 _II~_ -900nB 1(11 nno 0 lUll 14.1 (rum" H 1: b. T, detenmm' the n;mamm~ n: II. two the eqUlhbnum ~qu.:Iuon flg I·t r;F 0 ... ~1 0 , If, 10 IS () ... '0 ~ ~Jf 0 II If, In 15 IS _~OJ () If, "... Eumple 14.5 [Xh:nmne thr.: reddl)lh fOT the Io·...pan continuous beam shown ID b) the mt"thod of bJ t ork 245 B ... F o o By I £f To de1mn1 the cquau ,_se_ AB BC and CO The equauon are shown ID Fig 148 and of B are tabulated ID Table 14 I "the den With respect to B are naluated. These dcri t ofTabl.141 By ubsUtullng tbe expresstOD ~ r / nd , ,30 kN/m SOkN 1l ! D tB, I ~ xljD,=13' !--IOm--+I-Sm-+-Sm £1 =constant Solution The bl:am i urported b) four reaction!'>. A A 8 1 , and D Sioc:e onl) thTL'." cqullitmum equiltlon.... the degree of indetenmnacy of tbI equal to I Ld us !>elcd the re.H.tlon B, to be the redundant. The the redundant "III be detennined by mimmi/ing the strain enel1Y'-~-: with re p..:...1 to 8 The tram t'n rg)' of a beam !>ubJet:ted onl) to bending is e·xpI""'•• I L I 'E·- ih ,• I A,l,:C rdmg to the pnnclpk (lllt'ast "ork. A,=O~£ AI =245 -0.5 B) W AG.I4.8 d I'(,II) If ,8 ,-8 lid, ~O Bel re 'oe bt.un the equation lor bending moments M the feat:tlon at the UplX'Tl ... and n of the beam in term Appl 109 the three equlhbnum t:quallon v.e nle LF 0 ... - 0
  • 200. n 1 E. cOlUtanl £1 £ ....- 12m m fill. Pl4.4, Pl4.11 fill. Pl4.5, Pl4.12 15k lOk k , C , £ I ....A.a: :i/O OliO FB D tIOft~±21 10ft £.29000 .000 fill. PI4.&, Pl4.10 I&Pl4.7 £ Jl.. SOk I lklft sectkHll4.2 1••1 tfWouIli 14.1 Detennine the reaction and dra the ,hear and bending moment diagram fI r Ihe beam hO'll in Fig~ P14, I through PI48 usmg the three-moment equatIon PROBLEMS FIG. P14.1 A D E= consWit 2S kN/m ~c I 10 1 m I 20 2I m I £ =200 GPa I =SOD(10') nun' FIG. P14.2, P14.9 FIG. P14.3 o 41 ~ ORT T I 11' L lEfth " 4 ~ I~ d ·T loo, ( ( H0811 OST X) 12I I(T)(IO)(I2J' 12 08 ' 12flf2 41( 15, . 0.6T - 2.4TJ d,] 0 - 0.6( 400 4 T (B B< BD ",.:cordmg: to thl: pnnl,.'lrk of lea,t ork ~(iF..) fL . I:1(':If)~d' ~ 0 ....... (T 4£ (T £1 In this chapter. v.e hac studied 10 formulations of the force it}; method of analji... of statically indeterminate structures, the three-moment equation and the method of least work The three-moment equation represents. in a general form, patihilit) condition that the ~lope of the elastic curve be COD an inh:nor UpPOrl of the continuou~ beam This method whicla u~d for aniJl}zing (onlinuou~ beams subjected to external uppon ~ttlemenh. mvohc~ treatmg the bending moments rior and an} fixed suppon!'! of the beam as the redundanta moment equation i thcn applied at the location of each obtain a Sc:( 01 (:ompatihJlil} ClfualJons which can then be redundanr hcnding momenl The pnnciple of lea t ork tales that the magnllUtk Jallh oj un mdt'tammute true fure mllJl he me h Ihal lhe lured m Ih IrUl/urf" I a nummum To analyze an andete tUTe by the method 01 least v.ork the str-tin energy of the fir t expre sed In lenn of Ihe reoundants. Then the paruat The cpn: ion, llT the: tx-ndmg momcnt~f and the axial forces F m the redundant T and Ihl'IT dCmalic nlth respect 10 T are tabulated 1~_3 8~ ~ul:l {HUlin!! thc~(' cprC"lOn, and dcmatics into Eq (2 we TABLE 14.3 CHAPTER 14 Thl'ft-Moment Equafion and the Method of least Work608 SUMMARY
  • 201. - - - - - 1 5 f t - - _... £=~ '4.'4A beam IS upponed by a fixed Be a shown ID FIg. PI4 14 Dtl&_ cable by .... method ofleast wort. fIB. P14.14 .... method of least work See .... method of _ work See RKbOaI aad .... fon:e ID each mem. ID Fig PI4 13 _ !he method of .... Iooding shown ID Fig PI42 _ ...... IOmm.tA 65mmatB III' ='4 nR......~r.....?_........_of..-_ ft-+-IIDft--i--IOft -'----:20 ft---------< -.PMI
  • 202. 111 .... 15.1 ................ _ .. _ equill U Sb'UClUIII wIlIt IIJIlIph D.......InlIUnCi .... hodllliU""" J.. t JBBO - II COlhtder the COl1linuous beam shown in Fig. 15.I(a). Suppoae bh to dr.II. the mfluence line for the vertical reaction at the suppon 0 of the beam. The beam" subjected to a downWard concentrated load of unit magnitudc. the position of which 18 the coordinatc measured from the left end A of the beam a the figure. To deelop the mfluence line for the reaction B), we need to mine the e.pression for Or in terms of the variable position x oftbe load. Noting Ihat the beam is statically indeterminate to the first we !tClect the reaction 81 10 be the redundant. The roller suppon then removed from the actual indeterminate beam to obtain the cally determinate pnmary beam ,hown in Fig. 15.1(b). Next mar} beam IS ubje<.:tcd separatel) to the unit load positioned arbitrar} point ." at a dlstan<.:e x from the left end, and the O. as shown m fog. 15.1(b) and (c). respeclively. The expreSSlOll can no be detemlined b) usmg the compatibility condll1OD deflection of the pnm,,,)' beam at 0 due 10 the combined efti ex.ternal unit load and the unknon redundant B must be l.ere. Thus o _ la. fBB In hi4:h the llelbilny 4:oenlClcm /8' denotes the deflection mary beam at 0 due In the unll load at X (Fig 15 I b fl ;Ubillt) coetfklcnt IBN denotes the deOection at B due to the of the redundanl 0 fog. 15 IIe) . c :an use Eq (I:' I for 4:onstructing the inftuence I plaCing the unit load sU:cc:s hel)' at a number of pos1uons beam C'alual1ng ftHo lor ecI:h position of the unit load and alues of the ratio fB~ ISH Hoevcr a more efficlcnt B C ~~ CHAPTER 15 Influence Lines tor Stattc.lly Indelennlnate Structures !though ,lily of thc- mcthC'd, of ana.l) ~is of ind~tennInate pn: l'nlc-d In P.lrI fhl'1.' l'.lIl be u'c-d lor computmg the ORtiDlllI mfluenl'C Iinl.". C- "ill u,c the method of COfl'i.lstent deforma eu ~d m Ch.lpta I J for su("h purp~)-.e~. Once the innuence linea determin.ttc tructures h.I1.' lxen constructed. the} can be UIed. ,amI.' mannCT .1' IhlN." t<'lr detenninate structure~ discussed m In thIs ,..'harter. the pro("edun: for constructing innuence hnes Ii calh indC'tenmnatC' lx.lOh .lnd tru,~.. IS deeloped. and the a of fulkr-Bre".tu· principle for comtructmg qualitative loft for mdetenmn.ttc ~.Ims and frames is discussed. 612 -- --- - - - - - - - - - - ----------....; 15.1 INFLUENCE LINES FOR BEAMS AND TRUSSES r-x-l B C '!LL on: oa: D X t t tA B C la, Indetenmnale Beam dllnn + A.Jl X b Pnmary 8Qm SubJ«1ed (0 Urnt Load fIG. IS-I
  • 203. x t B, ._-- B I PlimorJ....Loodod willi b PriIU)''''' til ' ' .. + --------", t .s._&-...... 'IJ II'....'1117............
  • 204. CHAPnR 15 lnftuence lines for Statically Indetermln.te Structures I •. Ie Primary Beam loaded with Redundant B D..... Subjected", U.. LoodPrj _ILl ........UOOOllr_..' - 1S 8 C D eI Inftuencc Line Ii IkN ! 8 IA lIII c D I8 =J S 1.56 FIG. 15.3 contd D la) Indetenninate Beam EI =con..lant = c 3m x J kN 8 LR + R 8 (hI Primary Beam Subjected to Unit Load I kN x fBX 3m A 616 ... 11.3
  • 205. fDA fA 0 fOB JBD I G) 10£1 86 10 G Slllce ~ f D lD iU:COrdanc:e th M on the pnmary beam Fog 154(d and through F by USlnI the cooJUPte beam In Fig IS 4(e from whiCh v,e obtalo (he r. t from which ---======== ._cn__IOI-.::15~.,~..=n==:llMo:...:.::...:T:r:~-.:I1t:Example 15.2_ 1.5kNmk 8 1"8HI.. :tH." k' m' k. fBA I .. £/ :!.B k' m k. JBB EI 1:!6 k m' k.· fB' I•• £1 'HN m k:'< JBD lOB EI IB' ftB " II, IRl I.B Sm.. by 1.1 dl' b ~)f rl'.'lrrocal d..:tlcl.:ti(lIh. fBl - 'HI e place load at Btl" thl' rnnl.lf) Ix.lm f 1£ 15 ~(d .l~d compute the detleoliel" pt.'ml ~ through r t'l} U 109 thl.' bcJrn-Jdlel.:llon lormula glCJ1 mside .o<."r llth; bl,1{lk 1hu . J., 164J _ I 5 kN 'kN i.. 241 ' . The remaining ordinate of the mfluence Ime for 8 are calculated m manner Thc~ ordmates arc tabulated m Table 15.1. and the IOftuence 8 issho"n In hg. l'U(c) InflU! n((' '-me Jor .I( Ith the influence Ime for B known the of the mfluem:c line for the bcndin~ moment at C can now be placm~ the unit luad ,ucre Id) at points A through £ on the beam and by u~mg the com: pondmg alues of B computed "",""'lIlfI namplc. llS deph.:led m hg 15 1 1J "hen the umt load IS located t alue 01 the rtJlllOn .11 B IS B I 5 II. kN R) con",idenng the eq the Iree bod) 01 the purtlon of the beam to the left of ( 'C obtaUl The alue of the remammg ordmatcs of the mfluence line arc Imilar manner Tht:SC nrdmate for (( I hon m hg IS 3lg The ordinates of the influence line for B) can now be evaluated by a Eq. (I ,>ucl,;c,>"ic1y for each position of the unit load. For example w unit load is lOl.:atcd at A thc alue of 8 1 IS obtained as In "hlCh the nC~alle "go", indICate that Ihc'C deflectIons are in the direction. Note that the flexIbIlity coeflicient iSB in Eq. (I) denotes the Ipo,iliH' Jeflection of the primar} beam at B due to the unit value or dUDdant B, Fig. 15..~(c)J ,hcrca~ the del1cction f88 represent!> the (negative) deflectIon at B due to tile ntemal unit load at 8 (Fig. 153 d 243kN·m 3 kN + -£1 I , "1.6 lJ,,J4 " 1 kS m/k~) Influence Lme Ordmates I , I " o"'llJ " 148 " CHAPTER 15 Influence Unes for Statically Indeterminate Structures 618 A H ( n l lml L ad at TABLE 15.1
  • 206. • k k ko0603 fpB I. IECTJoIll11 ......... u.."" __..T_ SGIU11an The beam I IIIdetemunate 10 the nd D and G 81 the roller uppons D and G ...~«tlvcly The IIIftuence hne ordinat Ii II be ua lItrough G h",,~ on Fig 155 '''' Lin< fo, R-.Jundan, D and G The and G for an arhilrary posulon X of the uml I d l;: the compaubdny equation sec Fig 15 5 b Ihr ugh fo, Ip D G 0 f" f.DD GO Smce by Maxwell slaw fDx ltD loe place lit unlll beam (Fig IS 5 e and compute the deftectlon al pam A Ihr the beam-deftection fonnulas gnco IDSI(1e lite front c r fDA. (A.D 0 M The alue rthe d Imllar manner TI. ordog r I r II~ 1I1a are Ii lor the shear and bend tngm lie faPCCliv Draw the ml1uencc hnes for lhe ....~IOI" Fig IS Sa ~Example 15.3 0491 k k 816667 1680 0,492 +- f~ ~ () F -0.095 k k 108 Ipp ,1,(50) t 1(40) 0.492(20) - 0 A,=0603kk 0603 D tCL:II, () +'L:r 0 lmt Load r ,k k, .t n A 0 I 0 0 U B O.·N~ O.60.~ 0.095 0.397 C n. t''i o,~q 0.119 o746 (left 0.154 [right D 10 0 0 0 £ 0.079 O.07~ O.W3 O.O7:! F 0 0 10 0 The llrdmate, of the mfluence hne for n, l:dn no" be computed by Eq. I ...u~ce ... Iel~ for eal.h po Ition of the, unit load. For eumple unit load i... I~x:aled at B. the alue of D 1'1 glen b) The remamm~ ordmate... of the Influence line for D, are computed m I manner. ThcC" ordmate' are tabulated m Table 15.2. and the Influence D, l'1 ..ho.... n in F-ig, 15.4(t' Inl/IIt'f/( (' Lmn lor A. Will f With the mfluem:c line for D mfluence lines for the rem.llning reactions can no.... be detennined by the equatiom. of eqUlhbnurn. For example. for the position of the UDI point Bas ,ho....n In ....g. 15.4(g). the value of the reaction D" has been be 0.492 k:'k By applying lhe equilibrium equations. "e determme the the reaction, AI and F. to be TABLE 15.2 The alues (If the n:maming mfluenl.:c line ordmates are computed manner. Tht:'iC ordmatc!ii Me II ted 10 Table 15.2, and the influcDCle and f' afC hon in hg. 15.4 h and (I re"''''''i.:tie1). Intlunl/t I Inc lor .( unci I( The ordinates of the intluence hear dnd hendmg moment at C (an no he caluated by pl3ClDl llOCC 1e1~ at romt A throu~h F on the mdetenninate beam and corr~pondmg aluc of the rcactlnns computed prevIOusly F hown 111 • Ig 15.4~ .... hen the umt load i Ioc;ated al pomt B tbe reachon ue A fJ IlfJ3 k k· f) () 49:! k k. and F 009 Idenng Ih~ eljulhhnum of the frL"C bu~jy of the portion of the of ( C obtam 122 CHAPTER 15 Inftuence lines for Statically Indeterminate Structures
  • 207. ., '&11 .... II u.: far SIItIcIIIr In. I ........ 8trUCtWII I x I kN (~AI 5m 5m 5m 5m 5m 5m A D G. , 15m I J5m--'-' EI - ' ._Beam (1) Pn-rBeam 5ubje<1Iod 10 UIII Lood + foo --_oJ!. 1""'A-"'-:-~-9~--- G~D IkN (e) Primary Beam 5ubjeclod10 'edund... D + A'O'::';;'-fRlI...._."G~ x G I kN d) - ,.... SoihjorPd III • ' d 'G IkN A I C • II , G D ~....'. bd. ~.D
  • 208. Eo..... 15.4 10 O.K04 o.1Rb o -0.159 -O.J~7 o I + 0.228 0.032 - 0.1o l 011 l d cit /l , , (, 4 0 0 B o~~8 oO.l~ ( (I fl"" OIl6' J) 1.0 0 t {I Ij' I O.~.2~ f o ';;~5 1I.5l2 G 0 1.11 A, - 0.804 kN kN II, 1[5)· 0.22R( 15) 0.032(.10) 0 M, - 2.54 kN Th,.· n.'111ulnlng ardm,ih.' (If the Influence line, for the redundants are m .. Im.lar manner Tht''oC (Irdmalc are t.lhulated in Table 153 and enl'c hnl.' lor D ,IOJ G ,m.' hlmn In FIg. I S."lgJ and (h) respectJ {I/flu nd Lim lor ~ amI / t The ordinate!> of the influence n:mJmin~ reactIon 1;.111 no he dch:nnincd b) placing the unit load al POlO!'. ., through G on the indetcnnmatc beam and b} applying the of eqUlhbrium. "or eample. f()r the po"lltion of the unit load at 8 FII- the value, of lhe rC,H;llon, () and G, h,nc been found to be 0 228 kN -(J.()32 k'W kN. rC'J'lI.'(:liH~ly B) con,idering the equilibrium of the dt:h:nmnl,.' the alw.:, of the reactions A. and ./ ~ to be as follows TA8LE 15.3 Ora" the mfluenl.:e hne lnr the furee, in mcmber~ BC. BE and CE ho"n m • Ig 15€l1a l,v load arc tran..mith.'d to the top chord The .tlue of the remaimng influence line ordinates are computed m manner. Thc'oC ordinatr.'~ an: h~tcd in Tabk 15.3. and the influence I andt~ arc hO.El in Ilg. 15.5(J) and {k" re~pccti"ely. Solution The truss I tnlcrn<ally mdetermlllate to the fir t dcgree. We select tile Ff t In the diagonal mcmhcr ( F. to he the redundant. bljlu£nfe Lme lor R( JUtlJUIII ft. t: T{) determine the mftueoce .... pla a uml I ad Su!.:!.:e Idy at Jumh Band ( of the t po Ihon of the unit kldd "e apply the method of COllSlstenn1I~:: lmrut tht' alue 01 II The J'lnmitr) (russ obtained by 1'1 (t. I ubJt:Cted fl<trat Iy tn the URlt luad at Band C as shown CHAPTER 15 Influence lines for Statically Indeterminate Structures826
  • 209. _...-_.....,.....---._.... .
  • 210. IECTION 15.2 0 llUw 1nII_ LIlloo.,_ ....INoI......""'- 8 D £ Qual Innue l .. l 8 • 8 A_ l, U>od. 8 630 CHUTffl1S Influence Lines for Statically Indeterminate Structures On~c ,I qU.i1IL.ltI,.' Il1tlUl,'ll~e hne for a ...tructural response h.a IX.'l.'n l.'on IrUl.:tl.'d, It (,tn Ix: lI"Cd 10 dt..'X.'idc ""here to place load to m.Ilmize Ill: value 01 the re"'pOIl-.e function. A eli i."(t1on 9 ~ the alue' l,f .1 n;"poll"C function due 10 a undi tnbul.'d hH' 10.ld h m.1XlmUIll po ItlC or negatlc) when the plao:'d Oa Iho~ POrll("!J1 01 thl' tructure "here the ordinates rt." Pl"lJl '" fundlon lIl11ut."n(c Ime .Ire POHJC 'or negatic Beca ml1ul'nl..'l,'·linc Mdmall,' tend to dlmll1l...h rapidl} with di thr.: JX,int of .1rpIK<ltllll1 of thl' re"'ron~ function. Iie load p than Ihn.'l' '1'.10 kngth.. a"a~ from the location of the resPOIlle generJII} hae a negliglPle ctJeLl on the alue of the response ilh the 11 e·load p;.ltlern kml n. an indcterminale anal ...trudurc ('an he pc:rformcd to determine the maximum value or Pl"lJl..c funl:tlon. EulllJllB 15.5 • 8 D £ F 111 Dra quahtatle mflur.:l1l:e lin!':) IOf the Crll(;al reactions at suppon the o..:ndmg mllmr.:nt .It rom! B. •miJ the shc.:Jr .:Jnd hcnding moment of the four-...r,tn ulI1linUlIUs he.lm ,ho"n in Fi!!. 15.7(a) Also _L._.u r<lngr.:mr.:nl 01'. ul11lonnl) dl,lnl1uled iJo"""ard he load n I malmUm rChlllC' re.t<:llon.. at 'Urporh I and B. the: maximum nep lng moment at B thr.: lll,llmUm neg<ltiH.' ..hear al ( . and the maximum bending mon1l'nl at ( Amnaemem of Live Load ror Mnunum NcphYe M Cdl Quahtabvc InOuence LIM fur S Solution IIIJluc/J/( I inc lor I fo iJl.'tcfmlOe the qualilatle influence r.:rtk<ll re.actiun ~. at ?>uprnrt of. -c femme the -crtlcal restramt at A <tl'tu<ll ~am and gil' the rclc,I'oCd beam a ..mall dl'oplacement In Ibe • 8 C D E F H () E E F C Arronpmcnt of (j,", lood for MuimwD NepO 1<1 .:---:8- C~D:---:--"";F Qua1iIIIi... -Lme " II&. 157 Quaht<tllV Inl1uen.(' Lme (or A An n em nt of LIC' 1 ad for MaXimum Posiove A Ib, AG. 15.7 (I.:untd. C ~ofLI I.-dforM"'==
  • 211. - • Quahtab Influence line , W ~ ~ ~ ""'IlA L.. L.. ... ,5.8 Quahwsve Influence Line feX'MA • Arrangement of Uve 1.oId for MaJ,unum PoIibve MA (11 1111 I I I I .'( w( I I I I ,1111 W( W( I I I I 1111 A _L.. -~ .. d • - ~ A _L.. _L.. _L.. .. B A _.... - .... Ora.... quahtatle influence lmes lor the bending moment and shear at Ihe bUilding frame ho"n In I-Ig, 158(a Also. ~ho" the arra.~::: unifonnly dl InhUll:d do.... n.....arJ 111.' load II that 111 cause the 11 Itic bending moment and Iht: m.l~lmum ncgatic !lohear at A Solution InJiu nl( l.lII( /lIr 1" The quahtattve mflU('n","t' hne for the ment <1.1 -II o"n In t-Ig IS 8 h ute that IDt.:e the mernben connected tOgJ.:lhcr hy ngld Jomt th ongmal angles between mtersa:tmg I a l0lnl mu I he mamlamed to the deflected shape M,....1iII blam the hlaXUllum po lint: hendmg moment at A (he hve load oer th pan of the frame tht ordmat of Ihe mfluence CHAPTER 15 Influence Lines tor Statically Indeterminate Structures dir..."ct! n ,f I rill,: Jdkl;l:J sh·lI'll.: 01 ,Ihl: n.:lc.I'CJ beam th h' I"" t'l n;prc~nh Ih lo'Cn .11 h;j~ ~ll the 1IlI1ucn,,"C hoe 1e UC nllueru.:e ltoe 1M I l'l Ih.ll the Jdkdcd ,hare 1:0;, con I teat uppcrt onJltllm 01 the: rek d Ix••m: th.ll I pomts B D E and TI,; e ..1 beam "hlch an.: ,1I1....:hI;O III wlkr upr~lrt do notdisplace Tll max.lm. the roSltl~ ,.dUl'lll t _ the 11..: I{lad II IS placed f B oJ DE f till: o.:am "here thl' {lflhn,ltc'o of the mfluence line IX' III~ a shon In f-I l'i "7 I:l In. I III ,8 The quahtatlc mflucm:c Ime for 8 and the .uT'a11{!('lllcnt for ,he md lmum Pl)!<.lt!t' ollue l,r Bare determllled lD nunner and aT; hoo In 1"1£ l'i'" ( In U 'f I I , , , To determine the 4uahtatlc mfluence ~ndlng moment Jot B 't! msert a hm~(' at 8 in the aClual beam aad rdea..oo ~am I mall rotation In the JX' ItIC dlfct:llon of M. the rortl n to tht.: left (If B ulunh:rd'l(k" 1'< and the portion 10 the clcx:k"l ' a.. shenn In Ilg 15'" d Thl: ddk':ted ..hape of the rdeued. thu ('Obtamed f'pn:!oCnt 1hl: qU'Jhtatnc mf1uencc Ime for I" To cau the ma'lmum 1lI.'gatie 1x'nding moment at 8 we place load II <J ..r pan IB RD. llnd Ef ('If (h~ beam "here the ordina influence Ime lor /. Jf~ nq;..ltl.: as h(mn m Fig. 15.~ d. lnflu 'Ie, LUll fi T S( The quahl<ltlC mfluence line for S( IS de1..._ cultml! the auu,11 ocam at C !nd h) ~!I Ill!! the relea'Cd beam a dj,pla~ement In Ihl: po Itl1: dIrectIOn of S( b) moing end C of the of the be-...m dl) n ard and I:nd ( of the right portion uP'Aard Ilg, 15.7 e To obtam the nl<lXllnum ncgatle hear at C. the Jie load IS ~pan DI:. and the portion BC of the "pan Bn of lhe beam. where the the influence line for S .lfe neg,Hic.•I~ ~ho" n in fig 15.7(e). In/lu(/1C't' Lm!' lor /( Thc quahlatie influcm:e line for the moment at C and the 1Ic-Ioad arrangement for the maximum poIIh Ie arc !.hov.n In I ig. 15.7(f). 832 FIDPlll5.6
  • 212. -, B • ... "5.11 ... "5.10 ... ,,5.1 "'''5.1 E 401401 DC A B L-2paneI1l8m-16m 1- £.4=_ E F E con tun B C l----3 panels ailS ft - 4S ft £A:;: con tant 801---1-401 - 1--1----3 T601 _I 'Cr'=====~B~=~",,=~DF=JE FIG. P15.7 15•• Ora the influence hnes ~ the Be BF and CF of the truss shoWD ID F PI load, are tr.ln"mllted to the bottom h rd the AG. P15.6 15.7 Draw the inftuem;e lines for the forces tn memben Be and CD of the truss shown m Fig. PIS 7 live loads are transmitted to the top chord of the trus 15.' DrJw the IIlftuen 11 port .Ino the fOfl:C In m m ITU ho"n tn hg Plj 6 LI bf,llJllm choro 01 the truss FIG P15.5 B £/ = con~tant 15 It A A BCD .il ;;Y;; I :;, I--- 20 II --41-10 11-1-10 It £1 = constant hO11 III fig: PIS.3, Dctermme the influence hne at 5-ft mtcn ,i1... AG. P15.3 15.4 Draw the influence line... for the reaclion at POrt!. and the ~hcar and bending moment at pamt beam ,ho"n 111 Fig. PIS.4 Detcrmine the influence dinate... at lO·ft intenal..., 15.5 Dra the inftuenex Jine~ for Ihe reactJODI ports and the ~hear and bendmg moment at poiDl FIG. P15•• c8 11m £/:;: cun tanl A 601 SUMMARY 134 CHAPTER 15 Influence lines for Static.lly Indeterminate Structures In 11ll'S •.:h.lph:r t: lh,cu'...ed lIl11llf..'m:e linc~ for statlcall mlll.ll.. tru.. lUl"C' 'hl' rn'o..'dure for (.'on,trudmg such 1R8 b thl,.' Illl..'1lwd l,r con I lelll JcfMmalllll'" c"i'C'lltially lovol ;rudmg the IIl11UCIlCC: linc lor the n:Jundanb h) plaCIng a U(CL'~ Id~ ,It ,I nun~tx'r l,f r<'l11h along the length of the struc:tult tllf c:.Kh ~l ItJlln 11 the lIll1t lo,ld (omputmg the values or dund.lI1t h) lrrl~ Ill!): tht: melhod of con...hh:nt dcformatlo thl,: mllul,.'lKC hnc... for th.c rcdull~anh and. b} applYIng the 11011' of c:qUllit'tnum. detammll1g the mfluence Ime~ for other funclIon of the ...tructllre balu.lllon 01 the dclll'Ctlon Illohed in the application method Ill' Ll'n...I...(('l1t dcfl.'mlJlion can be con. Iderably expedited mg 1J.c1r... la ofrL't:ipn.:x:,l! denectlon~. The procedure for iog qualn.niH intlut:ncc line... for mdetenninate structures by Bre,lau· principle I... prc'Cnted in Section 15.2. SoclIon 15.1 PROBLEMS 15.1 Ora thl,.· II1flucnlc hoe for the reactiom at the ...up· port nd thc hear <tnd hendmg mllment at point B of the beam ho"" m Ilg. PI51 Determine the influence line or· dm t OIl '-m Hllcndl'i Sded tht" reaction at ~uprort C to he th redul1l,b.nt FIG. P15.1, P15.2 15.2 De ne the.: Influence Jines C r the real:tlOm: at the rt ~ r the f Pr. blem Ijib) !<Iccung the mo- l I rpon 4 t be the redundant. s....e Fig PI Ci I the mfluen lin r the rC"dl:Uon at uppon { d bendmg moment at pomt B of the beam
  • 213. C ~----- L ----+_ A 11I.,.5.15 11.'. Draw qualn.bYe inlI_ ment and shear at POlOt A of' Idle:'=== FIJ- PIS 16 Also show the .. dlstnbuted downward hve load posIbve bendmg moment at A and .. shear at If A 11.17 For the bwlelin. r..... """'" mine the arrangements of a unif'onaIy ward live load w that will cause the bendIng moment at pomt ..4 and till beneli.. mOlnent al po"nl B B 11I.'15.16 "''11.17 -+--L 000 ., ,& 15 .... LCI UI8II far S1I1Ic8IIJ' ............. ~ ... ,.5.12 ..... ,u 11.,. .... ,I.,. CIa ~ ....,. 1--- L --+-.-+-.-~- L --'+--L I~-"....;.-,,--,,--~ B C ..--:--:....;.-;;--1IIr
  • 214. 16.1 SLOPE-OEFLECTIOI EQUATIONS • r r _.... --CllAPI8l I. 51a11t-DoIIectIOn- the fundamentals of this method proVtdes a valuable intral~ matnx tltrues method hleh fonns the baSIS of m are currently used for slruclural analysis. We first derise the fundamental relallonsbips 1I01lII. apphcallon of the slope-deflection method and tbco concept of the slope-defleclton method We cOllSldcr the the method to the anall" of conltnuous beams and PI Dill of the frame In hleh joint ticlOslatIons are prevented sider the analy I of frames with joint translatlOns. When a continuous beam or a frame is subjected to CIl1le1llll temal moment generally deselop at the ends of Its inClividai The SIQpe-d(lIt!( llon equatlOnr relate Ihe moments at the 10 'he rolations and di.fplacemen1s of lis ends and lhe eo plied to the memher. To derive the slope-deflection equations, let us focus on an arbitrary member AB of the continuous beam 16.1 a). When the beam 's subjected to external loads and tlements. member AB deforms, as shown in the figure and ments are induced at its ends. The free-body diagram and curve for member AB are shown using an exaggerated 16.1(b). As indicated in this figure, double-subscript notabOO member end moments, with the first subscript identifyIng end at which the moment aclS and the second subscnpt other end of the member. Thus, MAS denotes the moment member AB, whereas MSA represents the moment at end B AB. Also. as shown in Fig. 16.I(b), OA and Os denote rotatIOns of ends A and B of the member with respect deformed honzontal) position of the member; <1 den translation between the two ends of the member m the pendicular to the undeformed axis of the member, and the notes the rotation of the member's chord (I.e the mm,1i necting the deformed poslt,ons of the member ends due translalton <1 Smce the deformat,ons are assumed to be rotation can be expressed a <1 L 111 member end moments end rotallOns QN/ hurd fUiallOn ar, po lllve wMn colUt/erd -
  • 215. T • lei ~~A:----------'"'i JW A
  • 216. _ ••71117_.-"5 By subotilUbllll Ihi cquabon mlo Eq 16 Sb ad cquabon (or AI we obtain 2EJ M. L1JJ and by subotilUlillll Eq 166a mID ather Eq J,...",IIUI! obtain the expmllllOll (or M••
  • 217. au . . . . . , DI I __ ,. C~A--z.. _ -'r_·~-J-:~==L-=-~'~~_-_-J,_ (I FIUd Bam I ,. I :(b) Simple-Bam 1IeadinI......."hI I (e) Jllud.Ead "=_.. _11 of the area IIIIdcr !be IIJDple-bam "Ii about !be eoda A lUId B are liven by
  • 218. • C 18 49 "-D lOk I 1613 I 8eIdillI- DiqdID (k-ft Jll,lU lOk 392k.~ pntrrmlrB:.....~~::--:ilr-----"'DI)4k ft , , , I IlJ8k 327Sk 1877k 49k Ie) Support Reactions 197 16.62 f)S_~(k IICnaIlt.2 -~oI"'111D7 • _ 1662 161 1.5k1ft II 3~rJ':~Jt)1S,,= 1338 B 32 5 d,_ElId M........... Shean FE~ ) EI c ;onslanl E =19.000 k I 1=500in" L - IFEM fB 4£1 L- 1FEM, ) 4t:1 ' II, , r/I , , JEt (L 1/ r/I + FEM 1t1 0. L (l 1/ 1/ 1/ f IS {l the m('lJ,fled !('1pt.'-d.:ftcct!on equations gIVen 16.1~ i:lrc similar in form .loJ Ih... modified slo[X'-Jc:tkl.:ti,)!1 ...quatlon~ can be expressed ( FEM••)"'I + f-E1s. - -~ BecJu It' J~ .md manzed .1 in hll..:h the "ub~npl , rcler" to the ritllillv connected end oftbc ~ here the moment / A ,Il'h and the ~ub~rjpt h idenlifies the II of the member lhe rotallon 01 the hinged end can now be wn To illU'"tr,tte the basic com:cpt of the !'ilope-dcflection method the three-span continuous beam sho".n in Fig. 16.3(a). Altho fIG. 16.3 1 ~ kilt III k A~4.1 (A~(_. 10---"" B ..... -'O:l~ 98 B(./" , --20ft ~1lI"-r-l/)ft+- 1511-----J 'bl CHAPTER 16 Slope-Deflection Method
  • 219. ...............,- FEM. L 12 50 k-ft FEM•• 50 k-fl or 50 k·ft For member BC FEMsc PL 3020 75 k-ft kT--8- r FEM • - 75 k·ft J or 75k ft by coosideriII8 Bas the near end Note that In accordance Wllh the l~llecl1on the counterclockWIse fixed-end moments are consIdered lO be JlOIIII Smc:e no external loads act on member CD Its fixed-end m men are thalli FEM D FEM 0 The fixed..,nd moments are shown on the diapam f trIIcture In Fig 163 c The s1ope-dellectioo equations for the lluee members unus beam can now be wnnen by usmg Eq I 9 rotariom upporlS f the _bDunUS beam transIa the rd 1hree members are zero I e 6 poria A and D are fixed. the rotabOOS 169 for member AB WIth A the obtam the ~ equa 2E/ O The unknown joint rotations arc determined by solving the equa equilibrium of the joinh thaI arc free to rotate. The free-body of the members and joints Band C of the continuous beam are s Fig. 16.3(b). In addItion 10 the extemal loads, each member I. IU to an mternal moment at each of its ends. Since the correct sen member end moments are not )oct knoiIl. it is assumed that 1M menh at the ends of all the members are positive (counte In accordance '-ith the slope·denection sign convention adopted preceding section. Note Ihat Ihe free-bod~ diagrdms of lhe JO the member end momcnh actmg in an opposite (clockWise diJllCIiili! accordance '-ith . c'-ton la'- of aClion and reaclion. Because the enllre structure is in equilibrium. each of Its and JOints mu't also be In equilibrium. By applying lhe mlOJlllSll.l hbnum c4u.tion, L: .11. () and L: At, _ () respectively to bodies of JOints Band C '-e oblam the equilibrium equatlODS II.A + II., = 0 .((8 t It I' 0 Degrees of Freedom With the loint l()(';i.Itions nOll c..tablished C identify the unla~ol dept.-odenl dl'.placement-. If,tn,kuion.. and rotations) of the J01DtI ,lruclure, Tht'.e unkno11 jOint displaccmenh are referred to U qrees of!rlrd(mr of the ..trueture. From the qualitatle deflected the continuou.. beam ..ho11 in hg. 16.3(a), e can see that nODe jomb can translate Furthermore. the fixed joints A and D aumaw tate. herea.. J0lnl'! Band C arc free to rotate. Thus the COD beam ha.. tO degree~ of freedom. 0B and Oc which represent know n rotation.. of joint~ Band C. respectively. The number of degrees of freedom IS sometimes called the kint:matic ind(!lt'rnlinac:r of the structure. Since the beam of FIg. 1 ha~ two degree~ of freedom. it I~ considered to be kinematicaUy mmale to the second degree. / structure without any degrees dom i~ termed kinemalically dt'laminate. In other words, if the placement~ of all the jOlO1S of a structure are either zero or kno structure is considered to be kinematically detcnninate. ,tru(lure .K1U.111~ !.:l)nlsb of ~l 'lI1glc l:(lnlmuous beam between Urr('1rt A .1Il0 D, for tht,.' purro",,, l~1 anal},is It ,is. consIdered: .:omfl4.'1scd (If thn.'C' m~mher ., 8. B( .InU CD. ngldly COD lOll1h f. B. C .InO D 1((,11"0 .11 111(,.' supporh of the structure the I.:Ol1lll1UOU 1x,Im h.... ~CI1 di, ulc:d 11110 members and j 0 1l1ta. the unknOI1" lerna I rc,ll:LJom, i.ll.:t onl} at the JOlOts. Equations of EqUilibrium MIl CHAPTER 16 Slop.-O.flection M.thod
  • 220. .. I kor o IIr -71 7 71 0 491 491 0 o 8, 000 rad 8 00018 rad Member End Shean The member end momen" )USI compuled diagnun of the memben and) 10" ID Fig 16 the ends of memben can now be de1enniDed ofequih"bnwn to the ftee bodies the mem M 0 392 S 20 Member End .....,11 The momen at lhe ends rth Ih mc:mbe • ~~'.~:~:''.:beam can now be determined by ubo!lIUltng the nn EJ and £18 rnto the s10JlHeflect equa Eq M. 0 I 08 46 SO 392 k-ft M,. 02 108 46 SO 71 k-ft MB( 02 10846 01132 M. 0218382 01 10846 5 49 1 k·ft or 49 I k-ft M D 026718382 491k.ft Moe 0133 18382 24 4k·ft Note that a positive answer for an end moment indICa thai counterclockwise whereas a negative answer for an end momen unpl a clockWise sense. To check that the solulton or Slmullaneous equalt"'" Eq 161 has been earned out correctly the numencal values of member end moments should be .ubslttuted into the )0101 equiJ.briwn equa on (Eqs. 16.17 If the solution IS correct then the eqwbbriwn equall should be satISfied. IICTJaIIIl2 ..... e....,.......,..,DD,IIIRIooI._ B UbollluUn lho: n me: I 01 r £ and I SIlO 'n SIlO 12' ft' de aode. be mu 2EI ., I 0 'L7£11- -Ie . •" I, 15 '21:.1 - 0" 0.133£/l/, 15 Io( ~EI :!.tJB I), 75 O~E/Il. + 0.1 EIII, + 75If 20 2FI :!.fJl +II. 75 O.~£fll, +0.1£///. - 75II • ~O and for mcmtx-r cn. These lope-ddledion equations automatu:all) satisf) the com condition, of the '1rUl.:lufC. Smce the member ends are rigidly to the adjacent joint. the rotatlon~ of member ends are equal to lations of the adjacent joints. Thu~. the (J terms in the slope equation... Eqs. 16.1R)) reprcCnt the rotations of the member '" ell as (hose of the jomb. Joint Rotations Smlll.lfl~ b) .Irrl~mg flj. 16.9) for member Be. c obtain 21:1"'0 0 01 50 O.'~EIf}. - 50 :0 - II 1.10. 1fJ8 46 k-ft UO, 18l~2 k.ft or 1I.4EIfI. +0 I Ellie ··25 and by substltutmg I'qs II ~.18d! and (16.1 Xc) into Eq. (16 I IO.~EIfI( ,IIIEIfI. 75) _ O.2~7EIO 0 or To dClcmline the unknown joint rotations 08 and tic. we SUbstl slope-dene'Clion equations (Eqs. (16.18)) inlo Ihe joint eq equalions (Eqs. (16.17)) and solve the resulling system of equat! multaneously for O. and II, Thus by substituting Eqs. (161811 116.18c) 1010 Fq. (16.17a), e obtam (O~EIO. 50)· (0.2EI0. ' O.IE///, + 75) 0 Olf.///. + O.4~7f.10( ·75 Soh 109 Eqs. I1>.1 ~d and (I h.1 % S1mulldneously for £/9. "-e obtam I5ll ClW'T£R 16 Slope-Deflection MelIIod
  • 221. --....,...- F Il I 5 20 127 JO 4 L" 0 '92 I 55 5 18 7 I 244 0 Chedcs S(n 49k So< 49 k ~91 II .Ioisc I' n k . .(I 'CR D87 k 13.38 k 1 .19.2 717 20 20 1 5120 " 1/ II _ 0 ~F II ~/ II The: fOfcgl1mg mcmlxr end ,he.u can. alternatil~I)I be evaluated b~ ~UJ"ll:rp..l"ltilm or ('no hear.. due: 10 the c:lcmalload and each of the cont! nKlm('lH~ .tl.:un~ ·rar.t1d~ tl l1 the ffiemocr. For example the shear at cnd .-1 of ffieml'ler -tB i, glcn b) CHAPTER 16 Slope-Deflection Method152 In "hleh the fin•• tcrm equah the ~hear due to the 1.5-k/ft unifonnly di..tribuh.-d Illad. "herea.. the ~ond and third terms arc the shears due to the ~9..:!-I..·ft and 71."'·k-ft moments. respectiH:I). at the ends A and 8 of the memher Support Reactions From the frce·body diagram of Joint B in Fig. 16.3(d), we can see that the erlical reaction at the roller support B is equal to the sum of the shears at ends B of members AD and BC; that is, B S.,+S" ~ 1662+1613~32.75kr Similar!} the crtlcal reaction at the roller support C equals the sum of Ihe ::hean. at end~ C of members BC and CD. Thus c - S" + S<o - 13.87 ,49 = 8.77 k f The reacuon:: at the fixed support A are equal to the shear and moment allhe end A of member AB; that is. A S.. -13.38k W. II .. - 392 k-ft ) Similarl} the reactions at the fixed support D equal the shear and m0- ment al end D of member CD. Thus D Soc 49 k I 10 10< 244k-ft) The suppon reacuon are shown m Fig 16 3 e . Shear and Bending Moment Diagrams With the suppon reacUons known the hear and bendlDg momtDt d~­ grams can now be constructed In the usual manner by U51Dg the btam sign com nlion descnbed In Section 5.1. Thr shear and bendmg m ment diagrams thus obtained for the continuous beam are shown 10 Fig 16.3 f and (g) respectively 16.3 ANALVSIS OF CONTINUOUS BEAMS Based on tbe d,SCUSSIon presented on tbe precedmg tI n .... procedure for Ihe analySIS ofconlmuOUS bealll> by the s1ope-<lcftecboo method be summarized as foDows I. Identify the degrees of freedom of lbe truewe F rolalio", beams. the degrees of freedom CODSI .... unknown theJOIDIS 1. Compute lixed-end moments For each member evaluate the lixed-end moments due to the the expemons g",en inside the bock ""' of the ten:Iock lixed-end momen "" I bee:~~:~:3. In the of support senJemen de the : chotds of memben adJ8COllI 10 the uppo I !be re e uanslabOD beWeCIl .... the member IengIh A L The ho
  • 222. -1fCl1lII1U ....,." C1C, _ I k 18k 2kill 3,n.lI( I B C )14 til A D t t I 81n 3741k 43k d~- 27.57 • It ,.,1U BeD 18k M~IA I MMC1B1) rm+mmM Bl ) ~ M (IB 1) 1 b 18k 984 27.57 3S6(lA I (lBl) rm+mmBl ).01.5 :It 10.5( IB Cl) 816 984 1 27.57 4 B =3141 (e) Member End MomenIs mel Shan 4 2 0.08£/0. 64 8 ISO 0133£/0, ISO 432 -0.16£/0.-432 ISO 00667£/0. - ISO 2£/ 20 30 ' 2£/ 0 30 ' 2£/ 25 0.)+648 2£/ 20 25 • M, FEM. Pa~ 648 k-ft , 64 8 k-llor L Pa ~ - 43.2 k·ft -43 H-llFEMs f or L FEMB( "L 2 30 - 150 k-fl , +150 k-ftor 12 12 FEMcs 150 k-ft ;J or -150 k-fl Equ,librQlnt Equal on The free-body d18graJD of JOint B 11 showD m 16 b thaI the member end momenb whICh .... asawned 10 be counte lock directiOD on the ends of the members must be .ppIied ID Sumlarl) by applymg Eq 169 for member Be. we obtam the lopc:-deftecliOD equation ole that in accordance with the slope-deflection sign convention the counter- clockWIse fixed-end moments are considered as positIVe, whereas the clockWiSe fixed-end moments are considered to be negative. Chord Rotations Smce no support settlements occur, the chord rotabons of both members are zero; that is'''AB "8e O. Slope-DeflectIOn EquatlOIH To relate the member end moments to the un- known Joint rotation, 08. we write the slope-deftection equations for the cwo members of the structure by applying Eq. (16.9). Note that smce the suppon A and C are fixed, the rotations 0.,4 - Oc O. Thus the slopc-deftectlon equation for member AB can be expressed as Dt:!emllne the rt3.l.:1lllnS mJ dnt the "hear and hendmg moment dUlgranu Ii tht 10- pan l,;(l111IRUOUS ~am shm" In ftg 165(a by the IOpe-deftceuon m hod Solution Ikq f',f dfmr From Fig. 16 'i(a. e can see that only Joml Sofdle beam IS free to rotate Thus the structure ha.. onl) one degree of freedom wtticb 1 the unkno,," Jomt rotation Os· Fi J.£nd"", ntf B) usmg the fixed-end moment ('xpres Ion given in_ SIde the back coer (If the book 'oe C'aluatc the fixed-end moments due to the e temalloads (or eal,:h member ClW'TBI 18 SI....·DeneeU.. MelIlod Ex-" 18.1 -
  • 223. - (CC.CI::1 ~D;')I02 Ie J kJft If BeamCoo £1_ h IECTtOII1U """"' .. Clnrnul __ B t4 k C -471 3kJft Member End Momen and Shean tb Free·Body Diagrams of J n Bonde 3klft ~1 201 02qprr~p7112 c 27 21 I ...1" B =477 63k A 18 fl 3 klft 20.7 27 2Jr~~026J 20.7 Ans. CheCkS I (j(,X o 1UE/OII • 150 I 0 EF 0 8.16 18 )741 -2 30) + 32.43 0 IIIMI II .p" /1B OIlX .~M :' . b-IX 356 k-ft ": 1H~ (I Ib '645 i 432 1015 k·ft or 101 5 k·ft; 1il( oU' .'M5J 1'0 1015 k·ft ) /l II O.ll6b71 JM 5) ISO I74J k·ft or 174Jk·ft; fwm hh:h The ..upport reaction acc hon in fig. Ib.5(dj. ~otc Ihat it J'llhllic an.....:r for an cnd moment indicates that ih o;ense is coun. h:n.:ln':.....I'oC. "herea.. a negatie an,,"er for an end moment imphc3 a c1ocklSC -.en....: Sinl:e the end mumenh .f8~ and .IBe are equal in magnitude but oppo. '>Ite In 'C'n~. the cquihl:lOum equation. 18 I +- .H8(' - 0, is indeed satisfied. .lo1/ha Em' Shc·t", The member end ,>hear~, obtained b} considering the equilibrium of each member. an: !'Iho"n in Fig. 16.5(c). Supporf RUJctionl The ceactionl> at the fix.ed l>UppUrb A and C are equal to the force" and momenh at the emh of the members connected to the3C jOints. To determine the reaction at the roller support B, c con!'lider the equJllbrium of the free bod) of joint B In the ertlcal direction (see Fi8. 16.5(c)) to obtain B SS4 SIIC 9.K4 + 27.57 - ~7.41 k 1 <; E 1/( 0 J5b 81b 55 1845 374 30 230 15 174 J 0.2 ",0 CheCkS Sh rand B ndrnq lomn" DUll/ram The shear and bendlDl moment di- agram can 00 be .,;on tructed b) usmg the Mum SJyn cona: "'10" delaibed II n S I Thne diagram are hon In Fig. 16 S c and f AJtS. J Inl R It" ToJ d.:lI:nmnc the unKno" n lomt rotalt..ln..". .1: SUhstllulc th.. Inp.:--dd'k-ctHln etjUJtllm f q ~ .Od .' I mto the cqUlhhnum equation (q 5 III (lbt.;,un fB~ 11l( 0 Eqwllhrlllm (ht'Ck To t:heck our l.:alt:ulation.. of member end shears and upport reactions e appl~ the equations of equilibnum to the free body of (be entire trUl:ture Thu !iCC hg 16.5(d OpJ'ltl II; .. kld! Jlrl'dll'll llll tho.: Ir~ Ixxl) ..,I thl,; Jl1lhl In uccl)rdance CUll'lI s lhmJ 1.1 0, .lprl)lllg thl' Ilwml'nt l.'qUlhhnum equation t: M. W1lh to thl: 11 ...'( bl~~ of l(llllt H .. oht.lIn thl: eqUlllhrlum equalll," It milt r Fnd I, III nlf The mcmlxr end momt'nh can no. be computed hy uh tltuUng the numl'nL.11 'alue 01 HOB bad mto the lopc-deftechon equa- tions hi I through 4 Thu CHAPTBl16 Stope-Deflection Method
  • 224. - CllAPlElt1. _alii 0............. 21 -21 (e) Shear Diagram (k) 51 3 ...............1 • _ •••.0.1 II IE/Ie 4 102 -102 M,1U (0 B,ndillll Moment Diagram (t II) Ioluaan IHgru of Futdom (JB and (Jc FI td End Momenls FEM. 324k-ln or 324k II 48 6 t-Il ~ or 48 6 k.fl 81 k-Il , or 81k II or 81 k.fl 486kll' or +48.6 k.fl 324 k-II or 34k
  • 225. - lOkN 20 kN m( :-:-_-:1D IC lOkN fb StabCally Detenninarc CanliIC'ct Portion 10kNim SECTION 11.3 AnaIyoiI 01 _ _ EI • constaftl (a) ConIlQUOUS Beam (d) Free-Body Diagrams of Joints Band C !--6m-+-9m_ _ 30 kl'l Ai B~120kNm (c) Stabcally lnde1mninate Part to be Analyzed 30 13 7~1-_.-J!4r1...!.l...!-!~l-L;jr--"ID I I ~127687 4159 OSupplldIleo<""'" 6873472 552730 lOkN 13(....,...---::-) (IBI) (DJi]j]m) CW) (--'~ IA BI 275;;; 275 I B CIllO 120 IC 687 I 3472 5527 I 30 6.87 . By' 41.59 C • 8527k (e) Member End Momemsllld Sbean ... lU 225 kN .mAns. Ans. or18. -OJ 125 1875- -225kN-m WBe> 06 125 +300= 225 kN m~ FI'C J·£nd Alomml FEM,." FEM6A 0 FEM -~ 675k m) or +675 k m ... 12 FEMe. 675k m) or 675 k m JOIllI R( tollOn To delcnmnc the unknown j~mt rotation Os, we substitute lhe slope-deflection equatIons Eqs. 1) and (2)) mto the equilibrium equation Eq. 3 to obtain 0.9E108 - -1125 '[em"" End Sht:tJn and Support ReQcI;om" See Fig. 16.7(c) and (d) Equlllhrrum Cht(k See fig:. 16.7(d) + I L F 0 52 5 - 15(20) + 225 - 60 + 82.5 ~ 0 Checks +(LMD~O -525(20) + 15(20)(10) - 225(10) +60(5) ~ 0 Checks 0.lE108 187.5 +(0.6EltJs --+-- 300) - 0 EU~B - - 125 kN - m~ from hu::h Detennine the member end moments and reactions for the continuous beam shown in Fig. 16.8(a) by the slope-deflection method. SOlution Since the moment and shear at end C of the cantilever member CD of the beam can be computed dlfcctly by applying the equations of equilibrium (see Fig. 16.8(b) It IS not necessary to include thiS member in the analysis. Thus, only the mdetennmate part AC of the beam shown in Fig. 16.8(c), needs to be analyud Note that as shown m thiS figure, the 120-kN . m moment and the 3O-kN fora: exerted at jOlOt C by the cantilever CD must be included in the analysis. lkg"e~ of Freedom From FIB 16.8(c) we can see that jOlOts B and C are free to rotate Thus the structure to be analyzed has two degrees of freedom hlCh are the unknown Jomt rotations 86 and 8e· or #emh , End fommt The member end moments can now be computed b) substituting: the numencal alue of £/88 into the slope-deftection equations Eqs. I and 2 Thus IEumplI1U
  • 226. 117 [J n and SECT.'•.3 ........ at ConInllll .... d Member End ttl (el free-Bod) Diagnutlll of Jomt Band C ~m £_7 CPa C""tinuous_ B fbi Owrd Rolan Due t Suppon Sdt ~ M.1U ~1~6::.5:.- ~=~~ -===~_ 6 (7 (8) Ans MS. Ans MS. m) m) 137 kor () I 111£108 , o222EIl1c - -675 It> Q) tl nwmhc:r "B lind Be 'EI tl n3F10sI 0 211 " llM~1I0.I 0 'F! ~ n 07 ' 0-U4£/88 o~22n!l{ • t>1 'iI 9 '1./ 0 (''' 5 (I ~~~£ln8 11444EW( I. " 9 l'tl S 4 lll~ () 667 412'; 275 k. m or 275 k It, 0444 412S 0222 97.62 675 27 S kN m I O'l"ll 412S () 444 9762 675 120 k m or 120 k m) Equl r WI! fi/I II n{ 8~ l..-on"lI.knng the mornen! I.."quihbriurn of the fl'C't I:o...(hc of l(llOts B .J.nd ( fig 16 Md ~c obtain the equilibnum ~quatllln~ ElOH 4125 kN m1 EIlJ{ -97.62 kN m2 ). Inl R 1011 n Suh<.tltutlln If the ..Iopc-deflection equatlonll Eq 12 through -t mill the equilibrium equatlom lEq". 15) and (6) }ic1d.. 8) ..ohing Fq 7) and II( ..imullaneousl), .....e detenninc the .. alues of ElfIs and EW, 10 llI.- te that the numemal ..aloe of I"... I" and Ie" do salls! the eqwbb- nurn equall n Eq S and 6 1 m EnJ ~a' and Support Rt'Q lrom Sec fig 168 e and f AnS f. ull , unr C Ii The eqwhbnum equations cha;k. I.mlta End lOlllt'fl/ The member end moments can no be computed by lIuhstltuting the numem:al .. alue.. of £/8. and EIlJ( into the slope-deflectton equation.. l-.q. I throug:h (4))' I~, (J 311 41.25, -131 kN m ClIAPmI'8 SIope·_ MolIlod..
  • 227. - AnI AnI Ana. An&. AnI An&. An&. o 9 M 2£1 See FI I I M. 98 kN m MIA 91 k m 101",- 91 k m or 91 k m M,= 56k m or 56 k m II SECTION 1U ..... of ewlt Til ..... O. 40 000 5 M. 2EI 8. o MD S6k m 101 -18k m M"""" EM _ and Suppor 1IIoc Efoi/ibr- c See F... 16 I: 0 236) 4 Member End Moments To compute the member end momen lute the numencal values of (J. (J and El 70 800 56 000 k the naht Sides of the slope-deftecllon equation Eq I throup 6 I By solving Eq 9 and 10 Simultaneously we detcrmme 8. 00005 rad o 0002 rod Jom, RotatIon. Substltutlon r the ~Olpe-<Icft<:cti'>n <qua,tio", through 6 mto the eqUIlibrium equall n -0.0025 0.02 8 -IN 7 (I) She-dT Diagram (kN) 91 (8) Bendmg Moment Diagram (kN· m) .>. 105 It' 1 A 8 C -9K ~'" 0 ~2 _ 0 0025 From Fig 169 b we can see that l/lco 0 Sl 'fN-Dejk "011 Equations Applymg Eq. 169 to members AS Be and CD WCinte 2EI 8 O. 00075 In whIch the negative Sign has been assigned to the value of "'AB to mdlcate that Its direction 15 clockWise as shown In Fig. 16.9(b . Slmilarly the chord rotation for member Be IS Chord Rotalwln The specified support settlement is depicted in Fig 16.9{b). using an exaggcr'dted scale. The inclined dashed lines in this figure indi- cate the chords (not the elastic curves) of the members in the deformed positions Because the length of member AB is 8 m the rotation of its chord is SolutIon Degrers of Freedom Os and fie . Filed-End Mom('nt~ Since no external loads act on the beam, the fixed-end moments are zero. C1W'TB116 $Il1jlt-Delloction Molhod - fI&. lU cold
  • 228. 1fl:IIOlI1U __ .. c.Coo"",, _ 1ft F : 1U: Determm the member end moment and reactions for the three-span contlnu ous beam hown In b 1610 a due to the umformly dlstnbuted load and due to the uppon nlem nt of In OIl B J In OIl ( and .. In at D. U~ the lope deflection method Solution lH r I Fr. "" Although all four Jomts of the beam are free to rotale " 111 ehmmate the rotations of the Imple supports at the ends If and D from the an h s b) usmg the modified slope-deftection equations for members AS and (D respecmel) Thu the analysl 111 mvohe ani) two unknon JOint rotatton liB and He FEM.. FEM. FEM/>( 66 7 k-ft J FI J-End I m nit 1Y ft C l!sc } to - -f~ c (b Cbord Rotati... Due to Support _ AI ~l...!....&.~br:J..~4.J...LJU.-! 0... B C ~20ft---l----20ft--1--- £ 29.000 1_ 7.800 lD 4 c.m;...... 1lcam 66 7k-ftor -66.7 k-ftor _ 66 7 k-ft ~ 220 12 FEMeFEM..FEM B in which the negative Sign has been assigned to the value of !/IA' to indicate that Its direction IS clockWise as shown m Fig 16.1O(b) The chord rotation for member BC can be computed In a Similar manner by usmg the settlement of upports Band C From Fig 16 10 b we observe that the relative settlement between the end of member BC IS I tn tn 0875 In. 00729 ft and (hord RllallOns The specified support settlements are depicted 10 FIg. 1610 b usmg an exaggerated scale The mcllned dashed hnes In thiS figure m· dlcate the chords not the elastic curves of the members In the defonned POSI- tions. It can be seen from thiS figure that since support A does not settle but upport B settles by In the relative settlement between the two ends of mem- ber AB is In 00521 ft Because the length of member AB IS 20 ft the rota· tlon of the chord of member A8 IS 00521 20 0.0026 (el Free Body DiBgmns of Jomls Band C 41 38 81 79 808 41 79 204 808 o::=IT=o) (IBI) ( o:TI=o)CW)(o:TI=olA BI4277~4277IB CI IC I 1 38 41 38 I 81 79 41 79 I 4 . B_I2317 C_6219 Cd MembcrEnd- andSheaB 000365 Similarl the hard rotation for member CD IS 000313 .[H II EqIlQI M o ,.,11.1' 0026 100 OI5£IB. 000039£1 100
  • 229. m L FEM HI II If HCTIOlIII.3 AIlIlyIlI ol _ _ (a> Continuous Beam ,., Ap-ib'leI--L I L ----j 111)One-Holf Beam with Symmetn< IIouDdUl' Condi- ... 11.11 , { , A P'l~B'-'--.,-tJ ';g:qE I--L ---- L ---- L --l--L--I £1 =constanl 6 PI (KI Ans. Ans. Ans, Ans -2.307.24 0.00149£1 +~:U 0.00063£1 J13 0' 427.7 k-ft ) or 808 k-ft ) EJOII 6.268.81 k·ft~ £If" - 1,131.57 k-fl' 0,1£/(11/ ().3SE/fJ( =- ~1.022,93 lilt -427 7 k·ft .IS( 421 7 k-f1 .... "'<'8 !10K k·ft , Ito KOK k-ft I :!I).lMXl 1",KUO) I:!' ~..rt~ into the right ~ides of the aboc o 'SElfl. 0.1 EIfI{ oI E1f/B +II 'SflfJ( 'f1 II, 1{ i)I)()lhS • t>tl 7'.,~u o"I'tl 111 flO, ( ("(111 f.I 66 i ~/.I .. J "- 3 n()().lbS I "" 720 _I oIt/II, n:'/HI n lMIII f.I "f>' In unOlll IOn OI5UOt,. OOlM~71.' Ion .:'11 0 • 0 Ans.II , Equ I J, 11m Equall" St.': hg Ib.101.: lBi 'fB( 0 fl' /1 0 - 0 II If If _ #l'mb.., E"J Shran and Support R('dcl;uns See hg. 16.IO{d and (e Ans EquIlibrium (nt" Ii The equilibrium equation~ check. '. e pre IOU I) analyzed lhe l,;onUnuou~ beam consuJered h~ m Exam- ple 13 14 by the method of L-on1 lent defonnatloos. Theoreticall) lhe slope- defla..1.lon method and the method of consistent defonnation~ should yteld IdentICal ults for a gIVen structure The small dlfferellCe$ between Ihe mulls determined he and those oblallled m Example I '.14 are due 10 the round-oft' errOR "'tn/her End .lOInol/ To compute the member end moments, c sub- 'ititute the numem:al alues of Elfl, and E/Oe back into the ~Iope·denection equation IEqs f1 throug.h 14)) to obtain By ~ohing Eq"'. 7) and (8) Imultaneou,Jy. we determine the "alue> of ElOs and EIfI( III be j, Int Ro at III 8~ uh tltutmg the ,lope-deflection equdtion Il:q through ~ 1010 the eqUilibrium «Iuall0n I Eq"_ 5. and (6). ",e obtam Sub~titullng EI equatton.. )' Il:h,h CHAPWl16 S1opo-_" Metllod172
  • 230. CltAPTfR 16 Slope-DenllC1lon Melllod "l. lL HL .... L nL "L ., ~ 2 T 1 2 " I) (181) (el', 1)(lcl)(, ,'o)(kb(c:ilJ) 81 ~ 18 cl ~'Ic DI' , I D E:I' L .e!!!... .L 1.Lo .L- 1.L- 12 L 12 r 12 "2L "i 12 12 "2L "2L 12 12 "21. ;1. 8="L c ..L D,="L Cd) Member End Moment and Shear.> 'I~ CI" • ~ .L',,18 ' ,I' , D E: )12C I I I I I .L .L .L .L .L T T (c) Support Reactions .L .L .L .L T T T 2 .L' .L' wL2 wL2 24 24 24 24 A D- 8 A c & D ~ E: E: V, / l, V ~.L .L .L .L .L- ..L ..L wL2 wL' T -T -T -T -If -If -If -If -12 (0 Shear Diagram (8) Bending Moment Diagram R6.16.11 coold The shean. at the ends of member AS are detenruned by considenng lhe eqUI- hbnum of the member The shear and moments at the ends of member Be can now be obtan~ by reflectmg the corn pandmg responses of member AB to the oght of the aXis and the member end moments and shears on the nght half of tbe beam can be ddenmned b) reftectmg the correspondmg responses on the lefl half to the other uJe of the au The mc:mber end moments and shears thus Obtained are !ilhOn In Fig 16 II d and the support reactions are given In Fig 1611 e The shear and bendmg moment diagrams for the beam are shown 1R Fil 16 II f and g respectl" I) AnI A thl eumple silo.... the uuhzatlon of structural symmetry can C()IlSider ably reduce the: mputatlonal effort reqUired In the analysiS The beam coosid"' ered 10 this mple Fig 16 11 a ha three degrees offrcedom, 8. 8 and Herby kiD' ad aDtase f the lrUcture s symmetry we were able IInma all the de of freedom from the analySIS RCT1OIlI6A .....,...., _ _ IIIlei_OfOf 16.• ANALYSIS OF FRAMES WITHOUT SIOESWAY PI
  • 231. ."1KT1CIII1U .... ,, _ _ '.111'_"11 B ,I I I I E £J=con tant , ---~-­, 1 Axis of ......- symmelr}',, (b) Frame with Sidesway D I I I, B B ~ E~ D'H~T-r.:~~-1I I I I I I I I I I (a) Frune without Sidesway (ej S)'IlUDCIIIC Frome SUbjcclallD S)'IlUDCIIIC Loading - No Sidesway I 80' c JOIDts prevents one JOlOt translation ID Its axial directioo. The Dumber of independent)OlDl translal10ns I then obtained by ubtractiDS from the total Dumber of _ble translal10ns of I f.... JOID" the Dumber of translalJons _ by the supports and members of the fnunc We ean .mfy our <:onclUl1Olll about the frames of FIgs. 16 12 a and (b ClW'IEIIl' ......._ - . " ... 11.12
  • 232. k k k t Itt nr B umg the fix~ nd m r fthe book II; bl In FE"', 40 20 100kft" HM 4 100 II. ft ~ 100 k- FEM FEM FEM FEM 2 )(, I k- FEI FEM k , ext vtC "nle the Ope~tl«h AC BD and CD and Eq It, .2E/ (1 100 oIE/U 100 20 114 2EI "l 100 - 0 2E/() 100 20 ., it'D 2EI 0 0,1 fJOp 20 (J fDB 2EI(2H 0.2£/fll) 20 D ltD 2£ 21 12l1( +HDI +1;0 0267£111 o )'3EI 30 2£ 2J 20D +Dc - 150 o I HEW 02 £1 It JO (ISO I~) 02£1 "It An. ltD 0 lib EqIMlI By appJ)lR the mOlllC'n '[; MliJtw0 ;~ em: bodies of on' C and D F 1 eqwlibnum eqUibODS IlCTIOIIIU ....,... 01 , .._ _ ......, f dEnd M In Id the back 23.14 B -145 9.7,+- 6921 (d) Support Rcacuons lhl Fr«·Bod,. Dlagram~of JOint!> C and 0 E=- !Y.OflO II. I (al Frame 1,4;- B 9*69.21 (C I Member End Momem... Shears and Alr.iaJ Force~ 18S_ A 276~ ~1I'i9 C - 21.2 CHAPTER 16 $IotIe-Ooftection Molhod - 6~ ! lifl '2 15 ~6_86 .2 klft C I~ 21 2 L I!'-'''':'!'-'''':!'-'''':'!'-'''':''-:D~'~ 21l LID I~ 2265 L I~D::-''------''------'-'-'--"-22 65 21._'/ .IC 1:1 .1::1."::1 IE_I~ 115.9 1864 -1.,205.7 I 27.65 J! '5 '-V -. 36.86 23.14 69.21 ---h193 1.45 - D 4011._ m "'- 11.13
  • 233. - 4 541.8 +54.8 M £1 M _,U ....,..... _ =~':ti;:;~';;'29.000)(8011)/(21' k 11.5 MALYSIS OF FIWIII WI11I (b) CbonI RIMa.ODS !lbe 10 SIIJIIIOR SettIcmco" MDC(:f;)MDE ..v MDt c) Free-Body Diapams of Jomts C and D 069- B 4~721 d Member IilId MomeIIII _1IId Axial FoI<:ea ...,..,. .:+ 4.6't' 4.6 7 1 SuppoR- a) Frame D E C 0 ~ tn. u-- B 1--30ft---+,--30 ft A tml_l_ E=29000ks. B' A -411 4'-Y 46 46 854 467 254 76.2 C • ~411 L C:- ::-: ' 4 II floh 48 f :-:----~£ -48 411-:-[;,1 54~1 De T_I JO t.,... 467 467 -...j..-I069 2.54 254 M 46 721 ~S48 --h92 C _411 0.69- o _ 11 .; Da. 1', _ -
  • 234. - and Next let us consider the dl placement diagram DDID It has been shown preVlou I tha'DD I and parallel to CC Therefore DO, DD A cos CC DD DD n 1ECnoI1U AooI,oio "' _ _ .'.'_ .. In whICh the ch rd 11 n be negative beca se the ar I k FI rotahan can be expressed In t nIlS f the S1denng the do placemen, diagram f) I 16 16 b SInce CC I perpendicular I AC hi b I fll wIth the vertleal CC must make ,he same ngl tal Thus from the diSplacement diagram fJ In C t can see that at ---L--- Re.l&.16 conld frame to an arbitrary horizontal displacement .6. and draw a qualitative deflected shape of the frame, which is consistent with its support con- ditions as well as with our assumption that the members of the frame are inextensibJe. To draw the deflected shape, which is shown in Fig. 16.16(b). we first imagine that the members BD and CD are discon- nected at JOint D. Since member AC is assumed to be inextenslble joint C can move only in an arc about point A. Furthermore since the translation ofjOint C is assumed to be small, we can consider the arc to be a straight hne perpendicular to member A C. Thus In order to move joint C horizontally by a distance A we must displace It in a direction perpendicular to member AC by a dlr tance CC Fig 16.16{b so that the horizontal component of CC equals A ote that although JOlOt C IS free to rotate Its rotation IS Ig· nored at thiS stage of the analySIS and the elastic curve AC of member AC IS drawn With the tangent at C parallel to the undefonned direction of the member. The member CD remains horizontal aDd translates as a ngId body IOto the position C D) with the displacement DDI equal to CC as hOol1 In the figure SlOce the hOrIZontal member CD II a sumed to be InexteDSJble and the translallon ofJomt D IS assumed to be small the nd D of thiS member can be moved from Its deformed poll. lion D only In the vertical direction. Smlllarly smce member SD IS also (dl ClW'TER 16 S/opt-Dellection Ml1!lod -
  • 235. Sm 8 -a°'4..Ju C....... AC .... 8Dd) F....Body o;.,r- (el Free-Body Diagram of the Entire Frame 8 '1BO 5m 8 I' I : I ' I : I ' I : I I lI'AC ,,/ J ',,,, ,, lb) QualilabVe Det1ected Shipe oftbe Frame FIlL 111.17 A ~C' D~,- ----, ,-- . ~ID' C '--;;00;:---,--...,-"""'1 SoIU11on Dr, "161 b 411 kN C I 401tN 0 IT c 0 7m Lf-'.~'.J£/ =confotant (a) Frame (1~.12b (16.12.) (1~.1I'1 (l6.Jlbl ~ "" 1-. l:""P. Ih lOot VBn I. co... p! 1~.JUh, - tan Pol1iI,0 I Itan P 1~ Jil<; L "I - co, Ii, (tan P, + tan /1,) i. a~ - ------ -- ------ cO,/I,(lanp" tan/I,) 01 cos II] = tI~ cos 11'1 al fli +0] ~in II:. = L By ~oling Eqs. (16.3la) and (16.3Ib) :)imultancously for tI] and d!. I.e obtain Once the equilibrium cquatioO have been e!<otabli!<ohed. the anal)sis can be completed m the u'ual manner. as diS(:ussed pre iousl). The: h1fcgomg c prc"lon, of l:hord rotations can he u-.ed to "Ole the I,,~-ddk'.'uon cqu.ttll'n lhcrcb~ relating member end nWmC:h1s to the thrt."C unkno n jOint di"placemcnh Oc OD, and .,. A.. In the case of the rcdangular frame... c"n'ldcn:d prciou"J}. the thn.."1: eqUlhbrium t."quatlon, nl..X"C' ,ar~ for the ..(~Iullon of the unkno n Join! dl'iplacl.'mcnts can he c..tabh hc.."d b} ...umming the moment'<. aCIlIlg on Joint" C and D and b} ...ummll1g the hMil'onlal fon;c" acting on the entire fr,lme. Ho"_ Cef. for framc!'l !th IIh':)IOCd kg..., it I' usuall) more comcnicni to e..tabli,h the third equihbrium equation b) ~umming the moment-. of all the force' and I.:ouple~ al.:tlllg on the entire frame about a moment I.:cnter O. "hKh i~ 1(l(.lled at the intcr-.cction of the longitudmal axe:.. of the t" 0 indined mcmbel', a~ ,ho n in Fig. 16.16(d). The location of the moment center () I.:an be determined b) using the condition~ ( I-ig. 1~.I6(dll Multistory Frames The foregomg method can be e:lended to the analysis of multistoJ) frame subj«ted to sidesay as illu!l.traled by Example 16.12. Ho- e',er bo.:ause of Ihe I,;onslderahle arnounl of compulallonal effort 10- ohed the analysis of such structures today IS perfonned on computers u)lng the matnll. fonnulalion of lhe displacement method presented 10 (hapter 18 H~ uh"lllUllO~ h.J' Ib,171 thn)ugh lh.2Q) inh rq fI6.:!h) "I.' (lhlctln the- l:h(lrd r,11.1I('11 III the three mcm~r... ltl {":Iln... 01 CHAPltR 16 Slope-DelIlCtion Method680
  • 236. - 40" lEI I "EI I £J~ o24E/~ o14E1~ o 08EI s 04EI and S 0"1 El s 2HD ~ 294 0 286EIH2£/ tJ 7 I , 2£1 "Inv 5 2EI 211 flo 392 OS'IEIfl 026£1 7 IECTI0II1U AnIIyIIa of _ _ •• Wi, 2£1 '1 5 II, IfDB If rEM 2k k fEM 411 4 294 k m H fEM fEM fEM HM (It rd IW ~ ~ In which S and S, repmcnt lhe shea stw-.11 m Fig 16 I £qulllbrlum EqUOtlo1U B} considering the m ment eqwbbnum f J n and D ..lie obtam the eqwlibnurn equations W'4 J 0 I, Y' 0 To estabb h the lhird eqwbbrium equation equatlon L FtOto the free bod of the ent ImD umns F'lI- lId I, d f.nd I Ide th ha k D B -5.8 '-f.-I77 16.47 40kN I <0 Suppon Reacti.... 5.8- B ~77 16.47 c Ie) Member End Momcnb. Shean. and Axial Forces 5.8_ 04 14.6~ 23Sl ... 16.17 amid A _5.8 146~ 23.53 -
  • 237. ., D B 2 - -f1j';"---1~ I)SupponIteoclioas ...,1." COlI'" leI Member End_ - 1IC1IIlI1U ......... _ _ ....' _ A 1764- 911 'Y88 x.:.'8841762 91.7 . rhM -SDITDI D I 16ft s_s,J'fMID D 16 ft MIDI ~i$BD - t o ",,,,,,,,,,,, :n3J ft / / 26.61 ft ,,,,,,,,,,,, c 1---20 f t - - - (c) Free-Body Diagram of the Entire Frame (d) ......Body Diqnms orCoIUlDD1 AC ODd BD CHAPTER II _ III ""1.,,
  • 238. -$' TE·••di/,,,urn ( 1lC'l1OII1U -,..."'_ , Det nmne the det1lXllon f J I F defteclt n m thod IExample 16.12 9 10 II 12 .1Hf} +.fIlS SHU ~ 16 and If{ - f( (53. B) • Jls/) S80(42.67). 10(26,67) 0TL 10 () B) !>ubstitutlOg thc~ expre~ IOns into the third equilibrium cquation. "'"c obtam FqUllthrwm f.quCJIWnI B) con ldcnng the momenl eqUilibrium tlf Joml dnd D "I: ~)hla," tht eqUilibrium cqu.Ullm.. Ilf "'! lrD 0 In ....hir.:h the ,hear.. al the lower end.. of the columns can be expre..sed in Icrm~ of column end moment a, ("oCc Hg. 16.18(d)) IllS ,toc 0 The third equilil:num equation i c'labh,hcd by ..umming the momcnh of all the fon.-.: and couple.. actmg on the fn:c body of the emire fr<lme abnul poml 0 "" hil:h j.. IOl:atcd .tl the inlc!"'C..'dion of the longilUdinal axe., of the two l"olumm. a.. ,h('l" " in fig. ItI.1X(L:}. Thu.. Jmn, DI pIal t'ffl'" Sub)titution 01 the slope-deftection equation) Eq~ through 6 IOto the «jUlJihrium cquation (EW" (7 through (9)) )Ield!i "4E/IIr 0 I E/llo l),lKJ"'5£/~ = 0 U 11JH( II 45J:..Jflo OOI2IEJ~ 0 1I711:JiI( 0877f:/Ho 0.183£h 800 8) IlOg Eq HI through (12 Imultaneou!il} ","e dctennine HOc 66 648 k-ft- t/flo 12"i 912 k·ft Elf> 5233 6k·ft .. CHAPTfR 16 Slope-Deflection Mertlod ./ f ,,1 II 'ti II 1)(,"5 t11l1fl 111118f1'/I '11 'ti II 1)6"<;; II ~II"I onlHIH/1/ '11 't/ II OblCi 1I1'J1"F/O 1/ I. '!.H oOt"''i II ~5111)/) o021-1U 1/ I. • 't/ ~ Q l (loY'5j 02t-W( 11.1 nfl,) 1111111//I 20 1/ 2fl , I II, .1 II oY', o2.ElOn +() 11:1fJ( O()1111l~ 10 ,
  • 239. 1M CHAPTER 16 Slope-Deflection Method P, III.PtI.11 A 8m-----l-- fI=""...... 2kift , ~ I A f!! I !III '1!;~ 1---15 It--I- III.Pl8.10 fI £ GPa / "'- Pt6.ll, Pl8.15 "'-PIli f 6m 8m 1= I.lOO IHi'l mm" 8m E=70GPa 25 kS/m 25 kNlm A~C I 10 , m I 2~m I £ =200 GPo 1=500( 1()6) mnr' fIG. Pt6.5 IOOkN UXH I I Af=~~~~( lJ _9m--+-om+6m - /---1- 2/ - - - - E. = constant AG. Pl6.4, P16.7 fIG. Pl6.3 169 20ft 1= 165010" 20 k 15ft 15fl 1. =29000kl ~EI _'II (I 1 FF'I.ljI ~ '''" 1 ... Pt8.2,PtU I 1... Solc Problem 16.:! for the loadmg loho"" ID FIg PI fl .2 and a settlement of I in. at support B £ 15 It-l In thl' ,h.tph:r. co stUJI..:J .1 d,t lI.:,d fonnul.t110n (If thc 1.11 pia ml;nt ..tlllncss Illdhod. ,.tlh.'d Ihe slop..:-ddll,.':tlon mcthod. fur thc anal ~'I of .tnd ff.Hnc Thc method I' ha-.ed (10 the: ..lope-deflection eqll.Hllln hu:h relate the ml,mt.'nh Jt the: emb of a member to the rotation and di~rla~cmenh of 11.. end, and the e'ternalload.. applied to the member The rrll:t.'durc for .tool!) ..I' <:-"entiall) im·ohe.. tl) idcntif)ing the unkno11 Joint di'plol,emcnb ldcgn.-c.. of freedom) of the structure, 2 for each memb..:r. riling slope-deflection equations relating member end moment to the unkm)n Joint dlspla,ements; (3) establi..hing the equation.. of equilibrium of the ..truclure in term.. of member end mo- ment..; 14) ub..titutin~ Ihe ..lope-dcflection equations into the equihb- rium equatiom and ,ohing the resulting s}stem of equations to deter. mine the unkno"n joint displ.tcements; and (5) computing member end mOffic:nt.. b} ..ub,tituting the alues of joint displacements back into the slope-deflection cquations. Once member end moments have been Cal· uatcd. memhcr end shear.. and axial forces. and support reactions. can be detennined through equilibrium considerations. D 10k I 15 It ( 10 It £1 I k I 8 SUMMARY ... Pl1.1 1"1 ...... IU Del "mne the re t.1lons dnd draw the r d be ng m men! dlagr m for the beams fooho"n f PI I PI 5 b) u ng the I pe-deftectlUn method PROBLEMS .-,1.3
  • 240. 11M CHAPTER 16 Stope-Deflection Method " 1 U kif. IIII I I I , k -t::='=:::!::~:=!='=:::!::9 c .1 D E=constant fill. PlB.24 11 kN/m • /) EI - lin tant IfIm ~ PlB.20 11.21 Sohe Problem 16.17 for the I dl f- Plo.I"7 and a -.ettlement of 50 mm at urpon n 18.12 Sohe Problem 16.18 for the loadmg h n I FI P16.lts and a 'iCulement of" tn. dl upron A 11.23 Determine the member end momenl and r II n for the frame in Fig. PI6.~J for the loadmg ho"n and t ,upport settlements of I in. at A and 1 IQ at J) ~ the ~lope-deftccllon method. . £1 =: ronslant f.=29.000 ko I = 3.500 in.4 II c I.:on lant f.• 20() Gp' 1 c 4fXl( 1061 ~ 2' kN/m 2 lift T B 1='~=!:~1;'C,=*===;> D 100k·ft 15ft J--'-----+--1-5f1~15 I. T 3m +N3m L FIG. PI6.17, Pl6.21 RG. P16.18, PI6.22 2 klft 10k D I A ~ '-...3IVft ---, B f. err-I I I I I I F- lO ft C ~ ~ 4Ok-/} I II&. PllI.25 101t IOU A 30 It -r--30 It B 15 + 0 It £1- comlanl H. f.= 1O.000k I 1-3oooin' A FIG. Pl6.23 2n1l +5 ft 21 ......,U I: =I.:on lant II.J4 ........ 11.31 Detenn'" the .....b<r end fill. Pl6.19 .tnd rCactlOn$ for the frames shown ID FI PI6 -4 P flI. PI&2I b} ~ng (he slope«ftectlon method I 169 for the loadmg ho"n m fig mot l ..'mmiu upponl 4Uk 15 .JfI 40" A 1~~em~=I=1;:ll=I=I=n~f.~=1 G B 2: J1O/} :A. F 1--1--+- 20 II + 201.--1--1--1 10ft 10ft 10ft IOfl E ~9nOOk I 1=:'.onOm" E =: confootanl - - 2I----!-- ~ III 1M""il"" alii II} B e / } 6m+ bm--i-6m-l-bm~ £1 =cetnstaJII IH ~tH 0 ~ I e I t IA..... B .... /l .... F ~J -I 'T- o 'r'51t~llItr~lOtrJ-~tr__ 1 ~ 1 1 fill. Pl6.12, Pl6.16 fIG Pl6.13 fill. Pl6.1. '-1... ' ..17 , ... 0.:
  • 241. - 17.1 Definitions and TImlIllOlogy 17.2 Ilasic Conceot of tile Moment ~ MetIlOll 17.3 AnaI'jIis or Continuous Beams 17A Analysis 01 Frames _ Sidesway 17.5 Analysis or Frames with Sidesway Summary Problems Moment-Distribution Method 17 The Empire State Building, Veil' York Proto eoortesv of Bethlehem Slee COfP(ll'al I15 ft -+15 ft 1 E= constant --3011-- I~ 2/ ~I C 2J 0 / / .1L _L.!! 18k- fl5. 'l6.31 3 kI(1C -rlllllllil :!Ok C 0 (,16~ J 81~4 ft 160 4ft-j £/ =conslanl fl5. 'l8.30 18 kNlm 1---2Oft--+-15 ft--l 1---5m---+-3 m--l i 25k - 12 It 1 I 5 kJft 8 £1= con lanl r-- ft -'+---12 ft----I £1 = constanl 8 ...L.. " fl5. 'l8.27 2m _PI... fl5. 'l1.211
  • 242. no CHAPTER t7 Moment-Distribution Method the "lllutUlIl III 1n1U1t mulUS el!u,IIIOlls. ,here.,s 111 the case of fram Ilh sid .H. Ihl.: llumllC'1 (11 slInult,Uleous equallons mvolved u ualt equals the num1"ll.:r 01 lI11kpc:ndl.:nt JOIIlI translations. The moml.:l1t-(,i!stnbulllll1 I11dhod " classified as a dlsplacem nt method and from.1 IheorcllLal ICfKlllll. It is er) similar to the slope defk'1.:tlOn method C(1l1sl0ereo 111 the prL"(.;eding chapter However unhk the lope-defleclloo methtlO III hu;h all the structure's equthbnum equa lion are sallshed Slmult.wellu...l} III the momenHJlstnbutlon method the moment cqUlhbnum equ.H1ons of the JOI~ts are solved Iteratively by uccessieh coosldenng the mome-ol eqUlhbnum at one JOint at a tune hlle the ~malllmg Jomts of the structure are assumed to be restramed agamst dlsplaL"'Cmenl e fir...t deriH~ the fundamental relations necessary for the apphca- tion of the moment-dIstribution method and then develop the ba IC concept of the method. We next conSIder the applicallon of the method to the anahsls (,If continuous beams and frames without sidesway and finall) di.-.c·uss the analp,is of frames ith sidesway. 17.1 DEFINITIONS AND TERMINOLOGY Before we can dcclop the moment-distribution method, it is necessary to adopt a sign comention and define the various terms used in the anal}sls, Sign Convention In applying the moment-distnbution method, we will adopt the same sign convention as used previously for the slope-deflection method: Counterclockwise member end moments are considered positive. Smce a counterclockWise moment at an end of a member must act m a clockWise directIOn on the adjacent JOInI. the foregomg sign convention imphes that clotli.ulse moment~ on JOlnH are considered po~it,ve _171 Dsne." .. fa I lSi "'fit opplied fit Q?- r if) t I fit oppIied Gtf-- sA t IL £1 fI6. 17.1 b Beam w FuEndHin M The bendmg tiffne K ofa mem r be applied at an end of the member , Thus by setttng 0 I rad ID Eq 17 I bending sttlfness of the beam of FIg I 4£1 K L Member Stiffness K K 4 1 L Con Ider a pn matlc beam AB which IS hmged at end A and fixed at end B a shon m Fig 171 a . If we apply a moment M at tbe end A the beam rotates by an angle 0 at the hmged end A and develops a m0- ment M. at the fixed end B a hnwn ID the figure The relationshiP bet"... the apphed moment M and the rotatton 0 can be estabhshed by
  • 243. 8IId FEM d M.17A FM -"17,2 ..... ~oI....._I...1II1117.7...__ on which FEM • and FEM 0 dono the!alive bon A between I the ma8Jllludeo weD the eliRlCbODS are the same It can be seen from F g I placement C8U1OS a chord rotabon on tbe~~~~~~~~~~then the two fuu:d.."d momen act 10 the to maonW. 31'0 s1DJlO1 t the IW ends the chord IOtabon due to relab ~:::a.: Fi J 4 b then both fuu:d-end ] (poaiu diroction to pmeal the 17.2 BASIC CONCEPT OF THE MOMEIl-llflTll1IUJIlIII ala 60 k·ft 60 k·ft 30 k-ftor or or x X 0051 o1251 - 0.2 OF 04150 -60k-ft 04150)- 6Ok-ft 02(150) 30k-fi Xac + X'D 005 XII< _ 0051 -04 Xac + X'D 0.1251 OF, OFac M,. OF..M M, -OF, M M,o - OF,oM X'DOF, KBA + KB( KBD These elislnbuuon facto" ondlcate that 40 po"""'t of the l50-k·ft 10 ment applied to joont B I exerted at end B of member AB 40 _ t a end B ofmember BC and the remaonong 20 percent at end B ofmember BD. Thus the moments at ends B of the three members are Based on the foregoing discussion, we can state that In general the eliSlnbutioo factor (OF) for an end of a member that I ngIdly COD n<ct<d 10 the adjacent joint equals the ratio of the ",Iau>e bendIng ~ ness of the member to the sum of the relative bending stilfnenes of B1l the members framtng 1010 the Jomt; that is, Funbermo", the moment dlStnbuted to or resISted by a rtgIdJ n<ct<d end of a member equal the dlstnbuuon factor for that end the D<g8uve of the moment apphed to the adjacent joont then the eli tnbuuon facto" for the ends B of membe" AB BC and BD are gIVen by FIud-End MomenII The fiud-end 10011I<II CXJlIIaoi, c:oudib weD Uve eli placcments of member - the k the book for con_t refeRIKlt. :::~:ibu:bon method the effects of ont tranIIab pan 8IId are aI taken onto lI<COUDI -Contliderthe .oetIIlomeat A au lSi 17 __-II •..,... .......n,
  • 244. 8111nc1n1 JDInI C __,,__ 8 1Il:1IDI17.2 .......II "II 1II" • _ AXed-End Moments DCCB COSA Be (8 Balanf>mg Jomt C ., I ,..-.... he106 ,L_~I 0 ~ ,01Ac106 B ,f Balanctng JOJnl 8 AS I os 05 I 10429 om I +50 50 +75 -75 • -125 -125 +322 +428 ,- 6.3- +161 ~- 63 • +21.4 8 1 I - 81 + 27 + 36 I - 41 + 14 • - 41 • + 1.8 • - 07 - 07 + 18 + 23 - 0.4 + 09 . - 04 • + 1.2 •- 0.5 - 0.5 + 02 + 02 - 0.3 + 0 I ~- 03 • + 0.1 - 0.05 1- OOS + 01 + 0.2 +38.9 -71.8 I +71.7 -491 +491 - +:10 ... 17.5 30k FEM -SOkft r 15k fi 1 245)("111111111) (B C) (C FEM. or k· 01911 A BI7187171 149 491 FEM Sk·ft r k FEM. or k·ft (h) Fmal Member End Moments (k-ft) FEM 0 _17----
  • 245. 10 10 0429 OS I oOM OM These dutnbuted moments are nIed eli tnbubOD table and baIf these m::::~~ ends B and D of memhen Be and CD .. table. JOIDI C then ped. 1ICT1OII17.2 -eooeo.t "_ _ _ M With JOint 8 n w balanced we can table hDe 3 that due 10 the carryo r elf< I balaDced momenl at Joml C Recall thaI the 10 zontal hne at Jomt C were balanced pre 1 Th again and dlstnbute the unbalanced moment t and CD a FIg 7 S OM s OM Rlllnclng Joint c +2 4 k-ft 2 (+42.8) COF,,(OM cs) =2(+32.2) +6. k-ft COFcD(OM<D) COMBe COMD, SImilarly the carryoer moment at the end D of member CD I com puted as IIaIInclng Jalnt B JOInl B 0 now "'leased The unbalanced momenl al this J tained by umDlIDg all the moments aCIIDg al the end B :::: AB and BC which '" ngidl connected 10 JOIDI B From the distribution table I 1 and 2 we caD see thaI there -:lO....<I~ lixed-end _ I I end B Df member AB whelal the member BC I subjected to a 75 k.n fixed-end 16 I k II carry r momenl Thus the unbalanced momen These carryover moments are recorded on the same hne of the moment distribution table a the distributed moments, with a honzontal arrow from each distributed moment to its carryover moment as bowa ID Fig 7.5a The total member end moments at this point m the anal depleted on Fog 17 5 e It can be seen from this figu", lhal JoiDI now 10 equlhbnum because It IS subjected to two equal but opposite. moments Joml B however IS not 10 equdibnum and It needs to balanced Before we release JOlDt B an lmagmary clamp I applild JOlDt C In Its rotated pautiOR as shown m Fig 17 5 e CIW'IEII17 _ tnhutetJ moments DMcIJ and DM cn to deeJop at ends C ofJncm. B( and ([) hll;h can be C.t1uated b) multlplymg the ncpU unbalanced moment I e 7~·k-ft b) the dlstnbuuon factors DF and DF rcspecmcl) Thus DM{. O"~9 +15 +322k-ft OM 0 571 75 - +42.8 k-ft These dlstnbuted moments are recorded on hne 2 of the mo dlstnbutlon table Fig 17 5 a and a hne is drawn beneath them mdll:ate that JOIOI ( IS no" balanced. ote that the sum of the three moments above the Ime at jomt C IS equal to zero Ie 75 32 41 0 The dlstnbutN moment at end C of member Be mduces a oer moment at the far end B fig:. 17.5(d)). which can be dctenniDad by muluplying the distributed moment by the carryover factor of the member. Smce JOint 8 remams clamped. the carryover factor of member Be IS Eq. 17.13)). Thus. the carryover moment at the end 8 member BCls
  • 246. m These carrymer moments are recordro on the nex.t hnl; Ime 3 01 Ih moment-distnbution table With an mdmed arrolo pomtmg trom each distributed moment to It! carr)o'er moment as shown m Fig I ~ I We can see from hne 3 of the moment-distnbulJon table that due to the carryover elfet:t, there are no +16.I-k-ft and 6 '-k·ft unbalanced moments at Joints Band C. respectiely. Thus Ihese jomt are balanced again. and the distnbuted moments thus obtamro arc recorded on hne 4- of the moment-dlstnbution lable One-half of the dl tnbuted m ments are Ihen carried o'er to the far ends of the members hne 5 and lhe pnx:t:ss IS continued unttl the unbalanced moments are n II Ibl m II The final member end moments obtamed b) algebralcall urnmm the entnes 10 each column of the moment-dlstnbutton tabl a rec rd on hne II of Ihe table Fig. 17 5 I ole that th final In In nt 10 agreement With those dctemuned prelou I)' m Fig I a nd 10 Section 16 2 by the ,Iope-deflcctlon method Th mall d ffi n t-cen the re ults obtamed b} different approaf,;h are d t the uDd olferrors Based on the d. n p ntal .. the p dure r. r the anal 0 nbDUOUS beam b mc:thod <an be ummariz<d a foil " SElmON 17.3 Anllylls of ContI_ _ Thu the balanCing of Jomt ( mdul:e th foUowm ment at end ( (f members BC and CD re ('ICCII I OM, 0429 75 l' 2 k f. OM 0571 , 4_ k-ft ~he four dl tnbuted moments are ret: rded on hn ., I tnbullon tabl and a hne I dm-n ben ath them- r Ioloth of Ihe table t mdlcate that all the Joant re n - Nt n In Ihe ne t lep of the anal) t the carl) 0 r m me t tha at lhe far ends of the members are computed b mullJpl 10 tnbuled m men b the carf)Oer fact rs 17.3 ANALYSIS DF CONTINUOUS BEAMS 125 k-fi 12 5 k-f1 05 '5 o., .25 m a ImJlar manner From hn I of the l,;an that the unbalanced momenl t UM, 75 k-ft DMB~ OM, Joml ( I then halanLeo mom n1-d1 tnbullon labl JOlOt ( I Practical Application of the Moment-Distribution Process rrocl,; ~ The.; Imal m~'J11 r ~'nJ nwmcn.... .Ire {)btained .by algcbralcall umm1l1g thl.: ('nlnI.:S 111 e,<llh ll,lumn III the m{lmc-nt-dl..tnbution table The hn tl mom!,;nt thu."i ollt.l1ned .tre n..-cordl,:d on hne R of the table d . h " ' an are hon on the fn..'t.'-bt.lJ) dlagr.m1 (,I l L' mcrnl'Crs In Ftg 175 h Uh.' that the final moml.'nt satl I) the equatlono,; 01 moment cqudlb- num at JOlOt 8 and ( nh the mc.:mhcr end momt.:nts "'no n mem~r end shears and upport reaLlIL.m t:an 00 he Jcolennmed h) cotl"ildenng the equlllbnum of the free hodu: of thl' mcmh..:r and JOlllt of the continuous beam a dtscu sed 10 Sl~t1iln 16..:! The ..hc,lr Jnd bendmg moment diagram ca. n then be construdl--J In the usu.1I manner b) the heum Uin (on n- I " see fig 1o.) In the ft.lTegomg dl"il,;u ion c determined the member end moments b SucLe....Ic!) halannng onc Joint of thc structure at a time. Although thi approaLh proldc a de.trI:r 111 19ht mto lhe basic concept of the mo- ment-Ji tnbutnm procl,; s Irom a practical ic-poinl. it is usually more f,;om enu:nt 10 usc an alternati c appro.lch in . hich all the joint of the ~tructure arc frr:t.' to rotatc arc h.t1anced simultaneousl} in the same !'Itep. All thc carr)ocr moments that arc induced at the tar end of the memlxrs <lTC then computed "'lIllUltancously in the folio. 109 step and lhe process of b<11ancing the Joinl"i and carl)ing oer moments I re pealed u.ntil the unbalanced momenb .It the joints arc negligibly small To Illustratc this .t1tcrn<1tic approach. consider agam the three· span conlinuous beam of Fig. 17.5(a). The moment-distribution table used. for carr}ing out lhc u)mpulations is sho.... " in Fig. 17.5 I Th prclllusl) computed di...trihulion factors .tIld fix.ed-end moments are re- ~orded on the top and the tirst line rC!'Ipcctic1y of the table as shown It1 the lIgure. The moment-dlstnbulion process is starled by balancmg Jomts Band ( I rom hne I 01 the moment-distnbutlon table Ftg 175 I .c can • that the unbalanced moment at Jomt 8 IS L M. 50 7' 25 k-f1 A dlscu sed pre IOU Iy the baJanLing of Joint 8 mduces dl tnbuted moment at ends B of Ih members 18 and BC -hll.;h can be eulu ted b multlpl}mg Ih nc:galle of the unbalanced moment b) the dl Inbu- lion factors Thu CHAPltA 17 Momen1-0ilbibution Method
  • 247. ClW'TER t7 n_od IECTIOtI 17.J ....,..." Cc:ll SJ•• __ 111 E' I'"17.1 IXlcmlln Ih m mtx-r nJ nHlmenl hlT lh(' IHl.~pal1 (ontmuoU5 beam h III nhlmcnHh Inhulll.m method anFgI <I l1ll 4 4 F M rIOft+-,5ft-! I f---- 1<; tt --+-JO ft---f lin 2 k. fI ~.JI==,B¥~bb!"J.J..!~CA _. I 0545 0.455 I +64 8 - 43.2 +150 -150 58.2 - 48.6, -29 1- .... - 243 +35.7 101.4 +101.4 -1743 Flxed-end moments 2 Balanc Joml 8 3 Carrymer J Fmal moments E/ con tant Di tnbuuon fa.:lo" (al Contmuoulio Beam and Moment-Distribution Table 18k Hft r: I :' rrllll 1I1T11- 357.r-,A--'----:Bl)101.4 101.4,IB ~)174 FIC. 17.7 (b) Fmal Member End Moments (koft) Solutio. This beam '" as pre lOusl)' analyzed in Example 16.1 by the slope--dcJlection method. Df ,,,hutiull Fm ton Only jomt B is free to rotate. The distribution ratto at thiS JOint are COM COM. OFB.. K.. 0.545 K.. KB< DFBc KB< K, .. + KB( Ie that the urn of the dlstnbution factors at joint B IS equal to I that DF., DF. 0545 0.455 C/IeCIII The d. lnbUllon factors are recorded In boxes beneath the corresponc:liDa ber ends on top of the moment-dlstnbutlon table as shown m Fig I FI End I ", nI A wmng that Jomt B IS clamped against rota calculate the fu.ed~nd m menh due to the external loads by USIDB the fiI<llklol moment presSion gJ n In de the back cover of the book FEM. 648 k·ft' or 648kft
  • 248. ClW'18I17 _ - _ - E 51117.2 Det rm e the member oJ momenl for the thn.':"-span I,;ontmuou In F 17 .a b lhe moment-t1t tnhullon mcthod OF OF 1= lllfl _"D ·iud··EndM 48 kfi FEM FEM 6 FEM 18 fi18 ft1 It I 05 0,1 I 05 05 I +l:!4 .... +81 81 +486 324 -16 ., -16 :! +16 :! +162 ---=8""1 +8.1 - 81 +8. · -4 1 -4 I +41 +41 --=IO +20 20 -+20 • -I 0 1() +10 +1.0 1-0' +05 • -0' +05 • ~O 2 : -02 _ +02 +02 _ -01 +01 01 +0 I -005 -005 +0.05 +005 +21.1 -70.2 +702 70.2 +70.2 -217 Di buonF I F 2 B C 4 B 80ndC 5 CaR} B Cony Balance J lOts Band C 9 Can) r I Balance J Inl 8 and C II Final moments (a) Contmuou!o-Beam and Momenl·Dlstnbullon Table 3 k1ft 3 klf1 3 klft r;.~:r:IJro) 70 {OIIIIIIIIJ)70.2(OJD:l::o::r:::=.'" mlA 8 1702 fB et702 Ic DIJI7 fbI Fmal Member End Moments (k-ftl OM. OMB(' OF UM. DFBC VMs 05 324 05 324 16 k fi I. a fi fI&. 17.8 Solution Tbl beam wa anal zed P IOU I)' In Eumple 16 2 by using the deftecb n m mod OM OM OF. UM OF o D Uti the beam re f ad 05 OF 05 IC
  • 249. 100 • m m DIF.,(-UJ;t.1 FEM DM OF tIMe
  • 250. III ;817 ..... II tl ........... 30kN 10k m IDB C mcp l20k 9m 4m-l6m (b) SlIIk:oU ConIinuous Beam CanIiJnw CB CD 1 615 +120 S2S oS + S -120 +120 ,.,1710 AI d FiJII1 MemberEnd M....... (kN m) DF FEM o m FEM m
  • 251. ""7.11 lOS tN m lOS tN m os os os os aodm pm'I usJy In Example 16 S by UIIDI At JOInt B OF OF. OF FEM. FEM", FEM FEM D FEM 0 te -n beam nal .. method Fi. MO#IWft A qualitatlic deftected shape of the con wnll all _ damped ....... rotabon aod ubjc<tod to the opeciljed .........., I depicted m FII 17 11(11 IISIIlI an exallF'lltod scale II ..... from hi figure that tbe mabvc settlements for the _ ....... 4 4 0 m aod4 0 By IUIDI the _-end moment prauOllS, we _ .. the 1IJe....... moments due to the uppon settlement to be Mommf D rribll/ron The moment diltnbution I carried out in the manner u shown on the momenl-di tribUhon table 1D FIS 17 11 FIMJ M""",n See the moment-di mllubon table aod flJ. I II d AM. .M ,.17 III A DII ............... -
  • 252. 7«1 A 4 H! FE! IfM OF 4_ 141 l' ," 1 til momcnt-<li,,,nbuloon OF anal cd 10 E'tampl AI tnt FE! FEM H! OF Delenntne the member end momcn f, r the ustng the momenl-dl Inbutl n me hod IICTlON 17.4 AINItpIs of Frt.-I ......... lld. i, SOlutian This frame w The procedure for the anal f fr that for th anal I of continuo so.:t1on HOVoecr unhke the : ntmu rna) be connectoo to a Jomt of a frame (0 record the computauon m uch a n Whereas some engmeers hke to r« rd th lations directly on a ketch of the fram the fonnat for such purposcs. e VoIII use a tabular illustrated b) the follo.... ing e:amplt: IExample 17.7 17.4 ANALYSIS OF FRAMES WITHOUT SIDESWAY 1718.) k-f, 66 7 k·fi tl5'"'1 2 20 2 12 66 7 k·ft 6(29IlOO)(7.800) G) (20)1( 12)' DF'I)( 'I ~n H sn ~(l 6(29(~10)(7.80()) G) (2f)~li)J - cO 11 . 0 ~o ~ 1.227.2 L·ft / :!11 0571DI, - - - 1/ XI' 1 :!O .1 !'If) O.4:!9Of :!Ol'I ~O + 1 - H~D -t JOlOt ( 1 luml R Dt IS rhus. the t tal fiAcd-end momenb dut to (he combmed elfC:( of the C' tcmaI I d and (he upron ttlemenh art J}I II' Ill" /,JI1 The fiAcd-end mom<nt due to the 2-.. It extemalload are At JOint n FHtd-l.nd lOIIIC1lt1 - quahl.ltie deflected hare of the continUOU~ beam "lth all Jomt, dJ.mpcd J.gam,t rol.llion and ubJCClcd to the JlL":iflL'd ~uppon IoCUkmenh I depICted in l i~ 171:!(b) u,ing an e,aggeraled "Calc. It 'an be 'en from thi hfure thc rdatic IoCttlements for the three member, are j.u In . ..//( I ~ ~ in and .'(1) 1~ -- ~ =-0 l in_ B) Uing the fhed· end·momenl c'prc"ion. c dctermine the licd·end ml)ments due 10 the U~ port ,<Hlement 10 he CHAPTER 17 Moment·Distribution Method
  • 253. _ ,&"" IDII ............ B - 1--301I--J-3O11--1 E 29.oooksi • F1ame I BD -0.6 ED Carry..... IDC DB DEC4 CD - r0.4 I..-. 10.429 o.s71 I 0.3 0.3 I 1 • I +100 100 +130 -130 +130 -130 • _ - 21.4 - 286, _+130 • 1- 10.7 ~ . - 143 + 75 ~ ~ 243 -182 - 182 - 122' + 52 + 7 .... + Z.6 ~+ H ,- 14 I - 1.1 I- I 1 1- 07 L_jowJ -+ 03 + 04 + 02~ -+ 02 - 0.1 - 01 - 01 I + 92.1 -115.9 +1159 -186.4 -194 +2056 0 I ~ - Ian
  • 254. hod of ,.aa!,'1is I II II An Important que tlon that an m Ih IS. ho 10 delenntne the member end m,.m,,,," the frame undt:rg~ s.tde a under Ih a Sml.'C the moment-dl tributlon method cann I be pUle Ihe momenls due to the knon fat rail ad R reel approach in hlCh Ihe frame I ubJ t d t joml trans.latlon!to !.:aused b) an unknon load Qa and in the direction of R as s.ho" n m fig 17 14 d joint tran lalion !to e detcrmme the rclatl c t an cnds of each member and c calculat the member in the same manner as done prc'IOU I. ID lh ments The fixed-end momen thu obi med ~~::::;~~:::momenl4slribuuon process to delerm n Ih /Q caused b) the )et-unk.non I d Q 0 I m MQ hae been dctemuned die magmlude Q c apphcatlon of equd bnum equal With the I d Q and the d Ired momen IR dUC' 10 lhe Ia eo" b) mol pi 109 1/ h the SEC110M 17.5 ...,... 01 ffIIIIII .... "'~I_'II' 8 I I I I I I I I I I I I (d I Frame Subjected 10 an Arbitrary Tram.lation.:1 ""Momenb A Ie) Frame Sl.lbJeCied loR- MRMomenh • Imaginary 1IIIIIIIIIrolkr --,-1 --~-I c - 8 C R ------1) D I , I I I , I I I + I I I I p - I I A 8 A 8 IbJ Frame ",ilh Sidc3)' Precnted ~ MuMomenh .... 17.14 8 aJ Actual Frarllt' - " Moment.. A ~~-I. r- -1 cl I I I I I I I liD - - ..I I I I I I I I I p_ I " Thu tlr 1.' ,.')n u.lcred the ,1ll.11),i" of "trw.:turc ,In hl,.h the tran latit)n, t1f Ihl' ItlI1h en: either 7ew or kno11 .1'0 In the (use of sUpJ'l)f( 'o(ttkmcnl In thl' 'oCl.:Ul)1l e appl) the morncnt-dl"tnbuhon mcth)d h) .l1la! ' Ir.lIlll..' h)'I: 10111" mil) undergo both rotations and IIdn I.ltll,.)n lh,lt h.lC: nnt l'x'1,,'n pre-.cnlxd. ·s dl,,"U'.~d in Stxtion 164 ,ul.:h frame arc l'()mnwnl) referred to.1 frames" IIh 'Ide'" a COIlMdcr lor e ample the rectangular frame ~hon III hg. 17,14 a ~ quahLJIIe detlC'h:d ,h,lpe of thl.' frame for an arbitrary loading I also ..hon III the figure u mg an c.ilggerated ",ale. While the fi cd Join" .f and B )1' the frame Me complete!) restralllcd again..t rotalion as cll a tran,I.1I1011. thc joint' C and D are free (0 rotate and trdn..late. H)cer. ,ml.:c tht' mcmhcr of the frame are assumed to be inctenslble and the defonnation, .tre "....umed to be small. the joints C and D dis- place b) the same .lmount. j, in the honLOntal direction onI) . as ..hon in the figure. 7. CHAPTER 17 Moment-Dlstnbution Method 17.5 ANALYSIS OF FRAMES WITH SIDESWAY
  • 255. fa 17.15 ""'....., BD IJl Mc= • B -45.4 + 02 Q 4.4 tD - 44 A lUod-Il!lId Moo_ Due '" Knowa Tru.I...... ~ .n_.s~.•IIIII••" " " t -
  • 256. f'-'lit 051 049 (f) Support _00 DF. C D 3Ok-·r----~ DF ~~)8S I k ft 172k-T 8S9k tJOlnt D FEM FE FEM FEM Mem/Nr End Momm' D.. ,o ... Arbllrory SitkSWQ A SiDoo loads at< applied to the: member.; of the: frame the member end 1IlC..... the: fnun< restrained apln t S1desway will be mo To detcrmino end moments M due to the 3O-k lateral load we subjOCt the liame tI8Iy known borizontal translabon A at jomt C. F'8'= 17 17(b tab.. deflected shape of the frame WIth all jOlOts clamped apIDat subj<cted to the honzontal dl placement 4 at j0101 C. Tbe JIIOllOltJN strucIIIl& such deftected hapes was discussed In Section 165 0IIl the frame mcmben are assumed to be mexteDSible and defmma umed to be _II an end ora member can tnmslate only 10 the pendicular to the member From thIS figure we can see that the Iabon 4. between the end of member AC 10 the diroctlon pe,rpeadilllllii member can be expressed 10 tenns of theJomt translaUOD A u 4.c CC ~4' I 254 4 Similarly the relab.. translabons for member.; CD and BD ... 4 D,D 24 34 14174 3 4 A.. 13 4 I 202A 3 Tbe II due to the: reIati.. _1IllDa.... .1' ,& 17 .1 Ip' DIILINItDn MIIbod -.1717
  • 257. !I.e.. "mnnp - _ " -R E --- Ff'I p_I'IIIIII'-R C-- -" D mtni~'1p- - j E I I I .1 1 , I " I p- J) C ---I I I I I A B 782 CHAPTER 17 Momenl-OIs1ribution Method Analysis of Multistory Frames The fl1r....g..'lI1g rnll.:l..'UUll..' L,ln Ix C'tcnul.'u to the analYSI"i of t Ilh l1luhirlc dc:gn:c-.. III fn:I.'J"'1ll of 'IUCS,I} ('on~ider the t~ n'(ungul.H fT.II111: ,11(1'11 in hg 171S(al. The momenl-dl tnbu an.II),! l'l Ihl IT.lIllC'" ,,:~rncJ, I,lul 10 three paris. In the fint PUt, sl"k,.ty 01 both 111lM of Ihe IT.IIllL·.• prccnted hy addmg Ima Tl,lIl..... at the: tlllN Ic:H:h.•1,h"ml1 10 hg. 17.J8{b) Member end mcnh In that dccll)P 10 th1" fr,lmc due to the ctcrnalloads are puh.~ b~ Ihe- monu.'nl-lh,tn!:luuoll pnxe,... and the restralnmg fo <lod R at the II1Mgmar) supporb arc Calualed b) appl)ing the Ulln' of cqUlhl:mum In the ~"('ond part of the anal}sis the lower of the frame I' al"'.1l.'o It) displace h) a kno,," amount AI while sldc,a) of the upper floor ., prccntcd. .a~ shOn In Fig. 1718 t1t."d-end momenl' l.:" b~ Ihi!'> di'placement are compUled and tribulcd to l)btam the member cnd momenb .fQI_ Wilh Ihe member E Ffi E A f" F~Q I , 1 " +C2 X I I +(I )( I I I ,~I C 011 C -Q I D I D I' I I I I I ~e.. !I. '-- 1,-- !I. I.- ","17.11 (a) Actual Frame - M Momenh (C) Frame SubjeCted 10 Known Tran lallon.11_ "'01 Moments (b) Frame with Sidesway Prevented - Mo Moments (d) Frame Subjected 10 Known TransIaliOll~ "'02 MomeRU SUMMARY
  • 258. - - IllL Pl7.14 IlLPl7.13 IlL Pl7.12, Pl7.1' o EI 10ft-+-- 8 1 kill -10ft FIG. P17.11 I I kill 20 kN/m 60kN A pnmwrn:l ~ E 1--8m---+-8m-+4m+4m [J- con tan, E=70GPa 1= 800(I ()6j mm' FIG. P17.9, P17.15 3lk 2;"i It 20ft EI= IlC. P17.10 ~ kllI A~D8 C 1--1111----+--1011-1--1011 £1 =constant FIG. P17.8 17.llIInItIIIl17.14 Dete"m "hear .lOd bendmg mom ot d. n fig P178 PI 14 h n the method 8m 1= 1.300 (10'1 rom' 8m E= 70GPa 100kN 100kN AI l ~ ~ + ~9m-+6m+6m+6m1 1--- I -----1- 21 E = constant 25 kN/m 1II.P17.5 A~ I 10m I 20m ~ I 21 E = 200 GPa 1= 500(1()6) IIIlII' FIG. P17.4, P17.7 FIG. P17.3 E ( o 10k I 20ft 1= 165010 4 c :c: IRk I Ok EJ =con ant 8 I--2U 11---1-IUII-l--Il 11-----11 II ~ I Hilt II 15 ft E _ 000 kn III. P17.2, P17.8 IlC. P17.1 184 CHAPTER 17 Moment·OlstribuUon Method ThL' .lI1al~sl' (ll If.llne, uh a ,'lOgle degree of freedom of!l I L.tfriL't.I (lut in tll r.lf". In the' hr,t p.ut the ~Ide!""ay IS pre'en the addilion of an Inla!!lOar) rllller to, the ,tructuce. Member en~ mcn" (hat de:d(lp in thl' Ir.lme:. due: to the external an: L:(lmpuh.~ b the mOnle:nt·dl"tnbutlon pnll'es,; and the re"llllilliJII forcc R .It the Imagmal') roller I' C'.lluah:d b) the apphcatlon l-quatton, of e:qutlihnunl. In the 'Ccond part of the anal} IS 10 calCU1lIo the memhc:r mllmcnh due 10 the I""lrce R applied in the 0PPOute lion the- ..tru,,:tun: i....llloC"d to displace b} an arbitrarily ...,...... J..:non" amount' and the member moments and the corre''P<1Ildi1ll force Q at the IOL:ati(ln of Rare ealuated as bc:fore. The actual mea. end moment an: detennined b} algebraicalJ) summing the m computed in the tiNt part and R Q times the moments computed m second part On...'(' mcmber cnd momenh arc known, member end shea ber aXial fl)rL"C,.•tnd 'mppon reactions can be evaluated throUgh librium con..ide:ration,. PROBLEMS .... 17.3 17.1 ....... 11.1 Detemune the reactillO and dra, the h rand bendmg moment diagram.. fl)T the beam.. lOho, n n FI PI I P175 b) the moment-di..tnbution method.
  • 259. D 30kNim c _ '.17 -,-DilL. Ian ........ Deocnninc the mombcr end mom :~~~~:.:::':: ~ PI I PI
  • 260. • . . " II . . . " ....... 9k- E 21 , 1 , lIk- e 21 16ft 1 A 30ft • E=C"lall.... ft ft ... Pl7.J1
  • 261. IT] 111 Fr IIC1lllIlll2 ....... ' - _ _CfW'TER 18 Introduction to Mltrix Structur.1 Anllysls In the matrl' stiffness method of anal}"is, the joint displacements of the structure are determined by solving a system of simultaneous equations hich i.. e,pressed in the form lal Aduul Tru RG.18.3 18.2 MEMBER STIFFNESS RELATIONS IN LOCAL COORDINATES (t'll Anal}tical Model and Degree.. of Freedom L + EJA=con b Fnme Member - l..ocal C,ooli....' • ",::;: 1 J: 1.,= ~-EI-bL..--~;.:r1 I =0 I =0 __--< I- L- 11Il.1U (18.P - Sd Frame Members To establish the !!otiffncss relationships for the members of plane frames. let us focus our attention on an arbitrary prismatic member m of the frame shm... n in fig 18.4 a . When the frdme is subjected to ext mal loads member In detorms and internal forces are induced at at ends. The undefonned and defomu."d positions of the member are sh In fig 184 b. As Indu;ated In this figure three dlsplacemen translations In the '( and ) directions and rotation about the _ axl needed to l;ompletel) speclf) the deformed position of each end in hich d denotes the joint displacement vector, as discussed pre- viously: P represents the effects of external loads at the joints of the structure: and S is called the structure .'itiffnes.' matrix. As will be diS-- cussed in Section 18.5, the stiffness matrix for the entire structure. S, is obtained by assembling the stiffness matrices for the individual members of the structure. Thl .1i!f;/l'S mlllrix !t)r a "U'mher is u.ecl to eJ:.prefiS the forte al the l'ncl.. orIhl' memha as jimctions oflhe t!i'iplacemt'nts of Ihe memher {'lid,. Note that the terms forces and diplac('menh are used here in the general sense to Include moments and rotations respectively In this section. we derie stiffness matrices for the members of plane frdmes. continuous beams. and plane trusses in the local coordinate systems of the members.
  • 262. • k, k k k ko4 ko k k k.,. 1be .um- ......eD1J, k ClIII be evaluated by -===I 'nher IlIPIJIIteIy to umt val_ of each of be IIX eDd d 1be _ber eDd fonleo required to ClUIIC be individual UDlt _ .... tIeD del mined by liliiii be prmctplcs of mc:dwmcw taiaII 8IId de slope ..........00 equaUODl Chapter 16 8IId by de eqationl of equilibrium 1be member eud fOlOllll t1lIII a:ptwwt the .,m..eodk:ienta Cor the member Let us evaluate be IIilfDeu coeIIicienll COliespond"'l to nIue ofde d.tpJ·ceml!ll '" at eud b of the member u I1toWIl II ~ ota that all otber ditplacemcnll of the IIIlIIIlber &Ill callilll from metIumIt: of"",,,riah that the axial dcfOl'lllllIioll '.IIber CIIUIed by an axial force (2, .. 1IW11 by '" QLuul..• IInliDe the force k, that mUll be applied at eud b ofthe -lfiij; II to caDle a dltplacament" I to be k Ell. L TIe at lhe far ead of lhe.memb.- eqailillrium: .. • ._....._ .........II1II,...
  • 263. 710 CHAPTER 18 Inlnlduction 10 Mltri. Slruc1Urll Anllysls Continuous Beam Members £,1 L Q Q k. ote that Eq 18.7 I obtaI ed fi is because the members oftru and. therefore the member fixed Truss Members FlG.18.6 k Q4' "4 0?-'Q,. "} I L £1 =,:onstanl Continu U'i beam member -local coordmate o jLl AL' 0 () I 0 () I () 12 oL 0 -12 oL EI 0 6L 4L' () -6L 2L L - i l .n' .1L' 0 0 -- 0 () I I 0 12 6L 0 12 -6L 0 6L 2L~ 0 -6L 4L' Pr("l4":,,cdmg In the "'illn~ m.lI111a. the ,~tiflnc!'l'" coefficients correa 109 10 the lImt JI'r1a":.·l11l'nl IIf' I .In: lound to be (fig, IRA h oF! 2F! '" - klL- " L' " L ote that the ,th column of the member stiffness matrix conSIsts of end rorce~ required to cau..-.e a unit alue of the displacement u while oth~r displaccmenh are zero. For example. the second column of. consl~ts of the ...ix end forces required to cause the displacement u a"ho"n in Fog. 18A(dl. and so on. From Eq. (18.5), we can see that the stiffness matrix k IS symmetric: that is. k,f = kii. It can be shown by ing Betti's law (Section 7.8) that stiffness matrices for linearly ew structures are always symmetric. Since the axial deformations of the members of continuous beams sub- jecled to lateral loads are zero, we do not need to consider the degroes of freedom in the direction of the member's centroidal axis in the anal Thus. onl} four degrees of freedom need to be considered for lhe man- be", of plane continuous beams. The degrees of freedom and the cone- sponding end forces for a continuous beam member are shown 10 F 18.5 '" 18.5
  • 264. • _ ate onented In dlne-t 10 tranlform tbc 111m relatioDa the am_ I«a1 coordinate ystem 10. Cl'llIii1_IIDIN : =~~1be:amber ...ftbeos relaUODI m aIobIl then CXlIIIbiDed 10 establilh tile ":::.::~In tbiJ IllCIIOIl, we diacusa tile 1 aDd end dilplacemenll from IccaJ to the _ben ofpIanc ftamea, '=::=::~ '::::: Coordiaal8 tranlformauon oftbc It Q Lid mthe r. DowiDIIICbOD • 1 ....
  • 265. 18 18 711 CIW'ltII18._ .. Mml _ " , Anl'ysl. The matrix T I l.:.tI1 _,boo ddlne the t.ransformati~'m of member displal:ements from !l')(:,tl hl glooal coordmates; that IS, end TTu &C1llII1U Cow. Ii TIn. _ Continuous Beam Members TN When anal~zing continuous beams. the member local coordinates oriented 0 that the positi e directions of the local x and .I axes are ~~ same as the posllic directions of the global X and Y axes, respective) Fig. 18. . This orientation enables us to avoid coordinate transfor_ mations tx~ause the member end forces and end displacements m th h · •global and local coordmates are 1 e same; that IS. Truss Members F-Q ,,. = u (18 19 ·~"'>':""_----x The tramfonnatlOD main by elpresslDg the local end ~ F as FIg. 189 band or to maW ~ on fll,1U (b) Member End _ aIld End Dispbcanents mLocal Coonlinata ) Y IQ"., Q"., Lx;') ,-x®t 0 ( 1)Q""I Q)." (bI Member End Force. and End Di p l - in Local Cooniinales o ) Y F II F4• l4 Lx bf 0 0Y-'F,.l, F v) ) Member End Force llRd End Displacements In Global Coordinate fl&.1U Consider an arbitrary member III of the truss shown in Fig. 18.9(a . The end forces and end displacements for the member, in local and global coordinates, are sho"n in I'ig. 18.9(b) and (cJ, respectively. Note that at each member end, two degrees of freedom and two end forces are needed in global coordinates to represent the components of the mem- ber axial displacement and axial force, respectively. Thus, in global co- ordmates. the truss member has a total of four degrees of freedom VI through .., and four end forces, F, through F" as shown in Fig. 18 9(c (a) Conunuous Beam b y,.....,......---....n x o
  • 266. 718 CltAPTEA 18 In1nHIuca.. 10 Motrla Structurel Molysls 18.4 MEMBER STIFFNESS RELATIONS iN GLOBAL COORDINATES IICl1OIIl" " -....__ f (18.24) (18.25)F - Kv +Ff Equation (IR.24) can be comcniently written as B "'lOg the: Ilh:I1ltx-1 t1tlnc: ... rd.l1iom. in h,xal coordinates (Section lX.2 and lhl.: tran h,ml.I!lllO rdatll)n~ (Section Ut-,). e can now de Hillp thl' '1JITnt: ... rd.Hh'" fllT memb:r.. In glohal coordinates Frame Members To c:~tahllsh tht- mcmocr tiITnC' , rdation In global coordinates, c first ,ub,tjtu!" thl:' ,uITne...' relation in local coordinates Q ku + Q Eq I .4 inti) the fon.:e transfom13tion n:latJons F ~ T 1Q (Eq. (18.17 ) to obtain F T'Q. TT ku+Q,) ~ TTku+T'Qr (1823 Then. h uh,tituting the displacement transformation relations u If E4 IS.14 Into E4 IIS2.1) "e detennine the desired relations be- t"ccn the member end force.... f, and end displacements. "', to be "here K .. T"kT F, .. TIQ, (18.26) (18.27) The matrix K is called the memhn .~·lifJneJS matrix in global coordinates and F is the memht" fIxed-end foret· rector in glohul coore/mates. Continuous Beam Members As slaled preiously. the local coordinates of Ihe members of coo- tinuous ~ams are oriented so that the positive directions of the local x and J axe are the same as the positive directions of the global X and } axes respectioely. Thus no trcmsformations of coordinates are needed. and the member stiffness relations in the local and global co- ordinatt'S are the same. Truss Members The uffness relations for truss members in global coordinates are ex. pressed a 18.5 STRUCTURE STIFFNESS RELATIONS
  • 267. from whicb we dotonD1DO tbo oxprosllODS for f_ at manbertobo F. --" ...........n_Ill""'" IlIIItiOlII eJIP- the ml loads P m termS of the:~OI=Dt~=~the member end ~ to .nd d,opl. ber trnou roIauons m s10bal coon!iDltel oodinI toll B wnlJtll Eq t8 25 ID oxp81lded fCll1ll obtIiD
  • 268. p p p D be convemClldy pm d m COII._ ............ I II AI.'"
  • 269. Slru lUI Stiffn WUI Member I A shown In Fig begiDlung JOlDt and J lOt 3 the 18 IJ we dete_;nnm<::::=--:-__--::--: L 1IECftOII1U "II r I fIr __ DctenDlDe the 1 Fig 18 I a by.he rna Computer PnIgram Example 18.1 In",",uctiDn 10 Mllrix Strocturel Anllpls 2. [,tIUctlC:- th~ ,trw.'tun: ,l1lrn..:" matn S ano fhed-joint force v h'T PlOT ~.Kh mcm~r (~f th..: ,lructlrc, pcrfonn the rollowlQ <,per.illOn'· . ") dOh . a. fllT (ru...~ go OITce11) to ,h:p -( l. ~ cmlsc. compute the mc:mba titlnc ... main 111 10c<11 coor~m<tles. k. Expre I of k 10f the l11t:'m~"" llf fr,lme' .-and continuous beams are given 10 Eq' 1:.5' and (I~.61. re'p<",'i,elj b. If the mcm~r j ... ,ubJected (0 cXh: d n.lal load Q '" then evaluate Its fixed-end hlft,..c l'ctor in k~, COllT mates. f b) using the pre"ll)n... l'tlf thed<nd ,moments Inside t~~ ~ck cover of the book and b~ .Iflrl) mg tlte equatIOns of equthbnum see Ex ampb 1~.2 ,md I~:I.. . . . c. For horizontal members Itlt the local -'" aXis positIVe to the right i.e.. 10 tlte "um: din.x:l1on a'S tlte global X axis) ,the mem- ber ,tilIne" relatJon~ 10 the local and global coordmates are the ",me i.e. Ii - k and F Q,): go to step 2(e). OtheFWI compute the member"~ tran~formation matrix T by using Eq 1~.12 d. Dctcrmmc the member slIffncss matrix in global coordinates Ii = TTkT IEq. 18.26)), and the corresponding fixed-end fon:e 'ector F, -- T'Q. (Eq. (18.27)). The matrix K must be sym· mctric. For trw~ses. it is usually more convenient to use the expheit form of Ii gen in Eq. (18.29). Also. for trusses. F 0 e. Identif, the member'> ,tructure degree of freedom numbers and store the pertinent elements of K and FJ in their proper pos'. tions in the structure stiffness matrix S and the fixed-joint force vector P,. rc,peeti,e1y. by using the procedure described 10 Section 18.5. The complete structure stifTness matrix S obtamed bj assembhng 'he ,tifTness coefficients of all the members of the structure must be s>mmctric. 3. Form the joint load ,eclor. P. 4. Determine the unkno n Jomt displacements. Substitute P P and S into the structure 'tiffn«s relation,. P - P, = Sd (Eq. 18.41 and sohe the resuhing s>stcm of simultaneous equations for th unkno n joint displacements d. S. Compute member end displacements and end fonoes. For each member do the follo ing: a. Obtain member end displacements in global coordmates, from the Joint di,placement'. d. by using the member s snlO" ture degree of freedom numbers. b. Detennine member end displacements in local coordlOates usmg the relallon hip u T, (Eq. 18.14)). For honzoD members ith the local " axis positie to the right u v e. Compute member end fore.., in local coordmates by USIng tIte relallonshipQ ku Q, Eq. (18.4 . Fortrusscs Q 0 d. Calculate member end forces in global coordmates by the trdnsfonnation relationship F TTQ (Eq. 1817 F CIW'TER 18
  • 270. • _,U ....._........... , T Q Q .. [ 50 ] 8b b P Sd p [ I.7H! 417.6][<11] 417b 1.b44.3 d, and S from Fig. 18.11 C IoC wnte Eq 50 ] 8b b Thus the Jomt load «IM l~ B} subsillulmg P from Eq 1 expanded fonn as 1, nt D/ m m ,,1 The "it1lfn("..~ rel,lIlons for the entire truss can be pressetlas Eq 41 lIhP 0 dJ 00434 In. B sohmg these equation.. simuitaneousl} ....e delcrmine the joint dlsplatemems to be ClW'TER 18 IntnJduc1lon 10 Ml1rix S1rVCluril Anllpl. - or lemher £"', Dlp!acem('nh alld End Forct'~ Member I The member end displacements in global coordinates, " can be obtained b} simply comparing the member's global degree of freedom numben ",ilh the structure degree of freedom numbers for the member. as follows. d [ 0.0434]. 0.0637 m. By 1liiO' !be ....1i""oIIip • ID 1ocI1 coordina be ....-.q" T Q, ..~ !be member end lOla pQ U: 4 o o in. 0.0434 -0.0637 ~0434]00637 06 08 0 o 0 06 '1 [:i] ~ [(~]14 2 d2 u Note that the structure degree of freedom numbers for the member 0 0 I are written on the nght side ofv as shown in Eq. (4). Since the structure de of freedom numbers corresponding to 1'1 and (.'2 are zero this mdlcates tha t 1 = l - O. Similarly. the numbers I and 2 corresponding to v and.. pee- uvel) mdicate that t dl and 14 d. II should be realized that these patJbdny equations could have been estabhshed alternatively SImply by a In pection of tbe hne dlagrdm of the structure Fig. 18.11 b . However the of the structure degree of freedom numbers enables us convemenlly to Ph.1ID thl procedure on a computer The member end dl placements m local coordinates can now be determi by usmg the relallonsh.p u T. Eq. 18 14 with T as defined ID Eq I B usmg Eq 18 1 we compute member end forces ID local coordin
  • 271. - 15338 7694 ~m?4616 Q (~) I IQ Q. _ l U "'''_no'''-'' 2 02 04+08 2 Member End 176.33 24 IINlm 15388 CW&HP223 177 m 04+04 02 5=£1 24kN1m 4593 30lIN 176114 ~<Deb@~ 891 6 09 Il J(-1I52+200~ I = r848 L -200 jJ 1:200 2 (d) Sb'UCQIIe Sbffncss Matrtl and fixed-JOint Force Vi tor Pd Pol, Q ITJ %-X(1) 4 (b AnoIytiaI Model 4m 10m EI ........ (a Contmuous Beam BOlIN IT] 6m y l -6~93] 6293 k o 629~] 62.93 lllnn(~ 111 0 0 0 lJ.I~4 - n0637 F T/Q F u T. u LF, 0 11 6293 100 (OS 60 007 ",0 Ch8dll Lf, 0 17 H 6927 100 In 60 0 Chldll It 11 6927 15 6293 20 100 eo 60 20 39", 0 C/lIdlI lXt omn the reactl nand tbe m mber end forces for the thnlMlplll llOUOU be m holo n In Fig 18 12 a hy usmg the matnx. tiff IlaIulIon D r f Fr dt nr Fr m th analyttc.lll model of the beam I I b '"" that th lnu;ture ha two degrees of freedom hi h are the unkn n r tall n of Jomt ., and 3 rnpecb I membcT I 1 rdmale tem are chosen that the poslU eli Support RUI(/IOIU As shO·n In hg 18 Il{c) the reactions at the joints 1,2 and 4 arc equal to the forces in global coordmates al the ends members wnnected to the jOints. Ana. EqUilthrlllm ChUA Appl)mg. the equations of equilibnum to th of th enllre sirudurc hg 18 life I.e obtain Q kg Q ~' 14<0, _: :][~0434]- [-:~:~ k Thus thealtlalf f'L'e1O membcr 'IS f-ig 18.11 d 629Jk T ClW'T£R 18 1_ _ 10 Mltril51rUCturol Anllysll ldI'Ibtr 3 h .,1111.2
  • 272. _ CHAPTER 18 IntrodUCtiOn 10 Matrix Structurll Anllylll 8~ U 109 the m~lOhe:r Illln~ rdatl~n~ Q ku Q. Eq. 184 we mem~r end ror~~ a~ 1IC1IIIII1U" ••tor.... • f, Q [ 0 1I1~ 00'; EI (UII.::! 00<> OOb 04 Ol),; O~ 0011 (106 0012 -006 006] [ (I ] [ 01 1 0 -006 EI 0 0-1 15409 [ lK91 K -1598 k m 6109. l"to ~-I ~ m Inn... 2 .. nr [~]~b[-I'~09] I 14 2: d 192:5 Q k. Q I [ 0012 0,06 -0012 006 1[ 0 1(~10,06 04 -OOb 0,2 154.09 The support react an: F. Q- ~OOI2 -0.06 0012 -0.06 0 0.06 0,2 -006 04 19235 [ I221kN 1176.83 kN m 1891 80 18339 117 HN AnI 153.8K kN m 4,98 Mnnbtr : • (:'1~ (f1 ~I(t'1I.. II Example'••3 Q k. Q, ( (J (~b 024 0096 024H 0 1F Q 024 08 024 04 19235 0096 024 00'lb 024 0 024 04 024 08 0
  • 273. ..-. ..-.-.- 4 I I
  • 274. II I 101 9 -60 but die 6nt ....- . ClDOIIIdo8l tam:
  • 275. Th IB.~lgl IB '1e then multiply " ]II XX~ 60 , -~]. 6 , [(~ ~ ~ o 0 I I .1 An 1 II X and ,ubtract it from ro I; then mul. 1.0-1'" Jnd ubtral:( it from TO 2. This yield... [ (1) (I ... b 'I' 15 .1Ild ,uhtr.ll.:t It from n.m 1 " ,Ind ,uhIT.ICI II lrom flm 1 lhi,) lc1d~ [ (II (1) ~.~~ I~.;~~] (I (I ~(lW ~~ .15~ 0" ldL~ rol .' h) ~ ~(lW [ (1) (I ~ I X -7705] 1 <>47 IXX~ 18.2111 0 0 I 6 tUltlpJ~ rlm rl) ~ in t tult1pl~ Tl' .' b) tirl~ Tl'" ~ b) -I., lIM API'£tlDIX B Review 0' Motri. Algebra 2. and .3 -- 6. 8 U Bll'cn .... D BA 4 4 B 1 ,6 4A IA Determine the prod_ C AI 2 5 5 C <aandD BA 1.3 Detennine the products - PROBLEMS SecIIon 1.3 1.1 Determine the matnx (' A JI If [ 12 -8 15] (:I A - -8 7 10 B 4 15 10 5 6 1.2 Determine the matrix C 2A B If (B 24 B250 3 and subtracting It from row 2 we -2 -4[~ Bj muillplj109 ro" I bj A" obtam The augmented malrix. is glen b) The Gau!)~·Jordan elimination method can also be used to detcnninc the imc~ ... of ...quare matrices. The procedure is similar to that described pre iOlisly for ...oh ing !'>imuhaneous equations, except that in the aug- mented matrix, the cocftlcicnt matrix is now replaced by the matrix A that is to be imerted and the vector of constants P is replaced by a unat matrix I of the same order as the matrix. A. Elementary TOY, operdtions are then performed on the augmented matrix to reduce the matrix. A to a unit matrix.. The matrix I y,hich y,as initially the unit matrix. no represents the imeNC of the original matrix A. To illustrate the foregoing procedure. let us compute the imerse of the 2 2 matrix Matrix Inversion
  • 276. - .....- • III.c.eM AG. CA JOlOt Coordmat s and Supports Screen JpM"Np D 8'_1 ReItrDjldl ........ E...... l -140 x..._ iii .......... iiix-euu& ... CHR .. ' ..... E...-...v-cw:&... · " v..... iii I--I _III -~- ...CoI
  • 277. _ _IX C Camp_ _1e ~ lift BIBLIOGRAPHY 20fl .. 1L Laible J P 198 Y, mston New York 11. H,b mlll •• Beer. ~.P. and Johnston E.R. Jr. (1981) Iedwniu IIJ .fcltt'riuls. McGraw-Hili. New York 5. Betti. E. (1872) II 'uom Cimt'l'lw. Series:! ols. 7 and 8. I. Boggs. R.G. (1(84) Elt'menlllfl Strudurl1l AnahJH Holt. Rinehart & Winston, New York. 7. Chaje!>, A. (1990) StrUt lural Anall I 2nd ed Prentice Hall. Englewood Clilfs. N.J. L Col/oqUIn! on /lillon' oj Structure 1982 Proccedm International AS!>OClatton for Rndge and Structural En- gmeenng. Cambndge England •• Cro~s. H. (1930) 'AnalySt of Contmuous Framn by Dt~tributing fixc:d-End Moments ,. Pr dill I Am('rlcun Soden of ell ,I En III r 56 919 928 11. Elias Z M. Anal, If. Wiley e York 11. Gere J.M and Weaver W Jr 196 AI A fi r Ullin ers Van ostrand Remhold t:W York 11. Glockner PG 19 I chamcs.· J urnal of 1M 5:"'..·...,1 89 1• .fSeE SlUndarJ f"wmml Dc I 1'1 Load ~ r 8lI ami 0Iht'1: ~tnuturt'. 2(KH Sfl AS( F ..Q2 Amer:an St'ICict) 01 CI II Engmeen. IfgtDla Z. Arbabi F ll991 StruilUTIJ! Anah, cuu/ &ha ~h:Gra,,·HiIL Nc York 3. Bathe. KJ. and WilOn. E,L 1~76 '"m" al llhJds In Finilt' Elmlem Anal.nif. Prentice Hall Englewood Chff ~J _,- __ -J'-_ fl 3()ft~ E I A are lOon tant E :!9000k.I.A=:Oan~
  • 278. M. Smith Ie 1988 S", r. A I ew York 1.1 II• WaDI CK I911J [n'mHldia StntcIllnJ &11McGraw-Hill ewYork loS III .t. Well HH 1989 AIIlII "0 SlI'II<lIIn An CIiuIIaJ/ aNi Modmo II, lJrotb 2Dd cd. Wi &11 York U &11 1.7 III U &11 UI 1.11 CIIAPTEIl3 1.1 U U U U Ion ioU ... Americ:oll .. Ho".,OOd, "."""'U_A nq , ,7, •
  • 279. -....-..."., - 5.n be I 1.1 U7 m k k k S. 120 k If II) k "' If 36k S 94 kfl 4 If S I .lOk 9n kN 711 k III k S ' "k S k·lt, I( t R " I , It n I 1'iO k m o~ B I ... R 0 L fII 8C1 13113 k-ft " I R. 11 L 12k. ~B Il .0 L 0; '0 R ') I I:! k. fB f 1"0 k·ft 1 . ~ It. - '8 L 211 k it : 1:!5 k·fL Is I( 75 kAt S, If Ss I US k Ss So 0; ./~ 2 7lMi k. 01 III -1.350 kN m.. I .1D II Ss L ....: -27 k; 58 '" 27k;18 I( at 21 ft from A S'R 1095 kN. S'L 375 k ·112.5 kN: SCI -1845k S So L =- 60 kN; Is 441 kN 01 -450 kN m St R - SSt =- -8 k; 5 sIl 2383 k' S - 21 17 k; Sf" '" - So 0; 18 -80 k·ft I( .10 :,,:;; -40 k·ft; +.Im;l~ 109 3 k·ft at 2589 ft from A .'" R - 30 k; Ss - S( L - -30 k; S( If So t ·-50 k; .If, 30Q k-ft: If" 800 k·fi h.llll<l =225 k-fL at 15 ft from A s.~ If ..= SSL - 33.33 kN; S8 If. Sf) L -66.67 kN; So If. Sc L 266 67 k S It 316.67 kN; SF L 1667 kN;.If. 3J3 3 k m .1(" _ 333.3 kN m. ifF -1666 7 k m S~ R = 7.5 k; S( L 22 S k. S( It 22 5 k S, L - -75 k; S.. 4 17 k 10 83 k SF R - 10 k; If, 225 k·fi If 50 + I~ 28.13 k-ft at 22 S ft 10 the left and t the right of ( S~ R - 125 k 1875 k ; SOL SF< 125 k If 3125k m .ISmfromA .... F aa3mbS .. S S, L -100 k S S 450 k m.lf 450k lS' U7 5.4lI U5 U' U7 m 66 f, It, l , 525 75. If hiL Moment :!()ll .. S 40 - ,- x s~ _214 k O. Ss w· 14 k I f t :!!(:!!(~(r.f, Itll~ l r. fRO 'h l ( ,. 20 f1: S .1.2) 16L); ( Ix(L~ .-.:') 6L J m: S 20; { 20. 6 m: S - 0; .I 60 9 m: S - -20: I -20,' + 180 7 m' S -60; I 60 14 m: S ~ IO( 1 - "( • I 1 110 ~ ..N 'i ~ 1 f o ,. ,) II'" ~ 'i-S ~ <. 116 2 L ~ ("I, ' H to "'4 k ,. ltoto'~ I fB( Rhl" 1 f If ,~ 1~ ~ T f( () 1' 81 II. ( f -S I~ k T f ~8 3' k IT <!, 4() kN. ,:!-S9 .... m. ON :tol -S~ .. m <!~ R6bk•. f 5U". If <!B .8 1B () <!t Os 44": kS. Sf -o-17.h k x"'5 ' ... , m; S8 . R944 kN, ~b!U 10., m O( Qs-O; St - -50 k~; 14 '0 kN 01; '8 h~5 kN; I" -- -150 kN m Q4 Qs S8 tl; S4 ~ 120 kN; 14 .'btl kN 01; .IB =- I:!O kN m Oc - 7 ~ kN; S, -=0 12.99 kN; .1, 77,94 kN 01; Q, S, 0: .f. - 8768 kN m forO (L 21:S- P12; M - P.'/2 Jor (I 2) : ... L: S - P12; .1 - P(L - rl 2 S II(L 2')/2: .1 - II".'(L - .')/2 S It L for 0 (L 2J. Bending Moment (-or (L 2) x J: Bending fIt' L)L S II(L (-or 0 "ur J m " (-or h m "orO " For 7 m 10 245 For 0 : nOI ,,' 2 For 20 fl : If For 0 3 J3 If For 0 .If u, U3 4.55 OIAPTER ~ .... 5.9 5.11 5.13 5.15 5.l7 5.25 5.11 5.2l ~ , T f' T f, C. F J C. I ( . F 204kTF 36 k-ft h C) 68.94 k (': Fj(,- 45 '4 kN T " 1625 k C' FHI T F. - 0 14 k C - 6 17 k C' FDJ r. H 4167 k C: 6 ~ C; FR<, - 5 k (TI; 511 k tT Ft.G 50 k Ie). ~ T F 100 ~ , b Indeh:mllnalC I I. t I. d [)eh.'lTlunalc ~ l n taNe I,. lktcmunalc It U, U5 ..,7 U3 Ul U3 .... U7 &A, ....
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  • 281. 936k 14 k 6HII 1&7 936k Q, 1 29 k 19k II 26k 4 k Q lOB k-fl 1U 26k Q k 21 k-fl 364k ISH ml....'nD 222 k II Q, 36 74 k IS k 249 k-ft