Transcription
FE Exam Review forStructural AnalysisProf. V. SaoumaOct. 2013 Structural Analysis is part of the afternoon exam.In the afternoon, you are to answer 60 questions, and Structural Analysis is about 10%of the test content (or about 6 questions).Each question is worth 2 points.You are expected to know:1. Structural analysis of statically determinate beams, trusses and frames.2. Deflection analysis of statically determinate beams, trusses and frames.3. Stability analysis of beams, trusses and frames.4. Column analysis (e.g. buckling, boundary conditions).5. Loads and load paths (e.g. dead, live, moving).6. Elementary statically indeterminate structures.The only page in the “Supplied-Reference Handbook” related to Structural Analysis(shown in the next page).Make sure that you know how to make best use of it, as it contains:1. Reminder of what do we mean by “Moving Loads”.2. Beam-Stiffness and moment carryover: to use for the analysis of staticallyindeterminate beams (unlikely that you get a SI frame).3. Equations for the calculations of the deflections of trusses and beams using thevirtual work method. Careful it is the virtual force/moment time the actualdisplacement (FL/AE for trusses, and M/EI for beams).4. Member fixed end actions for uniform and concentrated load.I strongly recommend that you also memorize:for the maximum deflection ofa uniformly loaded, simply supported beam.Careful with the SI units, GPa is 109 Pa or 109 N/m2 Many problems use the SI system.In most cases, you will be dealing with round numbers, which greatly simplify yourcalculations.Do not be tricked in believing that all triangles are 3-4-5.
26 ed tionmA吐Aせ十Aせ4パヨm一 m的町一つG AFL一A00一Aせ一一一一m eterminethetangentoffsetぅ仏 atthelowpoint.22‑ G1)Xy iG2L‑(0.016一 (‑0.02))(444m)2(2)(800m)ニ 4.44meleVlowpoint 742.12m十 4.44m 746.56m (747m)Al (0.08m)(0.15m) 0.012m2TheanswerisB.A2 (0.05m)(0.15m)2ェ �coordinate(meatむroぱidfortheT‑shapedbeamisUc �( (平十 叫))XJ08m比2(0.0075m )0.16m(C)0却m(D)0.38m mpressionnsion(D)2490N七e2pllss.comPPI・ 附W押 i
目di(eProblems ionatE.:Z MA 0RE，匂dEARH:" i2224N)(伊6m)ν N(D)16kN円U 'げ 1112NSolution8incesupportD isarollersupportthehorizontalre‑ac七ionforce hefree‑bodydiagramabou、.̲.EHDE !RE，νうRAy(15m)‑(20kN)(5m) 0kNRA" i20kN)(5m)15m 6.67kN esthehorizontalreactionatE ctionisFEH，υ‑RE，り 0lbf 1112N erEHiFEH，h (2)(1112N) 2224N七forceinmemberEHisTheresultanFEH V(Fi凹，出(Fi叫2ゾ (1112N)2十(山N)2ニ2487N compre回on (2490N ionatsupportA is6.67kN.Nextぅ 田 eshorizontalandverticalcompo‑m臨 hown.FAExPPI.附 w.ppi2p日sS.com
28 N.Summationofverticalforcesgiv 回う RAy 0kNFAByThiscanberearrangedtogiveFABy ‑RAy 七hefree‑bodydiagram zontalforce.Therefore，theforceinmember F isthesameastheforceinmemberAE.FEF FAE 8.33kN (8.3kN)TheanswerisD.(6.403m 一 I一一了一一一一 I.l'AB'U 4m hecross‑sectionalareaofeachmemberis1000m m2，themagni‑tudeoftheverticaldefl.ection抗 jointEismostnearlyν(6.403m (τ' --- )(‑6.67kN) ‑10.68kN(‑11kN)(A)0.70m Bforcew回 .1m m(C)1.6m m(D)2.8m )8.3悶SolutionUsetheprincipleofvi巾 toeachtrussjointandareshown回 follows. s.)τr'ussmemberlengthsinmetersareasfollows.B 5SolutionFrom80.141，FABy is-6.67 日 . Thehorizontalcom‑ponentofthememberABforceisAT 14 �トN︑ 1111/ 111I/Lιm一m m一m お一A住 民 一 笹 口 δ/ell‑ /1111 TheforceinmemberAEisFAE FAExD ‑FABx ‑(‑8.33kN) 8.33kNPPI.附 W押 i2p日ss.com6.6713.33
PradiceProblems凶 �� s44and45arebasedonthefollowingillustration.司町 5m 手 allvari‑ablesorganized. LFQkN2岨)memberkN) (m) ((kN) (6.4mノEB﹄t /vhuvhuThemodulusofelasticityofsteelisE 2.1X 1011 tEcanbefoundby引制I: 239.7凶一m︑ ︑ 6.4う二到 1.0 t. TheverticalreactionforceatsupportA RA約 kN(A)1.0kN(B)513(C) kN(D) rceatsup‑Themagnitudeoftheverport.A 420.53 0.83 0.8310.67‑16.67‑21.34十8.33十8.33 16.67十0.00‑6.67 A?QLPいM l/εvbzERl一Z 仁いEA2吋((1000mm):m)(品) 川 1Pa)誓)x(10005mThesupportA tsupportC.ThisresultsinaverticalreactionatsupportA onisalsodownward. (25kN)(2.5m) 0kN‑mRA"ニ (15kN)(5m)‑(25kN)(2.5m)吋TheanswerisB.5m10m 1.25kN (1.3kN)PPI.附 W押 i2p日sS.com
30 misunderreinforced.五 3000lbfjin22000lbfjinん 40，(A)1.3kN(B)14kN(C)25kN(D)39kN2As swertothisproblemistoconstructasheardiagram. う20in23inllF15kN LkN/mI/1c ιkips(C)200ft‑kips(D) C.Solutionん一Mω‑62.5Inan actualdesign and analysissituation， 乱 heckisnotrequiredtosolvethisspecificproblem.PPI・ 附W押i2p日sS.com
46 thecurveandroadismostnearlyPractice Exam y七hperpaverequirements?allthep釘 ametersarewithin組 acc叩七回(A)Yes，ablerange.(B) No，theVMAisexcessive(0) No，thedust‑toasphaltratioistoohighGm matNm istoohigh.(D) No，(A)28ft(B)36此(0)44氏(D)72ft田日36. gradeareasfollows.0kmjhdesignspeed 8coefficientoffriction 0.35.0sdriverreactiontime 2drivereyeheight 1.2mobject(tobe師 oided)height ismostnearly(A) xleload(ESAL)forthetrucktra伍 sectionofhighwayconsistsofaverticalcurvewitha1500m gnspeed 8drivereyeheight 1.2mobject(tobeavoided)height 0.2m七日oppingsightdistanceデ arlyProblems40and41町 ebasedonthefollowingi1lus七ra‑tion.ド 15kNi endingmomentinthebeamismostnearly(A)680m(B) 700m(0) 760m(D)840m38.A superpavedesignmixtureforahighwaywith7ESALs 10 h回 anominalmaximumaggregatesizeof19mm. istics:airvoids 4.0%VMAニ 13.2%VFA 70%.97dust‑to‑asphaltratio 0ニyrationsG87.1%atN 8gmm74gyrationsぅ Gm m 97.5%atN 1うPPI・ 附w.ppi2pass.com39.A ruckmakinganaverageof10trips40trucks，perday. loadequivalencyfactorforthefront配 torforsingleaxleis0.0877. .3kN‑m(B)14凶 ofbending08m m，4 themagnitudeoftheverticaldefl̲ec‑of2.0X 1tionatpointD ismostnearly(A)0.20m m(B)2.3m m(0)23m m(D)50m m
Prlldiceb:llm1 手ょ下予(A)‑0.71悶 /kN(B) ‑0.53kN/凶(C) ‑0却 kN/kN(D) veloadshearatsuppor七 )27kNだ 4000Ibf/in2(B)76kN(C)100kN(D)110凶fy irectionshownう the.ximumliveloadbendingmoment抗 supportC ismaymostne紅 l(A) 80kN‑m(B)90kN.m(C) 140kN.m(D) ussspanisshown. eforcesう veloadisdeadloadis150kips，350kips. alreinforcingbaris(A) no.7(B) 0kips.AssumingthaPPI.附 w.ppi2pass.com
Solution Practice Exam I56 �8，is1essthanthecurve1engthL:う札rL ‑A8 240097ESAL (000ESAL)240，;yう(100%)(VWi 伊豆(1%一 �南可 erthanthecurve1eng七h:L 2Sニ州240.Constructshearandmome凶 plied1oading，changeinmome凶 istheareaunder七hesheardiagram.一:‑1(200%)(VHi '2/AνIEq4JA‑IVAwhuM川ySince也 dvertica1curve1engthis‑qmill‑‑ 481.0m川一トL ﹄且晶ベー一A一h 一 円 円 VO(200%)(Vl五五万五:)21%‑(‑3%)A (2)(300m)一N L 756.4m natedasa19m ids 4.0%minimumVMλ 13%VFA 65‑75%dusιto‑aspha1tratio 0.6‑1.2yratiorおう maximumGm m 89%atNinit 8gatNmax 174gyrations，maximumGm m 98%伺M{kN.m) vespecifica七ions.The 1argest magnitude isA.TheanswerisD.39. ntD.ESAL七r叫(1sing1e砿 1e)(0.0877) (2tandemax1es)(0.1206) 0.3289ESALjtruck‑tripThe tota1daily ESAL for40trucks，each making10tripsaday，isE叫寄)0(印刷) 山cks)(10Xtruckωtrip 131.56ESALjdayPPI.州 w.ppi2pass.com江lOment 1S (J m(自白)ムD 10adingareshownonthefollowingmomentdiagram.同
Pr目cticeExam1Solufions 57ThemodulusofelasticityofsteelisE 2.1X defl.ec‑七Distionatpoin15ムD XX(‑EI Jイー云EI2呼判: 一山:1M(kN.m} 0IEIM evir‑tualunitloadingatpointD ��:1十年:123‑73‑26.042kN.m8.125kN2.m十 390.625 日 2.m33 286.458kN2.m2390.625kN.m十3(品)11(2.1X 10 Pa)/ 紅l立1 X IllJlJU‑一一ー 22.9m nregronfunctionA‑B0 x 5BC0 x 5CD0 x 2‑xEIEI同EI自PPI.附 w.ppi2poss.com
58 nofbeamatsuppo吋 mumverticalsheartt71.17kN17.79kN1 ̲ 1Theloadpositionshownresultsin出 emaximumresponseinthebeamatsuppor七 Cfor出 especi五eddi‑rectionoftrave.l �司日ttloadpositionformaximumbendingmoment(1.5 ) (7.117凶)x amatsuppor七 Cforthespeci五eddirec七ionoftrave.l �時max 1(‑5守)川 fmaximumbendingmomentatsupportCforth巴 tudeofmaximumver‑ticalshearatsupportC ncelineforbendingmo‑ment抗日uppo凶 PI.附 w.ppi2p日sS.com̲J.，"
Pr目di(eExam1Solutiolls ithspiralreinforcement，ゆ 0.70.8ettingゆPn teelgIVes叫E;fd務ゆiiλ哩 loadsappliedto出 (4000。主一 (0.85)(4000 ;)2As essiveloads，isgreaterthanAs 2.54inasedonPg 3kN/kNThecolumnhassixlongitudinalreinforcingbars. iA 一日主主一 orcings七eelrequiredis川(年) (0 aldesign/analysissituation，乱 d 1.6P!ivePuニ1.2ヱ(1.2)(150kips) (1.6)(350kips) ired内， eminimumareaofreinforcingsteelrequiredisAs pgAg yisgivenby凡 0.85P.。 んAs))(0.85五(Ag‑As) んAs) (0.85)(0.85f Aconcrete (0部2 elbased，alloadPu.onthefactoredaxiうde.2P吋 1.6P!ivePu 1 (1.2)(150kips) (1.6)(250kips) 580kipsPPI.附 w.ppi2p日sS.com
70 CivilDiscipli鵬闘 SpedficReviewfortheFE/EITEx脚Practice Exam IIThe七ra伍 cflowrelationshipisgivenbyq kv，inwhichqisthetra伍 o1000NpTheseparationismaintained抗 2m，but出 �﹁一助 (A)760veh/hr(B)880veh/hr(C)900veh/hr(D) is430ftforadesign38. Thesspeedof50mphonasectionofhighway. mostnearly蕩Z，10m(A)270氏(B) 380氏(C) 410ft(D) 450ft10m41. ThemaximumvalueforshearatsupportAismostnearly39. Aone‑laneruralroadhasa100 curveextendingfor230m alongitscenterline. Theroadis5m widewith3m wideshoulders. )0.00050(B)0.034(C)1.1(D)1.9(A)2000N(B) 2800N(C) 3000N(D) 3800N42. Themaximumvalueformomentatsuppor七A ‑m40. rof6.6. f0.3. area10incemeastrengthcoe血cientof0.20ぅ and阻 ficientforthesubbase?四43.A triangularpin‑connectedtrusscarriesaloadof4448N 15(D)0.20PN4448N4.6mPPI. 州w.ppi2p日sS.com
Pr目diceEx園田12 71ThetrussmemberpropertiesareE 200X 106 kPa2580.6m m.TheverticaldeflectionatpointandA ニ 2Pismostnearly(A)0.25m m(B) 0.48m m(C)0.51m m(D)0.75m forcedcon‑cretebeamareshown. steelpropertiesぽ e五 223000lbf/inう000lbf/in，andAs 3in2 ん 40，6ftl L 易 Ais18.75N，thereactionateachof出 eoutersupportsismostnearly(A)5.6N(B)7.2N(C)11N(D) 14N45. longthebeamismos(A)4.7N(B)9.4N(C) 14N(D)18N46. ly白う(A) 23，000lbf(B) 29，000脳(C)35，000lbf(D) 50，000lbf48. Thebeamsupports self‑weight. B)0.36in2(C) 0.67in2(D) 0.78in49. .043(C)0.051(D)0.058PPI.附 w.ppi2pass.com
Solution Practice Exam IIPradiceEx日m2501日tiol1S 81Thedensityis33.125veh/miformaximumtra缶 tionfactoroff 0，thesuperelevationisgivenby((75引1(吋(品)): ) (53::) (33.125品) 叫一(0.8(9.81 ) encebetweentheNCEESHandbook，gradesisA 1‑1% 3%1う40. rfaceDoriginalsurfaceSN αrecycleDrecycle十 α αbaseDbase十 αsubbaseDsubbase6.6 (0.42)(6in) (0.3)(3in)十 (0.2)(10in)十αsubbase(8in) 七heverticalcurvelength，Lう‑L ‑竺L400十 3.58(4)(430ft)2400十 (3.5)(430此)αsubbase 0.15TheanswerisC. pportA usingbasicbeamForexample fstatics，うう hecurveleng七h，400 3.58L 28一 一 A一一二400 (3.5)(430氏) (2)(430此) ‑ 4 h. Therefore，therequiredverticallengthofcurveisL 383.8抗 (380此)TheanswerisB.ト‑‑1一一一:51 20loadsplacedformaximumshearvalueatsupportA39. TheradiusofcurvatureisR 川寺)ThemaximumvalueforshearatsupportA isVA (1000N)(1.8) (1000N)(2.0) 3800NTheanswerisD.PPI.剛 w.ppi2p日sS.com
atpointM.一一2MRI 一弘一eU00一4でN一十一イ︑一勺G‑m 一十一﹀dlu一︐一円U41﹁ うM 伝山︑必RUv 剖42. nwithloadspositionedasshownformax‑凶 value. Asin五nding七heinfiuencelineimummomeforshear ingmomentatpointA ON3.0m一一 6mThemaximumvalueformoment atsupportA � rtualforces.MA (1000N)(8.0m) (1000N)(10.0m)二M.1.5N18000N.m (18kN.m)TheanswerisD.3O mI1.8N0N開43. 5NPN1.5NLMN ON RM，， (3.0m)十 (4448N)(4.6m)RM" 6820Nエ4.6m七[othe1eft]MM ON ‑RN"(3.0m) m)(N‑m)MPNPMN8142682001.81.50RN" 6820N [totheright]LFy 0N RMy ‑ 4448NRMν 4448NPPI.附 W押 i2p日sS.com[upw乱rd]5.49 804594.6470583.00total 127517
自ctic告Ex自m2Soluti開 5PrCalcu1atethevertica1de丑ection抗 nder七hesheardiagramuptothatpoint.ムp LFQ5L 玄 占 LFQFpLc 刷叫叫)べ中(や1 x(2附m m2)ル3m(t)(5.6叩 凡 1.12m(抗日2m)十(訪問問問 ニ(t)(5:)州‑18同?x 1.12m1.88m188m司TheanswerisA.45. eaunderthe10addiagram叩 5.6N‑ a1ueforshearupportA.occurs剖 sAtpoin七A，Vmax 9.4NV ‑9.4N 18.75N Mmax [‑5.7N.m[ 5.7N‑mTheanswerisD.PPIe附 w.ppi2p日sS.com
『."哩50AppendixAftemoonSa円lpleExamination42. OC1ispH‑dependentb. [HOCl]equals[OCllatpH7c. oncretesewerissetonaslopeof0.00l.The43. A61cflowwhenitishalιfullisajof伽 叫 制lb. 0.08m""/s3c.0.20m /s3d.0.28m /s・44. tBandbyacableACasshown.ThereactionforceatsupportB isAPractice Exam 9.70kN14.42kN45. f1ectionis20kNド3.53.m0mE 200GPa41 3x106m ma.b.c.d.16mm38m m22m m8mm
.SampleExamr‑‑r・一一一一r‑15.46. theoryis4kipsa. 13.6.7kipsb. 366kipsc. 3d. alueofthehorizontalthrustatthesupportsis[} []ー6m4m15mド10m当6オa. 282kNb. 197kNc. 218kNd. atB isE 29，000ksiforbothmembers2A 2inforAB2A 3inforBC10抗B20kipsドa. 1.2inb.0.05inc. 0.2ind. 0.1in8ft当
152AppendixAftemoonSampleExamination49. .3.5m replaceonedestroyedbya50. sedin出 issituationisa.unitpriceb. ilpile，if100ftbankcuydof51. 0 anda12%swellisdeposited?a.9.6b. 19.2c. 10.4d. 33.3i1lforawholestation(100ft)ofcutwith52. Determinethevolumeofftheendareasshownbelow. Al 120ft'/ ι が/ A' ，/150100ft3a. 820ydb.456.8ydjc.506.2yd4ydd. 569.JJ000onabuildingiddera53. ABCConstructionCompanyisthelowbt econd，tofthecontractequiredthesur ty . company ro000.Howmuchistlowestbidwas . 500，000b. 50，000c. 20，000d. 5000
SolutionsSolutions may be for problems with slightly differentdimensions;41. d.disjointedAllofthestatementsareco町ect.I was given two apparentlysets.One may conclude that problems are for the most part42. a.identical but with different dimensions.43. b.唱633Qf 0.2m /sQ/Qf 0.4内Q 0.08m' stake44. d. Thcouldbe脳 出 国ngthatcableACformsa3‑4‑5凶 anglewhenitd sn't.百leve同calandhorizontalcomponentsofT訂 eO.66Tand0.75，Trespectively.T10BxヱMB 0.66Tx2‑10x2.5 0司 T ω 4ヱぇ 久一 0.75x18.94 0弓ヱ久Bx 14.20π 0 x18.94‑10 0 By ‑2.50石弓 4RB ぽ型)X(3X1O‑6m4) 6 kNm2m‑J45. b. EI r2x Deflectionatrnidspan(x 2m)duetopointloadisgivenbyPb。20xO.5 ̲，̲6x4xωo 一一一 [‑x (Lーか )x] ̲- :. : [‑2 (4三j6LEr句l今j"l‑OS)x2] 0.016 Deflectionatrnidspan(x 2m)duetodistributedloadisgivenby44δ 5wL 5x4x4 0.022 一一 一一一一384EI 384x600Totalmidspandeflection 0.038m 38m mx)鉱 is:46. c. Bucklingaboutthestrong(K 2.0;L 32ft;rx 8.70in;KUr 88.3Bucklingabouttheweak(y)axis:K 0.7;L 32ft;ry 1.84in;KUr 146.1
唱64AppendixAftemoonSampleExaminationR π2EAπ2x29，000x273 ナーーす (子}内〆". jOO阻 p:ヲ47. b. ointemalx' xorCxissufficient. x'TakingmomentsaboutA(entirestructure):ヱMA 500x山 4ι ‑25C0y TakingmomentsaboutB (righthalfofstructure):ヱに一以 0MB.right 200x5‑6etC97.4kips.Solvingtheseequations，wegx ー148. d. Step1:UsingmethodofjointsatjointB:凡B 25.6k(T)FBC ー load(no
Structural Analysis Prof. V. Saouma Oct. 2013 Structural Analysis is part of the afternoon exam. In the afternoon, you are to answer 60 questions, and Structural Analysis is about 10% of the test content (or about 6 questions). Each question is worth 2 points. You are expected to know: 1.
