ME 211 Homework 4 Solutions

ME 211 Homework 4 Solutions - Problem#16.4...

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Unformatted text preview: Problem#16.4 ME211HomeworkSet#4Solutions 1032008 Locatethecentroidoftheshadedareaboundedbytheparabolaandtheliney=a. 1.)Theequationofthexcoordinateandycoordinateofacentroidareshown below. 2.)Tofind~xand~ytakeadifferentialcutanddeterminewherethecenterofgravity wouldbeinthatdifferentialcut.Asyoucanseeinthefigurebelow,thecenterof gravityinthexdirectionisx/2,andintheydirectionitisy. 3.)Determinethetotalareaoftheshadedregion.Todothis,youmusttakethe integralofthearea. x = ay1/ 2 2 dy = xdx a a a A= dA = xdy = A 0 0 a a 2 2 aydy = a a 3 / 2 = a 2 3 3 A a x x2 (ay) 1 ~ xdA = ( )xdy = dy = dy = a 3 2 4 0 2 0 2 0 xA = x dA = A ~ xdA A 2 1 x ( a2 ) = a3 3 4 3 x= a 8 A ~ ydA = a 2 2 (y)xdy = ay 3 / 2 dy = a a 5 / 2 = a 3 5 5 0 0 a yA = y dA = A ~ ydA = A 2 2 y a2 = a3 3 5 3 y= a 5 Problem#26.37 Replacetheloadingbyanequivalentresultantforceandspecifythelocationofthe forceonthebeam,measuredfrompointB. 1.) DrawtheFBDoftheloading. 2.) Determinethemagnitudeoftheresultantforcebybreakingupthedistribution intotwotrianglesandonerectangleasshownabove. F R 1 1 = (5)(3) + 5(3) + (3)(3) = 27kN 2 2 3.)Tofindthelocationoftheforce,takethemomentoftheforceaboutB. 1 1 M B = 27(d) = (5)(3)(4) + 5(3)(1.5) + (3)(3)(1) 2 2 d = 2.11m Problem#36.43 Determinethemagnitudeoftheresultantforceoftheloadingactingonthebeam andspecifywhereitactsfrompointO. 1.)Findthesumofforces.Treatthedistributedforcelikedifferentialarea 10 F Y = dA = [5(x - 8) 0 2 + 100] dx 5(10 - 8) 3 5(0 - 8) 3 - + 100(10) -100(0) 3 3 FR = 1867lb = 1.87kip 2.)FindthelocationofforcebyusingthecentroidformulalikewedidinProblem 6.4above. Themagnitudeoftheforceis1.87kipat3.66ftfromthebeginningofthexaxis. Problem#47.7 DeterminetheshearforceandmomentatpointsCandD. 1.)DrawtheFBDofwholebeamtosolvefortheexternalforcestothewholebeam. SupportReactions:Figure(a) 2.)Writetheequilibriumequationsnecessarytodeterminetheexternalforces. M F y B = 500lb(8 ft) - 300lb(8 ft) - AY (14 ft) = 0 AY = 114.29lb =AY + By - 500 - 200 - 300 By = 885.7lbs 3.)CutthebeamatCandDanddrawtheFBDs.Todeterminetheinternalshear forcesandmomentsatthosepoints,youmustcutthebeamtoexposethoseforces, andmakethemexternalforcesinasmallerbeamtosolve. InternalForces:Figure(b)andFigure(c) 4.)FindtheshearandmomentforcesatCusingFigure(b) 5.)FindtheshearandmomentforcesatDusingFigure(c) Problem#57.12 TheboomDFofthejibcraneandthecolumnDEhaveauniformweightof50lb/ft.If thehoistandloadweight300lb,determinethenormalforce,shearforce,and momentinthecraneatsectionspassingthroughpointsA,B,andC. 1.)DrawaFBDofthesectionA,andwritetheequilibriumequationstosolveforthe shearandmomentforce. F F X = NA = 0 = VA - 450 = 0 Y VA = 450lb M A = -150lb(1.5 ft) - 300lb(3 ft) = 0 M A = 1125lb * ft 3.)DrawaFBDofthesectionBandwritetheequilibriumequationstosolveforthe shearandmomentforce. F F X = NB = 0 = VB - 550 - 300 = 0 Y VB = 850lb M B - 550lb(5.5 ft) - 300lb(11 ft) = 0 M B = 6325lb * ft F F X = VC = 0 = NC - 650 - 300 - 250 = 0 Y NC = 1200lb MC - 650lb(6.5 ft) - 300lb(13 ft) = 0 MC = 8125lb * ft Problem#67.29 Thesemicirculararchissubjectedtoauniformdistributedloadalongitsaxisofwo perunitlength.Determinetheinternalnormalforce,shearforce,andmomentin thearchat=45. 1.)DrawtheFBD. 2.)Writetheequilibriumequations(Fx,Fy,andM) FRx = FRy = M Ro = w (rd )sin( ) = rw (-cos ) o o 0 0 = rw o (1- cos ) w (rd )cos( ) = rw (sin ) o o 0 0 = rw o (sin ) w (rd )r = r w o o 0 2 At = 45 F F x = 0 = -V + Frx cos( ) - FRy sin( ) V = 0.2929rw o cos(45) - 0.707rw o sin(45) V = -0.293rw o y = 0 = N + Fry cos( ) + FRx sin( ) N = -0.707rw o sin(45) - 0.2929rw o cos(45) N = -0.707rw o M o = 0 = -M + r 2wo ( 4 ) + (-0.707rwo )(r) M = -0.0783r 2 w o Problem#77.45 IfL=18ft,thebeamwillfailwhenthemaximumshearforceisVmax=800lb,orthe maximummomentisMmax=1200lb*ft.Determinethelargestintensitywofthe distributedloadingitwillsupport. 1.)DrawtheFBDofacutofthebeam.Ifyouanalyzethebeamfromtheleftside, youwillhavenoexternalforcestodetermine.Writetheequilibriumequations. 0xL Fy = 0 = -V - wx 2 V =- 2L wx 2 2L wx 3 M = 0 = M + 6L wx 3 M =- 6L 2.)Determinetheplaceofmaximumshearforce. Shearisthelargestattheendofthebeam. V =- wL2 wL =- 2L 2 wL -800 = - 2 w = 88.9lb / ft YoualsoknowthegreatestmomentoccursatLasshownonthegraphbelow.We mustverifythatthewaboveisthecorrectanswerbycheckingwhatwcorresponds tothemaxmomentthatthebeamcanwithstand. wx 2 wL2 =- 6 6 2 wL -1200 = - 6 w = 22.2lb / ft M max = - Thegreatestwthebeamcanwithstandisw=22.2lb/ft. Problem#87.56 Determinetheshearandmomentdiagramsforthebeam. 1.)DrawtheFBDofthewholebeamandwritetheequilibriumequationstofindthe supportreactions(aka.Theexternalforces). F Y = 80kip + BY - 20kip - 20kip - 4kip / ft(30 ft) = 0 2.)Findthemagnitudesofmoments.Utilizesymmetry.Wewillfirstfindthe momentatA,whichduetosymmetrywillbethesameatB. 0x<15 Moment M = M + 20kip(15 ft) = 0 M = -300kip * ft Shear F y = 0 = -V - 20kip V = -20kip 15x<45 M = M + 20kip(x) - 80kip(x -15) + 4kip / ft(x -15 ft) (x -15) 2 Fy = 0 = -V - 20kip + 80kip - 4kip / ft(x -15 ft) M = -20x + 80(x -15) - 4(x -15) V = -20 + 80 - 4(x -15) V = 120 - 4 x 45x<60 x -15 =0 2 M = M + 20kip(x) - 80kip(x -15) + 4kip / ft(30)(x - 30 ft) - 80(x - 45) = 0 M = -20x + 80(x -15) - 4(30)(x - 30) + 80(x - 45) F y = 0 = -V - 20kip + 80kip - 4kip / ft(30 ft) + 80 V = -20 + 80 - 4(30) + 80 V = 20 3.)Findthemagnitudeofthemomentatthecenterofthebeamduetosymmetry. M = M + 20kip(30 ft) - 80kip(15 ft) + 4kip / ft(15 ft)(7.5 ft) = 0 M = 150kipft 4.)Thedrawingsofthediagramsareshownbelow. Problem#97.58 Drawtheshearandmomentdiagramsfortheshaft.ThesupportatAisajournal bearingandatBitisathrustbearing. 1)DrawtheFBD 2.)Writetheshearandmomentequationsforthepiecesofthebeamanddrawthe correspondingdiagrams. 0<x<4 Shear 0 = -V - 400lbs V = -400lbs Moment 0 = M + 400x M = -400x 4x<16 Shear 0 = -V - 400 + Ay -100(x - 4) V = -100x + 1450 Moment M = -400x -100(x - 4) 16x<20 (x - 4) + 1450(x - 4) 2 Shear V = -950 Moment M = -400x -100(12)(x -10) + 1450(x - 4) - 800(x -16) Thesheardiagramisshownbelow Themomentdiagramisshownbelow. Asanalternativemethodtosolvingtheseproblems,youcanintegratefromwtoV toM. ...
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