ME 211 Homework 3 Solutions

ME 211 Homework 3 Solutions - Problem#15.4...

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Unformatted text preview: Problem#15.4 ME211HomeworkSet#3Solutions 9262008 Thetruss,usedtosupportabalcony,issubjectedtotheloadingshown.Approximate eachjointasapinanddeterminetheforceineachmember.Statewhetherthe membersareintensionorcompression.SetP1=800lbs,P2=0. Solution 1.) DrawtheFreeBodyDiagram(FBD)ateachjointandwritetheequilibrium equations.Continuetodrawthejointsuntilyouhaveansweredthequestion foralltheunknownsthatthequestionisasking. a. JointA Therearetwouknowns,soweneedtwoequilibriumequations. 1.) Fy = 0 = FAD sin(45) - P1 800 = FAD sin(45) FAD = 1131.1lbs(Compression) 2.) Fx = 0 = FAB - FAD cos(45) 0 = FAB -1131.1lbscos(45) FAB = 800lbs(Tension) b. JointB Therearetwouknownssoweneedtwoequilibriumequations. 1.) Fy = 0 = FBD - P2 FBD = 0lbs 2.) Fx = 0 = FBC - 800 FBC = 800lbs(Tension) c. JointD Therearetwouknowns,soweneedtwoequations. 1.) Fy = 0 = FDC sin(45) - P2 -1131.4 sin* (45) FDC = 113.1lbs(Tension) 2.) Fx = 0 = 1131.4 cos(45) + 1131.4 cos(45) - FDE FBC = 1600lbs(Compression) Problem#25.8 Determinetheforceineachmemberofthetrussandstateifthemembersarein tensionorcompression.SetP1=2kNandP2=1.5kN. 1.) DrawtheFBDateachjointoneatatime,andwritetheequilibrium equations.YoumightfindthatyoudonotneedtodrawaFBDforeveryjoint beforeyouhavesolvedforalltheforcesthatwereaskedfor.Therearepin connections,andyoucanlookupintable41thatapinintwodimensions hastwounknowns,FyandFx. JointC Therearetwouknowns,soweneedtwoequations. 1.) Fy = 0 = FBC sin(30) - P2 FBC = 3kN(Tension) 2.) Fx = 0 = FCD - FBC cos(30) FCD = 2.60kN(Compression) JointD y F F = 0 = FBD - P1 = 0 = FDE - FCD FBD = 2kN(Tension) x FDE = FCD = 2.60kN(Compression) JointB F y = 0 = FBE cos(30) - FBD cos* (30) FBE cos(30) = 2cos* (30) FBE = 2kN(Compression) F F x x = 0 = -FAB + FBE sin(30) + FBD sin(30) + FBC = 0 = -FAB + 2sin(30) + 2sin(30) + 3 FAB = 5kN(Tension) Problem#35.28 DeterminetheforceinmembersBC,HC,andHG.Afterthetrussissectionedusea singleequationofequilibriumforthecalculationofeachforce.Stateifthese membersareintensionorcompression. 1.)Ifyouareaskedtoonlydetermineacoupleofforcesinatrussopposedtothe forceineverybar,themethodofsectionsisagoodoptionsothatyoudonothaveto wastetimedeterminingtheforceineverybar. Tobegin,youmustfirstdrawtheFBDofthewholetrussbecauseyouwillneedthe reactionforceatA. 2.)Writetheequilibriumequationsnecessarytodeterminetheexternalforces. Therearetwounknownssoweneedtwoequations. 1.) M E = 0 = -Ay * 20 + 2 * 20 + 4 *15 + 4 *10 + 5 * 5 = 0 Ay = 8.25kN 2.) Fy = 0 = Ay + E y - 2 - 4 - 4 - 5 - 3 E y = 9.75kN 3.)DrawtheFBDofthesectionedtrussthatmakesalltheforcesthatarebeing askedforexternalforceslikeshownbelow.Solveforthoseunknownforces. 4.)Therearenowthreeuknowns,soweneedthreeequations. 1.) M H = 0 = -Ay * 5 + 2 * 5 + FBC * 3 FBC = 10.4kN(Compression) 2.) MC = 0 = -Ay *10 + 2 *10 + 4 * 5 + FHG = 9.16kN(Tension) 3.) MO = 0 = -2 * 2.5 + 8.25 * 2.5 - 4 * 7.5 + FHC = 2.24kN(Tension) 3 * FHC *12.5 34 5 * FHG * 5 29 Problem#45.33 DeterminetheforceinmembersGF,CF,andCDoftherooftrussandindicateifthe membersareintensionorcompression. 1.)Ifyouareaskedtoonlydetermineacoupleofforcesinatrussopposedtothe forceineverybar,themethodofsectionsisagoodoptionsothatyoudonothaveto wastetimedeterminingtheforceineverybar. Tobegin,youmustfirstdrawtheFBDofthewholetrussbecauseyoualwayshave tofindthereactionforcesexternaltothetrussinthemethodofsections. 2.)Writetheequilibriumequationsnecessarytodeterminetheexternalforces. M A = 0 = E y (4) - 2(0.8) -1.5(2.50) E y = 1.3375kN 3.)DrawtheFBDofthesectionedtrussthatmakesalltheforcesthatarebeing askedforexternalforceslikeshownbelow.Solveforthoseunknownforces. Therearethreeunknowns,soweneedthreeequations. 1.) MC = 0 = E y (2) - FGF (1.5) FGF = 1.78kN(Tension) 3 2.) M F = 0 = E y (1) - FCD ( )(1) 5 FCD = 2.23kN(Compression) 1.5 3.) M E = 0 = FCF ( )(1) 3.25 FCF = 0 Problem#55.36 DeterminetheforcePneededtosupportthe100lbweight.Eachpulleyhasaweight of10lbs.Also,whatarethecordreactionsatAandB? 1.)DrawaFBDofpulleyBandthelastpulley.PulleyCisnotneededbecauseP' incorporatestheimportantinformationthatwewouldbeabletogatherfromthe lastFBD. 2.)Writetheequilibriumforthepulleyfromfigure(a) F y = 0 = P'+2P -10 3.)Writetheequilibriumequationforthepulleyfromfigure(b) F y = 0 = 2P + P'-100 -10 4.)Usingthetwoequationsfromabove,wehavetwoequationsandtwounknowns, socansolveforbothunknowns P = 25lbs P'= 60lbs 5.)ThecordreactionsatAandBareequivalenttothecordreactionsatthatpoint. FA = P = 25lbs FB = P'= 60lbs Problem#65.43 DeterminethehorizontalandverticalcomponentsforceatpinsAandCofthetwo memberframe. 1.)DrawtheFBD.TheimportantpartofthisproblemisrecognizingthatmemberBC isatwoforcemember;therefore,theactionoftheforceactsfromonepointofthe twoforcemembertotheotherend.Alsoremembertoseparatethepartsifyou wanttoexaminereactionforcesatajointorinonebar. 2.)Writetheequilibriumequationsofthebar.(Fx,Fy,andM)Theunknownsare Ax,Ay,andFBC.Therearethreeunknowns;therefore,weneedthreeequations. 1.) M A = 0 = FBC cos(45)(3) - 600(1.5) FBC = 424.26N 2.) Fy = 0 = Ay + 424.26cos(45) - 600 Ay = 300N 3.) Fx = 0 = -Ax + 424.26sin(45) Ax = 300N 3.)NowthattheforceinbarFBChasbeendetermined,theresultantforceatChas beendeterminedbecausethesameforceactsthroughoutatwoforcemember.To findthereactionsforcesatC,breaktheresultantforceofthebarintoxandy components. Cx = FBC sin(45) = 424.26sin(45) = 300N Cy = FBC cos(45) = 424.26cos(45) = 300N Problem#75.53 Determinethehorizontalandverticalcomponentsofforce,whichthepinsexerton memberABC. 1.) DrawtheFBDandwritetheequilibriumequationsforpinA.Remember, fromTable41,ifitisonlyintwodimensions,thenthereareonlytwo unknownforcesatapin,xandy;therefore,weneedtwoequations. 1.) Fy = 0 = Ay - 80 Ay = 80lb 2.) Fx = 0 = Ax - 80 Ax = 80lb 2.)DrawtheFBDsofthetwobarsseparately. 3.)Therearefourunknownsthatweneedtosolvefor,Bx,By,Cx,Cy,soweneedfour equations. 1.) MC = 0 = 80(15 ft) - By (9 ft) By = 133.3lbs 2.) M D = 0 = -80(2.5 ft) + By (9) - Bx (3) Bx = 333lbs 3.) Fx = 0 = 80 + 333 - Cx Cx = 413lb 4.) Fy = 0 = -80 + 133.3 - Cy Cy = 53.3lb Problem#85.63 DeterminetheforcePonthecableifthespringiscompressed0.5incheswhenthe mechanismisinthepositionshown.Thespringhasastiffnessofk=800lb/ft. 1.)DrawtheFBDsofthetwobarsthatarenottwoforcemembers.Note,FCDisthe twoforcememberbecausethereareonlyforcesontheendsofthebar,sothesame forceactsuniformlythroughout. TodeterminetheforceatE,usethespringequation,kx=F,wherek=spring stiffness,andx=thechangeindistanceofthespring. FE = kx = 800( 0.5 ) ft = 33.33lbs 12 2.)Therearesixunknowns,P,FCD,Bx,By,Ay,Ax;thereforeweneedsixequations. Writethesummationofamoment,sumoftheforcesinthexdirection,andsumthe forcesintheydirectionforbothbars. 1.) M A = 0 = Bx (6) + By (6) - FE (30) M A = 0 = Bx (6) + By (6) - (33.3lbs)(30) Bx + By = 166.67lbs 2.) Fx = Ax + Bx - 33.35 = 0 Ax + Bx = 33.35 3.) Fy = -Ay + By = 0 Ay = By 4.) M D = 0 = By (6) - P(4) By = 0.667P 5.) Fx = 0 = -Bx + FCD cos(30) Bx = FCD cos(30) 6.) Fy = 0 = -By + FCD sin(30) - P By = FCD sin(30) - P 3.)Usingtheequationsabove,younowhave6equationsand6unknowns,soyou cansolveforPusingsubstitution. P = 46.9lbs Problem#95.69 IfaforceofP=6lbsisappliedperpendiculartothehandleofthemechanism, determinethemagnitudeofforceFforequilibrium.Themembersarepin connectedatA,B,C,andD. 1.)DrawtheFBDsofthebarsthatincorporatetheforcesthatweneedtodetermine. Note,FBCisatwoforcememberbecausethereareonlyforcesoneachendofthe bar;therefore,thesameforceactsthroughoutthewholebar. 2.)Therearesixunknowns,Ax,Ay,F,FBC,Dx,Dy;therefore,weneedsixequations. 1.) M A = 0 = FBC (4) - 6(25) FBC = 37.5lbs 2.) Fx = 0 = -Ax + 6 Ax = 6lbs 3.) Fy = 0 = -Ay + FBC Ay = 37.5lbs 4.) M D = 0 = -5(6) - 37.5(9) + 39(F) F = 9.42lbs Inthissituation,weonlyneededfourequations,becausewedidnotneedtosolve forDx,andDy.Sowesolvedfourequationsforfourunknowns. ...
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