ME 211 Homework 9 Solutions

ME 211 Homework 9 Solutions - Problem#16.20...

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Unformatted text preview: Problem#16.20 ME211HomeworkSet#9Solutions 1172008 Locatethecentroidybarofthechannel'scrosssectionalarea. 1.)DrawtheFBDandbreakthepiecedownintopartsofwhichyoucanfindthearea andthecentroid. 2.)Determinetheareaandthecentroidofeachindividualpiece(1,2,and1) Area1 = 6in(4in) y1 = 3in Area1 y = 72in 3 Area2 = 12in(2in) y 2 = 1in Area2 y 2 = 24in 3 3.)DeterminethesummationoftheareasandthesummationsofAreaxybar Area = Area + Area = 24in Area = 48in Area y = 72in + 24in Area y = 96in 1 2 2 3 3 x x x x 3 2 + 24in 2 4.)Solveforybarofthewholepiece Area y y= Area x x 96in 3 = = 2in 3 48in Thecentroidis2inchesfromthetopor4inchesfromthebottom. Problem#26.73 Determineybar,whichlocatesthecentroidalaxisx'forthecrosssectionalareaofthe Tbeam,andthenfindthemomentsofinertiaIx'andIy'. 1.) Tofindthecentroid,usetwosections 2.)Nowfindthemomentofinertiaaboutthisaxis 3.)Nowabouttheyaxis(Note:symmetrygetsridofAd^2term) 1 1 IY = 12 (250mm)(50mm) 3 + 12 (50mm)(300mm) 3 IY = 2604166.667mm 4 + 112500000mm 4 IY = 115104166.7mm 4 IY = 115(10 6 ) mm 4 Problem#36.83 Determinethemomentofinertiaofthebeam'scrosssectionalareawithrespectto thex'axispassingthroughthecentroidCofthecrosssection.Neglectthesizeofthe cornerweldsatAandBforthecalculation,ybar=104.3mm. 1.)DrawtheFBDandbreakthebarintosegmentstoallowtheuseofsectionsto determinethecentroid 2.)Themomentofinertiaaboutthex'axisforeachsegmentcanbedetermined usingtheparallelaxistheorem, Findthemomentofinertiaofeachsegement Segment 1 2 3 3.)Thetotalmomentofinertia Ai(mm2) (17.52) 15(150) (252) (dy)i(mm) 113.2 20.7 79.3 (Ibarx')i(mm4) (/4)(17.52) (Ady2)i(mm4) (Ix')i(mm4) 12.329(106) 12.402(106) 5.183(106) 12.654(106) . (1/12)15(1503) 0.964(106) (/4)(254) 12.347(106) Problem#412.9 ThealuminummachinepartissubjectedtoamomentofM=75N*m.Determinethe maximumtensileandcompressivebendingstressesinthepart. 1.)Firstfindthecentroidandtheareamoment 2.)Tensiononbottom,compressionontop Problem#512.21 Determinethesmallestallowablediameteroftheshaft,whichissubjectedtothe concentratedforces.ThesleevebearingsatAandBsupportonlyverticalforces, andtheallowablebendingstressisallow=22ksi. 1.)DrawtheFBD,theshearforceandmomentdiagramstodeterminethemaximum moment 2.)Findthemaximummomentfromthemomentgraphabove 3.)Findthediameterbyapplyingthebendingstressequationwiththeallowable bendingstressequalto22(103)psi. Problem#612.23 DeterminethemagnitudeofthemaximumloadPthatcanbeappliedtothebeamif thebeamismadeofamaterialhavinganallowablebendingstressof(allow)C=16ksi incompressionand(allow)t=18ksiintension. 1.)DrawtheFBDofthewholebeam,Figure(a),asectionthatyieldsthemax moment,andacrosssection. 2.)Firstfindthecentroidandtheareamoment 3.)Checkcompressionfirst Compression = M MAX c 5P(12in / ft)(2.75in) = = 16,000 psi I 124.33in 4 P = 12,056lbs = 12.1kip 4.)Checktension Tension = M MAX c 5P(12in / ft)(9in - 2.75in) = = 18,000 psi I 124.33in 4 P = 5968lb = 5.97kip Tensionwasthecriticalvalue. Problem#712.24 ThestrutCDontheutilitypolesupportsthecablehavingaweightof600lbs. DeterminetheabsolutemaximumbendingstressinthestrutifA,B,andCare assumedtobepinned. 1.)DrawtheFBD 2.)WritetheequilibriumequationsaboutthestrutCDtosolvefortheforces 4.)Usetheforcesjustfoundtodeterminetheshearforceandmomentdiagrams 5.)Usethemaxmomentfromthemomentgraphaboveinthebendingstress equationtosolveforthemaxbendingstress. Problem#814.11 Thescrewoftheclampexertsacompressiveforceof500lbsonthewoodblocks. Determinethemaximumnormalstressdevelopedalongsectionaa.Thecross sectionthereisrectangular,0.75in.by0.50in. 1.)DrawtheFBD 2.)Determinetheareaofthecrosssection 3.)Determinemaximumstressinthecrosssection.Inthissituation,thereis bothnormalstressandbendingstress Problem#914.17 Determinethemaximumandminimumnormalstressinthebracketatsectiona whentheloadisappliedatx=0. 1.)DrawtheFBDofthecontrolleverexposingthecrosssectionataa. 2.)Determinethesectionproperties 3.)Determinethestressinthecrosssection ...
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