ME 211 Homework 11 Solutions

ME 211 Homework 11 Solutions - Problem#115.5...

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Unformatted text preview: Problem#115.5 ME211HomeworkSet#11Solutions 11212008 Solveproblem15.4usingthestresstransformatioequaionsdevelopedinSec.15.2. Showtheresultonasketch. 1.)DrawtheFBDofthesectionAB 2.)Determinethenormalandshearstress:inaccordancewiththeestablishedsign convention, = +60 x = -3ksi y = 2ksi xy = -4ksi 3.)Applythestresstransformationequations:ApplyingEqs.91and92 x + y x -y + cos(2 ) + xy sin(2 ) 2 2 -3 + 2 -3 - 2 x' = + cos(120) + (-4 sin(120)) 2 2 x' = -2.71ksi -y x' y' = - x sin(2 ) + xy cos(2 ) 2 -3 - 2 x' y' = - sin(120) + (-4 cos(120)) 2 x' y' = 4.17ksi x' = Thenegativesignindicatesthat xisacompressivestress Problem#215.20 Thewoodenbeamissubjectedtoaloadof12kN.Ifgrainsofwoodinthebeamat pointAmakeanangleof25degreeswiththehorizontalasshown,determinethe normalandshearstressesthatactperpendicularandparalleltothegrainsdueto theloading. 1.) DrawtheFBD 2.)Writetheequilibriumequationintheydirection 3.)FindtheinternalreactionsatA 4.)Determinethestresses My 13714Nm(0.075m) = 1 I (0.2m)(0.3m 3 ) 12 A = 2.2857MPa(Tension) A = 0 A = 5.)Determinetheangletobeused,needthenormaltolineA 6.)Wewillusethe115degreeshere X = 2.2857MPa Y = 0 XY = 0.0MPa + Y X - Y X '= X + cos2 + XY sin2 2 2 2.2857 + 0 2.2857 - 0 X '= + cos2(115 ) + 0.0sin2(115 ) 2 2 X '= 0.408MPa(Tensile) + y x -y y '= x - cos2 - xy sin2 2 2 2.2857 + 0 2.2857 - 0 y '= - cos2(115) - (-0.0)sin2(115) 2 2 y '= 1.877MPa X - Y sin2 + XY cos2 2 2.2857 - 0 XY '= sin2(115 ) + -0.0cos2(115 ) 2 XY '= -0.875MPa XY '= Problem#315.31 DrawMohr'scirclethatdescribeseachofthefollowingstatesofstress. Part(a) 1.)Determinetheprincipalstresses 2.)Calculatetheaveragestressandradius Part(b) 1.)Determinetheprincipalstresses 2.)Calculatetheaveragestressandradius Part(c) 1.)Determinetheprincipalstresses 2.)Calculatetheaveragestressandradius Problem#415.32 Apointonathinplateissubjectedtotwosuccessivestatesofstressasshown. Determinetheresultingstateofstresswithreferencetoanelementorientedas shownattheright. 1.)FirstdrawtheFBD ELEMENT(a) Constructionofthecircle:Inaccordancewiththesignconvention,x'= y'=85MPa,andx'y'=0MPaforelement(a) 1.)Findthestressesonthesectionatthenewlydefinedcoordinatesystem x' = 85MPa y' = 85MPa x' y' = 0MPa 2.)Findtheaveragestress avg = x' + y' 85 + 85 = = 85.0MPa 2 2 3.)Determinetheradius A(85,0) C(85,0) R = A - C = (85,00) - (85,0) = 0MPa Theradiusis0;therefore,Mohr'scircleissimplyadotatC.Astheresult,the stateofstressisthesameregardlessoftheorientationoftheelement. ( x ) a = 85MPa ( y ) a = 85MPa ( xy ) a = 0MPa ELEMENT(b) Constructionofthecircle:Inaccordancewiththesignconvention,x'=y'=0MPa, andx'y'=60MPaforelement(b) 1.)Determinethestresses x' = 0MPa y' = 0MPa x' y' = 60MPa 2.)Determinetheaveragestress avg = x' + y' 0 + 0 = = 0MPa 2 2 3.)Determinetheradiusofthecircle A(0,60) C(0,0) R = A - C = (0,60) - (0,0) = 60MPa StressontheRotatedElement 6.)Determinethestressontherotatedelement:thenormalandshearstress components(x)b,(xy)barerepresentedbythecoordinatesofpointPonthecircle. (y)canbedeterminedbycalculatingthecoordinatesofQonthecircle. ( x ) b = 0 - 60.0cos(30) = -51.96MPa ( y ) b = 0 + 60.0cos(30) = 51.96MPa ( xy ) b = -60.0sin(30) = -30.0MPa 7.)Combinethestresscomponentsofthetwoelements x = ( x ) a + ( x ) b = 85 + (-51.96MPa) = 33.0MPa y = ( y ) a + ( y ) b = 85 + 51.96MPa = 137MPa xy = ( xy ) a + ( xy ) b = 0 + (-30.0MPa) = -30.0MPa Problem#515.39 Thethinwalledpipehasaninnerdiameterof0.5in,andathicknessof0.025in.Ifit issubjectedtoaninternalpressureof500psiandtheaxialtensionandtorsional loadingsshown,determinetheprincipalstressesatapointonthesurfaceofthe pipe 1.)Checktoseeifpressurevesselisvalid Yes,itisvalid. 2.)Determinethesectionproperties 3.)Determinethestresses 4.)Calculatetheaveragestressandtheradius 5.)DrawtheMohr'scircle 6.)Calculatetheprincipalstresses Problem#6SupplementalProblem#1 FigureP1showsasolidcylindricalshaftofdiameter10mmandlength1m,whichis builtintoawallatoneendA.AttheotherendB,aloadof10kNisappliedatthetop andperpendiculartothewall.Also,equalandoppositeloadsof5kNareapplied paralleltothewallatthetopandbottomasshown. Findthemaximumshearstressinthebeam. 1.)FindtheSectionProperties A = [(0.005m) 2 ] 1 1 J = r 4 = [(0.005) 4 ] 2 2 2.)Determinethestresses N My 10kN 10,000N(0.005)(0.005) + = + = 636.6MPa 2 A I (0.005) (0.005) 4 4 y = 0 x = = Tc 10kN(0.005)(0.005m) = = 254.65MPa 1 J (0.005) 4 2 3.)Determinetheaveragestressandtheradius avg = x + y 636.6MPa = = 318.3MPa 2 2 -y 2 2 R= ( x ) + xy = 407.6MPa 2 4.)Determinethemaximumshearstress max = 407.6MPa Problem#7SupplementalProblem#2 Asolidcircularsteelshaftofdiameter40mmissubjectedtoabendingmomentof 550Nmandatorqueof400Nm. Findthemagnitudeofthemaximumtensilestressintheshaft. 1.)FindtheSectionProperties A = [(0.02m) 2 ] 1 1 J = r 4 = [(0.02) 4 ] 2 2 1 1 I = r 4 = [(0.02) 4 ] 4 4 2.)Determinethestresses My 550Nm(0.02) = = 87.5MPa I 4 (0.02) 4 y = 0 x = Tc 400(0.02m) = = = 31.83MPa 1 J 4 (0.02) 2 3.)Determinetheaveragestressandtheradius avg = x + y 87.5MPa = = 43.75MPa 2 2 -y 2 2 R= ( x ) + xy = 54.10MPa 2 4.)Determinethemaximumtensilestress 1 = avg + R = 97.853MPa Problem#8SupplementalProblem#3 FigureP3showsathinwalledcylinderwithclosedendsthatisbuiltinatoneend andloadedbya50kNforceattheother.Ifthevesselalsocontainsagasata pressureof1MPa,findthelocationandmagnitudeof(i)themaximumtensile stress,and(ii)themaximumshearstress. 1.)FindtheSectionProperties A = [(0.25 2 - 0.24 2 )] 1 1 J = r 4 = [(0.25 4 - 0.24 4 )] 2 2 1 1 I = r 4 = [(0.25 4 - 0.24 4 )] 4 4 2.)Determinethestresses My Pr 200,000Nm(0.25) (10 6 )(0.24) x = + = + = 120.2MPa I 2t 2(0.01) (0.25 4 - 0.24 4 ) 4 Pr (10 6 )(0.24) y = = = 24 MPa t (0.01) =0 3.)Determinetheaveragestressandtheradius x = 120.2MPa y = 24 MPa xy = 0MPa avg = x + y 120.2MPa + 24 MPa = = 72.1MPa 2 2 -y 2 2 R= ( x ) + xy = 48.1MPa 2 4.)Determinethemaximumtensilestressandmaximumshearstress 1 = avg + R = 72.1+ 48.1 = 120.2MPa max = max = 60.1MPa 2 ...
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This note was uploaded on 05/12/2010 for the course MECHENG 211 taught by Professor ? during the Fall '07 term at University of Michigan.

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