ME 211 Homework 12 Solutions

ME 211 Homework 12 Solutions - Problem#115.44...

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Unformatted text preview: Problem#115.44 ME211HomeworkSet#12Solutions 11262008 Thestressatapointisshownontheelement.Determinetheprincipalstressesand theabsolutemaximumshearstress. 1.)ConstructionoftheCircle:Mohr'scirclefortheelementintheyzplaneis drawnfirst.Inaccordancewiththesignconvention,y=90psi,z=80psi,and yz=40psi.Hence, avg = 2.)ThecoordinatesforreferencepointsAandCareA(10,40)andC(5.00,0) y + z 90 + (-80) = = 5.00 psi 2 2 R = (90 - 5.00) 2 + 40 2 = 93.94 psi 3.)Inplaceprincipalstress:ThecoordinatesofpointAandBrepresent1and2, respectively 4.)ConstructionofThreeMohr'sCircles:Fromtheresultsobtainedabove, 1 = 5.00 + 93.94 = 98.94 psi 2 = 5.00 - 96.94 = -88.94 psi max = 98.94 psi int = -88.94 psi min = -100 psi 5.)AbsoluteMaximumShearStress:FromthethreeMohr'scirlces abs max = max - min 98.94 - (-100) = = 99.5 psi 2 2 Problem#215.70 The45degreestrainrosetteismountedonasteelshaft.Thefollowingreadingsare obtainedfromeachgauge:a=300(106),b=180(106),andc=250(106). Determinetheinplaneprincipalstrainsandtheirorientation. 1.) StrainRosettes(45):Applyingtheequationsinthetext b = 180(10-6 ) c = -250(10-6 ) a = -45 b = 0 c = 45 180(10-6 ) = x cos2 0 + y sin 2 0 + xy sin0cos0 x = 180(10-6 ) 300(10-6 ) = 180(10-6 )cos2 (-45) + y sin 2 (-45) + xy sin(-45)cos(-45) 210(10-6 ) = 0.5y - 0.5 xy -250(10-6 ) = 180(10-6 )cos 2 (45) + y sin 2 (45) + xy sin(45)cos(45) -340(10-6 ) = 0.5y + 0.5 xy y = -130(10-6 ) xy = -550(10-6 ) 2.)ConstructionoftheCircle:Withthestrainssolvedforabove,theaverage straincanbecalculated,andtheradiusdeterminedtocreatethecircle x = 180(10-6 ) y = -130(10-6 ) xy = -275(10-6 ) 2 + 180 + (-130) avg = x y = [ ](10-6 ) = 25.0(10-6 ) 2 2 A(180,-275) C(25.0,0)(10-6 ) R = ( (180 - 25) 2 + 275 2 (10-6 ) = 315.67(10-6 ) 3.)InplanePrincipalStrain:ThecoordinatesofpointsBandDrepresent12 respectively. x = (25.0 + 315.67)(10-6 ) = 341(10-6 ) y = (25.0 - 315.67)(10-6 ) = -291(10-6 ) 4.)OrientationofPrincipalStrain:Fromthecircle 275 = 1.7742 180 - 25.0 P1 = 30.3(Clockwise) tan(2 P1 ) = Problem#315.107 The60degreestrainrosetteismountedonthesurfaceofadome.Thefollowing readingsareobtainedforeachgauge:a=780(106),b=400(106),andc= 500(106).Determine(a)theprincipalstrainsand(b)themaximuminplaneshear strainandassociatedaveragestrain.Ineachcase,specifytheorientationofthe elementandshowhowthestraindeformstheelement. Part(a) 1.)Determinetheprincipalstrains a = 0 b = 60 c = 120 = x cos2 + y sin 2 + xy sin cos -780(10-6 ) = x cos 2 0 + y sin 2 0 + xy sin0cos0 x = -780(10-6 ) 400(10-6 ) = x cos2 120 + y sin 2 120 + xy sin120cos120 400(10-6 ) = 0.25x + 0.75y - 0.433 xy 500(10-6 ) = x cos2 60 + y sin 2 60 + xy sin60cos60 500(10-6 ) = 0.25x + 0.75y + 0.433 xy -100(10-6 ) = -0.866 xy xy = 115.47(10 ) x = -780(10-6 ) y = 860(10-6 ) xy = 57.74(10-6 ) 2 A(-780,57.74)(10-6 ) C(40,0)(10-6 ) Part(b) -6 max avg x - y 2 xy 2 -3 = 2* + = 1.641(10 ) 2 2 x + y -780(10-6 ) + 860(10-6 ) -5 = = = 4(10 ) 2 2 Problem#415.80 Thestraingaugeisplacedonthesurfaceofathinwalledsteelboilerasshown.Ifit is0.5inlong,determinethepressureintheboilerwhenthegaugeelongates 0.2(10^3)in.Theboilerhasathicknessof0.5in,andinnerdiameterof60in.Also, determinethemaximumx,yinplaneshearstraininthematerial.Est=29(10^3)ksi, vst=0.3. 1.)FirstdrawtheFBD 0.2(10-3 ) 2 = = 400(10-6 ) 0.5 1 2 = [ 2 - (1 + 3 )] E Pr 2 = 2t Pr 2 = t 3 = 0 1 Pr Pr 400(10-6 ) = - 0.3( ) 29(10 3 ) 2t t P = 0.967ksi 2 = 29ksi 1 = 58ksi 1 1 = 58 - 0.3(20 + 0)] = 1700(10-4 ) 3 [ 29(10 ) max 1 - 2 = = 1.30(10-3 ) 2 2 Problem#515.102 Thesmoothrigidbodycavityisfilledwithliquid6061T6aluminum.Whencooled itis0.012infromthetopofthecavity.Ifthetopofthecavityisnotcoveredandthe temperatureisincreasedby200degreesFahrenheit,determinethestrain componentsx,y,andzinthealuminum.Hint:UseEq.1532withanadditional straintermofT(Eq.104). 1.)Findthenormalstrain.Sincethealuminumisconfinedatitssidesbyarigid container,then x = y = 0 2.)UtilizetheHooke'sLawwiththeadditionalthermalstrain.Note:sinceitisnot constrainedinthezdirection,z=0. x = 1 [ x - v( y + z )] + T E 1 0= [ x - 0.35( y + 0)] + 13.1(10-6 )(200) 3 10.0(10 ) (1.)0 = x - 0.35 y + 26.2 1 [ y - v( x + z )] + T E 1 0= [ y - 0.35( x + 0)] + 13.1(10-6 )(200) 10.0(10 3 ) (2.)0 = y - 0.35 x + 26.2 y = 3.)SolvingEquations1and2above x = y = -40.31ksi 1 [ z - v( x + y )] + T E 1 z = [0 - 0.35(-40.31+ (-40.31))] + 13.1(10-6 )(200) 10.0(10 3 ) z = z = 5.44(10-3 ) Problem#615.93 Theshafthasaradiusof15mmandismadeofL2toolsteel.DeterminethetorqueT intheshaftifthetwostraingauges,attachedtothesurfaceoftheshaft,report strainsofx'=45(106)andy'=45(106).Also,computethestrainsactinginthex andydirections. 1.)Determinetheshearstress.Thetorsionalshearstresscanbeobtainedusing thetorsionformula, = Tc J xy = T(0.015) = 188628T 4 (0.015 ) 2 2.)StrainRosettes:Forpureshear,ofx=y=0. Applyinga=45(106)anda=30degrees, -45(10-6 ) = 0 + 0 + xy sin(30)cos(30) xy = -103.92(10-6 ) Applyinga=45(106)anda=120degrees, 45(10-6 ) = 0 + 0 + xy sin(120)cos(120) xy = -103.92(10-6 ) 3.)Shearstressandstrainrelationship:ApplyingHooke'sLaw, xy = G xy -188628T = 78.0(10 9 )[-103.92(10-6 )] T = 42.9N * m ...
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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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