3.11 Describethe Significanceof Structures Andmachinecomponents Madeof Twomaterialswithdifferent

Chapter 3

PhysicalPropertiesofMaterials

QUALITATIVEPROBLEMS

3.11 Describethe significanceof structures andmachinecomponents madeof twomaterialswithdifferent coefficientsofthermalexpansion.

Thestructuralfitofthemachinecomponents willdependonthethermalexpansioncoefficient.For instance, iftwomaterials with differentthermal expansion coefficientsare assembledtogether bysomemeansandthen heated, thestructurewilldevelopinternal stresses duetounevenexpansion. Ifthesestressesarehighenough,thestructurewillwarp,bend,orbuckleinordertobalanceorrelievethestresses;itwillpossiblyretain someinternal (residual) stressesaswell. Ifpreventedfromwarping, the structurewilldevelop highinternal stresses whichcanleadtocracks. This isnot alwaysdetrimental;shrink fitsaredesignedrecognizingthatmaterials may have differentcoefficientsofthermalexpansion, and somemachine elementssuchasthermocouplesandtemperatureprobesarebasedonamismatch ofthermal expansioncoefficients.

3.12 Whichofthepropertiesdescribedin thischapterareimportantfor (a)potsandpans,(b)cookiesheetsfor baking,(c) rulers,(d)paperclips,(e)musicwire,and(f)beveragecans?Explainyouranswers.

(a) Pots and pans: These require ahigh melting pointsothatthey don’t change phaseduring use;they should becorrosion resistant,cleanable inwater solutions, and haveahighthermal conductivity. The requirementsaresimilar toacookiesheet describednext.

(b) Cookiesheet: Requirescorrosionresistanceathightemperatures,thespecificheatshouldallowforrapid heating ofthe sheet, and ahighthermalconductivityshould allowforeven distributionofheat across the sheet. The melting temperatureshould behighenoughthatthesheetcansafelywithstandbaking temperatures.

(d) Paper clip: Should becorrosion resistant,with astiffnessthatholds papers togetherwithout requiring excessiveforce.

(e) Musicwire:Musicwire,asusedforguitars, ispreloaded toaveryhightension inordertoachievedesiredresonance. Assuch,itshouldhaveaveryhighstrengthandthepropercombinationofstiffnessanddensitytoachievetheproper acoustics.

(f) Beveragecan: Shouldhaveahighthermalconductivity,lowdensity, andgoodcorrosionresistance.

3.13 Notein Table3.1 thatthepropertiesof thealloysof metalshavea widerangecomparedwiththepropertiesofthepuremetals.Explainwhy.

Alloyingelements tend to disturbthe crystal lattice ofthe base metal, and they dosobydistortingthe lattice byoccupying lattice sites (substitutionalatoms), spacesbetween lat-tice sites (interstitials),orforming asecondphase (an intermetalliccompound ofthe twoelements). Lattice distortionwillreduceproperties thatdepend onarepeating lattice, suchasthermalconductivityand melting points. Propertiessuch asdensity and specificheatgenerally depend onthe propertiesofthe alloyingelements, andrangearound the valueforthe alloybasemetal. Also,‘alloys’isagenericterm,and caninclude averywiderange ofconcentrationand types ofalloying element, whereas pure metals have, bydefinition, onlyonechemistry.

3.14 Rankthefollowingin orderofincreasingthermalconductivity: aluminum,cop-per,silicon,titanium, ceramics,andplastics. Comment on theirapplicationsvis-`a-vis thesematerials.

Thermalconductivitydata iscontained in Table 3.1onp. 89. These materials,in orderofincreasing thermalconductivity,are ranked plastics (0.1-0.4W/mK), ceramics (10-17),titanium(17),aluminum (222),copper(393). Thisranking showswhymaterials suchasalu-minumandcopperareusedasheatsinkmaterials,andwhypolymersareusedasainsulator.Titaniumand ceramic materials,having anintermediatevalueofthermalconductivity,aresuitable forneither insulationorheat dissipation,and therefore donot have many thermalapplications.

3.15 Doescorrosionhaveanybeneficialeffects?Explain.

Corrosion isthoughtofmainlyasadetrimentalphenomenon. However,suchmanufacturingprocessesaschemicalmachining andchemicalmechanical polishingrelyoncorrosioneffects.Also,tosomeextent, cleaningofsurfacesreliesoncorrosion.

3.16 Explainhowthermalconductivitycanplayarolein thedevelopmentofresidualstressesin metals.

Thermalconductivityisoneofthemostimportantmaterial propertiesaffectingthermalstress(alongwiththermalexpansion). Interms ofresidual stresses,itismuchlessimportantthantheprocessinghistory. However,unevencoolingofcastings (PartII)orwelds(PartVI),forexample, cancausewarpage andresidual stresses.

3.17 Whatmaterialproperties aredesirablefor heatshieldssuchas thoseplacedonthespaceshuttle?

Material properties required forheat shieldsaresufficient strengthsothatthey donot failupontakeoff,reentry,andlanding; they must haveahighmelting pointsothatthey donotchangephaseordegrade at thehightemperaturesdeveloped during reentry,and theymustbeexceptionally highthermalinsulators sothattheshuttle cabindoesnotheat significantlyduring reentry.

3.18 Listexamplesofproductswherematerialsthataretransparentaredesired.Listapplicationsfor opaquematerials.

This isan open-ended problem, and studentsshould beencouraged to develop their ownexamples based ontheir insights and experiences. The followingare examples ofproductswheretransparencyisdesired: windowsand windshields, bottles, fluidcontainers(to allowdirectobservationofcontentvolumes), wrapping andpackaging, andglasses(eyewear). Thefollowingareexamplesofproducts whereopacityisdesired: windowsinrestrooms(ifpresent),glassinlight bulbs toproduce adiffuselight, foodpackaging toprotectthe contents fromlightradiationandassociated degradation,andproduct housingsforaestheticreasons.

3.19 RefertoFig.3.2 andexplainwhythetrendsseenaretobeexpected.

Thisisanopen-ended problem andstudentsshouldbeencouraged todeveloptheir ownob-stervations.Thetrends arenottoosurprising qualitatively,butthequantitativenatureofthetrends isatfirstverysurprising. Forexample,itisnotsurprising thathigh-modulus graphiteoutperformssteel,aspeopleareexposedtosporting equipmentsuchastennis racquets thataremadeoftheformerbutneverthelatter.However,peopledon’texpectgraphite tobe14timesbetterthansteel. Another surpriseinthetrendsisthepoorperformance ofglassfibersinanepoxymatrix. However,glassisprettydense,soweightsavingsarenot generally thereasonsforusingglassreinforcement.

QUANTITATIVEPROBLEMS

3.20 Ifwe assumethatall theworkdoneinplasticdeformationisconvertedintoheat,thetemperaturerisein aworkpieceis(1)directlyproportionaltotheworkdoneperunitvolumeand(2) inverselyproportionaltotheproductofthespecificheatandthedensityoftheworkpiece.UsingFig.2.6, andlettingtheareasunderthecurvesbetheunitworkdone,calculatethetemperaturerisefor (a)8650 steel,(b)304 stainlesssteel,and(c) 1100-H14aluminum.

Weusethefollowinginformation giveninChapters2and3:Theareaunder thetrue stress-true strain curveandthephysicalproperties foreachofthethree metals. Wethen followtheprocedure discussed onpp.63-64and useEq.(2.15)onp.82. Thus, for(a) 8650steel,the

areaunder thecurveinFig.2.6onp.63isaboutu=72,000in-lb/in3.Assumeadensityof

ρ=0.3lb/in3andaspecificheat c=0.12BTU/lb◦F.Therefore,

∆T = 72,000 =214◦F(0.3)(0.12)(778)(12)

For(b) 304stainless steel,wehaveu=175,000,ρ=0.3and c=0.12,hence∆T =520◦F.For(c)1100-H14aluminum,wehaveu=25,000in.-lb/in3,ρ=0.0975andc=0.215;hence

∆T =128◦F.

3.21 Thenaturalfrequency,f,ofacantileverbeamisgivenby

rEIgf=0.56

wL4

whereE isthemodulusofelasticity,Iisthemomentofinertia,gisthegravita-tionalconstant, wistheweight ofthebeamperunitlength,andListhelengthofthebeam.Howdoesthenaturalfrequencyofthebeamchange,ifatall,as itstemperatureisincreased?Assumethatthematerialissteel.

Let’sassumethatthebeamhasasquarecrosssectionwithasideoflengthh. Note,however,thatany crosssection willresult inthe sametrends, sostudentsshouldn’t bediscouragedfromconsidering, forexample, circular crosssections. The moment ofinertia forasquarecrosssectionis

I= h

Themomentofinertia willincrease astemperatureincreases, becausethe crosssectionwillbecomelargerduetothermalexpansion. Theweightperlength, w,isgivenby

w=WL

whereWistheweightofthebeam. SinceLincreaseswithincreasingtemperature,theweightperlengthwilldecreasewithincreasingtemperature.Alsonotethatthemodulusofelasticitywilldecreasewithincreasingtemperature(seeFig.2.7onp.64).Considertheratio ofinitialfrequency(subscript1)tofrequencyatelevated temperature(subscript2):

0.56

sE1 I1g

s E1I1

sE1I1

f1= w1L1 = (W/L1)L1

Simplifyingfurther,

0.56

sE2 I2g

w2L4

s E2I2

(W/L2)L4

sE2I2

s 3 s 4 3

f1=

E1 I1L2 =

E2I2L3

E1h1L2

E2h4L3

1 2 a

Letting αbethecoefficientofthermal expansion, wecanwrite

h2 =h1(1+α∆T)

Therefore, thefrequencyratio is

L2 =L1(1+α∆T)

s 4 3 s 4 3

3 s

f1=

E1h1L2 =

E1h1L1(1 +α∆T ) =

f2 E2h4L3

E h4

4 E (1+α∆T)

2 1 2

1(1+α∆T) 3 2

Tocompare theseeffects,considerthecaseofcarbon steel. Figure 2.7onp.64showsadropinelasticmodulus from190to130GPa overatemperatureincreaseof1000◦C.From Table

3.1onp.89,the coefficient ofthermal expansion forsteelis14.5µm/m◦C (average oftheextreme valuesgiveninthetable), sothatthechangeinfrequencyis:

f1=

E1 =

E2(1+α∆T)

s 190

130[1+(14.5 10−6)(1000)]=1.20

Thus, thenaturalfrequencyofthebeamdecreaseswhenheated. Thisisageneraltrend (andnotjustforcarbon steel),namelythatthethermalchangesinelasticmodulus playsalargerrolethan thethermalexpansion ofthebeam.

3.22 Itcanbeshownthatthermaldistortioninprecisiondevicesislow for highvaluesofthermalconductivitydividedbythethermalexpansioncoefficient.RankthematerialsinTable3.1 accordingtotheirabilitytoresistthermal distortion.

Thecalculations usingthedata inTable 3.1onp.89areasfollows.Whenarangeofvaluesisgivenforan alloy, the average value has been used. These materials have been rankedaccording totheratio ofthermalconductivitytothermalexpansion coefficient.

Material / Thermal
conductivity, / Thermal
expansion / k/α
k / coefficient,
α
Tungsten / 166 / 4.5 / 36.9
Molybdenum alloys / 142 / 5.1 / 27.8
Copper / 393 / 16.5 / 23.8
Silver / 429 / 19.3 / 22.2
Silicon / 148 / 7.63 / 19.4
Beryllium / 146 / 8.5 / 17.2
Gold / 317 / 19.3 / 16.4
Copper alloys / 234 / 18 / 13.0
Aluminum / 222 / 23.6 / 9.41
Tantalumalloys / 54 / 6.5 / 8.31
Aluminum alloys / 180 / 23 / 7.72
Columbium (niobium) / 52 / 7.1 / 7.32
Nickel / 92 / 13.3 / 6.92
Iron / 74 / 11.5 / 6.43
Magnesium / 154 / 26 / 5.92
Magnesium alloys / 106 / 26 / 4.10
Nickelalloys / 37 / 15.5 / 2.42
Steels / 33 / 14.5 / 2.28
Titanium / 17 / 8.35 / 2.04
Leadalloys / 35 / 29.1 / 1.20
Lead / 35 / 29.4 / 1.19
Titaniumalloys / 10 / 8.8 / 1.14

3.23 AddacolumntoTable3.1 thatliststhevolumetricheatcapacityofthematerialslisted,expressedinunitsofJ/cm3 K. Comparetheresultstothevaluefor liquidwater(4.184J/cm3 K).Notethatthevolumetricheatcapacity of a materialistheproductofitsdensity andspecificheat.

The additionalcolumn iscalculated as follows. Note thatoneneeds to be careful aboutkeepingconsistentunits.

Density / Specific
heat / Volumetric heat
capacity / Relative
volumetric
Material / (kg/m3) / (J/kgK) / (J/cm3 K) / heat capacity
Aluminum / 2700 / 900 / 2.43 / 0.58
Aluminum alloys / 2630-2820 / 880-920 / 2.31-2.59 / 0.55-0.62
Beryllium / 1854 / 1884 / 3.49 / 0.83
Columbium (niobium) / 8580 / 272 / 2.33 / 0.56
Copper / 8970 / 385 / 3.45 / 0.82
Copper alloys / 7470-8940 / 377-435 / 2.81-3.89 / 0.67-0.93
Gold / 19,300 / 129 / 2.49 / 0.59
Iron / 7860 / 460 / 3.61 / 0.86
Steels / 6920-9130 / 448-502 / 3.10-4.58 / 0.74-1.09
Lead / 11,350 / 130 / 1.47 / 0.35
Leadalloys / 8850-11,350 / 126-188 / 1.12-2.13 / 0.27-0.51
Magnesium / 1745 / 1025 / 1.79 / 0.43
Magnesium alloys / 1770-1780 / 1046 / 1.85 / 0.44
Molybdenum alloys / 10,210 / 276 / 2.83 / 0.68
Nickel / 8910 / 440 / 3.92 / 0.94
Nickelalloys / 7750-8850 / 381-544 / 2.95-4.81 / 0.70-1.15
Silicon / 2330 / 712 / 1.66 / 0.40
Silver / 10,500 / 235 / 2.47 / 0.59
Tantalumalloys / 16,600 / 142 / 2.36 / 0.56
Titanium / 4510 / 519 / 2.34 / 0.56
Titaniumalloys / 4430-4700 / 502-544 / 2.22-2.56 / 0.53-0.61
Tungsten / 19,290 / 138 / 2.66 / 0.64
Zinc / 7140 / 385 / 2.75 / 0.66
Zincalloys / 6640-7200 / 402 / 2.67-2.89 / 0.64-0.69
Ceramics / 2300-5500 / 750-950 / 1.72-5.22 / 0.41-1.25
Glasses / 2400-2700 / 500-850 / 1.2-2.3 / 0.29-0.55
Graphite / 1900-2200 / 840 / 1.60-1.85 / 0.38-0.44
Plastics / 900-2000 / 1000-2000 / 0.9-4.0 / 0.21-0.96
Wood / 400-700 / 2400-2800 / 0.96-1.96 / 0.23-0.47

3.24 Conducta literaturesearchandaddthefollowingmaterialstoTable3.1: cork,cement,ice, sugar,lithium,chromium,andplatinum.

Theadditions areasbelow.Notethatspecificvaluesmaychangedepending onsourcecited.

(a) Cork: density=193kg/m3,meltingpointisunavailable(corkdoesn’tmelt),specificheat

=2000J/kgK,thermal conductivity=0.05W/mK,coefficientofthermalexpansion

=30-50µm/m◦C,electrical resistivityupto1010 Ωcm.

(b) Cement: density =3120kg/m3, melting point isunavailable(cement doesn’t melt),specificheat=3300J/kgK,thermalconductivity=0.1W/mK,coefficientofthermalexpansion =7.4-13 µm/m◦C.

conductivity =2W/mK,coefficient ofthermalexpansion =50 µm/m◦C, electricalresistivity =182kΩ-m.

(d) Sugar: density=1587kg/m3,melting point=185◦C,specificheat =1250J/kgK,

(e) Lithium: density =535kg/m3,melting point =180◦,specificheat =3582J/kgK,thermal conductivity=84.8W/mK,electricalresistivity=92.8×10−9 Ω-m.

(f) Platinum:density =21.4g/cm3,melting point=1768◦C,specificheat =135J/kgK.thermal conductivity=70W/mK,coeffcient ofthermalexpansion =19µm/m◦C,electrical resistivity =105×10−9 Ω-m.

SYNTHESIS,DESIGNANDPROJECTS

3.25 Fromyourownexperience,makea listof parts,components,or productsthathavecorrodedandhavehadtobereplacedor discarded.

Bythestudent.Thisisanopen-ended problem thathavemany possibleanswers,andthesewillvarydepending onthebackgroundofthestudent.Therearemanyparts, usuallyassoci-ated with rusted steel,e.g.,automobile framesand bodies,bolts, bicyclepedals, etc. Otherparts thatarecommonlycorroded includeautomotivebatterycableterminals,marine partsofallkinds (especially ifocean going), nameplatesonoldmachinery, etc. Ifoneextendsthediscussiontocorrosion-assistedfailure,onecanincludejustaboutallparts whichfailbyfatigue, including shafts, and airplane fuselagesasshownbelow. This photographisadra-matic example ofcorrosion-assistedfatigue ofanaircraft fuselagethatoccurred mid-flight.(Source: FromHamrock,B.J., etal.,FundamentalsofMachineElements, 2nded.,NewYork,McGraw-Hill, 2005,p.265.).

3.26 Listapplicationswherethe followingproperties wouldbedesirable: (a) highdensity,(b)low density,(c) highmeltingpoint,(d)low meltingpoint,(e)highthermalconductivity, and(f)low thermalconductivity.

Bythe student. This isan open-ended problem, and many possible answers exist. Someexamplesare:

(a) High density: Adding weightto a part (like an anchor, bar bells or a boat), as aninertial elementinaself-windingwatch,andweightsforverticallyslidingwindows. Also,projectiles suchasbullets and shotgun particles areapplicationswherehighdensityisadvantageous.

(b) Lowdensity: Airplane components, aluminum tubing fortents, ladders, andhigh-speedmachinery elements. Mostsporting goodsgivebetterperformance ifdensity andhenceweightislow,suchastennis rackets, skis,etc.

(c) High melting point: Creep-resistantmaterials such asforgas-turbineblades oroveninsulation. Mold materials for die casting need to have high melting points, as dofilaments forlightbulbs.

(d) Lowmelting point: Soldering wire,fuseelements, waxforinvestmentcasting, and lu-bricantsthatdepend onaphasechangeareexamplesofsuchapplications.

(e) Highthermalconductivity:Rapid extractionofheat inradiatorsand heat exchangers,andcoolingfinsforelectricalcircuitsandtransformers.Cuttingtoolswithhighthermalconductivity can helpkeeptemperatureslowinmachining. Diesininjection moldingwithhighthermalconductivitycanextractheatmorequicklyallowinghigherproductionrates.

(f) Lowthermalconductivity:Coffeecups,winterclothing, andoveninsulation requirelowthermal conductivity.Inaddition,handles oncookware,lubricantsforhotforging,andthermos materials (unlessevacuated)needlowthermalconductivities.

3.27 Describeseveralapplicationsinwhichbothspecificstrengthandspecificstiffnessareimportant.

By the student. This problem isopen-ended and the studentsshould be encouraged todevelopanswersbasedontheir experienceandtraining.Twoexamplesare: (a)Tenttubing:requires lightweightmaterial foreaseofcarrying, whilepossessingsufficientlyhighstrengthandstiffnesstosupporttheweightofthetenttarp withoutexcessivebendingorbowing. (b)Racquetballortennis racquet: requires lightweightmaterialforcontrol overtheracquet’sdirection; also,highstrengthandstiffnessarerequired toefficientlytransfer theenergyoftheracquettotheball.

3.28 Designseveralmechanismsor instrumentsbasedon utilizingthe differencesinthermalexpansionofmaterials,suchasbimetallicstripsthatdevelopacurvaturewhenheated.

Bythestudent.Instrumentswillhaveacommonprincipleofmeasuring orregulating temper-atures suchasthermometersorbutterflyvalveswhichregulate fluidflowwhentemperaturesvary.

3.29 ForthematerialslistedinTable3.1, determinethespecificstrengthandspecificstiffness.Describeyourobservations.

Selected results are asfollows(the values whichgivehighest possiblequantitieshave beenused,e.g.,highstiffnessandlowdensity). Data istaken fromTable 2.2onp.59.

Material Y E Density Spec. strength Spec.stiffness

(MPa) (GPa) (kg/m3) (m ×103) (m ×106)

Aluminum / 35 / 69 / 2700 / 1.3 / 2.6
Alalloys / 550 / 79 / 2630 / 21.3 / 3.1
Copper / 76 / 105 / 8970 / 0.86 / 1.2
Cualloys / 1100 / 150 / 7470 / 15.0 / 2.05
Iron / 205 / 190 / 7860 / 2.66 / 2.5
Steels / 1725 / 200 / 6920 / 25.4 / 2.9
Lead / 14 / 14 / 11,350 / 0.13 / 0.126
Pballoys / 14 / 14 / 8850 / 0.161 / 0.16
Magnesium / 130 / 41 / 1745 / 7.6 / 2.4
Mgalloys / 305 / 45 / 1770 / 17.6 / 2.6
Moalloys / 2070 / 360 / 10,210 / 20.7 / 3.6
Nickel / 105 / 180 / 8910 / 1.2 / 2.06
Nialloys / 1200 / 214 / 7750 / 15.8 / 2.8
Titanium / 344 / 80 / 4510 / 7.8 / 1.8
Tialloys / 1380 / 130 / 4430 / 31.7 / 3.0
Tungsten / 550 / 350 / 19,290 / 2.9 / 1.8

3.30 Themaximumcompressiveforcethatalightweightcolumncanwithstandbeforebucklingdependson theratioof thesquarerootof thestiffnesstothedensityfor thematerial. ForthematerialslistedinTable2.2, determine(a)theratioof tensilestrengthtodensityand(b)theratioof elastic modulustodensity.Comment on thesuitabilityofeachfor beingmadeintolightweight columns.

Thisproblem usestheresults fromProblem 3.29.Tomakealightweightcolumn,onehastomaximizethespecificstrengthandthespecificstiffness. Reviewingthevaluesobtained,onecanobserve that:(a) Pure metals arenot usefulwhereas alloysaremuch morepreferable;(b)Titaniumalloyshavethehighest specificstrength(31,700m);(c)Aluminum alloyshavethe highestspecificstiffness(3.1); and (d) Amongthe lestdesirable materials areleadandcopper. Notethattheseresultsareconsistentwiththematerials ofchoiceformodernaircraft.

3.31 DescribepossibleapplicationsanddesignsusingalloysexhibitingtheInvareffectoflow thermalexpansion.

By the student. Ifthere isessentially no thermalexpansion, thematerial isexceptionalforsituationswhere thermalfatigue isaconsideration,orforprecision instruments wherenothermalexpansion would behighly desirable. Examples ofthe former include furnacesensorsandelectricalcomponents, andexamplesofthelatterincludemicromanipulatorsandmicro-electromechanicalsystems (MEMS); seep.908.

3.32 Collectsomepiecesof different metallicandnonmetallicmaterialslistedin Ta-ble 3.2. Usingsimpletestsand/or instruments,determinethevalidityof thedescendingorderofthephysicalpropertiesshownin thetable.

Bythe student.This isagoodprojectforstudents,and somedifferencesinthe trendscanbeobserved depending onthealloy, its source, and amountofcoldwork, heat treatingorannealing ithasundergone.

3.33 Designan actuatortoturnon a switchwhenthetemperaturedropsbelowacertainlevel.Use two materialswithdifferent coefficientsofthermalexpansionin yourdesign.

This isan open-ended problem with alarge number ofacceptable answers. The thermalexpansion effectcanbeusedtodeformacantilever, forexample, toactuateaswitch. Alter-natively, twomaterials thatarelongandthin canbeweldedattheir ends. Theycanthen bewrapped around amandrel, sothatafter elasticrecoverythey takeontheshapeofaspiral.The angular displacementofthe ends varies with temperature;apegattachedtooneendwhilefixingthe other willturn onaswitch assoonasthe pegtranslatestoanotherpeginaretaining fixture. This principal wasusedtogreat successincarburatorsinautomobilesbeforethe1980sinordertoachieveproper air/fuelratios asfunctions oftemperature.