Chapter 3

Structure and ManufacturingProperties of Metals

Questions

3.1 What is the difference between a unit cell anda single crystal?

A unit cell is the smallest group of atomsshowing the characteristic lattice structure ofa particular metal. A single crystal consistsof a number of unit cells; some examples arewhiskers, chips for semiconductor devices, andturbine blades.

3.2 Explain why we should study the crystal struc-ture of metals.

By studying the crystal structure of metals, in-formation about various properties can be in-ferred. By relating structure to properties, onecan predict processing behavior or select appro-priate applications for a metal. Metals withface-centered cubic structure, for example, tendto be ductile whereas hexagonal close-packedmetals tend to be brittle.

3.3 What effects does recrystallization have on theproperties of metals?

As shown in Figs. 3.17 on p. 96 and 3.18 onp. 97, strength and hardness are reduced, duc-tility is increased, and residual stresses are re-lieved.

3.4 What is the significance of a slip system?

The greater the number of slip systems, the

higher the ductility of the metal. Also, the slipsystem and the number of active slip systemsgive direct understanding of the material’s plas-tic behavior. For example, an hcp material hasfew slip systems. Thus, in a bulk material, fewgrains will be preferentially oriented with re-spect to a slip system and high stresses will berequired to initiate plastic deformation. On theother hand, fcc materials, have many slip sys-tems and thus a lower stress will be requiredfor plastic deformation. See also Section 3.3.1starting on p. 87.

3.5 Explain what is meant by structure-sensitiveand structure-insensitive properties of metals.

As described in Section 3.3.3 starting on p. 89,those properties that depend on the structure ofa metal are known as structure-sensitive proper-ties (yield and fracture strength, electrical con-ductivity). Those that are not (other physi-cal properties and elastic constants) are calledstructure-insensitive properties.

3.6 What is the relationship between nucleationrate and the number of grains per unit volumeof a metal?

This relationship is described at the beginningof Section 3.4 starting on p. 91. Generally, rapidcooling produces smaller grains, whereas slowcooling produces larger grains.

29

3.7 Explain the difference between recovery and re-crystallization.

These phenomena are described in Section 3.6on p. 96. Recovery involves relief of residualstresses, reduction in the number of disloca-tions, and increase in ductility. In recrystal-ization, new equiaxed and stress-free grains areformed, replacing the older grains.

3.8 (a) Is it possible for two pieces of the samemetal to have different recrystallization temper-atures? Explain. (b) Is it possible for recrys-tallization to take place in some regions of aworkpiece before other regions do in the sameworkpiece? Explain.

(a) Two pieces of the same metal can have dif-ferent recrystallization temperatures if thepieces have been cold worked to differentamounts. The piece that was cold workedto a greater extent will have more storedenergy to drive the recrystallization pro-cess, and hence its recrystallization tem-perature will be lower. See also Fig. 3.18on p. 97.

(b) Recrystallization may occur in some re-gions before others if

i. the workpiece was unevenly worked,as is generally the case in deformationprocessing of materials, since varyingamounts of cold work have differentrecrystallization temperatures, or

ii. the part has varying thicknesses; thethinner sections will heat up to the re-crystallization temperature faster.

3.9 Describe why different crystal structures ex-hibit different strengths and ductilities.

Different crystal structures have different slipsystems, which consist of a slip plane (the clos-est packed plane) and a slip direction (the close-packed direction). The fcc structure has 12 slipsystems, bcc has 48, and hcp has 3. The duc-tility of a metal depends on how many of theslip systems can be operative. In general, fccand bcc structures possess higher ductility thanhcp structures, because they have more slip sys-tems. The shear strength of a metal decreasesfor decreasing b/a ratio (b is inversely propor-tional to atomic density in the slip plane and a

is the plane spacing), and the b/a ratio dependson the slip system of the chemical structure.(See also Section 3.3.1 starting on p. 87.)

3.10 Explain the difference between preferred orien-tation and mechanical fibering.

Preferred orientation is anisotropic behavior ina polycrystalline workpiece that has crystalsaligned in nonrandom orientations. Crystalsbecome oriented nonrandomly in a workpiecewhen it is deformed, because the slip directionof a crystal tends to align along the generaldeformation direction. Mechanical fibering iscaused by the alignment of impurities, inclu-sions, or voids during plastic working of a metal;hence, the properties vary with the relative ori-entation of the stress applied to the orientationof the defect. (See also preferred orientation inSection 3.5 on p. 95.)

3.11 Give some analogies to mechanical fibering(such as layers of thin dough sprinkled withflour).

This is an open-ended problem with many ac-ceptable answers. Some examples are plywood,laminated products (such as countertops), win-ter clothing, pastry with layers of cream orjam, and pasta dishes with layers of pasta andcheese.

3.12 A cold-worked piece of metal has been re-crystallized. When tested, it is found to beanisotropic. Explain the probable reason forthis behavior.

The anisotropy of the workpiece is likely due topreferred orientation resulting from the recrys-tallization process. Copper is an example of ametal that has a very strong preferred orienta-tion after annealing. As shown in Fig. 3.19 onp. 97, no recrystallization occurs below a criti-cal deformation, being typically five percent.

3.13 Does recrystallization completely eliminate me-chanical fibering in a workpiece? Explain.

Mechanical fibering involves the alignment ofimpurities, inclusions, and voids in the work-piece during deformation. Recrystallizationgenerally modifies the grain structure, but willnot eliminate mechanical fibering.

30

3.14 Explain why we may have to be concerned withthe orange-peel effect on metal surfaces.

Orange peel not only influences surface appear-ance of parts, which may or may not be desir-able, but also affects their surface characteris-tics such as friction, wear, lubrication, and cor-rosion and electrical properties, as well as sub-sequent finishing, coating, and painting opera-tions. (See also surface roughness in practice inSection 4.3 on p. 137.)

3.15 How can you tell the difference between twoparts made of the same metal, one shaped bycold working and the other by hot working? Ex-plain the differences you might observe. Notethat there are several methods that can be usedto determine the differences between the twoparts.

Some of the methods of distinguishing hot vs.cold worked parts are:

(a) The surface finish of the cold-worked partwould be smoother than the hot-workedpart, and possibly shinier.

(b) If hardness values could be taken on theparts, the cold-worked part would beharder.

(c) The cold-worked part would likely containresidual stresses and exhibit anisotropicbehavior.

(d) Metallographic examination of the partscan be made: the hot-worked part wouldgenerally have equiaxed grains due to re-crystallization, while the cold-worked partwould have grains elongated in the generaldirection of deformation.

(e) The two parts can be subjected to mechan-ical testing and their properties compared.

3.16 Explain why the strength of a polycrystallinemetal at room temperature decreases as itsgrain size increases.

Strength increases as more entanglements ofdislocations take place with grain boundariesand with each other. Metals with larger grainshave less grain-boundary area per unit volume,and hence they are not be able to generate asmany entanglements at grain boundaries, thus

the strength will be lower. (See also Eq. (3.8)on p. 92.)

3.17 What is the significance of some metals, such aslead and tin, having recrystallization tempera-tures at about room temperature?

For these metals, room temperature is suffi-ciently high for recrystallization to occur with-out heating. These metals can be cold workedto large extent without requiring a recrystal-lization cycle to restore their ductility, henceformability. However, as the strain rate in-creases, their strength at room temperature in-creases because the metal has less time to re-crystallize, thus exhibiting a strain hardeningbehavior.

3.18 You are given a deck of playing cards heldtogether with a rubber band. Which of thematerial-behavior phenomena described in thischapter could you demonstrate with this setup?What would be the effects of increasing thenumber of rubber bands holding the cards to-gether? Explain. (Hint: Inspect Figs. 3.5 and3.7.)

The following demonstrations can be made witha deck of cards sliding against each other:

(a) Slip planes; permanent slip of cards withno rubber band, similar to that shown inFig. 3.5a on p. 86.

(b) Surface roughness that develops along theedges of the deck of cards, similar to thelower part of Fig. 3.7 on p. 88.

(c) Friction between the cards, simulating theshear stress required to cause slip, similarto Fig. 3.5 on p. 86. Friction between thecards can be decreased using talcum pow-der, or increased by moisture or soft glue(that has not set yet).

(d) Failure by slip, similar to Fig. 3.22b onp. 99.

(e) Presence of a rubber band indicates elasticbehavior and recovery when unloaded.

(f) The greater the number of rubber bands,the higher the shear modulus, G, which isrelated to the elastic modulus, E.

(g) The deck of cards is highly anisotropic.

31

3.19 Using the information given in Chapters 2 and3, list and describe the conditions that inducebrittle fracture in an otherwise ductile piece ofmetal.

Brittle fracture can be induced typically by:

(a) high deformation rates,

(b) the presence of stress concentrations, suchas notches and cracks,

(c) state of stress, especially high hydrostatictension components,

(d) radiation damage, and

(e) lower temperatures, particularly for met-als with bcc structure. In each case, thestress required to cause yielding is raisedabove the stress needed to cause failure,or the stress needed for crack propagationis below the yield stress of the metal (aswith stress concentrations).

3.20 Make a list of metals that would be suitablefor a (1) paper clip, (2) bicycle frame, (3) ra-zor blade, (4) battery cable, and (5) gas-turbineblade. Explain your reasoning.

In the selection of materials for these applica-tions, the particular requirements that are sig-nificant to these components are briefly out-lined as follows:

(a) Yield stress, elastic modulus, corrosion re-sistance.

(b) Strength, toughness, wear resistance, den-sity.

(c) Strength, resistance to corrosion and wear.

(d) Yield stress, toughness, elastic modulus,corrosion resistance, and electrical conduc-tivity.

(e) Strength, creep resistance, resistance tovarious types of wear, and corrosion resis-tance at high temperature.

Students are encouraged to suggest a variety ofmetals and discuss the relative advantages andlimitations with regard to particular applica-tions.

3.21 Explain the advantages and limitations of cold,warm, and hot working of metals, respectively.

These are explained briefly in Section 3.7 onp. 98. Basically, cold working has the advan-tages of refining the materials grain structurewhile increasing mechanical properties such asstrength, but it does result in anisotropy andreduced ductility. Hot working does not resultin strengthening of the workpiece, but the duc-tility of the workpiece is preserved, and thereis little or no anisotropy. Warm working is acompromise.

3.22 Explain why parts may crack when suddenlysubjected to extremes of temperature.

Thermal stresses result from temperature gra-dients in a material; the temperature will varysignificantly throughout the part when sub-jected to extremes of temperature. The higherthe temperature gradient, the more severe ther-mal stresses to which the part will be subjected,and the higher stresses will increase the proba-bility of cracking. This is particularly impor-tant in brittle and notch-sensitive materials.(See also Section 3.9.5 starting on p. 107 re-garding the role of coefficient of thermal expan-sion and thermal conductivity in developmentof thermal stresses.)

3.23 From your own experience and observations,list three applications each for the followingmetals and their alloys: (1) steel, (2) aluminum,(3) copper, (4) magnesium, and (5) gold.

There are numerous acceptable answers, includ-ing:

(a) steel: automobile bodies, structural mem-bers (buildings, boilers, machinery), fas-teners, springs, bearings, knives.

(b) aluminum: aircraft bodies, baseball bats,cookware, beverage containers, automo-tive pistons.

(c) copper: electrical wire, cookware, batterycable terminals, printed circuit boards.

(d) lead: batteries, toy soldiers, solders, glasscrystal.

(e) gold: jewelry, electrical connections, toothfillings, coins, medals.

3.24 List three applications that are not suitable foreach of the following metals and their alloys:

32

(1) steel, (2) aluminum, (3) copper, (4) magne-sium, and (5) gold.

There are several acceptable answers, including:

(a) steel: electrical contacts, aircraft fuselage,car tire, portable computer case.

(b) aluminum: cutting tools, shafts, gears, fly-wheels.

(c) copper: aircraft fuselage, bridges, subma-rine, toys.

(d) lead: toys, cookware, aircraft structuralcomponents, automobile body panels.

(e) gold: any part or component with a largemass and that requires strength and stiff-ness.

3.25 Name products that would not have been devel-oped to their advanced stages, as we find themtoday, if alloys with high strength and corrosionand creep resistance at elevated temperatureshad not been developed.

Some simple examples are jet engines and fur-naces. The student is encouraged to cite nu-merous other examples.

3.26 Inspect several metal products and componentsand make an educated guess as to what mate-rials they are made from. Give reasons for yourguess. If you list two or more possibilities, ex-plain your reasoning.

This is an open-ended problem and is a goodtopic for group discussion in class. Some ex-amples, such as an aluminum baseball bat orbeverage can, can be cited and students canbe asked why they believe the material is alu-minum.

3.27 List three engineering applications each forwhich the following physical properties wouldbe desirable: (1) high density, (2) low meltingpoint, and (3) high thermal conductivity.

Some examples are given below.

(a) High density: adding weight to a part(such as an anchor for a boat), flywheels,counterweights.

(b) Low melting point: Soldering wire, fuseelements (such as in sprinklers to sensefires).

(c) High thermal conductivity: cookware, carradiators, precision instruments that resistthermal warping. The student is encour-aged to site other examples.

3.28 Two physical properties that have a major in-fluence on the cracking of workpieces, tools, ordies during thermal cycling are thermal conduc-tivity and thermal expansion. Explain why.

Cracking results from thermal stresses that de-velop in the part during thermal cycling. Ther-mal stresses may be caused both by tempera-ture gradients and by anisotropy of thermal ex-pansion. High thermal conductivity allows theheat to be dissipated faster and more evenlythroughout the part, thus reducing the temper-ature gradient. If the thermal expansion is low,the stresses will be lower for a given tempera-ture gradient. When thermal stresses reach acertain level in the part, cracking will occur. Ifa material has higher ductility, it will be ableto undergo more by plastic deformation beforepossible fracture, and the tendency for crackingwill thus decrease.

3.29 Describe the advantages of nanomaterials overtraditional materials.

Since nanomaterials have fine structure, theyhave very high strength, hardness, andstrength-to-weight ratios compared to tradi-tional materials. The student is encouraged toreview relevant sections in the book; see, for ex-ample, pages 125-126, as well as nanoceramicsand nanopowders.

3.30 Aluminum has been cited as a possible substi-tute material for steel in automobiles. Whatconcerns, if any, would you have prior to pur-chasing an aluminum automobile?

By the student. Some of the main concernsassociated with aluminum alloys are that, gen-erally, their toughness is lower than steel alloys;thus, unless the automobile is properly designedand tested, its crashworthiness could suffer. Aperceived advantage is that weight savings withaluminum result in higher fuel efficiencies, butsteel requires much less energy to produce fromore, so these savings are not as high as initiallybelieved.

33

3.31 Lead shot is popular among sportsmen for hunt-ing, but birds commonly ingest the pellets(along with gravel) to help digest food. Whatsubstitute materials would you recommend forlead, and why?

Obviously, the humanitarian concern is asso-ciated with the waterfowl ingesting lead and,therefore, perishing from lead poisoning; theideal material would thus be one that is not poi-sonous. On the other hand, it is important forthe shot material to be effective for its purpose,as otherwise a bird is only wounded. Effectiveshot has a high density, thus a material with avery high density is desired. Referring to Table3.2 on p. 98, materials with a very high densitybut greater environmental friendliness are goldand tungsten, but obviously tungsten would bethe more logical choice.

3.32 What are metallic glasses? Why is the word“glass” used for these materials?

These materials are described in Section 3.11.9starting on p. 125. They are produced throughsuch processes as rapid solidification (describedin Section 5.10.8 starting on p. 235) so thatthe material has no grain structure or orien-

tation. Thus, none of the traditional metalliccharacteristics are present, such as deformationby slip, anisotropy, or grain effects. Becausethis is very similar to the microstructure andbehavior of glass, hence the term.

3.33 Which of the materials described in this chap-ter has the highest (a) density, (b) electri-cal conductivity, (c) thermal conductivity, (d)strength, and (e) cost?

As can be seen from Table 3.3 on p. 106, thehighest density is for tungsten, and the high-est electrical conductivity and thermal conduc-tivity in silver. The highest ultimate strengthmentioned in the chapter is for Monel K-500 at1050 MPa, and the highest cost (which variesfrom time to time) is usually is associated withsuperalloys.

3.34 What is twinning? How does it differ from slip?

This is illustrated in Fig. 3.5 on p. 86. In twin-ning, a grain deforms to produce a mirror-imageabout a plane of twinning. Slip involves slidingalong a plane. An appropriate analogy to dif-ferentiate these mechanisms is to suggest thattwinning is similar to bending about a plane,and slip is similar to shearing.

Problems

3.35 Calculate the theoretical (a) shear strength and(b) tensile strength for aluminum, plain-carbonsteel, and tungsten. Estimate the ratios of theirtheoretical strength to actual strength.

Equation (3.3) and Eq. (3.5) give the shear andtensile strengths, respectively, as

τ =G

2π

σ =E

10The values of E and ν are obtained from Table2.1 on p. 32, andG is calculated using Eq. (2.24)on p. 49,

G =E

2(1 + ν)

Thus, the following table can be generated:

Mat- E G τ σerial (GPa) ν (GPa) (GPa) (GPa)Al 79 0.34 29.5 4.6 7.9Steel 200 0.33 75.1 12.0 20W 400 0.27 157 25.1 40

3.36 A technician determines that the grain size ofa certain etched specimen is 6. Upon furtherchecking, it is found that the magnification usedwas 150, instead of 100 as required by ASTMstandards. What is the correct grain size?

To answer this question, one can either interpo-late from Table 3.1 on p. 93 or obtain the datafor a larger number of grain sizes, as well as thegrain diameter as a function of the ASTM No.

34

The following data is from the Metals Hand-book, ASM International:

ASTM Grains per Grains per Avg. grainNo mm2 mm3 dia., mm-3 1 0.7 1.00-1 4 5.6 0.500 8 16 0.351 16 45 0.252 32 128 0.183 64 360 0.1254 128 1020 0.0915 256 2900 0.0626 512 8200 0.0447 1024 23,000 0.0328 2048 65,000 0.022

Since the magnification ratio is 150/100=1.5,the diameter was magnified 1.5 times more thanit should have been. Thus, the grains appearedlarger than they actually are. Because the grainsize of 6 has an average diameter of 0.044 mm,the actual diameter is thus

d =0.044 mm

1.5= 0.0293mm

As can be seen from the table, this correspondsto a grain size of about 7.

3.37 Estimate the number of grains in a regular pa-per clip if its ASTM grain size is 9.

As can be seen in Table 3.1 on p. 93, an ASTMgrain size of 9 has 185,000 grains/mm3. An or-dinary paper clip (although they vary depend-ing on the size of paper clip considered) hass awire diameter of 0.80 mm and a length of 100mm. Therefore, the paper clip volume is

V =πd2l

4=π(0.80)2(100)

4= 50.5 mm3

The number of grains can thus be calculated as(50.5)(185,000)=9.34 million.

3.38 The natural frequency f of a cantilever beam isgiven by the expression

f = 0.56

√EIg

wL4,

where E is the modulus of elasticity, I is themoment of inertia, g is the gravitational con-stant, w is the weight of the beam per unitlength, and L is the length of the beam. How

does the natural frequency of the beam change,if any, as its temperature is increased?

Let’s assume that the beam has a square crosssection with a side of length h. Note, however,that any cross section will result in the sametrends, so students shouldn’t be discouragedfrom considering, for example, circular crosssections. The moment of inertia for a squarecross section is

I =h4

12The moment of inertia will increase as temper-ature increases, because the cross section willbecome larger due to thermal expansion. Theweight per length, w, is given by

w =W

L

where W is the weight of the beam. Since L in-creases with increasing temperature, the weightper length will decrease with increasing temper-ature. Also note that the modulus of elasticitywill decrease with increasing temperature (seeFig. 2.9 on p. 41). Consider the ratio of initialfrequency (subscript 1) to frequency at elevatedtemperature (subscript 2):

f1f2

=0.56

√E1I1gw1L4

1

0.56√

E2I2gw2L4

2

=

√E1I1

(W/L1)L41√

E2I2(W/L2)L4

2

=

√E1I1L3

1√E2I2L3

2

Simplifying further,

f1f2

=

√E1I1L3

2

E2I2L31

=

√E1h4

1L32

E2h42L

3a

Letting α be the coefficient of thermal expan-sion, we can write

h2 = h1 (1 + α∆T )

L2 = L1 (1 + α∆T )

Therefore, the frequency ratio is

f1f2

=

√E1h4

1L32

E2h42L

31

=

√E1h4

1L31 (1 + α∆T )3

E2h41 (1 + α∆T )4 L3

1

=

√E1

E2 (1 + α∆T )

35

To compare these effects, consider the case ofcarbon steel. Figure 2.9 on p. 41 shows a dropin elastic modulus from 190 to 130 GPa overa temperature increase of 1000◦C. From Table3.3 on p. 106, the coefficient of thermal expan-sion for steel is 14.5 µm/m◦C (average of theextreme values given in the table), so that thechange in frequency is:

f1f2

=

√E1

E2 (1 + α∆T )

=

√190

130 [1 + (14.5× 10−6) (1000)]

or f1/f2 = 1.20. Thus, the natural frequencyof the beam decreases when heated. This isa general trend (and not just for carbon steel),namely that the thermal changes in elastic mod-ulus plays a larger role than the thermal expan-sion of the beam.

3.39 A strip of metal is reduced in thickness by coldworking from 25 mm to 15 mm. A similar stripis reduced from 25 mm to 10 mm. Which oneof these strips will recrystallize at a lower tem-perature? Why?

In the first case, reducing the strip from 25 to15 mm involves a true strain (absolute value)of

ε = ln(

2515

)= 0.511

and for the second case,

ε = ln(

2510

)= 0.916

A review of Fig. 3.18 will indicate that, becauseof the higher degree of cold work and hencehigher stored energy, the second case will in-volve recrystallization at a lower temperaturethan the first case.

3.40 A 1-m long, simply-supported beam with around cross section is subjected to a load of 50kg at its center. (a) If the shaft is made fromAISI 303 steel and has a diameter of 20 mm,what is the deflection under the load? (b) Forshafts made from 2024-T4 aluminum, architec-tural bronze, and 99.5% titanium, respectively,what must the diameter of the shaft be for theshaft to have the same deflection as in part (a)?

(a) For a simply-supported beam, the deflec-tion can be obtained from any solid me-chanics book as

δ =PL3

48EI

For a round cross section with diameter of20 mm, the moment of inertia is

I =πd4

64=π(0.020)4

64= 7.85× 10−9 m4

From Table 2.1, E for steel is around 200GPa. The load is 50 kg or 490 N; there-fore, the deflection is

δ =PL3

48EI=

(490 N)(1 m)3

48(200 GPa)(7.85× 10−9 m4)

or δ = 0.00650 m = 6.5 mm.

(b) It is useful to express the diameter as afunction of deflection:

δ =PL3

48EI=

64PL3

48πEd4

Solving for d, we have

d =(

4PL3

3πEδ

)1/4

Thus, the following table can be constructed,with the elastic moduli taken from Table 2.1 onp. 32.

Material E (GPa) d (mm)2024-T4 Al 79 25.2Arch. bronze 110 23.299.5% Ti 80 25.1

3.41 If the diameter of the aluminum atom is 0.5 nm,estimate the number of atoms in a grain withan ASTM size of 5.

If the grain size is 5, there are 2900 grains permm3 of aluminum, and each grain has a volumeof 1/2900 = 3.45 × 10−4 mm3. Recall that foran fcc material there are four atoms per unitcell, with a total volume of 16πR3/3, and thatthe diagonal, a, of the unit cell is given by

a =(2√

2)R

36

Hence,

APFfcc =

(16πR3/3

)(2R√

2)3 = 0.74

Note that as long as all the atoms in the unitcell are of the same size, the atomic packingfactors do not depend on the atomic radius.Therefore, the volume of the grain taken up byatoms is (3.45×10−4)(0.74) = 2.55×10−4 mm3.(Recall that 1 mm=106 nm.) The diameter ofan aluminum atom is 0.5 nm, thus its radius is0.25 nm or 0.25× 10−6 mm. The volume of analuminum atom is

V =4πR3

3=

4π(0.25× 10−6)3

3

or 6.54 × 10−20 mm3. Dividing the volume ofaluminum in the grain by the volume of an alu-minum atom gives the total number of atomsin the grain as (2.55 × 10−4)/(6.54 × 10−20) =3.90× 1015.

3.42 Plot the following for the materials described inthis chapter: (a) yield stress versus density, (b)modulus of elasticity versus strength, and (c)modulus of elasticity versus relative cost. Hint:See Table 16.4.

The plots are shown below, based on the datagiven in Tables 2.1 on p. 32, 3.3 on p. 106, and16.4 on p. 971. Average values have been usedto obtain these plots.

0

200

400

600

800

1000

1200

0 5000 10,000 15,000 20,000

Yie

ld s

tres

s (M

Pa)

Density (kg/m3)

LeadMagnesium

Aluminum

CopperStainless steel

NickelTungsten

Titanium

MolybdenumSteel

0

50

100

150

200

250

300

350

400

0 5000 10,000 15,000 20,000

Ela

stic

mod

ulus

(GP

a) Tungsten

Molybdenum

Copper

NickelSteel

Titanium

Density (kg/m3)

Aluminum

Magnesium Lead

0

50

100

150

200

250

300

350

400

0.1 1 10 100 1000

Ela

stic

mod

ulus

(GP

a)

Molybdenum

Copper

Nickel

Steel

Titanium

Relative Cost

Aluminum

Magnesium

3.43 The following data is obtained in tension testsof brass:

Grain Size Yield stress(µm) (MPa)15 15020 14050 10575 90100 75

Does this material follow the Hall-Petch effect?If so, what is the value of k?

First, it is obvious from this table that the ma-terial becomes stronger as the grain size de-creases, which is the expected result. However,it is not clear whether Eq. (3.8) on p. 92 is ap-plicable. It is possible to plot the yield stressas a function of grain diameter, but it is betterto plot it as a function of d−1/2, as follows:

37

Yie

ld s

tren

gth

(MP

a) 160

60

80

100

120

140

0.05 0.3d-1/2

The least-squares curve fit for a straight line is

Y = 35.22 + 458d−1/2

with an R factor of 0.990. This suggests thata linear curve fit is proper, and it can be con-cluded that the material does follow the Hall-Petch effect, with a value of k = 458 MPa-

õm.

3.44 It can be shown that thermal distortion in pre-cision devices is low for high values of thermalconductivity divided by the thermal expansioncoefficient. Rank the materials in Table 3.3 ac-cording to their suitability to resist thermal dis-tortion.

The following table can be compiled, usingmaximum values of thermal conductivity andminimum values of thermal expansion coeffi-cient (to show optimum behavior for low ther-mal distortion):

Material k α k/αPlastics 0.4 72 0.00556Wood 0.4 2. 0.20Glasses 1.7 4.6 0.37Lead 35. 29.4 1.19Graphite 10. 7.86 1.27Ti alloys 12. 8.1 1.48Pb alloys 46 27.1 1.70Ti 17. 8.35 2.04Ceramics 17. 5.5 3.09Steels 52 11.7 4.44Ni alloys 63 12.7 4.96Mg alloys 138 26 5.31Mg 154. 26 5.92Iron 74. 11.5 6.43Nickel 92 13.3 6.91Columbium 52 7.1 7.3Tantalum 54 6.5 8.30Aluminum 222 23.6 9.40Al Alloys 239 23 10.3Cu alloys 234 16.5 14.18Gold 317. 19.3 16.4Berylium 146 8.5 17.1Si 148. 7.63 19.3Silver 429 19.3 22.2Copper 393 16.5 23.8Molybdenum 142 5.1 27.8Tungsten 166. 4.5 36.9

This data is shown graphically as follows:

TungstenMolybdenum

CopperSilver

Silver alloysBerylium

Cu-alloysAl-alloys

AluminumTantalum

ColumbiumNickel

MagnesiumMg-alloysNi-alloys

SteelCeramicsTitanium

Lead alloysTi-alloysGraphite

LeadGlasses

WoodPlastics

k/ (106 N/s)

0 4010 20 30

Increasingperformance

3.45 Assume that you are asked to give a quiz to stu-dents on the contents of this chapter. Preparethree quantitative problems and three qualita-tive questions, and supply the answers.

By the student. This is a challenging, open-ended question that requires considerable focusand understanding on the part of the students,and has been found to be a very valuable home-work problem.

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