manufacturing processes for engineering materials kalpakjian solution

Chapter 2

Fundamentals of the MechanicalBehavior of Materials

Questions

2.1 Can you calculate the percent elongation of ma-terials based only on the information given inFig. 2.6? Explain.

Recall that the percent elongation is defined byEq. (2.6) on p. 33 and depends on the originalgage length (lo) of the specimen. From Fig. 2.6on p. 37 only the necking strain (true and engi-neering) and true fracture strain can be deter-mined. Thus, we cannot calculate the percentelongation of the specimen; also, note that theelongation is a function of gage length and in-creases with gage length.

2.2 Explain if it is possible for the curves in Fig. 2.4to reach 0% elongation as the gage length is in-creased further.

The percent elongation of the specimen is afunction of the initial and final gage lengths.When the specimen is being pulled, regardlessof the original gage length, it will elongate uni-formly (and permanently) until necking begins.Therefore, the specimen will always have a cer-tain finite elongation. However, note that as thespecimen’s gage length is increased, the contri-bution of localized elongation (that is, necking)will decrease, but the total elongation will notapproach zero.

2.3 Explain why the difference between engineeringstrain and true strain becomes larger as strain

increases. Is this phenomenon true for both ten-sile and compressive strains? Explain.

The difference between the engineering and truestrains becomes larger because of the way thestrains are defined, respectively, as can be seenby inspecting Eqs. (2.1) on p. 30 and (2.9) onp. 35. This is true for both tensile and com-pressive strains.

2.4 Using the same scale for stress, we note that thetensile true-stress-true-strain curve is higherthan the engineering stress-strain curve. Ex-plain whether this condition also holds for acompression test.

During a compression test, the cross-sectionalarea of the specimen increases as the specimenheight decreases (because of volume constancy)as the load is increased. Since true stress is de-fined as ratio of the load to the instantaneouscross-sectional area of the specimen, the truestress in compression will be lower than the en-gineering stress for a given load, assuming thatfriction between the platens and the specimenis negligible.

2.5 Which of the two tests, tension or compression,requires a higher capacity testing machine thanthe other? Explain.

The compression test requires a higher capacitymachine because the cross-sectional area of the

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specimen increases during the test, which is theopposite of a tension test. The increase in arearequires a load higher than that for the ten-sion test to achieve the same stress level. Fur-thermore, note that compression-test specimensgenerally have a larger original cross-sectionalarea than those for tension tests, thus requiringhigher forces.

2.6 Explain how the modulus of resilience of a ma-terial changes, if at all, as it is strained: (1) foran elastic, perfectly plastic material, and (2) foran elastic, linearly strain-hardening material.

2.7 If you pull and break a tension-test specimenrapidly, where would the temperature be thehighest? Explain why.

Since temperature rise is due to the work input,the temperature will be highest in the neckedregion because that is where the strain, hencethe energy dissipated per unit volume in plasticdeformation, is highest.

2.8 Comment on the temperature distribution if thespecimen in Question 2.7 is pulled very slowly.

If the specimen is pulled very slowly, the tem-perature generated will be dissipated through-out the specimen and to the environment.Thus, there will be no appreciable temperaturerise anywhere, particularly with materials withhigh thermal conductivity.

2.9 In a tension test, the area under the true-stress-true-strain curve is the work done per unit vol-ume (the specific work). We also know thatthe area under the load-elongation curve rep-resents the work done on the specimen. If youdivide this latter work by the volume of thespecimen between the gage marks, you will de-termine the work done per unit volume (assum-ing that all deformation is confined betweenthe gage marks). Will this specific work bethe same as the area under the true-stress-true-strain curve? Explain. Will your answer be thesame for any value of strain? Explain.

If we divide the work done by the total volumeof the specimen between the gage lengths, weobtain the average specific work throughout thespecimen. However, the area under the true

stress-true strain curve represents the specificwork done at the necked (and fractured) regionin the specimen where the strain is a maximum.Thus, the answers will be different. However,up to the onset of necking (instability), the spe-cific work calculated will be the same. This isbecause the strain is uniform throughout thespecimen until necking begins.

2.10 The note at the bottom of Table 2.5 states thatas temperature increases, C decreases and mincreases. Explain why.

The value of C in Table 2.5 on p. 43 decreaseswith temperature because it is a measure of thestrength of the material. The value of m in-creases with temperature because the materialbecomes more strain-rate sensitive, due to thefact that the higher the strain rate, the less timethe material has to recover and recrystallize,hence its strength increases.

2.11 You are given the K and n values of two dif-ferent materials. Is this information sufficientto determine which material is tougher? If not,what additional information do you need, andwhy?

Although the K and n values may give a goodestimate of toughness, the true fracture stressand the true strain at fracture are required foraccurate calculation of toughness. The modu-lus of elasticity and yield stress would provideinformation about the area under the elastic re-gion; however, this region is very small and isthus usually negligible with respect to the restof the stress-strain curve.

2.12 Modify the curves in Fig. 2.7 to indicate theeffects of temperature. Explain the reasons foryour changes.

These modifications can be made by loweringthe slope of the elastic region and lowering thegeneral height of the curves. See, for example,Fig. 2.10 on p. 42.

2.13 Using a specific example, show why the defor-mation rate, say in m/s, and the true strain rateare not the same.

The deformation rate is the quantity v inEqs. (2.14), (2.15), (2.17), and (2.18) on pp. 41-46. Thus, when v is held constant during de-

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formation (hence a constant deformation rate),the true strain rate will vary, whereas the engi-neering strain rate will remain constant. Hence,the two quantities are not the same.

2.14 It has been stated that the higher the value ofm, the more diffuse the neck is, and likewise,the lower the value of m, the more localized theneck is. Explain the reason for this behavior.

As discussed in Section 2.2.7 starting on p. 41,with high m values, the material stretches toa greater length before it fails; this behavioris an indication that necking is delayed withincreasing m. When necking is about to be-gin, the necking region’s strength with respectto the rest of the specimen increases, due tostrain hardening. However, the strain rate inthe necking region is also higher than in the restof the specimen, because the material is elon-gating faster there. Since the material in thenecked region becomes stronger as it is strainedat a higher rate, the region exhibits a greater re-sistance to necking. The increase in resistanceto necking thus depends on the magnitude ofm. As the tension test progresses, necking be-comes more diffuse, and the specimen becomeslonger before fracture; hence, total elongationincreases with increasing values of m (Fig. 2.13on p. 45). As expected, the elongation afternecking (postuniform elongation) also increaseswith increasing m. It has been observed thatthe value of m decreases with metals of increas-ing strength.

2.15 Explain why materials with highm values (suchas hot glass and silly putty) when stretchedslowly, undergo large elongations before failure.Consider events taking place in the necked re-gion of the specimen.

The answer is similar to Answer 2.14 above.

2.16 Assume that you are running four-point bend-ing tests on a number of identical specimens ofthe same length and cross-section, but with in-creasing distance between the upper points ofloading (see Fig. 2.21b). What changes, if any,would you expect in the test results? Explain.

As the distance between the upper points ofloading in Fig. 2.21b on p. 51 increases, themagnitude of the bending moment decreases.

However, the volume of material subjected tothe maximum bending moment (hence to max-imum stress) increases. Thus, the probabilityof failure in the four-point test increases as thisdistance increases.

2.17 Would Eq. (2.10) hold true in the elastic range?Explain.

Note that this equation is based on volume con-stancy, i.e., Aolo = Al. We know, however, thatbecause the Poisson’s ratio ν is less than 0.5 inthe elastic range, the volume is not constant ina tension test; see Eq. (2.47) on p. 69. There-fore, the expression is not valid in the elasticrange.

2.18 Why have different types of hardness tests beendeveloped? How would you measure the hard-ness of a very large object?

There are several basic reasons: (a) The overallhardness range of the materials; (b) the hard-ness of their constituents; see Chapter 3; (c) thethickness of the specimen, such as bulk versusfoil; (d) the size of the specimen with respect tothat of the indenter; and (e) the surface finishof the part being tested.

2.19 Which hardness tests and scales would you usefor very thin strips of material, such as alu-minum foil? Why?

Because aluminum foil is very thin, the indenta-tions on the surface must be very small so as notto affect test results. Suitable tests would be amicrohardness test such as Knoop or Vickersunder very light loads (see Fig. 2.22 on p. 52).The accuracy of the test can be validated by ob-serving any changes in the surface appearanceopposite to the indented side.

2.20 List and explain the factors that you would con-sider in selecting an appropriate hardness testand scale for a particular application.

Hardness tests mainly have three differences:

(a) type of indenter,

(b) applied load, and

(c) method of indentation measurement(depth or surface area of indentation, orrebound of indenter).

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The hardness test selected would depend on theestimated hardness of the workpiece, its sizeand thickness, and if an average hardness or thehardness of individual microstructural compo-nents is desired. For instance, the scleroscope,which is portable, is capable of measuring thehardness of large pieces which otherwise wouldbe difficult or impossible to measure by othertechniques.

The Brinell hardness measurement leaves afairly large indentation which provides a goodmeasure of average hardness, while the Knooptest leaves a small indentation that allows, forexample, the determination of the hardness ofindividual phases in a two-phase alloy, as well asinclusions. The small indentation of the Knooptest also allows it to be useful in measuring thehardness of very thin layers on parts, such asplating or coatings. Recall that the depth of in-dentation should be small relative to part thick-ness, and that any change on the bottom sur-face appearance makes the test results invalid.

2.21 In a Brinell hardness test, the resulting impres-sion is found to be an ellipse. Give possibleexplanations for this phenomenon.

There are several possible reasons for thisphenomenon, but the two most likely areanisotropy in the material and the presence ofsurface residual stresses in the material.

2.21 Referring to Fig. 2.22 on p. 52, note that thematerial for indenters are either steel, tungstencarbide, or diamond. Why isn’t diamond usedfor all of the tests?

While diamond is the hardest material known,it would not, for example, be practical to makeand use a 10-mm diamond indenter because thecosts would be prohibitive. Consequently, ahard material such as those listed are sufficientfor most hardness tests.

2.22 What effect, if any, does friction have in a hard-ness test? Explain.

The effect of friction has been found to be mini-mal. In a hardness test, most of the indentationoccurs through plastic deformation, and thereis very little sliding at the indenter-workpieceinterface; see Fig. 2.25 on p. 55.

2.23 Describe the difference between creep andstress-relaxation phenomena, giving two exam-ples for each as they relate to engineering ap-plications.

Creep is the permanent deformation of a partthat is under a load over a period of time, usu-ally occurring at elevated temperatures. Stressrelaxation is the decrease in the stress level ina part under a constant strain. Examples ofcreep include:

(a) turbine blades operating at high tempera-tures, and

(b) high-temperature steam linesand furnacecomponents.

Stress relaxation is observed when, for example,a rubber band or a thermoplastic is pulled toa specific length and held at that length for aperiod of time. This phenomenon is commonlyobserved in rivets, bolts, and guy wires, as wellas thermoplastic components.

2.24 Referring to the two impact tests shown inFig. 2.31, explain how different the resultswould be if the specimens were impacted fromthe opposite directions.

Note that impacting the specimens shown inFig. 2.31 on p. 60 from the opposite directionswould subject the roots of the notches to com-pressive stresses, and thus they would not actas stress raisers. Thus, cracks would not propa-gate as they would when under tensile stresses.Consequently, the specimens would basicallybehave as if they were not notched.

2.25 If you remove layer ad from the part shown inFig. 2.30d, such as by machining or grinding,which way will the specimen curve? (Hint: As-sume that the part in diagram (d) can be mod-eled as consisting of four horizontal springs heldat the ends. Thus, from the top down, we havecompression, tension, compression, and tensionsprings.)

Since the internal forces will have to achieve astate of static equilibrium, the new part has tobow downward (i.e., it will hold water). Suchresidual-stress patterns can be modeled witha set of horizontal tension and compression

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springs. Note that the top layer of the mate-rial ad in Fig. 2.30d on p. 60, which is undercompression, has the tendency to bend the barupward. When this stress is relieved (such asby removing a layer), the bar will compensatefor it by bending downward.

2.26 Is it possible to completely remove residualstresses in a piece of material by the techniquedescribed in Fig. 2.32 if the material is elastic,linearly strain hardening? Explain.

By following the sequence of events depictedin Fig. 2.32 on p. 61, it can be seen that it isnot possible to completely remove the residualstresses. Note that for an elastic, linearly strainhardening material, σ′c will never catch up withσ′t.

2.27 Referring to Fig. 2.32, would it be possible toeliminate residual stresses by compression in-stead of tension? Assume that the piece of ma-terial will not buckle under the uniaxial com-pressive force.

Yes, by the same mechanism described inFig. 2.32 on p. 61.

2.28 List and explain the desirable mechanical prop-erties for the following: (1) elevator cable, (2)bandage, (3) shoe sole, (4) fish hook, (5) au-tomotive piston, (6) boat propeller, (7) gas-turbine blade, and (8) staple.

The following are some basic considerations:

(a) Elevator cable: The cable should not elon-gate elastically to a large extent or un-dergo yielding as the load is increased.These requirements thus call for a mate-rial with a high elastic modulus and yieldstress.

(b) Bandage: The bandage material must becompliant, that is, have a low stiffness, buthave high strength in the membrane direc-tion. Its inner surface must be permeableand outer surface resistant to environmen-tal effects.

(c) Shoe sole: The sole should be compliantfor comfort, with a high resilience. Itshould be tough so that it absorbs shockand should have high friction and wear re-sistance.

(d) Fish hook: A fish hook needs to have highstrength so that it doesn’t deform perma-nently under load, and thus maintain itsshape. It should be stiff (for better con-trol during its use) and should be resistantthe environment it is used in (such as saltwater).

(e) Automotive piston: This product musthave high strength at elevated tempera-tures, high physical and thermal shock re-sistance, and low mass.

(f) Boat propeller: The material must bestiff (to maintain its shape) and resistantto corrosion, and also have abrasion re-sistance because the propeller encounterssand and other abrasive particles when op-erated close to shore.

(g) Gas turbine blade: A gas turbine blade op-erates at high temperatures (depending onits location in the turbine); thus it shouldhave high-temperature strength and resis-tance to creep, as well as to oxidation andcorrosion due to combustion products dur-ing its use.

(h) Staple: The properties should be closelyparallel to that of a paper clip. The stapleshould have high ductility to allow it to bedeformed without fracture, and also havelow yield stress so that it can be bent (aswell as unbent when removing it) easilywithout requiring excessive force.

2.29 Make a sketch showing the nature and distribu-tion of the residual stresses in Figs. 2.31a and bbefore the parts were split (cut). Assume thatthe split parts are free from any stresses. (Hint:Force these parts back to the shape they werein before they were cut.)

As the question states, when we force back thesplit portions in the specimen in Fig. 2.31aon p. 60, we induce tensile stresses on theouter surfaces and compressive on the inner.Thus the original part would, along its totalcross section, have a residual stress distribu-tion of tension-compression-tension. Using thesame technique, we find that the specimen inFig. 2.31b would have a similar residual stressdistribution prior to cutting.

2.30 It is possible to calculate the work of plasticdeformation by measuring the temperature rise

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in a workpiece, assuming that there is no heatloss and that the temperature distribution isuniform throughout. If the specific heat of thematerial decreases with increasing temperature,will the work of deformation calculated usingthe specific heat at room temperature be higheror lower than the actual work done? Explain.

If we calculate the heat using a constant specificheat value in Eq. (2.65) on p. 73, the work willbe higher than it actually is. This is because,by definition, as the specific heat decreases, lesswork is required to raise the workpiece temper-ature by one degree. Consequently, the calcu-lated work will be higher than the actual workdone.

2.31 Explain whether or not the volume of a metalspecimen changes when the specimen is sub-jected to a state of (a) uniaxial compressivestress and (b) uniaxial tensile stress, all in theelastic range.

For case (a), the quantity in parentheses inEq. (2.47) on p. 69 will be negative, becauseof the compressive stress. Since the rest of theterms are positive, the product of these terms isnegative and, hence, there will be a decrease involume (This can also be deduced intuitively.)For case (b), it will be noted that the volumewill increase.

2.32 We know that it is relatively easy to subjecta specimen to hydrostatic compression, such asby using a chamber filled with a liquid. Devise ameans whereby the specimen (say, in the shapeof a cube or a thin round disk) can be subjectedto hydrostatic tension, or one approaching thisstate of stress. (Note that a thin-walled, inter-nally pressurized spherical shell is not a correctanswer, because it is subjected only to a stateof plane stress.)

Two possible answers are the following:

(a) A solid cube made of a soft metal has all itssix faces brazed to long square bars (of thesame cross section as the specimen); thebars are made of a stronger metal. The sixarms are then subjected to equal tensionforces, thus subjecting the cube to equaltensile stresses.

(b) A thin, solid round disk (such as a coin)and made of a soft material is brazed be-tween the ends of two solid round barsof the same diameter as that of the disk.When subjected to longitudinal tension,the disk will tend to shrink radially. Butbecause it is thin and its flat surfaces arerestrained by the two rods from moving,the disk will be subjected to tensile radialstresses. Thus, a state of triaxial (thoughnot exactly hydrostatic) tension will existwithin the thin disk.

2.33 Referring to Fig. 2.19, make sketches of thestate of stress for an element in the reducedsection of the tube when it is subjected to (1)torsion only, (2) torsion while the tube is in-ternally pressurized, and (3) torsion while thetube is externally pressurized. Assume that thetube is closed end.

These states of stress can be represented simplyby referring to the contents of this chapter aswell as the relevant materials covered in textson mechanics of solids.

2.34 A penny-shaped piece of soft metal is brazedto the ends of two flat, round steel rods of thesame diameter as the piece. The assembly isthen subjected to uniaxial tension. What is thestate of stress to which the soft metal is sub-jected? Explain.

The penny-shaped soft metal piece will tendto contract radially due to the Poisson’s ratio;however, the solid rods to which it attached willprevent this from happening. Consequently, thestate of stress will tend to approach that of hy-drostatic tension.

2.35 A circular disk of soft metal is being com-pressed between two flat, hardened circularsteel punches having the same diameter as thedisk. Assume that the disk material is perfectlyplastic and that there is no friction or any tem-perature effects. Explain the change, if any, inthe magnitude of the punch force as the disk isbeing compressed plastically to, say, a fractionof its original thickness.

Note that as it is compressed plastically, thedisk will expand radially, because of volumeconstancy. An approximately donut-shaped

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material will then be pushed radially out-ward, which will then exert radial compressivestresses on the disk volume under the punches.The volume of material directly between thepunches will now subjected to a triaxial com-pressive state of stress. According to yield cri-teria (see Section 2.11), the compressive stressexerted by the punches will thus increase, eventhough the material is not strain hardening.Therefore, the punch force will increase as de-formation increases.

2.36 A perfectly plastic metal is yielding under thestress state σ1, σ2, σ3, where σ1 > σ2 > σ3.Explain what happens if σ1 is increased.

Consider Fig. 2.36 on p. 67. Points in the in-terior of the yield locus are in an elastic state,whereas those on the yield locus are in a plas-tic state. Points outside the yield locus are notadmissible. Therefore, an increase in σ1 whilethe other stresses remain unchanged would re-quire an increase in yield stress. This can alsobe deduced by inspecting either Eq. (2.36) orEq. (2.37) on p. 64.

2.37 What is the dilatation of a material with a Pois-son’s ratio of 0.5? Is it possible for a material tohave a Poisson’s ratio of 0.7? Give a rationalefor your answer.

It can be seen from Eq. (2.47) on p. 69 that thedilatation of a material with ν = 0.5 is alwayszero, regardless of the stress state. To examinethe case of ν = 0.7, consider the situation wherethe stress state is hydrostatic tension. Equation(2.47) would then predict contraction under atensile stress, a situation that cannot occur.

2.38 Can a material have a negative Poisson’s ratio?Explain.

Solid material do not have a negative Poisson’sratio, with the exception of some composite ma-terials (see Chapter 10), where there can be anegative Poisson’s ratio in a given direction.

2.39 As clearly as possible, define plane stress andplane strain.

Plane stress is the situation where the stressesin one of the direction on an element are zero;plane strain is the situation where the strainsin one of the direction are zero.

2.40 What test would you use to evaluate the hard-ness of a coating on a metal surface? Would itmatter if the coating was harder or softer thanthe substrate? Explain.

The answer depends on whether the coating isrelatively thin or thick. For a relatively thickcoating, conventional hardness tests can be con-ducted, as long as the deformed region underthe indenter is less than about one-tenth ofthe coating thickness. If the coating thicknessis less than this threshold, then one must ei-ther rely on nontraditional hardness tests, orelse use fairly complicated indentation modelsto extract the material behavior. As an exam-ple of the former, atomic force microscopes us-ing diamond-tipped pyramids have been used tomeasure the hardness of coatings less than 100nanometers thick. As an example of the lat-ter, finite-element models of a coated substratebeing indented by an indenter of a known ge-ometry can be developed and then correlatedto experiments.

2.41 List the advantages and limitations of thestress-strain relationships given in Fig. 2.7.

Several answers that are acceptable, and thestudent is encouraged to develop as many aspossible. Two possible answers are: (1) thereis a tradeoff between mathematical complex-ity and accuracy in modeling material behaviorand (2) some materials may be better suited forcertain constitutive laws than others.

2.42 Plot the data in Table 2.1 on a bar chart, show-ing the range of values, and comment on theresults.

By the student. An example of a bar chart forthe elastic modulus is shown below.

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0 100 200 300 400 500

Aluminum

Copper

Lead

Magnesium

Molybdenum

Nickel

Steels

Stainless steels

Titanium

Tungsten

Elastic modulus (GPa)

Metallic materials

0 200 400 600 800 1000 1200

Ceramics

Diamond

Glass

Rubbers

Thermoplastics

Thermosets

Boron fibers

Carbon fibers

Glass fibers

Kevlar fibers

Spectra fibers

Elastic modulus (GPa)

Non-metallic materials

Typical comments regarding such a chart are:

(a) There is a smaller range for metals thanfor non-metals;

(b) Thermoplastics, thermosets and rubbersare orders of magnitude lower than met-als and other non-metals;

(c) Diamond and ceramics can be superior toothers, but ceramics have a large range ofvalues.

2.43 A hardness test is conducted on as-receivedmetal as a quality check. The results indicate

that the hardness is too high, thus the mate-rial may not have sufficient ductility for the in-tended application. The supplier is reluctant toaccept the return of the material, instead claim-ing that the diamond cone used in the Rockwelltesting was worn and blunt, and hence the testneeded to be recalibrated. Is this explanationplausible? Explain.

Refer to Fig. 2.22 on p. 52 and note that if anindenter is blunt, then the penetration, t, un-der a given load will be smaller than that usinga sharp indenter. This then translates into ahigher hardness. The explanation is plausible,but in practice, hardness tests are fairly reliableand measurements are consistent if the testingequipment is properly calibrated and routinelyserviced.

2.44 Explain why a 0.2% offset is used to determinethe yield strength in a tension test.

The value of 0.2% is somewhat arbitrary and isused to set some standard. A yield stress, repre-senting the transition point from elastic to plas-tic deformation, is difficult to measure. Thisis because the stress-strain curve is not linearlyproportional after the proportional limit, whichcan be as high as one-half the yield strength insome metals. Therefore, a transition from elas-tic to plastic behavior in a stress-strain curve isdifficult to discern. The use of a 0.2% offset isa convenient way of consistently interpreting ayield point from stress-strain curves.

2.45 Referring to Question 2.44, would the off-set method be necessary for a highly-strained-hardened material? Explain.

The 0.2% offset is still advisable whenever itcan be used, because it is a standardized ap-proach for determining yield stress, and thusone should not arbitrarily abandon standards.However, if the material is highly cold worked,there will be a more noticeable ‘kink’ in thestress-strain curve, and thus the yield stress isfar more easily discernable than for the samematerial in the annealed condition.

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Problems

2.46 A strip of metal is originally 1.5 m long. It isstretched in three steps: first to a length of 1.75m, then to 2.0 m, and finally to 3.0 m. Showthat the total true strain is the sum of the truestrains in each step, that is, that the strains areadditive. Show that, using engineering strains,the strain for each step cannot be added to ob-tain the total strain.

The true strain is given by Eq. (2.9) on p. 35 as

ε = ln(l

lo

)Therefore, the true strains for the three stepsare:

ε1 = ln(

1.751.5

)= 0.1541

ε2 = ln(

2.01.75

)= 0.1335

ε3 = ln(

3.02.0

)= 0.4055

The sum of these true strains is ε = 0.1541 +0.1335 + 0.4055 = 0.6931. The true strain fromstep 1 to 3 is

ε = ln(

31.5

)= 0.6931

Therefore the true strains are additive. Us-ing the same approach for engineering strainas defined by Eq. (2.1), we obtain e1 = 0.1667,e2 = 0.1429, and e3 = 0.5. The sum of thesestrains is e1+e2+e3 = 0.8096. The engineeringstrain from step 1 to 3 is

e =l − lolo

=3− 1.5

1.5=

1.51.5

= 1

Note that this is not equal to the sum of theengineering strains for the individual steps.

2.47 A paper clip is made of wire 1.20-mm in di-ameter. If the original material from which thewire is made is a rod 15-mm in diameter, calcu-late the longitudinal and diametrical engineer-ing and true strains that the wire has under-gone.

Assuming volume constancy, we may write

lflo

=(do

df

)2

=(

151.20

)2

= 156.25 ≈ 156

Letting l0 be unity, the longitudinal engineeringstrain is e1 = (156−1)/1 = 155. The diametralengineering strain is calculated as

ed =1.2− 15

15= −0.92

The longitudinal true strain is given byEq. (2.9) on p. 35 as

ε = ln(l

lo

)= ln (155) = 5.043

The diametral true strain is

εd = ln(

1.2015

)= −2.526

Note the large difference between the engineer-ing and true strains, even though both describethe same phenomenon. Note also that the sumof the true strains (recognizing that the radialstrain is εr = ln

(0.607.5

)= −2.526) in the three

principal directions is zero, indicating volumeconstancy in plastic deformation.

2.48 A material has the following properties: UTS =50, 000 psi and n = 0.25 Calculate its strengthcoefficient K.

Let us first note that the true UTS of this ma-terial is given by UTStrue = Knn (because atnecking ε = n). We can then determine thevalue of this stress from the UTS by follow-ing a procedure similar to Example 2.1. Sincen = 0.25, we can write

UTStrue = UTS(

Ao

Aneck

)= UTS

(e0.25

)= (50, 000)(1.28) = 64, 200 psi

Therefore, since UTStrue = Knn,

K =UTStrue

nn=

64, 2000.250.25

= 90, 800 psi

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2.49 Based on the information given in Fig. 2.6, cal-culate the ultimate tensile strength of annealed70-30 brass.

From Fig. 2.6 on p. 37, the true stress for an-nealed 70-30 brass at necking (where the slopebecomes constant; see Fig. 2.7a on p. 40) isfound to be about 60,000 psi, while the truestrain is about 0.2. We also know that the ratioof the original to necked areas of the specimenis given by

ln(

Ao

Aneck

)= 0.20

orAneck

Ao= e−0.20 = 0.819

Thus,

UTS = (60, 000)(0.819) = 49, 100 psi

2.50 Calculate the ultimate tensile strength (engi-neering) of a material whose strength coefficientis 400 MPa and of a tensile-test specimen thatnecks at a true strain of 0.20.

In this problem we have K = 400 MPa andn = 0.20. Following the same procedure as inExample 2.1, we find the true ultimate tensilestrength is

σ = (400)(0.20)0.20 = 290 MPa

andAneck = Aoe

−0.20 = 0.81Ao

Consequently,

UTS = (290)(0.81) = 237 MPa

2.51 A cable is made of four parallel strands of dif-ferent materials, all behaving according to theequation σ = Kεn, where n = 0.3 The materi-als, strength coefficients, and cross sections areas follows:

Material A: K = 450 MPa, Ao = 7 mm2;

Material B: K = 600 MPa, Ao = 2.5 mm2;

Material C: K = 300 MPa, Ao = 3 mm2;

Material D: K = 760 MPa, Ao = 2 mm2;

(a) Calculate the maximum tensile load thatthis cable can withstand prior to necking.

(b) Explain how you would arrive at an an-swer if the n values of the three strandswere different from each other.

(a) Necking will occur when ε = n = 0.3. Atthis point the true stresses in each cableare (from σ = Kεn), respectively,

σA = (450)0.30.3 = 314 MPa

σB = (600)0.30.3 = 418 MPa

σC = (300)0.30.3 = 209 MPa

σD = (760)0.30.3 = 530 MPa

The areas at necking are calculated as fol-lows (from Aneck = Aoe

−n):

AA = (7)e−0.3 = 5.18 mm2

AB = (2.5)e−0.3 = 1.85 mm2

AC = (3)e−0.3 = 2.22 mm2

AD = (2)e−0.3 = 1.48 mm2

Hence the total load that the cable cansupport is

P = (314)(5.18) + (418)(1.85)+(209)(2.22) + (530)(1.48)

= 3650 N

(b) If the n values of the four strands were dif-ferent, the procedure would consist of plot-ting the load-elongation curves of the fourstrands on the same chart, then obtain-ing graphically the maximum load. Alter-nately, a computer program can be writtento determine the maximum load.

2.52 Using only Fig. 2.6, calculate the maximumload in tension testing of a 304 stainless-steelround specimen with an original diameter of 0.5in.

We observe from Fig. 2.6 on p. 37 that neckingbegins at a true strain of about 0.1, and thatthe true UTS is about 110,000 psi. The origi-nal cross-sectional area is Ao = π(0.25 in)2 =0.196 in2. Since n = 0.1, we follow a proceduresimilar to Example 2.1 and show that

Ao

Aneck= e0.1 = 1.1

10






Thus

UTS =110, 000

1.1= 100, 000 psi

Hence the maximum load is

F = (UTS)(Ao) = (100, 000)(0.196)

or F = 19, 600 lb.

2.53 Using the data given in Table 2.1, calculate thevalues of the shear modulus G for the metalslisted in the table.

The important equation is Eq. (2.24) on p. 49which gives the shear modulus as

G =E

2(1 + ν)

The following values can be calculated (mid-range values of ν are taken as appropriate):

Material E (GPa) ν G (GPa)Al & alloys 69-79 0.32 26-30Cu & alloys 105-150 0.34 39-56Pb & alloys 14 0.43 4.9Mg & alloys 41-45 0.32 15.5-17.0Mo & alloys 330-360 0.32 125-136Ni & alloys 180-214 0.31 69-82Steels 190-200 0.30 73-77Stainless steels 190-200 0.29 74-77Ti & alloys 80-130 0.32 30-49W & alloys 350-400 0.27 138-157Ceramics 70-1000 0.2 29-417Glass 70-80 0.24 28-32Rubbers 0.01-0.1 0.5 0.0033-0.033Thermoplastics 1.4-3.4 0.36 0.51-1.25Thermosets 3.5-17 0.34 1.3-6.34

2.54 Derive an expression for the toughness of amaterial whose behavior is represented by theequation σ = K (ε+ 0.2)n and whose fracturestrain is denoted as εf .

Recall that toughness is the area under thestress-strain curve, hence the toughness for thismaterial would be given by

Toughness =∫ εf

0

σ dε

=∫ εf

0

K (ε+ 0.2)ndε

=K

n+ 1

[(εf + 0.2)n+1 − 0.2n+1

]

2.55 A cylindrical specimen made of a brittle mate-rial 1 in. high and with a diameter of 1 in. issubjected to a compressive force along its axis.It is found that fracture takes place at an angleof 45 under a load of 30,000 lb. Calculate theshear stress and the normal stress acting on thefracture surface.

Assuming that compression takes place withoutfriction, note that two of the principal stresseswill be zero. The third principal stress actingon this specimen is normal to the specimen andits magnitude is

σ3 =30, 000π(0.5)2

= 38, 200 psi

The Mohr’s circle for this situation is shownbelow.

2=90°

The fracture plane is oriented at an angle of45, corresponding to a rotation of 90 on theMohr’s circle. This corresponds to a stress stateon the fracture plane of σ = −19, 100 psi andτ = 19, 100 psi.

2.56 What is the modulus of resilience of a highlycold-worked piece of steel with a hardness of300 HB? Of a piece of highly cold-worked cop-per with a hardness of 150 HB?

Referring to Fig. 2.24 on p. 55, the value ofc in Eq. (2.29) on p. 54 is approximately 3.2for highly cold-worked steels and around 3.4for cold-worked aluminum. Therefore, we canapproximate c = 3.3 for cold-worked copper.However, since the Brinell hardness is in unitsof kg/mm2, from Eq. (2.29) we can write

Tsteel =H

3.2=

3003.2

= 93.75 kg/mm2 = 133 ksi

TCu =H

3.3=

1503.3

= 45.5 kg/mm2 = 64.6 ksi

11






From Table 2.1, Esteel = 30 × 106 psi andECu = 15 × 106 psi. The modulus of resilienceis calculated from Eq. (2.5). For steel:

Modulus of Resilience =Y 2

2E=

(133, 000)2

2(30× 106)

or a modulus of resilience for steel of 295 in-lb/in3. For copper,

Modulus of Resilience =Y 2

2E=

(62, 200)2

2(15× 106)

or a modulus of resilience for copper of 129 in-lb/in3.

Note that these values are slightly different thanthe values given in the text; this is due to thefact that (a) highly cold-worked metals such asthese have a much higher yield stress than theannealed materials described in the text, and(b) arbitrary property values are given in thestatement of the problem.

2.57 Calculate the work done in frictionless compres-sion of a solid cylinder 40 mm high and 15 mmin diameter to a reduction in height of 75% forthe following materials: (1) 1100-O aluminum,(2) annealed copper, (3) annealed 304 stainlesssteel, and (4) 70-30 brass, annealed.

The work done is calculated from Eq. (2.62) onp. 71 where the specific energy, u, is obtainedfrom Eq. (2.60). Since the reduction in height is75%, the final height is 10 mm and the absolutevalue of the true strain is

ε = ln(

4010

)= 1.386

K and n are obtained from Table 2.3 as follows:

Material K (MPa) n1100-O Al 180 0.20Cu, annealed 315 0.54304 Stainless, annealed 1300 0.3070-30 brass, annealed 895 0.49

The u values are then calculated fromEq. (2.60). For example, for 1100-O aluminum,where K is 180 MPa and n is 0.20, u is calcu-lated as

u =Kεn+1

n+ 1=

(180)(1.386)1.2

1.2= 222 MN/m3

The volume is calculated as V = πr2l =π(0.0075)2(0.04) = 7.069× 10−6 m3. The workdone is the product of the specific work, u, andthe volume, V . Therefore, the results can betabulated as follows.

u WorkMaterial (MN/m3) (Nm)1100-O Al 222 1562Cu, annealed 338 2391304 Stainless, annealed 1529 10,80870-30 brass, annealed 977 6908

2.58 A material has a strength coefficient K =100, 000 psi Assuming that a tensile-test spec-imen made from this material begins to neckat a true strain of 0.17, show that the ultimatetensile strength of this material is 62,400 psi.

The approach is the same as in Example 2.1.Since the necking strain corresponds to themaximum load and the necking strain for thismaterial is given as ε = n = 0.17, we have, asthe true ultimate tensile strength:

UTStrue = (100, 000)(0.17)0.17 = 74, 000 psi.

The cross-sectional area at the onset of neckingis obtained from

ln(

Ao

Aneck

)= n = 0.17.

Consequently,

Aneck = Aoe−0.17

and the maximum load, P , is

P = σA = (UTStrue)Aoe−0.17

= (74, 000)(0.844)(Ao) = 62, 400Ao lb.

Since UTS= P/Ao, we have UTS = 62,400 psi.

2.59 A tensile-test specimen is made of a materialrepresented by the equation σ = K (ε+ n)n.(a) Determine the true strain at which neckingwill begin. (b) Show that it is possible for anengineering material to exhibit this behavior.

12






(a) In Section 2.2.4 on p. 38 we noted thatinstability, hence necking, requires the fol-lowing condition to be fulfilled:

dσ

dε= σ

Consequently, for this material we have

Kn (ε+ n)n−1 = K (ε+ n)n

This is solved as n = 0; thus necking be-gins as soon as the specimen is subjectedto tension.

(b) Yes, this behavior is possible. Considera tension-test specimen that has beenstrained to necking and then unloaded.Upon loading it again in tension, it willimmediately begin to neck.

2.60 Take two solid cylindrical specimens of equal di-ameter but different heights. Assume that bothspecimens are compressed (frictionless) by thesame percent reduction, say 50%. Prove thatthe final diameters will be the same.

Let’s identify the shorter cylindrical specimenwith the subscript s and the taller one as t, andtheir original diameter as D. Subscripts f ando indicate final and original, respectively. Be-cause both specimens undergo the same percentreduction in height, we can write

htf

hto=hsf

hso

and from volume constancy,

htf

hto=(Dto

Dtf

)2

andhsf

hso=(Dso

Dsf

)2

Because Dto = Dso, we note from these rela-tionships that Dtf = Dsf .

2.61 A horizontal rigid bar c-c is subjecting specimena to tension and specimen b to frictionless com-pression such that the bar remains horizontal.(See the accompanying figure.) The force F islocated at a distance ratio of 2:1. Both speci-mens have an original cross-sectional area of 1

in2 and the original lengths are a = 8 in. andb = 4.5 in. The material for specimen a has atrue-stress-true-strain curve of σ = 100, 000ε0.5.Plot the true-stress-true-strain curve that thematerial for specimen b should have for the barto remain horizontal during the experiment.

F

2 1

a

b

cc

x

From the equilibrium of vertical forces and tokeep the bar horizontal, we note that 2Fa = Fb.Hence, in terms of true stresses and instanta-neous areas, we have

2σaAa = σbAb

From volume constancy we also have, in termsof original and final dimensions

AoaLoa = AaLa

andAobLob = AbLb

where Loa = (8/4.5)Lob = 1.78Lob. From theserelationships we can show that

σb = 2(

84.5

)Kσa

(Lb

La

)Since σa = Kε0.5

a where K = 100, 000 psi, wecan now write

σb =(

16K4.5

)(Lb

La

)√εa

Hence, for a deflection of x,

σb =(

16K4.5

)(4.5− x

8 + x

)√ln(

8 + x

8

)The true strain in specimen b is given by

εb = ln(

4.5− x

4.5

)By inspecting the figure in the problem state-ment, we note that while specimen a gets

13






longer, it will continue exerting some force Fa.However, specimen b will eventually acquire across-sectional area that will become infinite asx approaches 4.5 in., thus its strength mustapproach zero. This observation suggests thatspecimen b cannot have a true stress-true straincurve typical of metals, and that it will have amaximum at some strain. This is seen in theplot of σb shown below.

50,000

40,000

30,000

20,000

10,000

00 0.5 1.0 1.5 2.0 2.5

True

str

ess

(psi

)

Absolute value of true strain

2.62 Inspect the curve that you obtained in Problem2.61. Does a typical strain-hardening materialbehave in that manner? Explain.

Based on the discussions in Section 2.2.3 start-ing on p. 35, it is obvious that ordinary met-als would not normally behave in this manner.However, under certain conditions, the follow-ing could explain such behavior:

• When specimen b is heated to higher andhigher temperatures as deformation pro-gresses, with its strength decreasing as x isincreased further after the maximum valueof stress.

• In compression testing of brittle materials,such as ceramics, when the specimen be-gins to fracture.

• If the material is susceptible to thermalsoftening, then it can display such behav-ior with a sufficiently high strain rate.

2.63 In a disk test performed on a specimen 40-mmin diameter and 5 m thick, the specimen frac-tures at a stress of 500 MPa. What was theload on the disk at fracture?

Equation (2.20) is used to solve this problem.Noting that σ = 500 MPa, d = 40 mm = 0.04m, and t = 5 mm = 0.005 m, we can write

σ =2Pπdt

→ P =σπdt

2

Therefore

P =(500× 106)π(0.04)(0.005)

2= 157 kN.

2.64 In Fig. 2.32a, let the tensile and compressiveresidual stresses both be 10,000 psi and themodulus of elasticity of the material be 30×106

psi, with a modulus of resilience of 30 in.-lb/in3.If the original length in diagram (a) is 20 in.,what should be the stretched length in diagram(b) so that, when unloaded, the strip will befree of residual stresses?

Note that the yield stress can be obtained fromEq. (2.5) on p. 31 as

Mod. of Resilience = MR =Y 2

2E

Thus,

Y =√

2(MR)E =√

2(30)(30× 106)

or Y = 42, 430 psi. Using Eq. (2.32), the strainrequired to relieve the residual stress is:

ε =σc

E+Y

E=

10, 00030× 106

+42, 430

30× 106= 0.00175

Therefore,

ε = ln(lflo

)= ln

(lf

20 in.

)= 0.00175

Therefore, lf = 20.035 in.

2.65 Show that you can take a bent bar made of anelastic, perfectly plastic material and straightenit by stretching it into the plastic range. (Hint:Observe the events shown in Fig. 2.32.)

The series of events that takes place in straight-ening a bent bar by stretching it can be visu-alized by starting with a stress distribution asin Fig. 2.32a on p. 61, which would representthe unbending of a bent section. As we applytension, we algebraically add a uniform tensilestress to this stress distribution. Note that thechange in the stresses is the same as that de-picted in Fig. 2.32d, namely, the tensile stress

14






increases and reaches the yield stress, Y . Thecompressive stress is first reduced in magnitude,then becomes tensile. Eventually, the wholecross section reaches the constant yield stress,Y . Because we now have a uniform stress dis-tribution throughout its thickness, the bar be-comes straight and remains straight upon un-loading.

2.66 A bar 1 m long is bent and then stress re-lieved. The radius of curvature to the neutralaxis is 0.50 m. The bar is 30 mm thick andis made of an elastic, perfectly plastic materialwith Y = 600 MPa and E = 200 GPa. Cal-culate the length to which this bar should bestretched so that, after unloading, it will be-come and remain straight.

When the curved bar becomes straight, the en-gineering strain it undergoes is given by the ex-pression

e =t

2ρ

where t is the thickness and ρ is the radius tothe neutral axis. Hence in this case,

e =(0.030)2(0.50)

= 0.03

Since Y = 600 MPa and E = 200 GPa, we findthat the elastic limit for this material is at anelastic strain of

e =Y

E=

600 MPa200 GPa

= 0.003

which is much smaller than 0.05. Following thedescription in Answer 2.65 above, we find thatthe strain required to straighten the bar is

e = (2)(0.003) = 0.006

or

lf − lolo

= 0.006 → lf = 0.006lo + lo

or lf = 1.006 m.

2.67 Assume that a material with a uniaxial yieldstress Y yields under a stress state of principalstresses σ1, σ2, σ3, where σ1 > σ2 > σ3. Showthat the superposition of a hydrostatic stress, p,on this system (such as placing the specimen ina chamber pressurized with a liquid) does not

affect yielding. In other words, the material willstill yield according to yield criteria.

Let’s consider the distortion-energy criterion,although the same derivation could be per-formed with the maximum shear stress criterionas well. Equation (2.37) on p. 64 gives

(σ1 − σ2)2 + (σ2 − σ3)

2 + (σ3 − σ1)2 = 2Y 2

Now consider a new stress state where the prin-cipal stresses are

σ′1 = σ1 + p

σ′2 = σ2 + p

σ′3 = σ3 + p

which represents a new loading with an addi-tional hydrostatic pressure, p. The distortion-energy criterion for this stress state is

(σ′1 − σ′2)2 + (σ′2 − σ′3)

2 + (σ′3 − σ′1)2 = 2Y 2

or

2Y 2 = [(σ1 + p)− (σ2 + p)]2

+ [(σ2 + p)− (σ3 + p)]2

+ [(σ3 + p)− (σ1 + p)]2

which can be simplified as

(σ1 − σ2)2 + (σ2 − σ3)

2 + (σ3 − σ1)2 = 2Y 2

which is the original yield criterion. Hence, theyield criterion is unaffected by the superposi-tion of a hydrostatic pressure.

2.68 Give two different and specific examplesin which the maximum-shear-stress and thedistortion-energy criteria give the same answer.

In order to obtain the same answer for the twoyield criteria, we refer to Fig. 2.36 on p. 67 forplane stress and note the coordinates at whichthe two diagrams meet. Examples are: simpletension, simple compression, equal biaxial ten-sion, and equal biaxial compression. Thus, ac-ceptable answers would include (a) wire rope, asused on a crane to lift loads; (b) spherical pres-sure vessels, including balloons and gas storagetanks, and (c) shrink fits.

15






2.69 A thin-walled spherical shell with a yield stressY is subjected to an internal pressure p. Withappropriate equations, show whether or not thepressure required to yield this shell depends onthe particular yield criterion used.

Here we have a state of plane stress with equalbiaxial tension. The answer to Problem 2.68leads one to immediately conclude that boththe maximum shear stress and distortion energycriteria will give the same results. We will nowdemonstrate this more rigorously. The princi-pal membrane stresses are given by

σ1 = σ2 =pr

2t

andσ3 = 0

Using the maximum shear-stress criterion, wefind that

σ1 − 0 = Y

hencep =

2tYr

Using the distortion-energy criterion, we have

(0− 0)2 + (σ2 − 0)2 + (0− σ1)2 = 2Y 2

Since σ1 = σ2, then this gives σ1 = σ2 = Y , andthe same expression is obtained for pressure.

2.70 Show that, according to the distortion-energycriterion, the yield stress in plane strain is1.15Y where Y is the uniaxial yield stress of thematerial.

A plane-strain condition is shown in Fig. 2.35don p. 67, where σ1 is the yield stress of thematerial in plane strain (Y ′), σ3 is zero, andε2 = 0. From Eq. 2.43b on p. 68, we findthat σ2 = σ1/2. Substituting these into thedistortion-energy criterion given by Eq. (2.37)on p.64,(σ1 −

σ1

2

)2

+(σ1

2− 0)2

+ (0− σ1)2 = 2Y 2

and3σ2

1

2= 2Y 2

henceσ1 =

2√3Y ≈ 1.15Y

2.71 What would be the answer to Problem 2.70 ifthe maximum-shear-stress criterion were used?

Because σ2 is an intermediate stress and usingEq. (2.36), the answer would be

σ1 − 0 = Y

hence the yield stress in plane strain will beequal to the uniaxial yield stress, Y .

2.72 A closed-end, thin-walled cylinder of originallength l, thickness t, and internal radius r issubjected to an internal pressure p. Using thegeneralized Hooke’s law equations, show thechange, if any, that occurs in the length of thiscylinder when it is pressurized. Let ν = 0.33.

A closed-end, thin-walled cylinder under inter-nal pressure is subjected to the following prin-cipal stresses:

σ1 =pr

2t; σ2 =

pr

t; σ3 = 0

where the subscript 1 is the longitudinal di-rection, 2 is the hoop direction, and 3 is thethickness direction. From Hooke’s law given byEq. (2.33) on p. 63,

ε1 =1E

[σ1 − ν (σ2 + σ3)]

=1E

[pr

2t− 1

3

(prt

+ 0)]

=pr

6tESince all the quantities are positive (note thatin order to produce a tensile membrane stress,the pressure is positive as well), the longitudinalstrain is finite and positive. Thus the cylinderbecomes longer when pressurized, as it can alsobe deduced intuitively.

2.73 A round, thin-walled tube is subjected to ten-sion in the elastic range. Show that both thethickness and the diameter of the tube decreaseas tension increases.

The stress state in this case is σ1, σ2 = σ3 = 0.From the generalized Hooke’s law equationsgiven by Eq. (2.33) on p. 63, and denoting theaxial direction as 1, the hoop direction as 2, andthe radial direction as 3, we have for the hoopstrain:

ε2 =1E

[σ2 − ν (σ1 + σ3)] = −νσ1

E

16






Therefore, the diameter is negative for a tensile(positive) value of σ1. For the radial strain, thegeneralized Hooke’s law gives

ε3 =1E

[σ3 − ν (σ1 + σ2)] = −νσ1

E

Therefore, the radial strain is also negative andthe wall becomes thinner for a positive value ofσ1.

2.74 Take a long cylindrical balloon and, with a thinfelt-tip pen, mark a small square on it. Whatwill be the shape of this square after you blowup the balloon: (1) a larger square, (2) a rectan-gle, with its long axis in the circumferential di-rections, (3) a rectangle, with its long axis in thelongitudinal direction, or (4) an ellipse? Per-form this experiment and, based on your obser-vations, explain the results, using appropriateequations. Assume that the material the bal-loon is made of is perfectly elastic and isotropic,and that this situation represents a thin-walledclosed-end cylinder under internal pressure.

This is a simple graphic way of illustrating thegeneralized Hooke’s law equations. A balloonis a readily available and economical method ofdemonstrating these stress states. It is also en-couraged to assign the students the task of pre-dicting the shape numerically; an example of avaluable experiment involves partially inflatingthe balloon, drawing the square, then expand-ing it further and having the students predictthe dimensions of the square.

Although not as readily available, a rubber tubecan be used to demonstrate the effects of tor-sion in a similar manner.

2.75 Take a cubic piece of metal with a side lengthlo and deform it plastically to the shape of arectangular parallelepiped of dimensions l1, l2,and l3. Assuming that the material is rigid andperfectly plastic, show that volume constancyrequires that the following expression be satis-fied: ε1 + ε2 + ε3 = 0.

The initial volume and the final volume are con-stant, so that

lololo = l1l2l3 → l1l2l3lololo

= 1

Taking the natural log of both sides,

ln(l1l2l3lololo

)= ln(1) = 0

since ln(AB) = ln(A) + ln(B),

ln(l1lo

)+ ln

(l2lo

)+ ln

(l3lo

)= 0

From the definition of true strain given by

Eq. (2.9) on p. 35, ln(l1l0

)= ε1, etc., so that

ε1 + ε2 + ε3 = 0.

2.76 What is the diameter of an originally 30-mm-diameter solid steel ball when it is subjected toa hydrostatic pressure of 5 GPa?

From Eq. (2.46) on p. 68 and noting that, forthis case, all three strains are equal and all threestresses are equal in magnitude,

3ε =(

1− 2νE

)(−3p)

where p is the hydrostatic pressure. Thus, fromTable 2.1 on p. 32 we take values for steel ofν = 0.3 and E = 200 GPa, so that

ε =(

1− 2νE

)(−p) =

(1− 0.6

200

)(−5)

or ε = −0.01. Therefore

ln(Df

Do

)= −0.01

Solving for Df ,

Df = Doe−0.01 = (20)e−0.01 = 19.8 mm

2.77 Determine the effective stress and effectivestrain in plane-strain compression according tothe distortion-energy criterion.

Referring to Fig. 2.35d on p. 67 we note that,for this case, σ3 = 0 and σ2 = σ1/2, as canbe seen from Eq. (2.44) on p. 68. According tothe distortion-energy criterion and referring toEq. (2.52) on p. 69 for effective stress, we find

17






that

σ =1√2

[(σ1 −

σ1

2

)2

+(σ1

2

)2

+ (σ1)2

]1/2

=1√2

(14

+14

+ 1)1/2

σ1

=1√2

(√3√2

)σ1 =

√3

2σ1

Note that for this case ε3 = 0. Since volumeconstancy is maintained during plastic defor-mation, we also have ε3 = −ε1. Substitut-ing these into Eq. (2.54), the effective strainis found to be

ε =(

2√3

)ε1

2.78 (a) Calculate the work done in expanding a 2-mm-thick spherical shell from a diameter of 100mm to 140 mm, where the shell is made of a ma-terial for which σ = 200+50ε0.5 MPa. (b) Doesyour answer depend on the particular yield cri-terion used? Explain.

For this case, the membrane stresses are givenby

σ1 = σ2 =pt

2tand the strains are

ε1 = ε2 = ln(fr

fo

)Note that we have a balanced (or equal) biaxialstate of plane stress. Thus, the specific energy(for a perfectly-plastic material) will, accordingto either yield criteria, be

u = 2σ1ε1 = 2Y ln(rfro

)The work done will be

W = (Volume)(u)

=(4πr2oto

) [2Y ln

(rfro

)]= 8πY r2oto ln

(rfro

)Using the pressure-volume method of work, webegin with the formula

W =∫p dV

where V is the volume of the sphere. We inte-grate this equation between the limits Vo andVf , noting that

p =2tYr

and

V =4πr3

3so that

dV = 4πr2 dr

Also, from volume constancy, we have

t =r2otor2

Combining these expressions, we obtain

W = 8πY r2oto∫ rf

ro

dr

r= 8πY r2oto ln

(rfro

)which is the same expression obtained earlier.To obtain a numerical answer to this prob-lem, note that Y should be replaced with anaverage value Y . Also note that ε1 = ε2 =ln(140/100) = 0.336. Thus,

Y = 200 +50(0.336)1.5

1.5= 206 MPa

Hence the work done is

W = 8πY r2oto ln(rfro

)= 8π(206× 106)(0.1)2(0.001) ln(70/50)= 17.4kN-m

The yield criterion used does not matter be-cause this is equal biaxial tension; see the an-swer to Problem 2.68.

2.79 A cylindrical slug that has a diameter of 1in. and is 1 in. high is placed at the center ofa 2-in.-diameter cavity in a rigid die. (See theaccompanying figure.) The slug is surroundedby a compressible matrix, the pressure of whichis given by the relation

pm = 40, 000∆VVom

psi

wherem denotes the matrix and Vom is the orig-inal volume of the compressible matrix. Boththe slug and the matrix are being compressedby a piston and without any friction. The ini-tial pressure on the matrix is zero, and the slugmaterial has the true-stress-true-strain curve ofσ = 15, 000ε0.4.

18






Compressiblematrix

F

d1"

1"2"

Obtain an expression for the force F versus pis-ton travel d up to d = 0.5 in.

The total force, F , on the piston will be

F = Fw + Fm,

where the subscript w denotes the workpieceand m the matrix. As d increases, the matrixpressure increases, thus subjecting the slug totransverse compressive stresses on its circum-ference. Hence the slug will be subjected to tri-axial compressive stresses, with σ2 = σ3. Usingthe maximum shear-stress criterion for simplic-ity, we have

σ1 = σ + σ2

where σ1 is the required compressive stress onthe slug, σ is the flow stress of the slug mate-rial corresponding to a given strain, and givenas σ = 15, 000ε0.4, and σ2 is the compressivestress due to matrix pressure. Lets now deter-mine the matrix pressure in terms of d.

The volume of the slug is equal to π/4 and thevolume of the cavity when d = 0 is π. Hencethe original volume of the matrix is Vom = 3

4π.The volume of the matrix at any value of d isthen

Vm = π(1− d)− π

4= π

(34− d

)in3,

from which we obtain∆VVom

=Vom − Vm

Vom=

43d.

Note that when d = 34 in., the volume of the ma-

trix becomes zero. The matrix pressure, henceσ2, is now given by

σ2 =4(40, 000)

3d =

160, 0003

d (psi)

The absolute value of the true strain in the slugis given by

ε = ln1

1− d,

with which we can determine the value of σ forany d. The cross-sectional area of the workpieceat any d is

Aw =π

4(1− d)in2

and that of the matrix is

Am = π − π

4(1− d)in2

The required compressive stress on the slug is

σ1 = σ + σ2 = σ +160, 000

3d.

We may now write the total force on the pistonas

F = Aw

(σ +

160, 0003

d

)+Am

160, 0003

d lb.

The following data gives some numerical re-sults:

d Aw ε σ F(in.) (in2) (psi) (lb)0.1 0.872 0.105 6089 22,0700.2 0.98 0.223 8230 41,5900.3 1.121 0.357 9934 61,4100.4 1.31 0.510 11,460 82,0300.5 1.571 0.692 12,950 104,200

And the following plot shows the desired re-sults.

Forc

e (k

ip)

0

40

80

120

Displacement (in.)0 0.1 0.2 0.3 0.4 0.5

19






2.80 A specimen in the shape of a cube 20 mm oneach side is being compressed without frictionin a die cavity, as shown in Fig. 2.35d, where thewidth of the groove is 15 mm. Assume that thelinearly strain-hardening material has the true-stress-true-strain curve given by σ = 70 + 30εMPa. Calculate the compressive force requiredwhen the height of the specimen is at 3 mm,according to both yield criteria.

We note that the volume of the specimen is con-stant and can be expressed as

(20)(20)(20) = (h)(x)(x)

where x is the lateral dimensions assuming thespecimen expands uniformly during compres-sion. Since h = 3 mm, we have x = 51.6mm. Thus, the specimen touches the walls andhence this becomes a plane-strain problem (seeFig. 2.35d on p. 67). The absolute value of thetrue strain is

ε = ln(

203

)= 1.90

We can now determine the flow stress, Yf , ofthe material at this strain as

Yf = 70 + 30(1.90) = 127 MPa

The cross-sectional area on which the force isacting is

Area = (20)(20)(20)/3 = 2667 mm2

According to the maximum shear-stress crite-rion, we have σ1 = Yf , and thus

Force = (127)(2667) = 338 kN

According to the distortion energy criterion, wehave σ1 = 1.15Yf , or

Force = (1.15)(338) = 389 kN.

2.81 Obtain expressions for the specific energy fora material for each of the stress-strain curvesshown in Fig. 2.7, similar to those shown inSection 2.12.

Equation (2.59) on p. 71 gives the specific en-ergy as

u =∫ ε1

0

σ dε

(a) For a perfectly-elastic material as shown inFig 2.7a on p. 40, this expression becomes

u =∫ ε1

0

Eε dε = E

(ε2

2

)ε1

0

=Eε212

(b) For a rigid, perfectly-plastic material asshown in Fig. 2.7b, this is

u =∫ ε1

0

Y dε = Y (ε)ε10 = Y ε1

(c) For an elastic, perfectly plastic material,this is identical to an elastic material forε1 < Y/E, and for ε1 > Y/E it is

u =∫ ε1

0

σ dε =∫ Y/E

0

Eε dε+∫ ε1

Y/E

Y dε

=E

2

(Y

E

)2

+ Y

(ε1 −

Y

E

)=

Y 2

2E+ Y ε1 −

Y 2

E= Y

(ε1 −

Y

2E

)(d) For a rigid, linearly strain hardening ma-

terial, the specific energy is

u =∫ ε1

0

(Y + Epε) dε = Y ε1 +Epε

21

2

(e) For an elastic, linear strain hardening ma-terial, the specific energy is identical toan elastic material for ε1 < Y/E and forε1 > Y/E it is

u =∫ ε1

0

[Y + Ep

(ε− Y

E

)]dε

=∫ ε1

0

[Y

(1− Ep

E

)+ Epε

]dε

= Y

(1− Ep

E

)ε1 +

Epε21

2

2.82 A material with a yield stress of 70 MPa is sub-jected to three principal (normal) stresses of σ1,σ2 = 0, and σ3 = −σ1/2. What is the value ofσ1 when the metal yields according to the vonMises criterion? What if σ2 = σ1/3?

The distortion-energy criterion, given byEq. (2.37) on p. 64, is

(σ1 − σ2)2 + (σ2 − σ3)

2 + (σ3 − σ1)2 = 2Y 2

20






Substituting Y = 70 MPa and σ1, σ2 = 0 andσ3 = −σ1/2, we have

2(70)2 = (σ1)2 +

(−σ1

2

)2

+(−σ1

2− σ1

)2

thus,σ1 = 52.9 MPa

If Y = 70 MPa and σ1, σ2 = σ1/3 and σ3 =−σ1/2 is the stress state, then

2(70)2 =(σ1 −

σ1

3

)2

+(σ1

3− σ1

2

)2

+(−σ1

2− σ1

)2

= 2.72σ21

Thus, σ1 = 60.0 MPa. Therefore, the stresslevel to initiate yielding actually increases whenσ2 is increased.

2.83 A steel plate has the dimensions 100 mm × 100mm × 5 mm thick. It is subjected to biaxialtension of σ1 = σ2, with the stress in the thick-ness direction of σ3 = 0. What is the largestpossible change in volume at yielding, using thevon Mises criterion? What would this changein volume be if the plate were made of copper?

From Table 2.1 on p. 32, it is noted that forsteel we can use E = 200 GPa and ν = 0.30.For a stress state of σ1 = σ2 and σ3 = 0, thevon Mises criterion predicts that at yielding,

(σ1 − σ2)2 + (σ2 − σ3)

2 + (σ3 − σ1)2 = 2Y 2

or

(σ1 − σ1)2 + (σ1 − 0)2 + (0− σ1)

2 = 2Y 2

Resulting in σ1 = Y . Equation (2.47) gives:

∆ =1− 2νE

(σx + σy + σz)

=1− 2(0.3)200 GPa

[(350 MPa) + (350 MPa]

= = 0.0014

Since the original volume is (100)(100)(5) =50,000 mm3, the stressed volume is 50,070mm3, or the volume change is 70 mm3.

For copper, we have E = 125 GPa and ν = 0.34.Following the same derivation, the dilatationfor copper is 0.0006144; the stressed volume is50,031 mm3 and thus the change in volume is31 mm3.

2.84 A 50-mm-wide, 1-mm-thick strip is rolled to afinal thickness of 0.5 mm. It is noted that thestrip has increased in width to 52 mm. Whatis the strain in the rolling direction?

The thickness strain is

εt = ln(l

lo

)= ln

(0.5 mm1 mm

)= −0.693

The width strain is

εw = ln(l

lo

)= ln

(52 mm50 mm

)= 0.0392

Therefore, from Eq. (2.48), the strain in therolling (or longitudinal) direction is εl = 0 −0.0392 + 0.693 = 0.654.

2.85 An aluminum alloy yields at a stress of 50 MPain uniaxial tension. If this material is subjectedto the stresses σ1 = 25 MPa, σ2 = 15 MPa andσ3 = −26 MPa, will it yield? Explain.

According to the maximum shear-stress crite-rion, the effective stress is given by Eq. (2.51)on p. 69 as:

σ = σ1 − σ3 = 25− (−26) = 51 MPa

However, according to the distortion-energy cri-terion, the effective stress is given by Eq. (2.52)on p. 69 as:

σ =1√2

√(σ1 − σ2)

2 + (σ2 − σ3)2 + (σ3 − σ1)

2

or

σ =

√(25− 15)2 + (15 + 26)2 + (−26− 25)2

2

or σ = 46.8 MPa. Therefore, the effective stressis higher than the yield stress for the maximumshear-stress criterion, and lower than the yieldstress for the distortion-energy criterion. It isimpossible to state whether or not the mate-rial will yield at this stress state. An accuratestatement would be that yielding is imminent,if it is not already occurring.

2.86 A cylindrical specimen 1-in. in diameter and1-in. high is being compressed by dropping aweight of 200 lb on it from a certain height.After deformation, it is found that the temper-ature rise in the specimen is 300 F. Assuming

21






no heat loss and no friction, calculate the fi-nal height of the specimen, using the followingdata for the material: K = 30, 000 psi, n = 0.5,density = 0.1 lb/in3, and specific heat = 0.3BTU/lb·F.

This problem uses the same approach as in Ex-ample 2.8. The volume of the specimen is

V =πd2h

4=π(1)2(1)

4= 0.785 in3

The expression for heat is given by

Heat = cpρV∆T= (0.3)(0.1)(0.785)(300)(778)= 5500ft-lb = 66, 000 in-lb.

where the unit conversion 778 ft-lb = 1 BTUhas been applied. Since, ideally,

Heat = Work = V u = VKεn+1

n+ 1

= (0.785)(30, 000)ε1.5

1.5

Solving for ε,

ε1.5 =1.5(66, 000)

(0.785)(30, 000)= 4.20

Therefore, ε = 2.60. Using absolute values, wehave

ln(ho

hf

)= ln

(1 in.hf

)= 2.60

Solving for hf gives hf = 0.074 in.

2.87 A solid cylindrical specimen 100-mm high iscompressed to a final height of 40 mm in twosteps between frictionless platens; after the firststep the cylinder is 70 mm high. Calculate theengineering strain and the true strain for bothsteps, compare them, and comment on your ob-servations.

In the first step, we note that ho = 100 mm andh1 = 70 mm, so that from Eq. (2.1) on p. 30,

e1 =h1 − ho

ho=

70− 100100

= −0.300

and from Eq. (2.9) on p. 35,

ε1 = ln(h1

ho

)= ln

(70100

)= −0.357

Similarly, for the second step where h1 = 70mm and h2 = 40 mm,

e2 =h2 − h1

h1=

40− 7070

= −0.429

ε2 = ln(h2

h1

)= ln

(4070

)= −0.560

Note that if the operation were conducted inone step, the following would result:

e =h2 − ho

ho=

40− 100100

= −0.6

ε = ln(h2

ho

)= ln

(40100

)= −0.916

As was shown in Problem 2.46, this indicatesthat the true strains are additive while the en-gineering strains are not.

2.88 Assume that the specimen in Problem 2.87 hasan initial diameter of 80 mm and is made of1100-O aluminum. Determine the load requiredfor each step.

From volume constancy, we calculate

d1 = do

√ho

h1= 80

√10070

= 95.6 mm

d2 = do

√ho

h2= 80

√10040

= 126.5 mm

Based on these diameters the cross-sectionalarea at the steps is calculated as:

A1 =π

4d21 = 7181 mm2

A2 =π

4d22 = 12, 566 mm2

As calculated in Problem 2.87, ε1 = 0.357 andεtotal = 0.916. Note that for 1100-O aluminum,K = 180 MPa and n = 0.20 (see Table 2.3 onp. 37) so that Eq. (2.11) on p. 35 yields

σ1 = 180(0.357)0.20 = 146.5 MPa

σ2 = 180(0.916)0.20 = 176.9 Mpa

Therefore, the loads are calculated as:

P1 = σ1A1 = (146.5)(7181) = 1050 kN

P2 = (176.9)(12, 566) = 2223 kN

22






2.89 Determine the specific energy and actual energyexpended for the entire process described in thepreceding two problems.

From Eq. (2.60) on p. 71 and using εtotal =0.916, K = 180 MPa and n = 0.20, we have

u =Kεn+1

n+ 1=

(180)(0.916)1.2

1.2= 135 MPa

2.90 A metal has a strain hardening exponent of0.22. At a true strain of 0.2, the true stressis 20,000 psi. (a) Determine the stress-strainrelationship for this material. (b) Determinethe ultimate tensile strength for this material.

This solution follows the same approach as inExample 2.1. From Eq. (2.11) on p. 35, andrecognizing that n = 0.22 and σ = 20, 000 psifor ε = 0.20,

σ = Kεn → 20, 000 = K(0.20)0.22

or K = 28, 500 psi. Therefore, the stress-strainrelationship for this material is

σ = 28, 500ε0.22 psi

To determine the ultimate tensile strength forthe material, realize that the strain at neckingis equal to the strain hardening exponent, orε = n. Therefore,

σult = K(n)n = 28, 500(0.22)0.22 = 20, 400 psi

The cross-sectional area at the onset of neckingis obtained from

ln(

Ao

Aneck

)= n = 0.22

Consequently,

Aneck = Aoe−0.22

and the maximum load is

P = σA = σultAneck.

Hence,

P = (20, 400)(Ao)e−0.22 = 16, 370Ao

Since UTS= P/Ao, we have

UTS =16, 370Ao

Ao= 16, 370 psi

2.91 The area of each face of a metal cube is 400 m2,and the metal has a shear yield stress, k, of 140MPa. Compressive loads of 40 kN and 80 kNare applied at different faces (say in the x- andy-directions). What must be the compressiveload applied to the z-direction to cause yield-ing according to the Tresca criterion? Assumea frictionless condition.

Since the area of each face is 400 mm2, thestresses in the x- and y- directions are

σx = −40, 000400

= −100 MPa

σy = −80, 000400

= −200 MPa

where the negative sign indicates that thestresses are compressive. If the Tresca criterionis used, then Eq. (2.36) on p. 64 gives

σmax − σmin = Y = 2k = 280 MPa

It is stated that σ3 is compressive, and is there-fore negative. Note that if σ3 is zero, then thematerial does not yield because σmax − σmin =0 − (−200) = 200 MPa < 280 MPa. There-fore, σ3 must be lower than σ2, and is calculatedfrom:

σmax − σmin = σ1 − σ3 = 280 MPa

or

σ3 = σ1 − 280 = −100− 280 = −380 MPa

2.92 A tensile force of 9 kN is applied to the ends ofa solid bar of 6.35 mm diameter. Under load,the diameter reduces to 5.00 mm. Assuminguniform deformation and volume constancy, (a)determine the engineering stress and strain, (b)determine the true stress and strain, (c) if theoriginal bar had been subjected to a true stressof 345 MPa and the resulting diameter was 5.60mm, what are the engineering stress and engi-neering strain for this condition?

First note that, in this case, do = 6.35 mm, df

= 5.00 mm, P=9000 N, and from volume con-stancy,

lod2o = lfd

2f → lf

lo=d2

o

d2f

=6.352

5.002= 1.613

23






(a) The engineering stress is calculated fromEq. (2.3) on p. 30 as:

σ =P

Ao=

9000π4 (6.35)2

= 284 MPa

and the engineering strain is calculatedfrom Eq. (2.1) on p. 30 as:

e =l − lolo

=lflo− 1 = 1.613− 1 = 0.613

(b) The true stress is calculated from Eq. (2.8)on p. 34 as:

σ =P

A=

9000π4 (5.00)2

= 458 MPa

and the true strain is calculated fromEq. (2.9) on p. 35 as:

ε = ln(lflo

)= ln 1.613 = 0.478

(c) If the final diameter is df = 5.60 mm, thenthe final area is Af = π

4 d2f = 24.63 mm2.

If the true stress is 345 MPa, then

P = σA = (345)(24.63) = 8497 ≈ 8500 N

Therefore, the engineering stress is calcu-lated as before as

σ =P

Ao=

8500π4 (6.35)2

= 268 MPa

Similarly, from volume constancy,

lflo

=d2

o

d2f

=6.352

5.602= 1.286

Therefore, the engineering strain is

e =lflo− 1 = 1.286− 1 = 0.286

2.93 Two identical specimens 10-mm in diameterand with test sections 25 mm long are madeof 1112 steel. One is in the as-received condi-tion and the other is annealed. What will bethe true strain when necking begins, and whatwill be the elongation of these samples at thatinstant? What is the ultimate tensile strengthfor these samples?

This problem uses a similar approach as for Ex-ample 2.1. First, we note from Table 2.3 onp. 37 that for cold-rolled 1112 steel, K = 760MPa and n = 0.08. Also, the initial cross-sectional area is Ao = π

4 (10)2 = 78.5 mm2.For annealed 1112 steel, K = 760 MPa andn = 0.19. At necking, ε = n, so that the strainwill be ε = 0.08 for the cold-rolled steel andε = 0.19 for the annealed steel. For the cold-rolled steel, the final length is given by Eq. (2.9)on p. 35 as

ε = n = ln(l

lo

)Solving for l,

l = enlo = e0.08(25) = 27.08 mm

The elongation is, from Eq. (2.6),

Elongation =lf − lolo

× 100 =27.08− 25

25× 100

or 8.32 %. To calculate the ultimate strength,we can write, for the cold-rolled steel,

UTStrue = Knn = 760(0.08)0.08 = 621 MPa

As in Example 2.1, we calculate the load atnecking as:

P = UTStrueAoe−n

So that

UTS =P

Ao=

UTStrueAoe−n

Ao= UTStruee

−n

This expression is evaluated as

UTS = (621)e−0.08 = 573 MPa

Repeating these calculations for the annealedspecimen yields l = 30.23 mm, elongation =20.9%, and UTS= 458 MPa.

2.94 During the production of a part, a metal witha yield strength of 110 MPa is subjected to astress state σ1, σ2 = σ1/3, σ3 = 0. Sketch theMohr’s circle diagram for this stress state. De-termine the stress σ1 necessary to cause yieldingby the maximum shear stress and the von Misescriteria.

For the stress state of σ1, σ1/3, 0 the followingfigure the three-dimensional Mohr’s circle:

24






123

For the von Mises criterion, Eq. (2.37) on p. 64gives:

2Y 2 = (σ1 − σ2)2 + (σ2 − σ3)

2 + (σ3 − σ1)2

=(σ1 −

σ1

3

)2

+(σ1

3− 0)2

+ (0− σ1)2

=49σ2

1 +19σ2

1 + σ21 =

149σ2

1

Solving for σ1 gives σ1 = 125 MPa. Accordingto the Tresca criterion, Eq. (2.36) on p. 64 onp. 64 gives

σ1 − σ3 = σ1 = 0 = Y

or σ1 = 110 MPa.

2.95 Estimate the depth of penetration in a Brinellhardness test using 500-kg load, when the sam-ple is a cold-worked aluminum with a yieldstress of 200 MPa.

Note from Fig. 2.24 on p. 55 that for cold-worked aluminum with a yield stress of 200MPa, the Brinell hardness is around 65kg/mm2. From Fig. 2.22 on p. 52, we can esti-mate the diameter of the indentation from theexpression:

HB =2P

(πD)(D −√D2 − d2)

from which we find that d = 3.091 mm forD = 10mm. To calculate the depth of pene-tration, consider the following sketch:

5 mm

3 mm

Because the radius is 5 mm and one-half thepenetration diameter is 1.5 mm, we can obtainα as

α = sin−1

(1.55

)= 17.5

The depth of penetration, t, can be obtainedfrom

t = 5− 5 cosα = 5− 5 cos 17.5 = 0.23 mm

2.96 The following data are taken from a stainlesssteel tension-test specimen:

Load, P (lb) Extension, ∆l (in.)1600 02500 0.023000 0.083600 0.204200 0.404500 0.604600 (max) 0.864586 (fracture) 0.98

Also, Ao = 0.056 in2, Af = 0.016 in2, lo = 2in. Plot the true stress-true strain curve for thematerial.

The following are calculated from Eqs. (2.6),(2.9), (2.10), and (2.8) on pp. 33-35:

A σ∆l l ε (in2) (ksi)0 2.0 0 0.056 28.50.02 2.02 0.00995 0.0554 45.10.08 2.08 0.0392 0.0538 55.70.2 2.2 0.0953 0.0509 70.70.4 2.4 0.182 0.0467 90.0.6 2.6 0.262 0.0431 1040.86 2.86 0.357 0.0392 1170.98 2.98 0.399 0.0376 120

The true stress-true strain curve is then plottedas follows:

True

str

ess,

(k

si) 160

0

40

80

120

0 0.2 0.4

True strain,

25






2.97 A metal is yielding plastically under the stressstate shown in the accompanying figure.

50 MPa

40 MPa

20 MPa

(a) Label the principal axes according to theirproper numerical convention (1, 2, 3).

(b) What is the yield stress using the Trescacriterion?

(c) What if the von Mises criterion is used?(d) The stress state causes measured strains

of ε1 = 0.4 and ε2 = 0.2, with ε3 not beingmeasured. What is the value of ε3?

(a) Since σ1 ≥ σ2 ≥ σ3, then from the figureσ1 = 50 MPa, σ2 = 20 MPa and σ3 = −40MPa.

(b) The yield stress using the Tresca criterionis given by Eq. (2.36) as

σmax − σmin = Y

So that

Y = 50 MPa− (−40 MPa) = 90 MPa

(c) If the von Mises criterion is used, thenEq. (2.37) on p. 64 gives

(σ1−σ2)2 +(σ2−σ3)2 +(σ3−σ1)2 = 2Y 2

or

2Y 2 = (50− 20)2 +(20+40)2 +(50+40)2

or2Y 2 = 12, 600

which is solved as Y = 79.4 MPa.(d) If the material is deforming plastically,

then from Eq. (2.48) on p. 69,

ε1 + ε2 + ε3 = 0.4 + 0.2 + ε3 = 0

or ε3 = −0.6.

2.98 It has been proposed to modify the von Misesyield criterion as:

(σ1 − σ2)a + (σ2 − σ3)

a + (σ3 − σ1)a = C

where C is a constant and a is an even inte-ger larger than 2. Plot this yield criterion fora = 4 and a = 12, along with the Tresca andvon Mises criteria, in plane stress. (Hint: SeeFig. 2.36 on p. 67).

For plane stress, one of the stresses, say σ3, iszero, and the other stresses are σA and σB . Theyield criterion is then

(σA − σB)a + (σB)a + (σA)a = C

For uniaxial tension, σA = Y and σB = 0 sothat C = 2Y a. These equations are difficultto solve by hand; the following solution wasobtained using a mathematical programmingpackage:

Y

Y

B

A

Tresca

von Mises

a=12a=4

Note that the solution for a = 2 (von Mises)and a = 4 are so close that they cannot bedistinguished in the plot. When zoomed intoa portion of the curve, one would see that thea = 4 curve lies between the von Mises curveand the a = 12 curve.

2.99 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 student,and has been found to be a very valuable home-work problem.

26






27






28






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 60 9.5 7.9Steel 200 0.33 149 23.7 20W 400 0.27 274 43.6 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 1000C. From Table3.3 on p. 106, the coefficient of thermal expan-sion for steel is 14.5 µm/mC (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-

√µm.

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


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.

38






39






40






Chapter 4

Surfaces, Tribology, DimensionalCharacteristics, Inspection, andProduct Quality Assurance

Questions

4.1 Explain what is meant by surface integrity.Why should we be interested in it?

Whereas surface roughness describes the geo-metric features of a surface, surface integrityconsists of not only the geometric descriptionbut also the mechanical and metallurgical prop-erties and characteristics. As described in Sec-tion 4.2 starting on p. 132, surface integrity hasa major effect on properties, such as fatiguestrength and resistance to corrosion, and hencethe service life of a product.

4.2 Why are surface-roughness design requirementsin engineering so broad? Give appropriate ex-amples.

As described in Section 4.3 starting on p. 134,surface-roughness design requirements for typ-ical engineering applications can vary by asmuch as two orders of magnitude for differentparts. The reasons and considerations for thiswide range include the following:

(a) Precision required on mating surfaces,such as seals, gaskets, fittings, and toolsand dies. For example, ball bearingsand gages require very smooth surfaces,whereas surfaces for gaskets and brakedrums can be quite rough.

(b) Tribological considerations, that is, the ef-fect of surface roughness on friction, wear,and lubrication.

(c) Fatigue and notch sensitivity, becauserougher surfaces generally have shorter fa-tigue lives.

(d) Electrical and thermal contact resistance,because the rougher the surface, the higherthe resistance will be.

(e) Corrosion resistance, because the rougherthe surface, the more the possibility thatcorrosive media may be entrapped.

(f) Subsequent processing, such as paintingand coating, in which a certain degree ofroughness can result in better bonding.

(g) Appearance, because, depending on theapplication, a rough or smooth surfacemay be preferred.

(h) Cost considerations, because the finer thefinish, the higher is the cost.

4.3 We have seen that a surface has various layers.Describe the factors that influence the thicknessof each of these layers.

These layers generally consist of a work-hardened layer, oxides, adsorbed gases, and var-ious contaminants (see Fig. 4.1 on p. 132). The

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thickness of these layers is influenced by thenature of surface-generation process employed(casting, forming, machining, grinding, polish-ing, etc.) and the environment to which the sur-face is exposed during and after its generation.Thus, for example, dull cutting tools or severesurface deformation during metalworking oper-ations produce a relatively thick work-hardenedlayer. In addition to production methods andchoice of processing parameters, an equally im-portant factor is the effect of the environmentand temperature on the workpiece material.

4.4 What is the consequence of oxides of metals be-ing generally much harder than the base metal?Explain.

The consequences are numerous, and the oxidecan be beneficial as well as detrimental. In slid-ing contact, the oxide is a hard surface that, as aresult, is wear resistant [see Eq. (4.6) on p. 145],and it can also protect the substrate from fur-ther chemical attack. However, if an oxide wearparticle spalls from the surface, a detrimentalthree-body wear situation can result. Also, asdiscussed in Chapter 2, the hard surface layersmay be detrimental from a fatigue standpointif their ductility is compromised. Finally, if amaterial is plastically deformed, as in the pro-cesses described in Chapters 6 and 7, the oxidelayer may crack or even break off, resulting in asurface finish that may be unacceptable for theparticular application.

4.5 What factors would you consider in specifyingthe lay of a surface?

Specifying the lay of a surface requires consid-erations such as the nature of the mating sur-faces and their application, direction of rela-tive sliding, frictional effects, lubricant entrap-ment, and optical factors such as appearanceand reflectivity of the surface. Physical proper-ties such as thermal and electrical conductivitymay also be significant.

4.6 Describe the effects of various surface defects(see Section 4.3 starting on p. 134) on the per-formance of engineering components in service.How would you go about determining whetheror not each of these defects is important for aparticular application?

Surface defects can have several effects on theperformance of engineering components in ser-vice. Among these effects are premature failureunder various types of loading, crevice corro-sion, adverse effects on lubrication, and whetherthe components will function smoothly or therewill be vibration or chatter.

The manner in which their importance for aparticular operation can be assessed is by ob-serving the defect type and its geometry, andhow these defects would relate to componentperformance. The direction and depth of acrack, for example, should be reviewed with re-spect to the direction of tensile stresses or direc-tion of relative movement between the surfaces.Another example is the possibility of crevicecorrosion in the presence of a hostile environ-ment.

4.7 Explain why the same surface roughness valuesdo not necessarily represent the same type ofsurface.

As can be seen in Eqs. (4.1) and (4.2) on p. 134,there is an infinite range of values for a, b, c, d,etc. that would give the same arithmetic meanvalue Ra or the root-mean-square average valueRq. This can be seen graphically, as the sur-faces shown below are examples that result inthe same Ra values but are very different ge-ometrically and have different tribological per-formance (from Bhushan, B., Introduction toTribology, Wiley, 2002).

(a)

(b)

(c)

(d)

(e)

(f)

4.8 In using a surface-roughness measuring instru-ment, how would you go about determining thecutoff value? Give appropriate examples.

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The determination depends on various factors.For example, if waviness is repeatable, the cut-off need not be longer than the waviness cycle.Also, if there is poor control of processing pa-rameters during manufacturing, such that thesurface produced is highly irregular, then cutoffshould be long enough to give a representativeroughness value. If the quality of the workpiecematerial is poor, with numerous flaws, inclu-sions, impurities, etc., the cutoff must be longenough to be representative of the surface ingeneral. The lay could also play a significantrole in cutoff selection. The cutoff should berelated to the spacing of asperities, which hasbeen found to be about an order of magnitudelarger than the roughness for most surfaces.Thus, several recommendations can be foundin the technical literature for different methodsof surface preparation. For example, the fol-lowing data is recommended by a profilometermanufacturer:

Ra Cut-off length(µm) mm0.025 0.080.05 0.250.1 0.250.2 0.250.4 0.250.8 0.81.6 0.83.2 2.56.3 2.512.5 2.5

4.9 What is the significance of the fact that the sty-lus path and the actual surface profile generallyare not the same?

This situation indicates that profilometer tracesare not exact duplicates of actual surfaces andthat such readings can be misleading for pre-cise study of surfaces. (Note, however, that theroughness in Fig. 4.4 on p. 137 is highly exag-gerated because of the differences between thehorizontal and vertical scales.) For example,surfaces with deep narrow valleys will be mea-sured smoother than they really are. This canhave significant effects on the estimating the fa-tigue life, corrosion, and proper assessment ofthe capabilities of various manufacturing pro-cesses.

4.10 Give two examples each in which waviness of asurface would be (1) desirable and (2) undesir-able.

Suggested examples are, for desirable: sug-gested examples are aesthetic reasons, appear-ance, beneficial effects of trapping lubricantsbetween two surfaces. For undesirable: uneven-ness between mating surfaces, difficulty of pro-viding a tight seal, sliding is not smooth. Thestudent is encouraged to give other examples.

4.11 Explain why surface temperature increaseswhen two bodies are rubbed against each other.What is the significance of temperature rise dueto friction?

This topic is described in Section 4.4.1 start-ing on p. 138. When bodies rub against eachother, friction causes energy dissipation whichis in the form of heat generation at the sur-faces. If the rubbing speed is very slow, and thethermal conductivity of the workpiece is veryhigh, then the temperature rise may be negli-gible. More commonly, there can be a majortemperature rise at the surface. The signifi-cance of this temperature rise is that surfacesmay be more chemically active or may be de-velop higher thermal stresses and possibly re-sult in heat checking. Note that this is not nec-essarily detrimental because chemical reactiv-ity is required for many boundary and extreme-pressure lubricants to bond to a surface.

4.12 To what factors would you attribute the factthat the coefficient of friction in hot working ishigher than in cold working, as shown in Ta-ble 4.1?

The factors that have a significant influence onfriction are described in Section 4.4.1 startingon p. 138. For hot working, specifically, im-portant factors are the tendency for increasedjunction strength (due to greater affinity), ox-ide formation, strength of oxide layers, and theeffectiveness of lubricants at elevated tempera-tures.

4.13 In Section 4.4.1, we note that the values of thecoefficient of friction can be much higher thanunity. Explain why.

This phenomenon is largely a matter of defi-nition of the coefficient of friction µ, and also

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indicates the desirable feature of the concept offriction factor, m, which can range between 0and 1; see Eq. (4.5) on p. 140. Consider, forexample, Eq. (4.3) on p. 139. If, for a variety ofreasons, the mating surfaces have developed ex-tensive microwelds during sliding, then the loadis removed and the two surfaces are allowed toslide against each other, there would be con-siderable friction force, F , required. Since forthis case the load N is now negligible, the co-efficient of friction, µ = F/N , would indeed bevery high. This can also be seen if the matingsurfaces consisted of adhesive tape or VelcroR©.

4.14 Describe the tribological differences betweenordinary machine elements (such as meshinggears, cams in contact with followers, and ballbearings with inner and outer races) and ele-ments of metalworking processes (such as forg-ing, rolling, and extrusion, which involve work-pieces in contact with tools and dies).

The tribological differences are due to signifi-cant differences in parameters such as contactloads and stresses, relative speeds between slid-ing members, workpiece temperatures, temper-ature rise during application, types of materialsinvolved, types of lubricants used, and the par-ticular environment. Also, referring to Fig. 4.6on p. 140, note that the manufacturing pro-cesses all take place at very high normal stressesand as a result, non-linear relationships be-tween friction and normal force are uncommon.With machine elements such as gears, cams,and bearings, however, normal forces are not ashigh, and a Coulomb friction law as stated inEq. (4.3) on p. 139 generally applies. Studentsare encouraged to develop a list with severalspecific examples.

4.15 Give the reasons that an originally round spec-imen in a ring-compression test may becomeoval after deformation.

The specimen may flow more easily in one di-rection than another for reasons such as:

(a) anisotropy of the workpiece material,

(b) the lay of the specimen surfaces, thus af-fecting frictional characteristics,

(c) the lay on the surface of the flat dies em-ployed,

(d) uneven lubricant layer over the matingsurfaces, and

(e) lack of symmetry of the test setup, such asplatens that are not parallel.

4.16 Can the temperature rise at a sliding interfaceexceed the melting point of the metals? Ex-plain.

When the heat generated due to friction andthat due to work of plastic deformation exceedsthe rate of heat dissipation from the surfacesthrough conduction and convection, the sur-faces will soften and even melt, and the heatinput will be dissipated as heat of fusion neces-sary for changing from a solid to a liquid phase.This heat represents a high amount of energy,thus the surface temperature will not exceed themelting point.

4.17 List and briefly describe the types of wear en-countered in engineering practice.

This topic is discussed in Section 4.4.2 onp. 144. Basically, the types of wear are:

• Adhesive wear, where material transfer oc-curs because one material has bonded tothe other and relative motion shears thesofter material; see Fig. 4.10 on p. 145.

• Abrasive wear, where a hard asperityplows into a softer material, producing achip, as shown in Fig. 4.10 on p. 145. Thiscan be a two-body or a three-body phe-nomenon.

• Corrosive wear, which occurs when chemi-cal or electrochemical reactions take place,thereby removing material from surfaces.

• Fatigue wear, common in bearings andgears, is due to damage associated withcyclic loading, where cracks propagate andcause material loss through spalling.

• Erosion, caused by the abrasive action ofloose hard particles.

• Impact wear, refers to spalling associatedwith dynamic loading of a surface.

4.18 Explain why each of the terms in the Archardformula for adhesive wear, Eq. (4.6) on p. 145,should affect the wear volume.

The following observations can be made regard-ing this formula:

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(a) The wear coefficient, k, indicates the affin-ity of the two contacting surfaces to de-velop microwelds, as shown in Table 4.2on p. 146. The greater the affinity, thegreater the probability of forming strongmicrowelds, hence the higher the adhesivewear.

(b) The greater the distance traveled, L, ob-viously the higher the amount of wear.

(c) The greater the normal load, W , thegreater the tendency to form strong mi-crowelds, hence the greater the wear.

(d) The higher the hardness of the softer body,the lower the possibility of forming strongjunctions at the interface, hence the lowerthe wear. Note also the significant effectof lubrication on the magnitude of k, as tobe expected.

4.19 How can adhesive wear be reduced? How canfatigue wear be reduced?

Adhesive wear can be reduced by studying theeffects outlined in the answer to Problem 4.18above. Fatigue wear can be reduced by:

(a) reducing the load and sliding distance andincreasing the hardness, consistent withProblem 4.18 above;

(b) improving the quality of the contactingmaterials, such as eliminating inclusions,impurities, and voids;

(c) improving the surface finish and integrityduring the manufacturing process;

(d) surface working, such as shot peening orother treatments;

(e) reducing contact stresses; and

(f) reducing the number of total cycles. (Seealso Section 2.7 starting on p. 56.)

4.20 It has been stated that as the normal load de-creases, abrasive wear is reduced. Explain whythis is so.

For abrasive wear to occur, the harder orrougher surface must penetrate the softer sur-face to some depth. Thus, this phenomenonbecomes similar to a hardness test, wherebythe harder the surface, the less the penetrationof the indenter. Consequently, as the normal

load decreases, the surfaces do not penetrateas much and, hence, the groove produced (byan abrasive particle sliding against a surface) ismore shallow, thus abrasive wear is lower.

4.21 Does the presence of a lubricant affect abrasivewear? Explain.

Although it is not readily apparent fromEq. (4.6) on p. 145, the presence of a lubri-cant can affect abrasive wear by virtue of thefact that a lubricant can have some effect (al-though to a very minor extent) on the depthof penetration, as well as the manner in whichthe slivers are produced and their dimensions(as described in Chapter 8). It should also benoted that the presence of a lubricant will causethe wear particles to stick to the surfaces, thusinterfering with the operation. This topic hasnot been studied to any extent, thus it wouldbe suitable for literature search on the part ofstudents.

4.22 Explain how you would estimate the magnitudeof the wear coefficient for a pencil writing onpaper.

Referring to Eq. (4.6) on p. 145, since the wearvolume, the force on the pencil, and the slidingdistance can be determined, we can then cal-culate the dimensionless wear coefficient, k/H.The hardness of the pencil material can be mea-sured through a microhardness test. The tri-bology of pencil on paper is an interesting areafor inexpensive experimentation. Note also thatdifferent types of paper will result in differentwear coefficients as well (e.g., rough construc-tion paper vs. writing paper vs. newspaper, oreven wax paper). This topic can easily be ex-panded into a design project to encourage stu-dents to develop wear tests to determine k.

4.23 Describe a test method for determining thewear coefficient k in Eq. (4.6). What wouldbe the difficulties in applying the results fromthis test to a manufacturing application, suchas predicting the life of tools and dies?

Several tests have been developed for evaluat-ing wear coefficients, and this topic would besuitable as a student project. The following areamong the more commonly used:

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• The pin-on-disk test uses a pin sliding overa rotating disk; wear volume is obtainedfrom the change in the length and geome-try of the pin or by using profilometry onthe disk.

• The pin-on-flat test uses a pin that recip-rocates against a flat surface.

• A ring test uses rotating rings and is espe-cially used to evaluate fatigue wear.

• An abrasive wear test that is commonlyperformed uses a rubber wheel to pressloose abrasives against a workpiece.

All of these tests have significant drawbackswhen applied to manufacturing processes. Mostimportantly, it is difficult to reproduce the con-tact stresses encountered in a manufacturingenvironment, such as temperature and strainrate), which will then have a major effect onwear. Furthermore, there is a need to maintainthe same surface condition as encountered inmanufacturing operations.

4.24 Why is the abrasive wear resistance of a mate-rial a function of its hardness?

Higher hardness indicates greater resistance topenetration, hence less penetration of the abra-sive particles or hard protrusions into surfaces,and the grooves produced are not as deep.Thus, abrasive wear is a function of hardness.

4.25 We have seen that wear can have detrimen-tal effects on engineering components, tools,dies, etc. Can you visualize situations in whichwear could be beneficial? Give some examples.(Hint: Note that writing with a pencil is a wearprocess.)

Consider, for example:

(a) running-in periods of machinery,

(b) burnishing, involving improvements insurface finish and appearance due to asmall amount of controlled wear.

(c) using sandpaper to remove splinters fromwood,

(d) using a scouring pad on cookware to re-move dried or burnt food particles.

(e) grinding and other manufacturing opera-tions, as described in Chapter 9, where

fine tolerances and good surface finishesare achieved through basically controlledwear mechanisms.

The student is encouraged to think of more ex-amples.

4.26 On the basis of the topics discussed in this chap-ter, do you think there is a direct correlation be-tween friction and wear of materials? Explain.

The answer is no, not directly. Consider, forexample, the fact that ball and roller bearingshave very low friction yet they do undergo wear,especially by surface fatigue. Also, ceramicshave low wear rate, yet they can have signifi-cant frictional resistance. The following data,obtained from J. Halling, Principles of Tribol-ogy, 1975, p. 9, clearly demonstrates that highfriction does not necessarily correspond to highwear:

Wear rateMaterials µ (cm3/cm ×10−12)Mild steel on mild 0.62 157,000

steel60/40 leaded brass 0.24 24,000PTFE 0.18 2000Stellite 0.60 310Ferritic stainless steel 0.53 270Polyethylene 0.65 30Tungsten carbide on 0.35 2

itself

4.27 You have undoubtedly replaced parts in variousappliances and automobiles because they wereworn. Describe the methodology you would fol-low in determining the type(s) of wear thesecomponents have undergone.

This is an open-ended problem, and the stu-dent should be asked to develop a methodol-ogy based on Section 4.4.2 starting on p. 144.The methodology should include inspection ata number of levels, for example, visual determi-nation of the surface, as well as under a lightmicroscope and scanning electron microscope.Surface scratches, for instance, are indicativeof abrasive wear; spalling would suggest fatiguewear; and a burnished surface suggests adhesivewear. The wear particles must also be investi-gated. If the particles have a bulky form, theyare likely to be adhesive wear particles, includ-ing surface oxides. Flakes are indicative of ad-hesive wear of metals that do not have an oxide

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surface. Abrasive wear results in slivers or wearparticles with a larger aspect ratios.

4.28 Why is the study of lubrication regimes impor-tant?

The reason is primarily due to the fact that eachregime, ranging from full-fluid film to sliding ofdry surfaces, has its own set of variables thataffect performance, load-bearing capacity, fric-tion, wear, temperature rise, and surface dam-age. Consequently, in the event of poor perfor-mance, one should concentrate and further in-vestigate those particular parameters. Also, thelubricant film plays a major role in the ultimateworkpiece surface roughness that is produced.The student may elaborate further, based onthe topics covered in Section 4.4.3 starting onp. 149.

4.29 Explain why so many different types of metal-working fluids have been developed.

The student may discuss this topic, based onvarious topics listed in Section 4.4.4 startingon p. 151; thus, for example, end result ex-pected (to reduce friction or wear), the partic-ular manufacturing process employed, the ma-terials used, the temperatures that will be en-countered, costs involved, etc.

4.30 Differentiate between (1) coolants and lubri-cants, (2) liquid and solid lubricants, (3) directand indirect emulsions, and (4) plain and com-pounded oils.

The answers can be found in Section 4.4.4 start-ing on p. 151. Basically,

(a) a coolant is mainly intended to removeheat, whereas a lubricant has friction andwear reduction functions as well. For ex-ample, water is an excellent coolant but isa poor lubricant (unless used in hydrody-namic lubrication), whereas water-solubleoils generally serve both functions.

(b) The difference between liquid and solid lu-bricants is that they have different phases.However, although solid lubricants aresolid at room temperature, they may notbe so at operating temperatures.

(c) Direct emulsions have oil suspended in wa-ter; indirect (invert) emulsions have waterdroplets suspended in oil.

(d) Plain oils contain the base oil only,whereas compounded oils have various ad-ditives in the base oil to fulfill special crite-ria such as lubricity and workpiece surfacebrightening.

4.31 Explain the role of conversion coatings. Basedon Fig. 4.13, what lubrication regime is mostsuitable for application of conversion coatings?

Conversion coatings provide a rough and poroussurface on workpieces. The porosity is infil-trated by the lubricant, thus aiding in entrain-ment and retention of the lubricant in the met-alworking process. Considering the regimes oflubrication, it is clear that conversion coatingsare not useful in full-film lubrication, since athick lubricant film already exits at the inter-faces without the need for a rough surface. Itis, however, beneficial for boundary or mixed-lubrication regimes.

4.32 Explain why surface treatment of manufacturedproducts may be necessary. Give several exam-ples.

This topic is described at the beginning of Sec-tion 4.5 on p. 154. Examples are:

• Some surfaces may be coated with ahard material for wear resistance, such asceramic-coated cutting or forming tools.

• Jewelry and tableware are electroplatedwith gold or silver, for aesthetic and somefunctional reasons.

• Bolts, nuts, and other fasteners are zinccoated for corrosion resistance.

• Some automotive parts are plated withdecorative chrome (although not used asoften now) for aesthetic reasons.

The student is encouraged to develop additionalspecific applications, based on the materialscovered in this section of the text.

4.33 Which surface treatments are functional, andwhich are decorative? Give several examples.

A review of the processes described indicatesthat most surface treatments are functional. Afew, such as electroplating, anodizing, porce-lain enameling, and ceramic coating, are gener-ally regarded as both functional and decorative.

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The students are encouraged to give specific ex-amples from their personal experience and ob-servations.

4.34 Give examples of several typical applications ofmechanical surface treatment.

The applications are described in Section 4.5.1starting on p. 154. Some examples are:

• The shoulders of shafts can be roller bur-nished to impart a compressive residualstress and thus improve fatigue life.

• Crankshafts, rotors, cams, and other simi-lar parts are shot peened in order to in-crease surface hardness and wear resis-tance, as well as improve fatigue life.

• Railroad rails can be hardened by explo-sive hardening.

The student is encouraged to give additionalexamples.

4.35 Explain the difference between case hardeningand hard facing.

Case hardening is a heat treatment process (de-scribed in Section 5.11.3 on p. 241) performedon a manufactured part; hard facing involvesdepositing metal on a surface using varioustechniques described in the text.

4.36 List several applications for coated sheet metal,including galvanized steel.

There are numerous applications, ranging fromgalvanized sheet-steel car bodies for corrosionprotection, to sheet-metal television cabinets,office equipment, appliances, and gutters anddown spouts. Polymer-coated steels are typi-cally used for food and beverage containers andalso for some sheet-metal parts. The student isencouraged to develop lists for specific applica-tions.

4.37 Explain how roller-burnishing processes induceresidual stresses on the surface of workpieces.

Roller burnishing, like shot peening, inducesresidual surface compressive stresses due tolocalized plastic deformation of the surface.These stresses develop because the surface layertends to expand during burnishing, but the

bulk prevents these layers from expanding later-ally freely. Consequently, compressive residualstresses develop on the surface.

4.38 List several products or components that couldnot be made properly, or function effectively inservice, without implementation of the knowl-edge involved in Sections 4.2 through 4.5.

This is an open-ended problem that can be an-swered in many ways. Some examples of com-ponents that require the knowledge in Sections4.2 through 4.5 include:

• Brake drums, rotors, and shoes could notbe designed properly without an under-standing of friction and wear phenomena.

• Crankshaft main bearings and pistonbearings require an understanding of lu-brication.

• A wide variety of parts have functionalcoatings, such as galvanized sheet metalfor automotive body panels, zinc coatingson bolts and nuts, and hard chrome coat-ings for wear resistance, and cutting toolscan have nitride coatings through chemicalvapor deposition.

• Aircraft fuselage components have a strictsurface roughness requirement.

4.39 Explain the difference between direct- andindirect-reading linear measurements.

In direct reading, the measurements are ob-tained directly from numbers on the measuringinstruments, such as a rule, vernier caliper, ormicrometer. In indirect reading, the measure-ments are made using calipers, dividers, andtelescoping gages. These instruments do nothave numbers on them and their setting is mea-sured subsequently using a direct-measuring in-strument.

4.40 Why have coordinate-measuring machines be-come important instruments in modern manu-facturing? Give some examples of applications.

These machines are built rigidly and are veryprecise, and are equipped with digital readoutsand also can be linked to computers for on-lineinspection of parts. They can be placed close tomachine tools for efficient inspection and rapid

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feedback for correction of processing parame-ters before the next part is made. They arealso being made more rugged to resist environ-mental effects in manufacturing plants, such astemperature variations, vibration, and dirt.

4.41 Give reasons why the control of dimensional tol-erances in manufacturing is important.

This topic is described in Section 4.7 starting onp. 170. Generally, for instance, products per-form best when they are at their design spec-ification, so dimensional tolerances should becontrolled to obtain as good of performance asis possible. The student is encouraged to giveseveral examples.

4.42 Give examples where it may be preferable tospecify unilateral tolerances as opposed to bi-lateral tolerances in design.

By the student. For example, when a shrinkfit is required, it may be beneficial to specify ashaft and hole tolerance with a unilateral tol-erance, so that a minimum contact pressure isassured.

4.43 Explain why a measuring instrument may nothave sufficient precision.

A caliper, for example, that can only measureto the nearest 0.001 in. is not precise enoughto measure a 0.0005 in. press-fit clearance be-tween two mating gears.

4.44 Comment on the differences, if any, between (1)roundness and circularity, (2) roundness and ec-centricity, and (3) roundness and cylindricity.

(a) The terms roundness and circularity areusually interchangeable, with the term outof roundness being commonly used. Circu-larity is defined as the condition of a sur-face of revolution where all points of thesurface intersected by any plane perpen-dicular to an axis or passing through a cen-ter are equidistant from the center. Also,we usually refer to a round shaft as beinground, whereas there are components andparts in which only a portion of a surfaceis circular. (See, for example, circular in-terpolation in numerical control, describedin Fig. 14.11c on p. 882).

(b) Eccentricity may be defined as not havingthe same center, or referring to concentric-ity in which two or more features have acommon axis. Thus, a round shaft may bemounted on a lathe at its ends in such amanner that its rotation is eccentric.

(c) Cylindricity is defined similarly to circu-larity, as the condition of a surface of rev-olution in which all points of the surfaceare equidistant from a common axis. Astraight shaft with the same roundnessalong its axis would possess cylindricity;however, in a certain component, round-ness may be confined to only certain nar-row regions along the shaft, thus it doesnot have cylindricity over its total length.

4.45 It has been stated that dimensional tolerancesfor nonmetallic stock, such as plastics, are usu-ally wider than for metals. Explain why. Con-sider physical and mechanical properties of thematerials involved.

Nonmetallic parts have wider tolerances be-cause they often have low elastic modulus andstrength, are soft, have high thermal expansion,and are therefore difficult to manufacture withhigh accuracy. (See also Chapter 10).

4.46 Describe the basic features of nondestructivetesting techniques that use electrical energy.

Nondestructive testing techniques that use elec-trical energy are magnetic particle, ultrasonic,acoustic emission, radiography, eddy current,and holography. Their basic features are de-scribed in Section 4.8.1.

4.47 Identify the nondestructive techniques that arecapable of detecting internal flaws and thosethat only detect external flaws.

Internal flaws: ultrasonic, acoustic emission, ra-diography, and thermal. External flaws: liq-uid penetrants, magnetic particle, eddy current,and holography. Some of these techniques canbe utilized for both types of defects.

4.48 Which of the nondestructive inspection tech-niques are suitable for nonmetallic materials?Why?

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By the student. Since nonmetallic materials arecharacterized by lack of electrical conductivity,techniques such as magnetic particle and eddycurrent would not be suitable.

4.49 Why is automated inspection becoming an im-portant aspect of manufacturing engineering?

As described throughout the text, almost allmanufacturing equipment is now automated(see also Chapters 14 and 15). Consequently,inspection at various stages of productionshould also be automated in order to improveproductivity, by keeping the flow of materialsand products at an even, rapid pace.

4.50 Describe situations in which the use of destruc-tive testing techniques is unavoidable.

Destructive testing techniques will be neces-sary for determining, for example, the mechan-ical properties of products being made, becausenondestructive techniques generally cannot doso. Such property determination requires testsamples (as described throughout Chapter 2),such as from different regions of a forging or acasting, before proceeding with large-scale pro-duction of the product. This approach is partic-ularly important for parts that are critical, suchas jet-engine turbine components and medicalimplants.

4.51 Should products be designed and built for a cer-tain expected life? Explain.

Product life cycle and cradle-to-cradle designare discussed in Section 16.4. The students areencouraged to review this material, as well asdescribe their own thoughts and cite their expe-riences with purchasing various products. Thisis an important topic, and includes several tech-nical as well as economic considerations andpersonal choices.

4.52 What are the consequences of setting lower andupper specifications closer to the peak of thecurve in Fig. 4.23?

In statistical process control, setting the speci-fications closer to the center of the distribution

will cause more of the sample points to fall out-side the limits, as can be seen in Fig. 4.21c onp. 177, thus increasing the rejection rate.

4.53 Identify factors that can cause a process to be-come out of control. Give several examples ofsuch factors.

This situation can occur because of various fac-tors, such as:

(a) the gradual deterioration of coolant or lu-bricant,

(b) debris interfering with the manufacturingoperation,

(c) an increase or decrease in the temperaturein a heat-treating operation,

(d) a change in the properties of the incomingraw materials, and

(e) a change in the environmental conditions,such as temperature, humidity, and airquality.

The student is encouraged to give other exam-ples.

4.54 In reading this chapter, you will have noted thatthe specific term dimensional tolerance is oftenused, rather than just the word tolerance. Doyou think this distinction is important? Ex-plain.

As a general term, tolerances relate not onlyto dimensions but to parameters such as themechanical, physical, and chemical propertiesof materials, including their compositions. Forexample, in the electronics industry, there aretolerances with respect to part dimensions, butalso with respect to electrical properties. Inmost mechanical engineering design applica-tions, however, the distinction is not significant.

4.55 Give an example of an assignable variation anda chance variation.

This topic is defined and described in Section4.9.1 starting on p. 176, with an example onbending of beams to determine their strength.The students are encouraged to describe an ex-ample of their own.

50






Problems

4.56 Referring to the surface profile in Fig. 4.3, givesome numerical values for the vertical distancesfrom the center line. Calculate the Ra and Rq

values. Then give another set of values for thesame general profile, and calculate the samequantities. Comment on your results.

As an example, two students took the same fig-ure and enlarged it, one with a copy machine,the other by scanning it into a computer andzooming in on the figure. The first studentprinted the figure on graph paper and inter-polated numbers for points a through l. Thesecond sketched a grid over the drawing andinterpolated numbers as well. Their results aregiven as:

Point Student 1 Student 2 Modified 2a 3.8 5.0 3.79b 2.5 3.0 2.27c 4.0 5.5 4.17d 5.5 7.5 5.69e 2.0 2.5 1.90f -3.5 -4.5 -3.41g -5.0 -6.5 -4.93h -4.0 -5.5 -4.17i -4.0 -5.0 -3.79j -5.5 -7.0 -5.31k -3.5 -4.5 -3.41l -1.0 -1.0 -0.76

Note that the scales are slightly off, due to thefact that the grids used were different. To havethe same peak-to-peak value as Student 1, thevalues of Student 2 were scaled as in the Modi-fied 2 column. The Ra and Rq values, as calcu-lated from Eqs. (4.1) and (4.2) on p. 134, are:

Source Ra Rq

Student 1 3.69 5.45Student 2 4.79 5.12Modified 2 3.63 3.88

Note that the roughness values depend on thescales used for the plots. When normalized lin-early, so that the data has the same peak-to-peak value, theRa roughnesses match very well.However, the Rq roughness values do not match

well; a second-order mapping of data pointswould improve the performance, however.

4.57 Calculate the ratio of Ra/Rq for (a) a sine wave,(b) a saw-tooth profile, (c) a square wave.

This solution uses the continuous forms ofroughness given by Eqs. (4.1) and (4.2) onp. 134.

(a) The equation of a sine wave with amplitudea is

y = a sin2πxl

Thus, Eq. (4.1) gives

Ra =1l

∫ l

0

∣∣∣∣a sin2πxl

∣∣∣∣ dx =2al

∫ l/2

0

sin2πxldx

Integrating,

Ra = −2al

l

2π

(cos

2πxl

)l/2

0

= −aπ

(cosπ − cos 0) = −aπ

(−1− 1)

=2aπ

To evaluate Rq for a sine wave, recall that∫sin2 u du =

u

2− 1

4sin 2u

which can be obtained from any calculus bookor table of integrals. Therefore, from Eq. (4.2),

R2q =

1l

∫ l

0

y2 dx =1l

∫ l

0

a2 sin2 2πxldx

Evaluating the integral,

R2q =

a2

l

l

2π

[2πx2l

− 14

sin4πxl

]l

0

=a2

2π[(π − 0)− (0− 0)] =

a2

2So that Rq = a√

2, and

Ra

Rq=

2a/πa/√

2=

2√

2π

≈ 0.90

(b) For a saw-tooth profile, we can use symme-try to evaluate Ra and Rq over one-fourth of

51






the saw tooth. The equation for the curve overthis range is:

y =4alx

so that, from Eq. (4.1),

Ra =4l

∫ l/4

0

4axldx


Ra =16al2

(12x2

)l/4

0

=16al2

12

(l2

16− 0)

=a

2

From Eq. (4.2),

R2q =

4l

∫ l/4

0

y2 dx =4l

∫ l/4

0

16a2

l2x2 dx


R2q =

64a2

l3

(13x3

)l/4

0

=64a2

l313l3

64=a2

3

Therefore, Rq = a√3

and

Ra

Rq=

a/2a/√

3=√

32≈ 0.866

(c) For a square wave with amplitude a,

Ra =1l

∫ l

0

a dx =a

l(x)l

0 = a

and

R2q =

1l

∫ l

0

a2 dx =a2

l(x)l

0 = a2

so that Rq = a. Therefore,

Ra

Rq=a

a= 1.0

4.58 Refer to Fig. 4.7b and make measurements ofthe external and internal diameters(in the hor-izontal direction in the photograph) of the fourspecimens shown. Remembering that in plas-tic deformation the volume of the rings remainsconstant, calculate (a) the reduction in heightand (b) the coefficient of friction for each of thethree compressed specimens.

The volume of the original specimen is (seeFig. 4.8 on p. 142):

V =πh

4(d2

o − d2i

)=π(0.25)

4(0.752 − 0.3752

)or V = 0.0828 in3. Note that the volume mustremain constant, so that

hπ

4(d2

o − d2i

)= 0.0828 in3

or

h =0.105d2

o − d2i

Specimen 1 has not been deformed, so its di-mensions are taken from Fig. 4.8. The re-maining dimensions are scaled to be consistentwith these values. The height value cannot bedirectly measured because of the angle of viewin the figure; so these are calculated from vol-ume constancy. Sample measurements are asfollows:

di do hID di (in.) (in.) (in.)1 0.375 0.75 0.252 0.477 0.97 0.1473 0.282 1.04 0.1044 0.1757 1.04 0.100

The reduction in height is calculated from

% Reduction in height =ho − hf

ho× 100

and the values of the coefficient of friction arethen obtained from Fig. 4.8a on p. 142 to obtainthe following:

% Reduction % ReductionID in height in di µ1 0 0 —2 41.2 -27.2 0.013 58.4 24.8 0.104 60 53.1 0.20

Note that the fricton coefficient increases as thetest progresses. This is commonly observed inlubricated specimens, where the lubricant filmis initially thick, but breaks down as the contactarea between the ring and die increases.

52






4.59 Using Fig. 4.8a, make a plot of the coefficient offriction versus the change in internal diameterfor a reduction in height of (1) 25%, (2) 50%,and (3) 60%.

Typical data obtained from Fig. 4.8a on p. 142are summarized in the table below. Note thatparticular results may vary.

% Change in Internal Diameter20% Red. 40% Red. 60% Red.

µ in height in height in height0 -12 -30 —0.02 -7 -16 -400.03 -4 -10 -240.04 -3 -5 -120.055 0 0 00.1 4 10 250.2 8 22 530.3 11 28 700.4 13 34 800.577 15 38 —

The plot follows:

0.5

0.4

0.3

0.2

0.1

0

Fric

tion

Coe

ffici

ent,

-50 0 50 100

Reduction in internal diameter, %

Red. = 0.20

Red = 0.40

Red = 0.60

4.60 In Example 4.1, assume that the coefficient offriction is 0.20. If all other initial parametersremain the same, what is the new internal di-ameter of the ring specimen?

If µ=0.20, Fig. 4.8a on p. 142 shows that thereduction in inner diameter is about 34% for areduction in height of 50%. Since the originalID is 15 mm, we therefore have

15 mm− IDfinal

15 mm= 0.34

or IDfinal = 9.9 mm.

4.61 How would you go about estimating forces re-quired for roller burnishing? (Hint: Considerhardness testing.)

The procedure would consist of first determin-ing the contact area between the roller and thesurface being burnished. The force is then theproduct of this area and the compressive stresson the workpiece material. Because of the con-strained volume of material subjected to plasticdeformation, the level of this stress is on the or-der of the hardness of the material, as describedin Section 2.6.8 on p. 54, or about three timesthe yield stress for cold-worked metals; see alsoFig. 2.24 on p. 55.

4.62 Estimate the plating thickness in electroplatinga 50-mm solid metal ball using a current of 1A and a plating time of 2 hours. Assume thatc = 0.08.

Note that the surface area of a sphere is A =4πr2, so that the volume of the plating is V =4πr2h, where h is the plating thickness. FromEq. (4.7) on p. 159,

V = cIt = 4πr2h

Solving for the plating thickness, and usingproper units, we find

h =cIt

4πr2=

(0.08)(1)(7200)4π(25)2

= 0.073 mm

4.63 Assume that a steel rule expands by 1% be-cause of an increase in environmental tempera-ture. What will be the indicated diameter of ashaft whose actual diameter is 50.00 mm?

The indicated diameter will be 50.00 -0.01(50.00) = 49.50 mm.

4.64 Examine Eqs. (4.2) and (4.10). What is therelationship between Rq and σ? What wouldbe the equation for the standard deviation of acontinuous curve?

The two equations are very similar. Note thatif the mean is zero, then Eq. (4.10) on p. 178 isalmost exactly the same as Eq. (4.2) on p. 134.When a large number of data points are consid-ered, the equations are the same. Rq roughnesscan actually be thought of as the standard de-viation of a curve about its mean line. Thestandard deviation of a continuous curve can

53






simply be expressed by the analog portion ofEq. (4.2):

σ =

√1l

∫ l

0

y2 dx

or, if the mean is µ,

σ =

√1l

∫ l

0

(y − µ)2 dx

4.65 Calculate the control limits for averages andranges for the following: number of samples =7; ¯x = 50; R = 7.

From Table 4.3, we find that for a sample sizeof 7, we have A2 = 0.419, D4 = 1.924 andD3 = 0.078. Equations (4.11) and (4.12) givethe upper and lower control limits for the aver-ages as

UCLx = ¯x+A2R = 50 + (0.419)(7) = 52.933

LCLx = ¯x−A2R = 50− (0.419)(7) = 47.067

For the ranges, Eqs. (4.13) and (4.14) yield

UCLR = D4R = (1.924)(7) = 13.468

LCRR = D3R = (0.078)(7) = 0.546

4.66 Calculate the control limits for the following:number of samples = 7; ¯x = 40.5; UCLR =4.85.

From Table 4.3, we find that for a sample sizeof 7, we have A2 = 0.419, D4 = 1.924 andD3 = 0.078. If the UCLR is 4.85, then fromEq. (4.14),

UCLR = D4R

solving for R,

R =UCLR

D4=

4.851.924

= 2.521

Therefore, Eqs. (4.11) and (4.12) give the upperand lower control limits for the averages as

UCLx = ¯x+A2R = (40.5)+(0.419)(2.521) = 41.556

LCLx = ¯x−A2R = (40.5)−(0.419)(2.521) = 39.444

Equation (4.14) gives:

LCLR = D3R = (0.078)(2.521) = 0.197

4.67 In an inspection with a sample size of 10 anda sample number of 40, it was found that theaverage range was 10 and the average of aver-ages was 75. Calculate the control limits foraverages and ranges.

From Table 4.3, we find that for a sample sizeof 10, we have A2 = 0.308, D4 = 1.777 andD3 = 0.223. Equations (4.11) and (4.12) onp. 180 give the upper and lower control limitsfor the averages as

UCLx = ¯x+A2R = (75)+(0.308)(10) = 78.080

LCLx = ¯x−A2R = (75)− (0.308)(10) = 71.920


UCLR = D4R = (1.777)(10) = 17.77

LCRR = D3R = (0.223)(10) = 2.23

4.68 Determine the control limits for the data shownin the following table:

x1 x2 x3 x4

0.65 0.75 0.67 0.650.69 0.73 0.70 0.680.65 0.68 0.65 0.610.64 0.65 0.60 0.600.68 0.72 0.70 0.660.70 0.74 0.65 0.71

Since the number of samples is 4, from Table 4.3on p. 180 we find that A2 = 0.729, D4 = 2.282,and D3 = 0. We calculate averages and rangesand fill in the chart as follows:

x1 x2 x3 x4 x R0.65 0.75 0.67 0.65 0.6800 0.100.69 0.73 0.70 0.68 0.7000 0.050.65 0.68 0.65 0.61 0.6475 0.070.64 0.65 0.60 0.60 0.6225 0.050.68 0.72 0.70 0.66 0.6900 0.060.70 0.74 0.65 0.71 0.7000 0.09

Therefore, the average of averages is ¯x =0.6733, and the average range is R = 0.07.Eqs. (4.11) and (4.12) give the upper and lowercontrol limits for the averages as

UCLx = ¯x+A2R = (0.6733) + (0.729)(0.07)

54






or UCLx=0.7243.

LCLx = ¯x−A2R = (0.6733)− (0.729)(0.07)

or LCLx=0.6223. For the ranges, Eqs. (4.13)and (4.14) yield

UCLR = D4R = (2.282)(0.07) = 0.1600

LCRR = D3R = (0)(0.07) = 0

4.69 Calculate the mean, median and standard de-viation for all of the data in Problem 4.68.

The mean is given by Eq. (4.8) on p. 178 as

x =0.65 + 0.75 + 0.67 + . . .+ 0.71

24= 0.6733

The median is obtained by arranging the dataand finding the value that defines where 50% ofthe data is above that value. For the data inProblem 4.68, the median is between 0.67 and0.68, so it is reported as 0.675. The standarddeviation is given by Eq. (4.10) on p. 178 as

σ =

√(0.65− 0.6733)2 + . . .+ (0.71− 0.6733)2

23

which in this case is determined as σ = 0.0411.

4.70 The average of averages of a number of sam-ples of size 7 was determined to be 125. Theaverage range was 17.82, and the standard de-viation was 5.85. The following measurementswere taken in a sample: 120, 132, 124, 130, 118,132, 121, and 127. Is the process in control?

For a sample size of 7, we note from Table 4.3on p. 180 that A2 = 0.419, D4 = 1.924, andD3 = 0.078. Equations (4.11) and (4.12) givethe upper and lower control limits for the aver-ages as

UCLx = ¯x+A2R = (125)+(0.419)(17.82) = 132.46

LCLx = ¯x−A2R = (125)−(0.419)(17.82) = 117.53


UCLR = D4R = (1.924)(17.82) = 34.28

LCRR = D3R = (0.078)(17.82) = 1.390

For the sample shown, the average is x = 125.3and the range is R = 132 − 118 = 14. Theseare both within their respective control limits,therefore the process is in control. Note thatthe standard deviation is 5.85, and Eq. (4.14)on p. 180 allows an alternative method of cal-culation of the average range.



55






56






Chapter 5

Metal-Casting Processes andEquipment; Heat Treatment

Questions

5.1 Describe the characteristics of (1) an alloy, (2)pearlite, (3) austenite, (4) martensite, and (5)cementite.

(a) Alloy: composed of two or more elements,at least one element is a metal. The al-loy may be a solid solution or it may formintermetallic compounds.

(b) Pearlite: a two-phase aggregate consistingof alternating lamellae of ferrite and ce-mentite; the closer the pearlite spacing oflamellae, the harder the steel.

(c) Austenite: also called gamma iron, it hasa fcc crystal structure which allows for agreater solubility of carbon in the crys-tal lattice. This structure also possesses ahigh ductility, which increases the steel’sformability.

(d) Martensite: forms by quenching austen-ite. It has a bct (body-centered tetrag-onal) structure, and the carbon atoms ininterstitial positions impart high strength.It is hard and very brittle.

(e) Cementite: also known as iron-carbide(Fe3C), it is a hard and brittle intermetal-lic phase.

5.2 What are the effects of mold materials on fluidflow and heat transfer?

The most important factor is the thermal con-ductivity of the mold material; the higher theconductivity, the higher the heat transfer andthe greater the tendency for the fluid to solid-ify, hence possibly impeding the free flow of themolten metal. Also, the higher the cooling rateof the surfaces of the casting in contact withthe mold, the smaller the grain size and hencethe higher the strength. The type of surfacesdeveloped in the preparation of mold materi-als may also be different. For example, sand-mold surfaces are likely be rougher than thoseof metal molds whose surfaces can be preparedto varying degrees of roughness, including thedirections of roughness (lay).

5.3 How does the shape of graphite in cast iron af-fect its properties?

The shape of graphite in cast irons has the fol-lowing basic forms:

(a) Flakes. Graphite flakes have sharp edgeswhich act as stress raisers in tension.This shape makes cast iron low in tensilestrength and ductility, but it still has highcompressive strength. On the other hand,the flakes also act as vibration dampers,a characteristic important in damping ofmachine-tool bases and other structures.

(b) Nodules. Graphite can form nodules orspheroids when magnesium or cerium is

1






added to the melt. This form has in-creased ductility, strength, and shock re-sistance compared to flakes, but the damp-ing ability is reduced.

(c) Clusters. Graphite clusters are much likenodules, except that they form from thebreakdown of white cast iron upon anneal-ing. Clusters have properties that are ba-sically similar to flakes.

(d) Compacted flakes. These are short andthick flakes with rounded edges. This formhas properties that are between nodularand flake graphite.

5.4 Explain the difference between short and longfreezing ranges. How are they determined?Why are they important?

Freezing range is defined by Eq. (5.3) on p. 196in terms of temperature difference. Referringto Fig. 5.6 on p. 197, note that once the phasediagram and the composition is known, we candetermine the freezing range, TL − TS . As de-scribed in Section 5.3.2 on p. 196, the freezingrange has an important influence on the for-mation and size of the mushy zone, and, conse-quently, affects structure-property relationshipsof the casting.

5.5 We know that pouring molten metal at a highrate into a mold has certain disadvantages.Are there any disadvantages to pouring it veryslowly? Explain.

There are disadvantages to pouring metalslowly. Besides the additional time needed formold filling, the liquid metal may solidify orpartially solidify while still in the gating systemor before completely filling the mold, resultingin an incomplete or partial casting. This canhave extremely detrimental effects in a tree ofparts, as in investment casting.

5.6 Why does porosity have detrimental effects onthe mechanical properties of castings? Whichphysical properties are also affected adverselyby porosity?

Pores are, in effect, internal discontinuities thatare prone to cracking and crack propagation.Thus, the toughness of a material will decreaseas a result of porosity. Furthermore, the pres-ence of pores in a piece of metal under tension

indicates that the material around the pores hasto support a greater load than if no pores werepresent; thus, the strength is also lowered. Con-sidering thermal and electrical conductivity, aninternal defect such as a pore decreases both thethermal and electrical conductivity, nting thatair is a very poor conductor.

5.7 A spoked hand wheel is to be cast in gray iron.In order to prevent hot tearing of the spokes,would you insulate the spokes or chill them?Explain.

Referring to Table 5.1 on p. 206, we note that,during solidification, gray iron undergoes an ex-pansion of 2.5%. Although this fact may sug-gest that hot tearing cannot occur, considera-tion must also be given to significant contrac-tion of the spokes during cooling. Since the hot-tearing tendency will be reduced as the strengthincreases, it would thus be advisable to chill thespokes to develop this strength.

5.8 Which of the following considerations are im-portant for a riser to function properly? (1)Have a surface area larger than that of the partbeing cast. (2) Be kept open to atmosphericpressure. (3) Solidify first. Explain.

Both (1) and (3) would result in a situationcontrary to a riser’s purpose. That is, if a risersolidifies first, it cannot feed the mold cavity.However, concerning (2), an open riser has someadvantages over closed risers. Recognizing thatopen risers have the danger of solidifying first,they must be sized properly for proper func-tion. But if the riser is correctly sized so that itremains a reservoir of molten metal to accom-modate part shrinkage during solidification, anopen riser helps exhaust gases from the moldduring pouring, and can thereby eliminate someassociated defects. A so-called blind riser thatis not open to the atmosphere may cause pock-ets of air to be trapped, or increased dissolu-tion of air into the metal, leading to defects inthe cast part. For these reasons, the size andplacement of risers is one of the most difficultchallenges in designing molds.

5.9 Explain why the constant C in Eq. (5.9) de-pends on mold material, metal properties, andtemperature.

2






The constant C takes into account various fac-tors such as the thermal conductivity of themold material and external temperature. Forexample, Zircon sand has a higher thermal con-ductivity than basic silica sand, and as a result,a casting in Zircon (of equal volume and surfacearea) will require less time to solidify than thatcast in silica.

5.10 Explain why gray iron undergoes expansion,rather than contraction, during solidification.

As gray cast iron solidifies, a period ofgraphitization occurs during the final stages,which causes an expansion that counteracts theshrinkage of the metal. This results in an over-all expansion.

5.11 How can you tell whether a cavity in a castingare due to porosity or to shrinkage?

Evidence of which type of porosity is present,i.e., gas or shrinkage, can be gained by study-ing the location and shape of the cavity. If theporosity is near the mold surface, core surface,or chaplet surface, it is most likely to be gasporosity. However, if the porosity occurs in anarea considered to be a hot spot in the casting(see Fig. 5.37 on p. 249), it is most likely tobe shrinkage porosity. Furthermore, gas poros-ity has smooth surfaces (much like the holes inSwiss cheese) and is often, though not always,generally spherical in shape. Shrinkage poros-ity has a more textured and jagged surface, andis generally irregular in shape.

5.12 Explain the reasons for hot tearing in castings.

Hot tearing is a result of tensile stresses that de-velop upon contraction during solidification inmolds and cores if they are not sufficiently col-lapsible and/or do not allow movement underthe resulting pressure during shrinkage.

5.13 Would you be concerned about the fact thata portion of an internal chill is left within thecasting? What materials do you think chillsshould be made of, and why?

The fact that a part of the chill remains withinthe casting should be a consideration in the de-sign of parts to be cast. The following factorsare important:

(a) The material from which the chill is madeshould be compatible with the metal be-ing cast (it should have approximatelythe same composition of the metal beingpoured).

(b) The chill must be clean, that is, withoutany lubricant or coating on the surface,because any gas evolved when the moltenmetal contacts the chill may not readilyescape.

(c) The chill may not fuse with the casting,developing regions of weakness or stressconcentration. If these factors are under-stood and provided for, the fact that apiece of the chill remains within the cast-ing is generally of no significant concern.

5.14 Are external chills as effective as internal chills?Explain.

The effectiveness will depend on the location ofthe region to be chilled in the mold. If a regionneeds to be chilled (say, for example, to direc-tionally solidify a casting), an external chill canbe as effective as an internal chill. Often, how-ever, chilling is required at some depth beneaththe surface of a casting to be effective. For thiscondition an internal chill would be more effec-tive.

5.15 Do you think early formation of dendrites in amold can impede the free flow of molten metalinto the mold? Explain.

Consider the solidification of an alloy with avery long freezing range. The mushy zone forthis alloy will also be quite large (see Fig. 5.6).Since the mushy condition consists of interlac-ing dendrites surrounded by liquid, it is appar-ent that this condition will restrict fluid flow,as also confirmed in practice.

5.16 Is there any difference in the tendency forshrinkage void formation for metals with shortfreezing and long freezing ranges, respectively?Explain.

In an alloy with a long freezing range, the pres-ence of a large mushy zone is more likely tooccur, and thus the formation of miocroporos-ity. However, in an alloy with a short freezingrange, the formation of gross shrinkage voids ismore likely to occur.

3






5.17 It has long been observed by foundrymen thatlow pouring temperatures (that is, low super-heat) promote equiaxed grains over columnargrains. Also, equiaxed grains become finer asthe pouring temperature decreases. Explain thereasons for these phenomena.

Equiaxed grains are present in castings near themold wall where rapid cooling and solidificationtake place by heat transfer through the rela-tively cool mold. With low pouring tempera-ture, cooling to the solidification temperatureis faster because there is less heat stored in themolten metal. With a high pouring tempera-ture, cooling to the solidification temperatureis slower, especially away from the mold wall.The mold still dissipates heat, but the metal re-mains molten for a longer period of time, thusproducing columnar grains in the direction ofheat conduction. As the pouring temperature isdecreased, equiaxed grains should become finerbecause the cooling is more rapid and largegrains do not have time to form from the moltenmetal.

5.18 What are the reasons for the large variety ofcasting processes that have been developed overthe years?

By the student. There are several acceptableanswers depending on the interpretation of theproblem by the student. Students may ap-proach this as processes that are applicationdriven, material driven, or economics driven.For example, while investment casting is moreexpensive than sand casting, closer dimensionaltolerances and better surface finish are possi-ble. Thus, for certain parts such as barrelsfor handguns, investment casting is preferable.Consider also the differences between the hot-and cold-chamber permanent-mold casting op-erations.

5.19 Why can blind risers be smaller than open-toprisers?

Risers are used as reservoirs for a casting in re-gions where shrinkage is expected to occur, i.e,areas which are the last to solidify. Thus, risersmust be made large enough to ensure that theyare the last to solidify. If a riser solidifies beforethe cavity it is to feed, it is useless. As a result,an open riser in contact with air must be larger

to ensure that it will not solidify first. A blindriser is less prone to this phenomenon, as it isin contact with the mold on all surfaces; thus ablind riser may be made smaller.

5.20 Would you recommend preheating the molds inpermanent-mold casting? Also, would you re-move the casting soon after it has solidified?Explain.

Preheating the mold in permanent-mold castingis advisable in order to reduce the chilling effectof the metal mold which could lead to low metalfluidity and the problems that accompany thiscondition. Also, the molds are heated to re-duce thermal damage which may result fromrepeated contact with the molten metal. Con-sidering casting removal, the casting should beallowed to cool in the mold until there is nodanger of distortion or developing defects dur-ing shakeout. While this may be a very shortperiod of time for small castings, large castingsmay require an hour or more.

5.21 In a sand-casting operation, what factors deter-mine the time at which you would remove thecasting from the mold?

This question is an important one for any cast-ing operation, not just sand casting, becausea decrease in production time will result in adecrease in product cost. Therefore, a castingideally should be removed at the earliest pos-sible time. Factors which affect time are thethermal conductivity of the mold-material andof the cast metal, the thickness and the over-all size of the casting, and the temperature atwhich the metal is being poured.

5.22 Explain why the strength-to-weight ratio of die-cast parts increases with decreasing wall thick-ness.

Because the metal die acts as a heat sink for themolten metal, the metal chills rapidly, develop-ing a fine-grain hard skin with higher strength.As a result, the strength-to-weight ratio of die-cast parts increases with decreasing wall thick-ness.

5.23 We note that the ductility of some cast alloysis very low (see Fig. 5.13). Do you think thisshould be a significant concern in engineeringapplications of castings? Explain.

4






The low ductility of some cast alloys should cer-tainly be taken into consideration in the engi-neering application of the casting. The low duc-tility will:

(a) affect properties, such as toughness and fa-tigue,

(b) have a significant influence on furtherprocessing and finishing of the casting,i.e., machining processes, such as milling,drilling, and tapping, and

(c) possibly affect tribological behavior.

It should be noted that many engineering ap-plications do not require high ductility; for ex-ample, when stresses are sufficiently small toensure the material remains elastic and whereimpact loads do not occur.

5.24 The modulus of elasticity of gray iron variessignificantly with its type, such as the ASTMclass. Explain why.

Because the shape, size, and distribution ofthe second-phase (i.e., the graphite flakes) varygreatly for gray cast irons, there is a large cor-responding variation of properties attainable.The elastic modulus, for example, is one prop-erty which is affected by this factor.

5.25 List and explain the considerations involved inselecting pattern materials.

Pattern materials have a number of importantmaterial requirements. Often, they are ma-chined, thus good machinability is a require-ment. The material should be sufficiently stiffto allow good shape development. The materialmust have sufficient wear and corrosion resis-tance so that the pattern has a reasonable life.The economics of the operation is affected alsoby material cost.

5.26 Why is the investment-casting process capableof producing fine surface detail on castings?

The surface detail of the casting depends onthe quality of the pattern surface. In invest-ment casting, for example, the pattern is madeof wax or a thermoplastic poured or injectedinto a metal die with good surface finish. Con-sequently, surface detail of the casting is verygood and can be controlled. Furthermore, the

coating on the pattern (which then becomes themold) consists of very fine silica, thus contribut-ing to the fine surface detail of the cast product.

5.27 Explain why a casting may have a slightly dif-ferent shape than the pattern used to make themold.

After solidification, shrinkage continues untilthe casting cools to room temperature. Also,due to surface tension, the solidifying metalwill, when surface tension is high enough, notfully conform to sharp corners and other intri-cate surface features. Thus, the cast shape willgenerally be slightly different from that of thepattern used.

5.28 Explain why squeeze casting produces partswith better mechanical properties, dimensionalaccuracy, and surface finish than expendable-mold processes.

The squeeze-casting process consists of a com-bination of casting and forging. The pressureapplied to the molten metal by the punch, orupper die, keeps the entrapped gases in solu-tion, and thus porosity is generally not foundin these products. Also, the rapid heat transferresults in a fine microstructure with good me-chanical properties. Due to the applied pres-sure and the type of die used, i.e., metal, gooddimensional accuracy and surface finish are typ-ically found in squeeze-cast parts.

5.29 Why are steels more difficult to cast than castirons?

The primary reason steels are more difficult tocast than cast irons is that they melt at a highertemperature. The high temperatures compli-cate mold material selection, preparation, andtechniques involved for heating and pouring ofthe metal.

5.30 What would you recommend to improve thesurface finish in expendable-mold casting pro-cesses?

One method of improving the surface finish ofcastings is to use what is known as a facingsand, such as Zircon. This is a sand having bet-ter properties (such as permeability and surfacefinish) than bulk sand, but is generally moreexpensive. Thus, facing sand is used as a first

5






layer against the pattern, with the rest of themold being made of less expensive (silica) sand.

5.31 You have seen that even though die casting pro-duces thin parts, there is a limit to the mini-mum thickness. Why can’t even thinner partsbe made by this process?

Because of the high thermal conductivity thatmetal dies exhibit, there is a limiting thicknessbelow which the molten metal will solidify pre-maturely before filling the mold cavity. Also,the finite viscosities of the molten metal (whichincreases as it begins to cool) will require higherpressures to force the metal into the narrow pas-sages of the die cavities.

5.32 What differences, if any, would you expect inthe properties of castings made by permanent-mold vs. sand-casting methods?

As described in the text, permanent-moldcastings generally possess better surface fin-ish, closer tolerances, more uniform mechani-cal properties, and more sound thin-walled sec-tions than sand castings. However, sand cast-ings generally can have more intricate shapes,larger overall size, and lower in cost (dependingupon the alloy) than permanent-mold castings.

5.33 Which of the casting processes would be suit-able for making small toys in large numbers?Explain.

This is an open-ended problem, and the stu-dents should give a rationale for their choice.Refer also to Table 5.2 and note that die castingis one of the best processes for this application.The student should refer to the application re-quiring large production runs, so that toolingcost per casting can be low, the sizes possiblein die casting are suitable for such toys, and thedimensional tolerances and surface finish are ac-ceptable.

5.34 Why are allowances provided for in making pat-terns? What do they depend on?

Shrinkage allowances on patterns are correc-tions for the shrinkage that occurs upon solidifi-cation of the casting and its subsequent contrac-tion while cooling to room temperature. Theallowance will therefore depend on the amountof contraction an alloy undergoes.

5.35 Explain the difference in the importance ofdrafts in green-sand casting vs. permanent-mold casting.

Draft is provided to allow the removal of thepattern without damaging the mold. If themold material is sand and has no draft, themold cavity is likely to be damaged upon pat-tern removal, due to the low strength of thesand mold. However, a die made of high-strength steel, which is typical for permanent-mold castings, is not at all likely to be damagedduring the removal of the part; thus smallerdraft angles can be employed.

5.36 Make a list of the mold and die materials usedin the casting processes described in this chap-ter. Under each type of material, list the cast-ing processes that are used, and explain whythese processes are suitable for that particularmold or die material.

This is an open-ended problem, and studentsshould be encouraged to develop an answerbased on the contents of this chapter. An ex-ample of an acceptable answer would, in a briefform, be:

• Sand: Used because of its ability to resistvery high temperatures, availability, andlow cost. Used for sand, shell, expanded-pattern, investment, and ceramic-moldcasting processes.

• Metal: Such as steel or iron. Results inexcellent surface finish and good dimen-sional accuracy. Used for die, slush, pres-sure, centrifugal, and squeeze-casting pro-cesses.

• Graphite: Used for conditions similar tothose for metal molds; however, lowerpressures should be employed for this ma-terial. Used mainly in pressure- andcentrifugal-casting.

• Plaster of paris: Used in plaster-mold cast-ing for the production of relatively smallcomponents, such as fittings and valves.

5.37 Explain why carbon is so effective in impartingstrength to iron in the form of steel.

Carbon has an atomic radius that is about 57%of the iron atom, thus it occupies an interstitialposition in the iron unit cell (see Figs. 3.2 on

6






p. 84 and 3.9 on p. 90). However, because itsradius is greater than that of the largest holebetween the Fe atoms, it strains the lattice,thus interfering with dislocation movement andleading to strain hardening. Also, the size ofthe carbon atom allows it to have a high solu-bility in the fcc high-temperature phase of iron(austenite). At low temperatures, the structureis bcc and has a very low solubility of carbonatoms. On quenching, the austenitic structuretransforms to body-centered tetragonal (bct)martensite, which produces high distortion inthe crystal lattice. Because it is higher, thestrength increase is more than by other elementadditions.

5.38 Describe the engineering significance of the ex-istence of a eutectic point in phase diagrams.

The eutectic point corresponds to a compo-sition of an alloy that has a lowest meltingtemperature for that alloy system. The lowmelting temperature associated with a eutec-tic point can, for example, help in controllingthermal damage to parts during joining, as isdone in soldering. (See Section 12.13.3 startingon p. 776).

5.39 Explain the difference between hardness andhardenability.

Hardness represents the material’s resistance toplastic deformation when indented (see Section2.6 starting on p. 51), while hardenability isthe material’s capability to be hardened by heattreatment. (See also Section 5.11.1 starting onp. 236).

5.40 Explain why it may be desirable or necessary forcastings to be subjected to various heat treat-ments.

The morphology of grains in an as-cast struc-ture may not be desirable for commercial appli-cations. Thus, heat treatments, such as quench-ing and tempering (among others), are carriedout to optimize the grain structure of castings.In this manner, the mechanical properties canbe controlled and enhanced.

5.41 Describe the differences between case hardeningand through hardening insofar as engineeringapplications are concerned.

Case hardening is a treatment that hardensonly the surface layer of the part (see Table5.7 on p. 242). The bulk retains its toughness,which allows for blunting of surface cracks asthey propagate inward. Case hardening gener-ally induces compressive residual stresses on thesurface, thus retarding fatigue failure. Throughhardened parts have a high hardness across thewhole part; consequently, a crack could propa-gate easily through the cross section of the part,causing major failure.

5.42 Type metal is a bismuth alloy used to cast typefor printing. Explain why bismuth is ideal forthis process.

When one considers the use of type or for preci-sion castings such as mechanical typewriter im-pressions, one realizes that the type tool musthave extremely high precision and smooth sur-faces. A die casting using most metals wouldhave shrinkage that would result in the distor-tion of the type, or even the metal shrinkingaway from the mold wall. Since bismuth ex-pands during solidification, the molten metalcan actually expand to fill molds fully, therebyensuring accurate casting and repeatable type-faces.

5.43 Do you expect to see larger solidification shrink-age for a material with a bcc crystal structureor fcc? Explain.

The greater shrinkage would be expected fromthe material with the greater packing efficiencyor atomic packing factor (APF) in a solid state.Since the APF for fcc is 0.74 and for bcc it is0.68, one would expect a larger shrinkage fora material with a fcc structure. This can alsobeen seen from Fig. 3.2 on p. 84. Note, how-ever, that for an alloy, the answer is not as sim-ple, since it must be determined if the alloy-ing element can fit into interstitial positions orserves as a substitutional element.

5.44 Describe the drawbacks to having a riser thatis (a) too large, or (b) too small.

The main drawbacks to having a riser too largeare: the material in the riser is eventuallyscrapped and has to be recycled; the riser hassto be cut off, and a larger riser will cost more

7






to machine; an excessively large riser slows so-lidification; the riser may interfere with solidifi-cation elsewhere in the casting; the extra metalmay cause buoyancy forces sufficient to sepa-rate the mold halves, unless they are properlyweighted or clamped. The drawbacks to hav-ing too small a riser are mainly associated withdefects in the casting, either due to insufficientfeeding of liquid to compensate for solidifica-tion shrinkage, or shrinkage pores because thesolidification front is not uniform.

5.45 If you were to incorporate lettering on a sandcasting, would you make the letters protrudefrom the surface or recess into the surface?What if the part were to be made by invest-ment casting?

In sand casting, where a pattern must be pre-pared and used, it is easier to produce lettersand numbers by machining them into the sur-face of a pattern; thus the pattern will have re-cessed letters. The sand mold will then haveprotruding letters, as long as the pattern isfaithfully reproduced. The final part will thenhave recessed letters.

In investment casting, the patterns are pro-duced through injection molding. It is easier toinclude recessed lettering in the injection mold-ing die (instead of machining protruding let-ters). Thus, the mold will have recessed let-ters and the pattern will have protruding let-ters. Since the pattern is a replica of the finalpart, the part will also have protruding letters.

In summary, it is generally easier to produce re-cessed letters in sand castings and protrudingletters in investment casting.

5.46 List and briefly explain the three mechanismsby which metals shrink during casting.

Metals shrink by:

(a) Thermal contraction in the liquid phasefrom superheat temperature to solidifica-tion temperature,

(b) Solidification shrinkage, and

(c) Thermal contraction in the solid phasefrom the solidification temperature toroom temperature.

5.47 Explain the significance of the “tree” in invest-ment casting.

The tree is important because it allows simul-taneous casting of several parts. Since signifi-cant labor is involved in the production of eachmold, this strategy of increasing the number ofparts that are poured per mold is critical to theeconomics of investment casting.

5.48 Sketch the microstructure you would expect fora slab cast through (a) continuous casting, (b)strip casting, and (c) melt spinning.

The microstructures are as follows:

Continuouscast

Strip cast Melt spunP

roce

ssin

gd

irect

ion

Note that the continuous cast structure showsthe columnar grains growing away from themold wall. The strip-cast metal has been hotrolled immediately after solidification, and isshown as quenched, prior it is annealed to ob-tain an equiaxed structure. The melt-spunstructure solidifies so rapidly that there are noclear grains (an amorphous metal).

5.49 The general design recommendations for a wellin sand casting are that (a) its diameter shouldbe twice the sprue exit diameter, and (b) thedepth should be approximately twice the depthof the runner. Explain the consequences of de-viating from these rules.

Refer to Figure 5.10 for terminology used in thisproblem.

(a) Regarding this rule, if the well diameter ismuch larger than twice the exit diameter,liquid will not fill the well and aspiration ofthe molten metal may result. On the otherhand, if the diameter is small compared tothe sprue exit diameter, and recognizingthat wells are generally not tapered, thenthere is a fear of aspiration within the well(see the discussion of sprue profile in Sec-tion 5.4 starting on p. 199.

8






(b) If the well is not deeper than the runner,then turbulent metal first splashed intothe well is immediately fed into the cast-ing, leading to aspiration and associateddefects. If the well is much deeper, thenthe metal remains in the well and can so-lidify prematurely.

5.50 Describe the characteristics of thixocasting andrheocasting.

Thixocasting and rheocasting involve castingoperations where the alloy is in the slushy stage.Often, ultrasonic vibrations will be used to en-sure that dendrites remain in solution, so thatthe metal is a slurry of molten continuous phaseand suspended particles. In such casting op-erations, the molten metal has a lower super-heat and, therefore, requires less cycle time,and shrinkage defects and porosity can be de-creased. This is further described in Section5.10.6 starting on p. 233.

5.51 Sketch the temperature profile you would ex-pect for (a) continuous casting of a billet, (b)sand casting of a cube, (c) centrifugal castingof a pipe.

This would be an interesting finite-element as-signment if such software is made availableto the students. Consider continuous casting.The liquid portion has essentially a constanttemperature, as there is significant stirring ofthe liquid through the continuous addition ofmolten metal. The die walls extract heat, andthe coolant spray at the die exterior removesheat even more aggressively. Thus, a sketch ofthe isotherms in continuous casting would be asfollows:

MoldMolten metal

Isotherms

Conductionboundary

Convectionboundary

The cube and the pipe are left to be completedby the student.

5.52 What are the benefits and drawbacks to hav-ing a pouring temperature that is much higherthan the metal’s melting temperature? Whatare the advantages and disadvantages in hav-ing the pouring temperature remain close to themelting temperature?

If the pouring temperature is much higher thanthat of the mold temperature, there is less dan-ger that the metal will solidify in the mold,and it is likely that even intricate molds canbe fully filled. This situation makes runners,gates, wells, etc., easier to design because theircross sections are less critical for complete moldfilling. The main drawback is that there isan increased likelihood of shrinkage pores, coldshuts, and other defects associated with shrink-age. Also there is an increased likelihood of en-trained air since the viscosity of the metal willbe lower at the higher pouring temperature. Ifthe pouring temperature is close to the melt-ing temperature, there will be less likelihoodof shrinkage porosity and entrained air. How-ever, there is the danger of the molten metalsolidifying in a runner before the mold cavityis completely filled; this may be overcome withhigher injection pressures, but clearly has a costimplication.

5.53 What are the benefits and drawbacks to heatingthe mold in investment casting before pouringin the molten metal?

Heating the mold in investment casting is advis-able in order to reduce the chilling effect of themold, which otherwise could lead to low metalfluidity and the problems that accompany thiscondition. Molds are usually preheated to someextent. However, excessive heating will com-promise the strength of the mold, resulting inerosion and associated defects.

5.54 Can a chaplet also act as a chill? Explain.

A chaplet is used to position a core. It has ageometry that can either rest against a moldface or it can be inserted into a mold face. Ifthe chaplet is a thermally conductive material,it can also serve as a chill.

9






5.55 Rank the casting processes described in thischapter in terms of their solidification rate.For example, which processes extract heat thefastest from a given volume of metal and whichis the slowest?

There is, as expected, some overlap between thevarious processes, and the rate of heat transfercan be modified whenever desired. However, ageneral ranking in terms of rate of heat extrac-tion is as follows: Die casting (cold chamber),squeeze casting, centrifugal casting, slush cast-

ing, die casting (hot chamber), permanent moldcasting, shell mold casting, investment casting,sand casting, lost foam, ceramic-mold casting,and plaster-mold casting.

5.56 The heavy regions of parts typically are placedin the drag in sand casting and not in the cope.Explain why.

A simple explanation is that if they were to beplaced in the cope, they would develop a highbuoyancy force that would tend to separate themold and thus develop flashes on the casting.

Problems

5.57 Referring to Fig. 5.3, estimate the followingquantities for a 20% Cu-80% Ni alloy: (1) liq-uidus temperature, (2) solidus temperature, (3)percentage of nickel in the liquid at 1400C(2550F), (4) the major phase at 1400C, and(5) the ratio of solid to liquid at 1400C.

We estimate the following quantities from Fig.5.3 on p. 192: (1) The liquidus temperature is1400C (2550F). (2) The solidus temperatureis 1372C (2500F). (3) At 2550F, the alloy isstill all liquid, thus the nickel concentration is80%. (4) The major phase at 1400C is liquid,with no solids present since the alloy is not be-low the liquidus temperature. (5) The ratio iszero, since no solid is present.

5.58 Determine the amount of gamma and alphaphases (see Fig. 5.4b) in a 10-kg, AISI 1060steel casting as it is being cooled to the follow-ing temperatures: (1) 750C, (2) 728C, and(3) 726C.

We determine the following quantities fromFig. 5.6 on p. 197: (a) At 750C, the alloy isjust in the single-phase austenite (gamma) re-gion, thus the percent gamma is 100% (10 kg),and alpha is 0%. (b) At 728C, the alloy isin the two-phase gamma-alpha field, and theweight percentages of each is found by the lever

rule (see Example 5.1):

%α =(xγ − xo

xγ − xα

)× 100%

=(

0.77− 0.600.77− 0.022

)× 100%

= 23% or 2.3 kg

%γ =(xo − xα

xγ − xα

)× 100%

=(

0.60− 0.0220.77− 0.022

)× 100%

= 77% or 7.7 kg

(c) At 726C, the alloy is in the two-phase al-pha and Fe3C field. No gamma phase is present.Again the lever rule is used to find the amountof alpha present:

%α =(

6.67− 0.606.67− 0.022

)×100% = 91% or 9.1 kg

5.59 A round casting is 0.3 m in diameter and 0.5 min length. Another casting of the same metalis elliptical in cross section, with a major-to-minor axis ratio of 3, and has the same lengthand cross sectional area as the round casting.Both pieces are cast under the same conditions.What is the difference in the solidification timesof the two castings?

10






For the same length and cross sectional area(thus the same volume), and the same castingconditions, the same C value in Eq. (5.11) onp. 205 on p. 205 should be applicable. The sur-face area and volume of the round casting is

Around = 2πrl + 2πr2 = 0.613 m2

Vround = πr2l = 0.0353 in2

Since the cross-sectional area of the ellipse isthe same as that for the cylinder, and it has amajor and minor diameter of a and b, respec-tively, where a = 3b, then

πab = πr2

3b2 = r2 → b =

√(0.15)2

3or b = 0.0866 m, so that a = 0.260 m. The sur-face area of the ellipse-based part is (see a ba-sic geometry text for the area equation deriva-tions):

Aellipse = 2πab+ 2π√a2 + b2l = 1.002 m2

The volume is still 0.0353 in2. According toEq. (5.11) on p. 205, we thus have

Tround

Tellipse=

(V/Around)2

(V/Aellipse)2=(Aellipse

Around

)2

= 2.67

5.60 Derive Eq. (5.7).

We note that Eq. (5.5) on p. 200 gives a rela-tionship between height, h, and velocity, v, andEq. (5.6) on p. 201 gives a relationship betweenheight, h, and cross sectional area, A. With thereference plate at the top of the pouring basin(and denoted as subscript 0), the sprue top isdenoted as 1, and the bottom as 2. Note thath2 is numerically greater than h1. At the topof the sprue we have vo = 0 and ho = 0. As afirst approximation, assume that the pressurespo, p1 and p2 are equal and that the frictionalloss f is negligible. Thus, from Eq. (5.5) wehave

ho +po

ρg+v2

o

2g= h1 +

p1

ρg+v21

2g+ f

or, solving for v1,

0 = h1 +v21

2g→ v1 =

√2gh1

Similarly,

ho +po

ρg+v2

o

2g= h2 +

p2

ρg+v22

2g+ f

orv2 =

√2gh2

Substituting these results into the continuityequation given by Eq. (5.6), we have

A1v1 = A2v2

A1

√2gh1 = A2

√2gh2

A1

A2=√

2gh2√2gh1

=√h2

h1

which is the desired relationship.

5.61 Two halves of a mold (cope and drag) areweighted down to keep them from separat-ing due to the pressure exerted by the moltenmetal (buoyancy). Consider a solid, spheri-cal steel casting, 9 in. in diameter, that is be-ing produced by sand casting. Each flask (seeFig. 5.10) is 20 in. by 20 in. and 15 in. deep. Theparting line is at the middle of the part. Esti-mate the clamping force required. Assume thatthe molten metal has a density of 500 lb/ft3 andthat the sand has a density of 100 lb/ft3,

The force exerted by the molten metal is theproduct of its cross-sectional area at the partingline and the pressure of the molten metal due tothe height of the sprue. Assume that the spruehas the same height as the cope, namely, 15 in.The pressure of the molten metal is the prod-uct of height and density. Assuming a densityfor the molten metal of 500 lb/ft3, the pres-sure at the parting line will be (500)(15/12) =625 lb/ft2, or 4.34 psi. The buoyancy force isthe product of projected area and pressure, or(625)(π)(9/12)2 = 1100 lb. The net volume ofthe sand in each flask is

V = (20)(20)(15)− (0.5)(

4π3

)(9)3

or V = 4473 in3 = 2.59 ft3. For a sand den-sity of 100 lb/ft3, the cope weighs 454 lb. Un-der these circumstances, a clamping force of1100− 259 ≈ 850 lb is required.

11






5.62 Would the position of the parting line in Prob-lem 5.61 influence your answer? Explain.

The position of the parting line does have an in-fluence on the answer to Problem 5.54, because(a) the projected area of the molten metal willbe different and (b) the weight of the cope willalso be different.

5.63 Plot the clamping force in Problem 5.61 as afunction of increasing diameter of the casting,from 10 in. to 20 in.

Note in this problem that as the diameter ofthe casting increases, the cross-sectional areaof the molten metal increases, hence the buoy-ancy force also increases. At the same time,the weight of the cope decreases because of thelarger space taken up by the molten metal. Us-ing the same approach as in Problem 5.61, theweight of the casting as a function of diameteris given by

Fb = ρV = (500)(π

6d3)

=(261 lb/ft3

)d3

The volume of sand in the cope is given by:

V = (20)(20)(15)(

112

)3

− (0.5)(π

6

)d3

= 3.47 ft3 − 0.261d3

Therefore, the weight of the sand is given by:

Fw = ρsandV

=(100 lb/ft3

) (3.47 ft3 − 0.261d3

)= 347 lb−

(26.1 lb/ft3

)d3

The required clamping force is given by equilib-rium as

Fc = Fb − Fw

=(261 lb/ft3

)d3 − 347 lb

+(26.1 lb/ft3

)d3

=(287 lb/ft3

)d3 − 347 lb

This equation is plotted below. Note that for asmall diameter, no clamping force is needed, asthe weight of the cope is sufficient to hold thecope and drag together.

0 5 10 15 20

Casting diameter, in.

Cla

mp

forc

e, lb

-500

0

500

1000

1500

5.64 Sketch a graph of specific volume vs. temper-ature for a metal that shrinks as it cools fromthe liquid state to room temperature. On thegraph, mark the area where shrinkage is com-pensated for by risers.

The graph is as follows. See also Fig. 5.1b onp. 189.

Sp

ecifi

c d

ensi

ty

Time

Shrinkage of liquid

Shrinkage of solid

Shrinkage compensated

by riser

Shrinkage compensated bypatternmaker's rule

Solidificationshrinkage

5.65 A round casting has the same dimensions asin Problem 5.59. Another casting of the samemetal is rectangular in cross-section, with awidth-to-thickness ratio of 3, and has the samelength and cross-sectional area as the roundcasting. Both pieces are cast under the sameconditions. What is the difference in the solid-ification times of the two castings?

The castings have the same length and cross-sectional area (thus the same volume) and thesame casting conditions, hence the same Cvalue. The total surface area of the round cast-ing, with l = 500 mm and r = 150 mm, is

Around = 2πrl + 2πr2

= 2π(150)(500) + 2π(150)2

= 6.13× 105 mm2

12






The cross-sectional area of the round casting isπr2 = π(150)2 = 70, 680 mm2. The rectangularcross section has sides x and 3x, so that

70, 680 = 3x2 → x = 153 mm

hence the perimeter of the rectangular castingwith the same cross-sectional area and axes ra-tio of 3 is 1228 mm. The total surface area is

Arect = 2(70, 680) + (1228)(500)

or Arect = 7.55 × 105 mm2. According toChvorinov’s rule, cooling time for a constantvolume is inversely proportional to surface areasquared. Therefore,

trecttround

=(

6.13× 105

7.55× 105

)2

= 0.66

5.66 A 75-mm thick square plate and a right circu-lar cylinder with a radius of 100 mm and heightof 50 mm each have the same volume. If eachis to be cast using a cylindrical riser, will eachpart require the same size riser to ensure properfeeding of the molten metal? Explain.

Recall that it is important for the riser to so-lidify after the casting has solidified. A castingthat solidifies rapidly would most likely requirea smaller riser than one which solidifies over alonger period of time. Lets now calculate therelative solidification times, using Chvorinov’srule given by Eq. (5.11) on p. 205 on p. 205.For the cylindrical part, we have

Vcylinder = πr2h = π(0.1 m)2(0.050 m)

or Vcylinder = 0.00157 m3. The surface area ofthe cylinder is

Acylinder = 2πr2 + 2πrh= 2π(0.1)2 + 2π(0.1)(0.05)= 0.0942 m2

Thus, from Eq. (5.11) on p. 205 on p. 205,

tcylinder = C

(0.001570.0942

)2

=(2.78× 10−4

)C

For a square plate with sides L and heighth = 0.075 m, and the same volume as the cylin-der, we have

Vplate = 0.00157 m3 = L2h = L2(0.075 m)

Solving for L yields L = 0.144 m. Therefore,

Aplate = 2L2 + 4Lh= 2(0.144)2 + 4(0.144)(0.075)

or Aplate = 0.0847 m2. From Eq. (5.11) onp. 205 on p. 205,

tplate = C

(0.001570.0847

)2

=(3.43× 10−4

)C

Therefore, the cylindrical casting will takelonger to solidify and will thus require a largerriser.

5.67 Assume that the top of a round sprue has a di-ameter of 4 in. and is at a height of 12 in. fromthe runner. Based on Eq. (5.7), plot the pro-file of the sprue diameter as a function of itsheight. Assume that the sprue has a diameterof 1 in. at its bottom.

From Eq. (5.7) on p. 201 and substituting forthe area, it can be shown that

d21

d2=√

h

h1

Therefore,

d =

√d21

√h1

h→ d = Ch−0.25

The difficulty here is that the reference locationfor height measurements is not known. Oftenchokes or wells are used to control flow, but thisproblem will be solved assuming that properflow is to be attained by considering hydrody-namics in the design of the sprue. The bound-ary conditions are that at h = ho, d = 4 (whereho is the height at the top of the sprue from thereference location) and at h = ho +12 in., d = 1in. The first boundary condition yields

4 = C(ho)−0.25 or C = 4h0.25o

The second boundary condition yields

1 = C(ho + 12)−0.25 =(4h0.25

o

)(ho + 12)−0.25

This equation is solved as ho = 0.047 in., sothat C = 1.863. These values are substitutedinto the expression above to obtain

d = 1.863(h+ 0.047)−0.25

13






Note that ho is the location of the bottom ofthe sprue and that the sprue is axisymmetric.The sprue shape, based on this curve, is shownbelow.

4 in

1 in

12 in

5.68 Estimate the clamping force for a diecastingmachine in which the casting is rectangular,with projected dimensions of 75 mm x 150 mm.Would your answer depend on whether or notit is a hot-chamber or cold-chamber process?Explain.

The clamping force is needed to compensate forthe separating force developed when the metalis injected into the die. When the die is full,and the full pressure is developed, the separat-ing force is F = pA, where p is the pressure andA is the projected area of the casting. Note thatthe answer will depend on whether the opera-tion is hot- or cold-chamber, because pressuresare higher in the cold-chamber than in the hot-chamber process.

The projected area is 11,250 mm2. In the hot-chamber process, an average pressure is takenas 15 MPa (see Section 5.10.3), although thepressure can range up to 35 MPa. If we use anaverage pressure, the required clamping force is

Fhot = pA = (35)(11, 250) = 394 kN

For the cold-chamber process and using a mid-range pressure of 45 MPa, the force will be

Fcold = pA = (45)(11, 250) = 506 kN

5.69 When designing patterns for casting, pattern-makers use special rulers that automatically in-corporate solid shrinkage allowances into theirdesigns. Therefore, a 12-in. patternmaker’sruler is longer than a foot. How long should apatternmaker’s ruler be for the making of pat-terns for (1) aluminum castings (2) malleablecast iron and (3) high-manganese steel?

It was stated in Section 5.12.2 on p. 248 thattypical shrinkage allowances for metals are 1

8to 1

4 in./ft, so it is expected that the ruler bearound 12.125-12.25 in. long. Specific shrink-age allowances for these metals can be obtainedfrom the technical literature or the internet.For example, from Kalpakjian, ManufacturingProcesses for Engineering Materials, 3rd ed.,p. 280, we obtain the following:

ShrinkageMetal allowance, %Aluminum 1.3Malleable cast Iron 0.89High-manganese steel 2.6

From the following formula,

Lruler = Lo(1 + shrinkage)

We find that for aluminum,

LAl = (12)(1.013) = 12.156 in.

For malleable cast iron,

Liron = (12)(1.0089) = 12.107 in.

and for high-manganese steel,

Lsteel = (12)(1.026) = 12.312 in.

Note that high-manganese steel and malleablecast iron were selected for this problem becausethey have extremely high and low shrinkageallowances, respectively. The aluminum rulerfalls within the expected range, as do mostother metals.

5.70 The blank for the spool shown in the accompa-nying figure is to be sand cast out of A-319, analuminum casting alloy. Make a sketch of thewooden pattern for this part. Include all nec-essary allowances for shrinkage and machining.

14






0.50 in.

0.45 in.

4.00 in.

3.00 in.

The sketch for a typical green-sand casting pat-tern for the spool is shown below. A cross-sectional view is also provided to clearly in-dicate shrinkage and machining allowances, aswell as the draft angles. The important ele-ments of this pattern are as follows (dimensionsin inches):

(a) Two-piece pattern.(b) Locating pins will be needed in the pattern

plate to ensure that these features alignproperly.

(c) Shrinkage allowance = 5/32 in./ft.(d) Machining allowance = 1/16 in.(e) Draft = 3.

1.50 in.4.58 in.

3° (typical)

5.71 Repeat Problem 5.70, but assume that the alu-minum spool is to be cast using expendable-pattern casting. Explain the important differ-ences between the two patterns.

A sketch for a typical expandable-pattern cast-ing is shown below. A cross-sectional view isalso provided to clearly show the differencesbetween green-sand (from Problem 5.70) andevaporative-casting patterns. There will besome variations in the patterns produced bystudents depending on which dimensions are as-signed a machining allowance. The importantelements of this pattern are as follows (dimen-sions in inches):

(a) One-piece pattern, made of polystyrene.

(b) Shrinkage allowance = 5/32 in./ft

(c) Machining allowance = 1/16 in.

(d) No draft angles are necessary.

0.56 in.

0.52 in.

3.04

in.

4.05 in.

5.72 In sand casting, it is important that the copemold half be held down with sufficient force tokeep it from floating when the molten metalis poured in. For the casting shown in theaccompanying figure, calculate the minimumamount of weight necessary to keep the copefrom floating up as the molten metal is pouredin. (Hint: The buoyancy force exerted by themolten metal on the cope is related to the effec-tive height of the metal head above the cope.)

15






A A

Section A-A

3.00

2.00 2.50

1.00 0.50

2.00 1.00

1.00

2.50

4.00

5.00

R = 0.75

Material: Low-carbon steel Density: 0.26 lb/in3

All dimensions in inches

3.00

The cope mold half must be sufficiently heavyor be weighted sufficiently to keep it from float-ing when the molten metal is poured into themold. The buoyancy force, F , on the cope is ex-erted by the metallostatic pressure (caused bythe metal in the cope above the parting line)and can be calculated from the formula

F = pA

where p is the pressure at the parting line andA is the projected area of the mold cavity. Thepressure is

p = wh = (0.26 lb/in3)(3.00 in.) = 0.78 psi

The projected mold-cavity area can be calcu-lated from the dimensions given on cross sectionAA in the problem, and is found to be 10.13 in2.Thus, the force is

F = (0.78)(10.13) = 7.9 lb

5.73 The optimum shape of a riser is spherical toensure that it cools more slowly than the cast-ing it feeds. Spherically shaped risers, however,are difficult to cast. (1) Sketch the shape of ablind riser that is easy to mold, but also hasthe smallest possible surface area-to-volume ra-tio. (2) Compare the solidification time of theriser in part (a) to that of a riser shaped like aright circular cylinder. Assume that the volumeof each riser is the same, and that for each theheight is equal to the diameter (see Example5.2).

A sketch of a blind riser that is easy to cast isshown below, consisting of a cylindrical and ahemispherical portions.

Hemisphere

h=r

r

Note that the height of the cylindrical portionis equal to its radius (so that the total height ofthe riser is equal to its diameter). The volume,V , of this riser is

V = πr2h+12

(4πr3

3

)=

5πr3

3

Letting V be unity, we have

r =(

35π

)1/3

The surface area of the riser is

A = 2πrh+ πr2 +12(4πr2

)= 5πr2

Substituting for r, we obtain A = 5.21. There-fore, from Eq. (5.11) on p. 205, the solidificationtime, t, for the blind riser will be

t = C

(V

A

)2

= C

(1

5.21

)2

= 0.037C

16






From Example 5.2, we know that the solidifica-tion time for a cylinder with a height equal toits diameter is 0.033C. Thus, this blind riserwill cool a little slower, but not much so, and iseasier to cast.

5.74 The part shown in the accompanying figureis a hemispherical shell used as an acetabular(mushroom shaped) cup in a total hip replace-ment. Select a casting process for this partand provide a sketch of all patterns or toolingneeded if it is to be produced from a cobalt-chrome alloy.

R = 28

557

3

2520

Dimensions in mm

This is an industrially-relevant problem, as thisis the casting used as acetabular cups for totalhip replacements. There are several possibleanswers to this question, depending on the stu-dent’s estimates of production rate and equip-ment costs. In practice, this part is producedthrough an investment casting operation, wherethe individual parts with runners are injectionmolded and then attached to a central sprue.The tooling that would be required include: (1)a mold for the injection molding of wax into thecup shape. (2) Templates for placement of thecup shape onto the sprue, in order to assureproper spacing for even, controlled cooling. (3)Machining fixtures. It should be noted that thewax pattern will be larger than the desired cast-ing, because of shrinkage as well as the incor-poration of a shrinkage allowance.

5.75 A cylinder with a height-to-diameter ratio ofunity solidifies in four minutes in a sand cast-ing operation. What is the solidification timeif the cylinder height is doubled? What is thetime if the diameter is doubled?

From Chvorinov’s rule, given by Eq. (5.11) on

p. 205, and assuming n = 2 gives

t = C

(V

A

)2

= C

[ (πd2h/4

)(πd2/2 + πdh)

]

= C

(dh

2d+ 4h

)= 4 min

Solving for C,

C = (4 min)(

2d+ 4hdh

)If the height is doubled, then we can use d2 = dand h2 = 2h to obtain

t = C

(d2h2

2d2 + 4h2

)= (4 min)

(2d+ 4hdh

)(d(2h)

2d+ 4(2h)

)= (4 min)

(4d+ 8h2d+ 8h

)If d = h, then

t = (4 min)(

12h10h

)= 4.8 min

If the diameter is doubled, so that d3 = 2d andh3 = h, then

t = C

(d3h3

2d3 + 4h3

)= (4 min)

(2d+ 4hdh

)((2d)(h)

2(2d) + 4(h)

)= (4 min)

(4d+ 8h4d+ 4h

)or, for d = h,

t = (4 min)(

12h8h

)= 6 min

5.76 Steel piping is to be produced by centrifugalcasting. The length is 12 feet, the diameter is 3ft, and the thickness is 0.5 in. Using basic equa-tions from dynamics and statics, determine therotational speed needed to have the centripetalforce be 70 times its weight.

The centripetal force can be obtained from anundergraduate dynamics textbook as

Fc =mv2

r

17






where m is the mass, v is the tangential veloc-ity, and r is the radius. It is desired to havethis force to be 70 times its weight, or

70 =Fc

W=mv2/r

mg=v2

rg

since r is the mean radius of the casting, or 1.25ft, v can be solved as

v =√

(70)rg =√

(70)(1.25)(32.2)

or v = 53 ft/sec or 637 in./sec. The rotationalspeed needed to obtain this velocity is

ω =v

r=

53 ft/sec1.25 ft

= 42.4 rad/sec

This is equivalent to 405 rev/min.

5.77 A sprue is 12 in. long and has a diameter of 5in. at the top, where the metal is poured. Themolten metal level in the pouring basin is takenas 3 in. from the top of the sprue for design pur-poses. If a flow rate of 40 in3/s is to be achieved,what should be the diameter of the bottom ofthe sprue? Will the sprue aspirate? Explain.

Assuming the flow is frictionless, the velocityof the molten metal at the bottom of the sprue(h = 12 in. = 1 ft) is

v =√

2gh =√

2(32.2)(1)

or v = 8.0 ft/s = 96 in./s. For a flow rate of 40in3/s, the area needs to be

A =Q

v=

40 in3/s96 in./s

= 0.417 in2

For a circular runner, the diameter would thenbe 0.73 in., or roughly 3

4 in. Compare this tothe diameter at the bottom of the sprue basedon Eq. (5.7), where h1 = 3 in., h2 = 15 in., andA1 = 19.6 in2. The diameter at the bottom ofthe sprue is calculated from:

A1

A2=√h2

h1

A2 =A1√h2/h1

=19.6√15/3

= 8.8 in2

d =

√4πA2 = 3.34 in

Thus, the sprue confines the flow more than isnecessary, and it will not aspirate.

5.78 Small amounts of slag often persist after skim-ming and are introduced into the molten metalflow in casting. Recognizing that the slag ismuch less dense than the metal, design moldfeatures that will remove small amounts of slagbefore the metal reaches the mold cavity.

There are several dross-trap designs in use infoundries. (A good discussion of trap designis given in J. Campbell, Castings, 1991, ReedEducational Publishers, pp. 53-55.) A conven-tional and effective dross trap is illustrated be-low:

It designed on the principle that a trap at theend of a runner will take the metal through therunner and keep it away from the gates. Thedesign shown is a wedge-type trap. Metal en-tering the runner contacts the wedge, and theleading front of the metal wave is chilled andattaches itself to the runner wall, and thus it iskept out of the mold cavity. The wedge must bedesigned so as to avoid reflected waves that oth-erwise would recirculate the dross or the slag.

The following design is a swirl trap:

Top view

Inlet Outlet

Swirlchamber

Inlet

Outlet

DrossMolten metal

Side view

This is based on the principle that the dross orslag is less dense than the metal. The metal en-ters the trap off of the center, inducing a swirl inthe molten metal as the trap fills with moltenmetal. Because it is far less dense than themetal, the dross or slag remains in the center of

18






the swirl trap. Since the metal is tapped fromthe outside periphery, dross or slag is excludedfrom the casting.

5.79 Pure aluminum is being poured into a sandmold. The metal level in the pouring basin is10 in. above the metal level in the mold, andthe runner is circular with a 0.4 in. diameter.What is the velocity and rate of the flow of themetal into the mold? Is the flow turbulent orlaminar?

Equation (5.5) on p. 200 gives the metal flow.Assuming the pressure does not change appre-ciably in the channel and that there is no fric-tion in the sprue, the flow is

h1 +v21

2g= h2 +

v22

2g

Where the subscript 1 indicates the top of thesprue and 2 the bottom. If we assume that thevelocity at the top of the sprue is very low (aswould occur with the normal case of a pouringbasin on top of the sprue with a large cross-sectional area), then v1 = 0. The velocity atthe bottom of the sprue is

v22 = 2g(h1 − h2)

or

v2 =√

2g∆h =√

2(32.2 ft/s2)(12 in/ft)(10 in)

or v2 = 87.9 in./s. If the opening is 0.4-in. indiameter, the area is

A =π

4d2 =

π

4(0.4)2 = 0.126 in2

Therefore, the flow rate is

Q = v2A = (87.9)(0.126) = 11.0 in3/s.

Pure aluminum has a density of 2700 kg/m3

(see Table 3.3) and a viscosity of around 0.0015Ns/m2 around 700C. The Reynolds number,from Eq. (5.10) on p. 202, is then (using v =87.9 in/s = 2.23 m/s and D = 0.4 in.=0.01016m),

Re =vDρ

η

=(2.23 m/s)(0.01016 m)(2700 kg/m3)

0.0015 Ns/m2

or Re= 40,782. As discussed in Section 5.4.1starting on p. 199, this situation would rep-resent turbulence, and the velocity and/or di-ameter should be decreased to bring Re below20,000 or so.

5.80 For the sprue described in Problem 5.79, whatrunner diameter is needed to ensure a Reynoldsnumber of 2000? How long will a 20 in3 castingtake to fill with such a runner?

Using the data given in Problem 5.79, aReynolds number of 2000 can be achieved byreducing the channel diameter, so that

Re = 2000 =vDρ

η=

(2.23)(2700)0.0015

D

or D = 0.000498 m = 0.0196 in.

For this diameter, the initial flow rate would be

Q = v2A =πv24D2 =

π

4(87.9)(0.0196)2

= 0.0266 in3/sec

This means that a 20 in3 casting would take 753s (about 12 min) to fill and only if the initialflow rate could be maintained, which is gener-ally not the case. Such a long filling time isnot acceptable, since it is likely that metal willsolidify in runners and thus not fill the moldcompletely. Also, with such small a small run-ner, additional mechanisms need to be consid-ered. For example, surface tension and fric-tion would severely reduce the velocity in theReynolds number calculation above.

This is generally the case with castings; to de-sign a sprue and runner system that maintainslaminar flow in the fluid would result in exces-sively long fill times.

5.81 How long would it take for the sprue in Problem5.79 to feed a casting with a square cross-sectionof 6 in. per side and a height of 4 in.? Assumethe sprue is frictionless.

Note that the volume of the casting is 144in3, with a constant cross-sectional area of 36in2. The velocity will change as the mold fills,because the pouring basin height above themolten metal will decrease. The velocity willvary according to:

v = c√

2gh =√

2gh

19






The flow rate is given by

Q = vA =vπd2

4=πd2

√2gh

4

The mold cavity fills at a rate of Q/(36 in2), or

dh

dt=Q

A= −πd

2√

2gh4A

where the minus sign has been added so thath refers to the height difference between themetal level in the mold and the runner, whichdecreases with respect to time. Separating thevariables,

dh√h

= −πd2√

2g4A

dt

Integrating,(2√h)6in

10in= −πd

2√

2g4A

(t)t0

From this equation and using d=0.4 in. andA = 36 in2, t is found to be 14.7 s. As a compar-ison, using the flow rate calculated in Problem5.79, the mold would require approximately 13s to fill.

5.82 A rectangular mold with dimensions 100 mm ×200 mm × 400 mm is filled with aluminum withno superheat. Determine the final dimensionsof the part as it cools to room temperature. Re-peat the analysis for gray cast iron.

Note that the initial volume of the box is(0.100)(0.200)(0.400)=0.008 m3. From Table5.1 on p. 206, the volumetric contraction foraluminum is 6.6%. Therefore, the box volumewill be

V = (1− 0.066)(0.008 m3) = 0.007472 m3

Assuming the box has the same aspect ratio asthe mold (1:2:4) and that warpage can be ig-nored, we can calculate the dimensions of thebox after solidification as 97.7 mm × 195.5 mm× 391 mm. From Table 3.3 on p. 106, the melt-ing point of aluminum is 660C, with a coef-ficient of thermal expansion of 23.6 µm/mC.Thus, the total strain in cooling from 660C toroom temperature (25C) is

ε = α∆t = (23.6µm/mC)(660C− 25C)

or ε = 0.0150. The final box dimensions aretherefore 96.2 × 192.5 × 385 mm.For gray cast iron, the metal expands uponsolidification. Assuming the mold will allowfor expansion, the volume after solidification isgiven by

V = (1.025)(0.008 m3) = 0.0082 m3

If the box has the same aspect ratio as the ini-tial mold cavity, the dimensions after solidifica-tion will be 100.8 × 201.7 × 403.3 mm. Usingthe data for iron in Table 3.3, the melting pointis taken as 1537C and the coefficient of ther-mal expansion as 11.5 µm/mC. Therefore,

ε = α∆t = (11.5µm/mC)(1537C− 25C)

or ε = 0.0174. Hence, the final dimensions are99.0 × 198.1 × 396 mm. Note that even thoughthe cast iron had to cool off from a higher initialtemperature, the box of cast iron is much closerto the mold dimensions than the aluminum.

5.83 The constant C in Chvorinov’s rule is given as3 s/mm2 and is used to produce a cylindricalcasting with a diameter of 75 mm and a heightof 125 mm. Estimate the time for the cast-ing to fully solidify. The mold can be brokensafely when the solidified shell is at least 20 mm.Assuming the cylinder cools evenly, how muchtime must transpire after pouring the moltenmetal before the mold can be broken?

Note that for the cylinder

A = 2(π

4d2)

+ πdh

= 2[π4

(75)2]

+ π(75)(125)

= 38, 290 mm2

V =π

4d2h =

π

4(75)2(125) = 5.522× 105 mm3

From Chvorinov’s rule given by Eq. (5.11) onp. 205,

t = C

(V

A

)2

= (3 s/mm2)(

5.522× 105

38, 290

)2

or t = 624 s, or just over 10 min to solidify.The second part of the problem is far more dif-ficult, and different answers can be obtaineddepending on the method of analysis. Thesolution is not as straightforward as it mayseem initially. For example, one could say that

20






the 20 mm wall is 53.3% of the thickness, sothat 0.533(624)=333 s is needed. However, thiswould not be sufficient because an annular sec-tion at an outside radius has more material thanone closer to the center. It is thus reasonableand conservative to consider the time requiredfor the remaining cylinder to solidify. Usingh = 85 mm and d = 35 mm, the solidificationtime is found to be 21.8 s. Therefore, one stillhas to wait 602 s before the mold can be broken.

5.84 If an acceleration of 100 g is necessary to pro-duce a part in true centrifugal casting and thepart has an inner diameter of 10 in., a meanouter diameter of 14 in., and a length of 25 ft.,what rotational speed is needed?

The angular acceleration is given by α = ω2r.Recognizing that the largest force is experi-enced at the outside radius, this value for r isused in the calculation:

α = ω2r = 100 g = 3220 ft/s2

Therefore, solving for ω,

ω =√α/r =

√(3220 ft/s2

)/(0.583 ft)

or ω = 74 rad/s = 710 rpm.

5.85 A jeweler wishes to produce twenty gold ringsin one investment-casting operation. The waxparts are attached to a wax central sprue ofa 0.5 in. diameter. The rings are located infour rows, each 0.5 in. from the other on thesprue. The rings require a 0.125-in. diameterand 0.5-in. long runner to the sprue. Estimate

the weight of gold needed to completely fill therings, runners, and sprues. The specific gravityof gold is 19.3.

The particular answer will depend on the geom-etry selected for a typical ring. Let’s approxi-mate a typical ring as a tube with dimensionsof 1 in. outer diameter, 5/8 in. inner diameter,and 3/8 in. width. The volume of each ring isthen 0.18 in3, and a total volume for 20 rings of3.6 in3. There are twenty runners to the sprue,so this volume component is

V = 20(π

4d2)L = 20

(π4

(0.125 in.)2)

(0.5 in.)

or V = 0.123 in3. The central sprue has alength of 1.5 in., so that its volume is

V =π

4d2L =

π

4(0.5 in.)2(1.5 in.) = 0.29 in3

The total volume is then 4.0 in3, not includingthe metal in the pouring basin, if any. The spe-cific gravity of gold is 19.3, thus its density is19.3(62.4 lb/ft3) = 0.697 lb/in3. Therefore, thejeweler needs 2.79 lb. of gold.



Design

5.87 Design test methods to determine the fluidity ofmetals in casting (see Section 5.4.2 starting onp. 203). Make appropriate sketches and explainthe important features of each design.

By the student. The designs should allow somemethod of examining the ability of a metalto fill the mold. One example, taken fromKalpakjian and Schmid, Manufacturing Engi-

neering and Technology, 5th ed, Prentice-Hall,2001, is shown below.

21






Pouring cup

SprueFluidity index

5.88 The accompanying figures indicate various de-fects and discontinuities in cast products. Re-view each one and offer design solutions to avoidthem.

(a)

Fracture

(b)

Riser

Gate

Casting

(c)

Sink mark

(d)

Cold tearing

By the student. Some examples are for (a) frac-ture is at stress raiser, a better design would uti-lize a more gradual filet radius; (b) fracture atthe gate indicates this runner section is too nar-row and thus it solidified first, hence the gateshould be larger.

5.89 Utilizing the equipment and materials avail-able in a typical kitchen, design an experimentto reproduce results similar to those shown inFig. 5.12.

A simple experiment can be performed withmelted chocolate and a coffee cup. If a part-ing agent is sprayed into the cup, and moltenchocolate is poured, after a short while the still

molten center portion can be poured from thecup, leaving a solidified shell. This effect canbe made more pronounced by using chilling thecups first.

5.90 Design a test method to measure the perme-ability of sand for sand casting.

Permeability suggests that there is a potentialfor material to penetrate somewhat into theporous mold material. The penetration can bemeasured through experimental setups, such asusing a standard-sized slug or shape of sand, ap-plying a known pressure to one side, and thenmeasuring the flow rate through the sand.

5.91 Describe the procedures that would be involvedin making a bronze statue. Which casting pro-cess or processes would be suitable? Why?

The answer depends on the size of the statue.A small statue (say 100 mm tall) can be diecast if the quantities desired are large enough,or it can be sand cast for fewer quantities. Thevery large statues such as those found in publicparks, which typically are on the order of 1 to 3m tall, are produced by first manufacturing orsculpting a blank from wax and then using theinvestment-casting process. Another option fora large casting is to carefully prepare a ceramicmold.

5.92 Porosity developed in the boss of a casting isillustrated in the accompanying figure. Showthat by simply repositioning the parting line ofthis casting, this problem can be eliminated.

Cope

Drag

Boss

Riser Part

Core

Note in the figure that the boss is at some dis-tance from the blind riser. Consequently, theboss can develop porosity (not shown in the fig-ure, but to be added by the instructor) becauseof a lack of supply of molten metal from the

22






riser. The sketch below shows a repositionedparting line that would eliminate porosity inthe boss. Note in the illustration below that theboss can now be supplied with molten metal asit begins to solidify and shrink.

5.93 For the wheel illustrated in the accompanyingfigure, show how (a) riser placement, (b) coreplacement, (c) padding, and (d) chills may beused to help feed molten metal and eliminateporosity in the isolated hob boss.

Rim

Hub boss

Four different methods are shown below.

(a) Riser

(b) Core

(c) Pads

(d) Chills

5.94 In the figure below, the original casting designshown in (a) was changed to the design shown

in (b). The casting is round, with a vertical axisof symmetry. As a functional part, what advan-tages do you think the new design has over theold one?

1 in.(25 mm)

1.5 in.(38 mm)

1 in.(25 mm)

1 in.(25 mm)

Ribs or brackets

(a)

(b)

By the student. There are several advantages,including that the part thickness is more uni-form, so that large shrinkage porosity is lesslikely, and the ribs will control warping due tothermal stresses as well as increasing joint stiff-ness.

5.95 An incorrect and a correct design for castingare shown, respectively, in the accompanyingfigure. Review the changes made and commenton their advantages.

Outside core

(a) incorrect

(b) correct

Outside core

By the student. The main advantage of thenew design is that it can be easily cast withoutthe need for an external core. The original partwould require two such cores, because the ge-ometry is such that it cannot be obtained in asand mold without cores.

23






5.96 Three sets of designs for die casting are shownin the accompanying figure. Note the changesmade to original die design (number 1 in eachcase) and comment on the reasons.

(1)

(2)

Parting line

(a)

(1) (2)

(b)

Parting line

Parting line

(3)

(1) (2)

(c)

By the student. There are many observations,usually with the intent of minimizing changesin section thickness, eliminating inclined sur-faces to simplify mold construction, and to ori-ent flanges so that they can be easily cast.

5.97 It is sometimes desirable to cool metals moreslowly than they would be if the molds weremaintained at room temperature. List and ex-plain the methods you would use to slow downthe cooling process.

There can be several approaches to this prob-lem, including:

• Heated molds will maintain temperatureshigher than room temperature, but willstill allow successful casting if the moldtemperature is below the melting temper-ature of the metal.

• The mold can be placed in a container;heat from the molten metal will then warmthe local environment above room temper-ature.

• The mold can be insulated to a greater ex-tent, so that its steady-state temperatureis higher (permanent-mold processes).

• The mold can be heated to a higher tem-perature.

• An exothermic jacket can be placedaround the molten metal.

• Radiation heat sources can be used to slowthe rate of heat loss by conduction.

5.98 Design an experiment to measure the constantsC and n in the Chvorinov’s Rule [Eq. (5.11)].

The following are some tests that could be con-sidered:

• The most straightforward tests involveproducing a number of molds with a familyof parts (such as spheres, cubes or cylin-ders with a fixed length-to-diameter ratio),pouring them, and then breaking the moldperiodically to observe if the metal has so-lidified. This inevitably results in spilledmolten metal and may therefore a difficulttest procedure to use.

• Students should consider designing moldsthat are enclosed but have a solidificationfront that terminates at an open riser; theycan then monitor the solidification timescan then be monitored, and then deter-mine fit Eq. (5.11) on p. 205 to their data.

• An alternative to solidifying the metal isto melt it within a mold specially designedfor such an experiment.

5.99 The part in the accompanying figure is to becast of 10% Sn bronze at the rate of 100 partsper month. To find an appropriate casting pro-cess, consider all the processes in this chapter,then reject those that are (a) technically in-admissible, (b) technically feasible but too ex-pensive for the purpose, and (c) identify the

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most economical process. Write a rationale us-ing common-sense assumptions about productcost.

4.0 in.

10.0 in.

1.0 in.

10 in.

0.45±0.05 in.

Ra=125 in.

Ra=60 in.

The answers could be somewhat subjective, be-cause the particular economics are affected by

company capabilities and practices. The follow-ing summary is reasonable suggestion:

Process Note Cost rationaleSand casting This is probably

best.Shell-mold casting (a)Lost Foam Need tooling to

make blanks. Toolow of productionrate to justify.

Plaster mold (a)Ceramic mold (b)Lost Wax Need to make

blanks. Too low ofproduction rate tojustify, unless rapidtooling is used.

Vacuum casting (b)Pressure casting (b)Die casting (b)Centrifugal casting (b)CZ Process (b)Notes: (a) technically inadmissible; (b) Too ex-pensive.

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26






Chapter 6

Bulk Deformation Processes

Questions

Forging

6.1 How can you tell whether a certain part isforged or cast? Describe the features that youwould investigate to arrive at a conclusion.

Numerous tests can be used to identify cast vs.forged parts. Depending on the forging temper-ature, forged parts are generally tougher thancast parts, as can be verified when samples fromvarious regions of the part are subjected to atensile test. Hardness comparisons may alsobe made. Microstructures will also indicateforged vs. cast parts. Grain size will usuallybe smaller in forgings than in castings, and thegrains will undergo deformation in specific di-rections (preferred orientation). Cast parts, onthe other hand, will generally be more isotropicthan forged parts. Surface characteristics androughness are also likely to be different, depend-ing on the specific casting processes used andthe condition of the mold or die surfaces.

6.2 Why is the control of volume of the blank im-portant in closed-die forging?

If too large of a blank is placed into the diesin a closed-die forging operation, presses will(a) jam, (b) not complete their stroke, and (c)subject press structures to high loads. Nu-merous catastrophic failures in C-frame presseshave been attributed to such excessive loads. If,on the other hand, the blank is too small, thedesired shape will not be completely impartedonto the workpiece.

6.3 What are the advantages and limitations of acogging operation? Of die inserts in forging?

Because the contact area in cogging is muchsmaller (incremental deformation) than in aregular forging operation, large sections of barscan be reduced at much low loads, thus requir-ing lower-capacity machinery, which is an im-portant economic advantage. Furthermore, var-ious cross sections can be produced along thelength of the bar by varying the strokes duringcogging. A corresponding disadvantage is thetime and large number of strokes required tocog long workpieces, as well as the difficulty incontrolling deformation with sufficient dimen-sional accuracy and surface finish.

6.4 Explain why there are so many different kindsof forging machines available.

Each type of forging machine has its own advan-tages and limitations, each being ideally suitedfor different applications. Major factors in-volved in equipment selection may be summa-rized as follows:

(a) Force and energy requirements,

(b) Force-stroke characteristics,

(c) Length of ram travel,

(d) Production rate requirements,

(e) Strain-rate sensitivity of the workpiecematerials,

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(f) Cooling of the workpiece in the die in hotforging, and its consequences regarding diefilling and forging forces,

(g) Economic considerations.

6.5 Devise an experimental method whereby youcan measure the force required for forgingonly the flash in impression-die forging. (SeeFig. 6.15a.)

An experimental method to determine theforces required to forge only the flash (for anaxisymmetric part) would involve making thedie in two concentric pieces, each with its ownload cell to measure the force. The die in thecenter would only cover the projected area ofthe part itself, and the outer die (ring shaped)would cover the projected area of the annularflash. During forging, the load cells are moni-tored individually and, thus, the loads for thepart and the flash, respectively, can be mea-sured independently. Students are encouragedto devise other possible and practical methods.

6.6 A manufacturer is successfully hot forging a cer-tain part, using material supplied by CompanyA. A new supply of material is obtained fromCompany B, with the same nominal composi-tion of the major alloying elements as that ofthe material from Company A. However, it isfound that the new forgings are cracking eventhough the same procedure is followed as be-fore. What is the probable reason?

The probable reason is the presence of impu-rities, inclusions, and minor elements (such assulfur) in the material supplied by Company B.Note that the question states that both materi-als have the “same nominal composition of themajor alloying elements”. No mention is maderegarding minor elements or impurity levels.

6.7 Explain why there might be a change in thedensity of a forged product as compared to thatof the cast blank.

If the original material has porosity, such asfrom a poor casting with porosity due to gasesor shrinkage cavities, its density will increaseafter forging because the pores will close underthe applied compressive stresses. On the otherhand, the original blank may be free of anyporosity but due to adverse material flow and

state of stress during plastic deformation, cavi-ties may develop (similar to voids that developin the necked region of a tensile-test specimen,see Fig. 3.24 on p. 100). Thus, the density willdecrease after forging due to void formation.

6.8 Since glass is a good lubricant for hot extrusion,would you use glass for impression-die forgingas well? Explain.

Glass, in various forms, is used for hot forgingoperations. However, in impression-die forging,even thin films (because glass is incompressible)will prevent the part from producing the die ge-ometry, and thus develop poor quality, and mayprevent successful forging of intricate shapes. Ifthe glass lubricant solidifies in deep recesses ofthe dies, they will be difficult and costly to re-move.

6.9 Describe and explain the factors that influencespread in cogging operations on square billets.

A review of the events taking place at the die-workpiece interface in cogging indicates thatthe factors that influence spreading are:

(a) Friction: the lower the friction, the morethe spreading because of reduced lateralresistance to material flow.

(b) Width-to-thickness ratio of the workpiece:the higher this ratio, the lower the spread-ing.

(c) Contact length (in the longitudinaldirection)-to-workpiece ratio); the higherthis ratio, the higher the spreading. Recallthat the material flows in the direction ofleast resistance.

6.10 Why are end grains generally undesirable inforged products? Give examples of such prod-ucts.

As discussed in Section 6.2.5 starting onp. 283, end grains are generally undesirable be-cause corrosion occurs preferentially along grainboundaries. Thus end grains present manygrain boundaries at the surface for corrosionto take place. In addition, they may result inobjectionable surface appearance, as well as re-ducing the fatigue life of the component becauseof surface roughness that results from corrosion.

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6.11 Explain why one cannot produce a finishedforging in one press stroke, starting with ablank.

Forgings are typically produced through a seriesof operations, such as edging, blocking, etc., asdepicted in Fig. 6.25 on p. 285. This is done fora number of reasons:

(a) The force and energy requirements on thepress are greatly reduced by performingthe operations sequentially;

(b) The part may have to be subjected to in-termediate annealing, thus allowing lessductile materials to be forged to compli-cated shapes.

(c) Reviewing the Archard wear law given byEq. (4.6) on p. 145, it can be seen thatlow die wear rates can be achieved if thesliding distance and/or the force is low ina stroke. Reviewing Fig. 6.25 on p. 285,it can be seen that each operation willinvolve a large sliding distance betweenthe workpiece and dies, thus causing morewear.

6.12 List the advantages and disadvantages of usinga lubricant in forging operations.

The advantages include:

(a) a reduction in the force and energy re-quired;

(b) less localization of strain, resulting in im-proved forgeability;

(c) the lubricant acts as a thermal barrier, sothat the part can remain hotter longer andthus have more ductility;

(d) the lubricant can protect the workpiecefrom the environment, especially in hotforging, and also act as a parting agent.

The disadvantages include:

(a) The lubricant adds cost to the operation;

(b) a thick film can result in orange-peel effecton the workpiece;

(c) lubricants may be entrapped in die cavi-ties, thus part dimensions my not be ac-ceptable;

(d) the lubricant must subsequently be re-moved from the part surface, an additionaland difficult operation;

(e) disposal of the lubricant can present envi-ronmental shortcomings.

6.13 Explain the reasons why the flash assists in diefilling, especially in hot forging.

The flash is excess metal which is squeezed outfrom the die cavity into the outer space betweenthe two dies. The flash cools faster than the ma-terial in the cavity due to the high a/h ratio andthe more intimate contact with the relativelycool dies. Consequently, the flash has higherstrength than the hotter workpiece in the diecavity and, with higher frictional resistance inthe flash gap, provides greater resistance to ma-terial flow outward through the flash gap. Thus,the flash encourages filling of complex die cav-ities.

6.14 By inspecting some forged products (such asa pipe wrench or coins), you can see that thelettering on them is raised rather than sunk.Offer an explanation as to why they are madethat way.

By the student. It is much easier and economi-cal to machine cavities in a die (thus producinglettering on a forging that are raised from itssurface) than producing protrusions (thus pro-ducing lettering that are like impressions on theforged surface). Note that to produce a protru-sion on the die, material surrounding the lettersbe removed, a difficult operation for most let-tering. Recall also similar consideration in castproducts.

Rolling

6.15 It was stated that three factors that influ-ence spreading in rolling are (a) the width-to-thickness ratio of the strip, (b) friction, and (c)the ratio of the radius of the roll to the thick-ness of the strip. Explain how each of thesefactors affects spreading.

These parameters basically all contribute to thefrictional resistance in the width direction of thestrip by changing the aspect ratio of the con-tact area between the roll and the strip (see alsoAnswer 6.9 above).

29






6.16 Explain how you would go about applyingfront and back tensions to sheet metals duringrolling.

Front tensions are applied and controlled by thetake-up reel of a rolling mill; the higher thetorque to this reel or the higher the rotationalspeed, the greater the front tension. Back ten-sion is applied by the pay-off reel by increasingthe braking torque on the pay-off reel or reduc-ing its rotational speed.

6.17 It was noted that rolls tend to flatten under rollforces. Which property(ies) of the roll materialcan be increased to reduce flattening? Why?

Flattening is elastic deformation of the origi-nally circular roll cross section, and results ina larger contact length in the roll gap. There-fore, the elastic modulus of the roll should beincreased.

6.18 Describe the methods by which roll flatteningcan be reduced.

Roll flattening can be reduced by:

(a) decreasing the reduction per pass,

(b) reducing friction, and/or

(c) increasing the roll stiffness (for example,by making it from materials with highmodulus of elasticity, such as carbides).

6.19 Explain the technical and economic reasons fortaking larger rather than smaller reductions perpass in flat rolling.

Economically, it is always beneficial to reducethe number of operations involved in manufac-turing of products. Reducing the number ofpasses in rolling achieves this result by lower-ing the number of required operations. Thisallows less production time to achieve the finalthickness of the rolled product. Of course, anyadverse effects of high reductions per pass mustalso be considered.

6.20 List and explain the methods that can be usedto reduce the roll force.

In reviewing the mechanics of a flat rolling op-eration, described in Section 6.3.1 starting onp. 290, it will be apparent that the roll force,F , can be reduced by:

(a) using smaller-diameter rolls,

(b) taking lower reduction per pass,

(c) reducing friction,

(d) increasing strip temperature, and

(e) applying front and/or back tensions, σf

and σb.

6.21 Explain the advantages and limitations of usingsmall-diameter rolls in flat rolling.

The advantages of using smaller diameter rollsin flat rolling are the following:

(a) compressive residual stresses are devel-oped on the workpiece surface,

(b) lower roll forces are required,

(c) lower power requirements,

(d) less spreading, and

(e) the smaller diameter rolls are less costlyand easier to replace and maintain.


(a) larger roll deflections, possibly requiringbackup rolls, and

(b) lower possible drafts; see Eq. (6.46) onp. 298.

6.22 A ring-rolling operation is being used suc-cessfully for the production of bearing races.However, when the bearing race diameter ischanged, the operation results in very poor sur-face finish. List the possible causes, and de-scribe the type of investigation you would con-duct to identify the parameters involved andcorrect the problem.

Surface finish is closely related to lubricant filmthickness, thus initial investigations should beperformed to make sure that the film thicknessis maintained the same for both bearing races.Some of the initial investigations would involvemaking sure, for example, that the lubricantsupply is not reduced with a larger race size.Also, the higher the rolling speed, the greaterthe film thickness, so it should be checked thatthe rolling speed is the same for both cases. For-ward slip should be measured and the rollingspeeds adjusted accordingly.

30






6.23 Describe the importance of controlling rollspeed, roll gap, temperature, and other relevantprocess variables in a tandem-rolling operation.

Control of tandem rolling is especially impor-tant because the conditions at a particularstand can affect the those at another stand. Forexample, with roll gap and rolling speeds, theeffect of poor control is the application of toomuch or too little front and/or back tension.As is clear from the description on p. 301, thismay result in larger roll forces and torques, orcan lead to chatter.

6.24 Is it possible to have a negative forward slip?Explain.

It is possible to have negative forward slip, butonly in the presence of a large front tension.Consider that it is possible to apply a largeenough front tension so that the rolls slip. Aslightly lower front tension will have significantslippage, so the workpiece velocity will be muchlower than the roll velocity. From Eq. (6.24)onp. 291, the forward slip will be negative. Notethat it is not possible to have negative forwardslip if there are no front or back tensions.

6.25 In addition to rolling, the thickness of platesand sheets can also be reduced by simplystretching. Would this process be feasible forhigh-volume production? Explain.

Although stretching may first appear to be afeasible process, there are several significantlimitations associated with it, as compared torolling:

(a) The stretching process is a batch operationand it cannot be continuous as in rolling.

(b) The reduction in thickness is limited bynecking of the sheet, depending on itsstrain-hardening exponent, n.

(c) As the sheet is stretched, the surface finishbecomes dull due to the orange-peel effect,and thickness and width control becomesdifficult.

(d) Stretching the sheet requires some meansof clamping at its ends which, in turn, willleave marks on the sheet, or even causetearing.

(e) There would be major difficulties in-volved in applying high temperature dur-ing stretching of less ductile materials.

6.26 In Fig. 6.33, explain why the neutral pointmoves towards the roll-gap entry as friction in-creases.

The best way to visualize this situation is toconsider two extreme conditions. Let’s first as-sume that friction at the roll-strip interface iszero. This means that the roll is slipping withrespect to the strip and as a result, the neutral(no-slip) point has to move towards the exit.On the other hand, if we assume that friction isvery high, the roll tends to pull the strip withit; in this case, the neutral point will tend tomove towards the entry of the roll gap.

6.27 What typically is done to make sure the prod-uct in flat rolling is not crowned?

There are a number of strategies that can befollowed to make sure that the material in flatrolling is not crowned, that is, to make surethat its thickness is constant across the width.These include:

(a) Use work rolls that are crowned.(b) Use larger backing rolls that reduce elastic

deformation of the work rolls.(c) Apply a corrective moment to the shafts

of the work rolls.(d) Use a roll material with high stiffness.

6.28 List the possible consequences of rolling at (a)too high of a speed and (b) too low of a speed.

There are advantages and disadvantages toeach. Rolling at high speed is advantageous inthat production rate is increased, but it has dis-advantages as well, including:

• The lubricant film thickness entrained willbe larger, which can reduce friction andlead to a condition where the rolls slipagainst the workpiece. This can lead to adamaged surface finish on the workpiece.

• The thicker lubricant film associated withhigher speeds can result in significantorange-peel effect, or surface roughening.

• Because of the higher speed, chatter mayoccur, compromising the surface quality orprocess viability.

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• There is a limit to speed associated withthe power source that drive the rolls.

Rolling at low speed is advantageous becausethe surface roughness of the strip can matchthat of the rolls (which can be polished). How-ever, rolling at too low a speed has conse-quences such as:

• Production rate will be low, and thus thecost will be higher.

• Because a sufficiently thick lubricant filmcannot be developed and maintained,there may be a danger of transferringmaterial from the workpiece to the roll(pickup), thus compromising surface fin-ish.

• The strip may cool excessively before con-tacting the rolls. This is because a longbillet that is rolled slowly will lose someof its heat to the environment and also byconduction through the roller conveyor.

6.29 Rolling may be described as a continuous forg-ing operation. Is this description appropriate?Explain.

This is a good analogy. Consider the situationof forging a block to a thinner cross sectionthrough increments (as in incremental form-ing). As the number of stages increases, the op-eration eventually approaches that of the stripprofile in rolling.

6.30 Referring to appropriate equations, explain whytitanium carbide is used as the work roll inSendzimir mills, but not generally in otherrolling mill configurations.

The main reason that titanium carbide is usedin a Sendzimer mill is that it has a high elas-tic modulus, and thus will not flatten as much;see Eq. (6.48) on p. 299 and the text immedi-ately after this equation. Titanium carbide isnot used for other roll configuration because ofthe size of the rolls required and the high costof TiC rolls.

Extrusion

6.31 It was stated that the extrusion ratio, die ge-ometry, extrusion speed, and billet temperatureall affect the extrusion pressure. Explain why.

Extrusion ratio is defined as the ratio of billet(initial) area to final area. If redundant workis neglected, the absolute value of true strain isε = ln(Ao/Af ). Thus, the extrusion ratio af-fects the extrusion force directly in an ideal sit-uation. Die geometry has an effect because itinfluences material flow and, thus, contributesto the redundant work of deformation. Extru-sion speed has an effect because, particularlyat elevated temperatures, the flow stress willincrease with increasing strain rate, dependingon the strain-rate sensitivity of the workpiecematerial. On the other hand, higher temper-atures lower the yield stress and thus, reduceforces.

6.32 How would you go about preventing centerburstdefects in extrusion? Explain why your meth-ods would be effective.

Centerburst defects are attributed to a state ofhydrostatic tensile stress at the centerline of thedeformation zone in the die. The two majorvariables affecting hydrostatic tension are thedie angle and extrusion ratio. These defects canbe reduced or eliminated by lowering the dieangle, because this increases the contact lengthfor the same reduction and thereby increasesthe deformation zone. Similarly, a higher ex-trusion ratio also increases the size and depthof the deformation zone, and thus will reduce oreliminate the formation of these cracks. Theseconsiderations are also relevant to strip, rod,and wire drawing.

6.33 How would you go about making a steppedextrusion that has increasingly larger cross-sections along its length? Is it possible? Wouldyour process be economical and suitable forhigh production runs? Explain.

If the product has a stepped profile, such as around stepped shaft with increasing diameter,the smaller diameter is extruded first. The dieis then changed to one with a larger openingand the part is extruded further. A still largerthird, and further larger cross sections, can beproduced by changing the die to a larger diam-eter opening. The process would obviously notbe economical at all for high production runs.For shorter pieces, it is possible to make a diewith a stepped profile, as shown in Fig. 6.57 onp. 317, where the length of the stroke is small.

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6.34 Note from Eq. (6.54) that, for low values of theextrusion ratio, such as R = 2, the ideal ex-trusion pressure p can be lower than the yieldstress, Y , of the material. Explain whether ornot this phenomenon is logical.

Equation (6.54) on p. 310 is based on the energyprinciple and is correct. Note that the extru-sion pressure, p, acts on the (undeformed) billetarea. Consequently, it is not necessary that itsmagnitude be at least equal to the yield stressof the billet material.

6.35 In hydrostatic extrusion, complex seals are usedbetween the ram and the container, but not be-tween the extrusion and the die. Explain why.

The seals are not needed because the leadingend of the workpiece, in effect, acts as a sealagainst the die. The clearance between theworkpiece and the die is very small, so that thehydraulic fluid in the container cannot leak sig-nificantly. This may present some startup prob-lems, however, before the workpiece becomeswell-conformed to the die profile.

6.36 List and describe the types of defects that mayoccur in (a) extrusion and (b) drawing.

Recognizing that a defect is a situation that cancause a workpiece to be considered unsuitablefor its intended operation, several defects canoccur. Extrusion defects are discussed in Sec-tion 6.4.4 starting on p. 318. Examples includepoor surface finish or surface cracking (suchas bamboo defect), tailpipe or fishtailing, andchevron cracking. In drawing, defects includepoor surface finish and chevron cracking. Bothextrusion and drawing also can have a loss in di-mensional accuracy, particularly as attributedto die wear.

6.37 What is a land in a die? What is its function?What are the advantages and disadvantages tohaving no land?

The land is shown in Fig. 6.60 on p. 320 fordrawing, but is too small to be seen for thefigures illustrating extrusion. The land is theportion of a die that is parallel to the work-piece travel that bears against the workpiece.The land is needed to ensure that workpiece di-mensions are controlled and that die wear does

not affect dimensions, since die wear mainly oc-curs on the inlet side of the die. The disadvan-tage to the land is that the workpiece surfacecan be damaged by scratching against the land;generally, the smaller the land, the better theworkpiece surface.

6.38 Under what circumstances is backwards extru-sion preferable to direct extrusion? When ishydrostatic extrusion preferable to direct extru-sion?

Comparing Figs. 16.47a and 16.47b on p. 309it is obvious that the main difference is thatin backward extrusion the billet is stationary,and in direct extrusion it is moving relative tothe container walls. The main advantage be-comes clear if a glass pillow is used to providehot-working lubricant between the workpieceand the die. On the other hand, if there issignificant friction between the workpiece andthe chamber, energy losses associated with fric-tion are avoided in backwards extrusion (be-cause there is no movement between the bodiesinvolved).

6.39 What is the purpose of a container liner in di-rect extrusion (see Fig. 6.47a)? Why is there nocontainer liner used in hydrostatic extrusion?

The container liner is used as a sacrificial wearpart, similar to the pads used in an automotivedisk brake. When worn, it is far less expen-sive to replace a liner than to replace the entirecontainer. In hydrostatic extrusion, the billetdoesn’t contact the container, and thus wear isnot a concern.

Drawing

6.40 We have seen that in rod and wire drawing, themaximum die pressure is at the die entry. Why?

The reason is that at the die entry, the stateof stress is plane stress with equal biaxial com-pression (in the radial direction). Thus, accord-ing to yield criteria the state of stress is in thethird quadrant of Fig. 2.36 on p. 67 and hencethe pressure has a value of Y . At the die exit,however, we have longitudinal tension and bi-axial (radial) compression due to contact withthe die. According to the yield criteria, because

33






of the tensile stress present, the die pressure islower than that at the entry to the die (see alsoAnswer 6.43 below).

6.41 Describe the conditions under which wet draw-ing and dry drawing, respectively, are desirable.

Wet drawing would be suitable for large coilsof wire that can be dipped fully in the lubri-cant, whereas dry drawing would be suitablefor straight short rods.

6.42 Name the important process variables in draw-ing, and explain how they affect the drawingprocess.

These are described in Section 6.5 starting onp. 320. The important variables include:

• Yield stress, Y ; it directly affects the drawstress and die life.

• Die angle, α. The die angle in the defor-mation zone affects the redundant work; inthe entry area, the die angle is importantfor encouraging lubricant entrainment.

• Friction coefficient, µ. The friction coef-ficient affects the frictional component ofwork and, hence, the draw stress. See alsoEq. (6.68) on p. 322.

• Reduction in area. As described, there isa limit to the reduction in area that canbe achieved in drawing.

• Lubrication condition. Effective lubrica-tion reduces friction, but also may lead toa rough surface due to the orange peel ef-fect.

6.43 Assume that a rod drawing operation can becarried out either in one pass or in two passesin tandem. If the die angles are the same andthe total reduction is the same, will the drawingforces be different? Explain.

The drawing forces will be the same, unless thesurface of the rod is undergoing some changeswhile it is between the two dies, due to externaleffects such as the environment or additionallubrication. The reason why the forces are notdifferent is that the drawing process can be re-garded as consisting of a series of incrementalreductions taking place in one die. Ideally, wecan slice the die into a number of segments

and, thus, make it a tandem process. Note thatas the distance between the individual die seg-ments decreases, we approach the one-die con-figuration. Also note that in a tandem opera-tion, the front tension of one segment becomesthe back tension of an adjacent segment.

6.44 Refer to Fig. 6.60 and assume that reduction inthe cross section is taking place by pushing arod through the die instead of pulling it. As-suming that the material is perfectly plastic,sketch the die-pressure distribution, for the fol-lowing situations: (a) frictionless, (b) with fric-tion, and (c) frictionless but with front tension.Explain your answers.

Note that the mathematical models developedfor drawing and extrusion predict the drawstress or extrusion pressure, but do not showthe die pressure. A quantitative relationshipcould be derived for the die pressure, recogniz-ing that p−σx = Y ′ based on yield criteria, andthen examining Eqs. (6.63) through (6.67) onp. 321. However, a qualitative sketch of the diepressure can be generated based on the physi-cal understanding of the friction hill and asso-ciated pressure plots in forging and rolling inSections 6.2 and 6.3. A qualitative sketch ofthe die pressures is given below. Note that theactual position of the curve for the frictionlesscase with front tension depends on the level offront tension provided.

Frictionless

With frictionFrictionless with front tension

Position, x

Dimensionless pressure, p/Y

6.45 In deriving Eq. (6.74), no mention was maderegarding the ductility of the original materialbeing drawn. Explain why.

The derivation of Eq. (6.77) on p. 326 is basedon the fact that, at failure, the tensile stress inthe wire or rod has reached the uniaxial yieldstress of the material. Thus, it is implicitlyassumed that the material is able to undergothe reduction in cross-sectional area and that itis ultimately failing under high tensile stresses.

34






Note that a less ductile material will fail prema-turely because of lack of ductility but not lackof strength.

6.46 Why does the die pressure in drawing decreasestoward the die exit?

We refer to Eq. (6.71) on p. 322 which rep-resents yield criteria in the deformation zone.Note that as we approach the die exit, the draw-ing stress, σ, increases; consequently, the diepressure, p, must decrease, as also shown inFig. 6.62 on p. 322.

6.47 What is the magnitude of the die pressure atthe die exit for a drawing operation that is be-ing carried out at the maximum reduction perpass?

The die pressure at the exit in this case willbe zero. This is because of the condition setby Eq. (6.73) on p. 324 which deals only withuniaxial stress. Note that there is a finite diepressure in a normal drawing operation, as de-picted in Fig. 6.62 on p. 322, and that the draw-ing stress at the exit is lower than the uniaxialyield stress of the material, as it should for asuccessful drawing operation to take place.

6.48 Explain why the maximum reduction per passin drawing should increase as the strain-hardening exponent, n, increases.

The reason is that the material is continuouslystrain hardening as it reaches the die exit. Con-sequently, at the exit it is stronger and, thus,can resist higher stresses before it yields. Con-sequently, a strain-hardening material can un-dergo higher reductions per pass, as can also beseen in Example 6.8.

6.49 If, in deriving Eq. (6.74), we include friction,will the maximum reduction per pass be thesame (that is, 63%), higher, or lower? Explain.

If we include friction, the drawing stress willbe higher. As a result, the maximum reductionper pass will be lower than 63%. In other words,the cross-sectional area of the exiting materialhas to be larger than the ideal case in orderto support the increased drawing stress due tofriction, without yielding.

6.50 Explain what effects back tension has on thedie pressure in wire or rod drawing, and dis-cuss why these effects occur.

The effect of back pressure is similar to that ofback tension in rolling (see Figs. 6.35 on p. 295and 6.62 on p. 322), namely, the pressure drops.This satisfies yield criteria, in that, as tensionincreases, the apparent compressive yield stressof the material decreases.

6.51 Explain why the inhomogeneity factor, φ, in rodand wire drawing depends on the ratio, h/L, asplotted in Fig. 6.12.

By observing Figs. 6.12 on p. 276 and 6.13bon p. 277, we note that the higher the h/L ra-tio, the more nonuniform the deformation ofthe material. For example, keeping h constant(hence the same initial and final diameters), wenote that as L decreases, the die angle has tobecome larger. This, in turn, indicates higherredundant work (see Fig. 6.51 on p. 311).

6.52 Describe the reasons for the development of theswaging process.

The major reasons include:

(a) variety of parts that can be produced withrelatively simple tooling,

(b) capacity to produce internal profiles onlong workpieces,

(c) compact equipment,(d) good surface finish and dimensional accu-

racy, and(e) improved workpiece properties due to cold

working of the material.

6.53 Occasionally, wire drawing of steel will takeplace within a sheath of a soft metal, such ascopper or lead. Why would this procedure beeffective?

The main reason that steel wire drawing takesplace in a sheath of a softer metal is to reducethe frictional stress. Recall from Eq. (4.5) onp. 140 that, for the same friction factor, m, thefrictional stress is lower if the workpiece hard-ness is lower. By placing the sheath in contactwith the die, the soft metal acts as a solid lubri-cant and reduces the frictional stresses. This, inturn, reduces forces and hence increases drawa-bility.

35






6.54 Recognizing that it is very difficult to manufac-ture a die with a submillimeter diameter, howwould you produce a 10 µm-diameter wire?

The most common method of producing verysmall wires is through bundle drawing, whereina large number of wires (up to hundreds) aresimultaneously drawn through one die. Specialcare must be taken to provide good lubrication;otherwise, the wires will weld together duringdrawing. The student should be encouraged tosuggest additional techniques.

6.55 What changes would you expect in the strength,hardness, ductility, and anisotropy of annealedmetals after they have been drawn throughdies? Why?

We would expect that the yield stress of thematerial is higher, assuming that the opera-tion is performed at room temperature. Since itis directly related to strength, hardness is alsohigher. The ductility is expected to decrease,as the material has been strain hardened. Be-cause of preferred orientation during deforma-tion, some anisotropy is also to be expected incold-drawn rods.

General

6.56 With respect to the topics covered in this chap-ter, list and explain specifically two exampleseach where friction (a) is desirable and (b) isnot desirable.

The student is encouraged to provide severalspecific examples. For example, friction is de-sirable in rolling and controlling material flowin forging. It is undesirable in rod and wiredrawing (except to obtain a burnished surface)and extrusion.

6.57 Choose any three topics from Chapter 2 andwith a specific example for each, show their rel-evance to the topics covered in this chapter.

By the student. For example, a student coulddiscuss yield criteria, and then show how theyare used to develop pressure and force equationsfor specific operations.

6.58 Same as Question 6.57 but for Chapter 3.

By the student. For example, a student couldselect thermal effects on mechanical properties,as discussed in Section 3.7 starting on p. 98,and apply it to a discussion of cold versus hotforging.

6.59 List and explain the reasons that there are somany different types of die materials used forthe processes described in this chapter.

Among several reasons are the level of stressesand type of loading involved (such as static ordynamic), relative sliding, temperature, ther-mal cycling, dimensional requirements, size ofworkpiece, frictional considerations, wear, andeconomic considerations.

6.60 Why should we be interested in residual stressesdeveloped in parts made by the forming pro-cesses described in this chapter.

Residual stresses and their significance are dis-cussed in detail in Section 2.10 starting on p. 59.The student should elaborate further with spe-cific references to the processes discussed in thischapter.

6.61 Make a summary of the types of defects foundin the processes described in this chapter. Foreach type, specify methods of reducing or elim-inating the defects.

By the student; see also Sections 3.8, 4.2, and4.3.

Problems

Forging 6.62 In the free-body diagram in Fig. 6.4b, the in-

36






cremental stress dσx on the element was shownpointing to the left. Yet it would appear that,because of the direction of frictional stresses,µp, the incremental stress should point to theright in order to balance the horizontal forces.Show that the same answer for the forging pres-sure is obtained regardless of the direction ofthis incremental stress.

We will derive the pressure using the same ap-proach as described in Section 6.2.2 starting onp. 269. The equivalent version of Fig. 6.4 onp. 269 is shown below.

dxx

a

h

(a) (b)

y

x

y

σy

x + dx

y

Using the stresses as shown in part (b), we have,from equilibrium and assuming unit width,

(σx + dσx)h− 2µσydx− σxh = 0

ordσx −

2µσy

hdx = 0

For the distortion-energy criterion, it should berecognized that σx is now tensile, whereas inthe text it is compressive. Therefore, Eq. (6.11)becomes

σy + σy =2√3Y = Y ′

Thusdσx = −dσy

When substituted into the equilibrium equa-tion, one obtains

σy = Ce−2µx/h

Using the boundary conditions that σx = 0(and therefore σy = Y ′) at x = 0, gives thevalue of C as

C = Y ′e2µa/h

Therefore, substituting into the expression forσy,

σy = Ce−2µx/h = Y ′e2µa/he−2µx/h

= Y ′e2µ(a−x)/h

which is the same as Eq. (6.13) on p. 270.

6.63 Plot the force vs. reduction in height curve inopen-die forging of a solid cylindrical, annealedcopper specimen 2 in. high and 1 in. in diam-eter, up to a reduction of 70%, for the casesof (a) no friction between the flat dies and thespecimen, (b) µ = 0.25, and (c) µ = 0.5. Ignorebarreling and use average-pressure formulas.

For annealed copper we have, from Table 2.3 onp. 37, K = 315 MPa = 46,000 psi and n = 0.54.The flow stress is

Yf = (46, 000 psi)ε0.54

where the absolute value of the strain is

ε = ln(ho

h

)From volume constancy, we have

π

4r2oho =

π

4r2h

or

r =

√r2o

(ho

h

)Note that ro = 0.5 in and ho = 2 in. The forg-ing force is given by Eqs. (6.18) and (6.19) onp. 272 as:

F = Yf

(1 +

2µr3h

)(πr2)

Some of the points on the curves are the follow-ing:

Forging Force, kip% Red. µ = 0 µ = 0.25 µ = 0.510 11.9 12.4 13.120 20.1 21.3 22.430 29.6 31.7 33.840 41.9 45.6 49.450 59.3 66.3 73.660 86.1 100. 114.70 133.1 167. 201.

37






The curve is plotted as follows:Fo

rgin

g fo

rce,

kip

0

50

100

150

200

% Reduction

0 20 40 60 7010 30 50

=0=0.25=0.5

6.64 Use Fig. 6.9b to provide the answers to Problem6.63.

The force required for forging is the productof the average pressure and the instantaneouscross-sectional area. The average pressure isobtained from Fig. 6.9b on p. 272. Note thatfor µ = 0, pave/Yf = 1, and thus the an-swer is the same as that to Problem 6.63 givenabove. The following table can be developedwhere pave/Yf is obtained from Fig 6.9b, and hand r are calculated as in Problem 6.63. Notethat Fig. 6.9b does not give detailed informa-tion for 2r/h < 10, which is where the data forthis problem lies. However, the µ = 0.25 values(interpolated between the µ = 0.2 and µ = 0.3curves) are noticeably above 1 by 2r/h = 5 orso, so we give the value 1.25, and all interme-diate values are linearly interpolated from thisreading. Similarly, for µ = 0.5, a value betweenµ = 0.3 and sticking suggests pave/Y is around1.6 or so by 2r/h = 3. This is the basis for thenumbers below.

% pave/Yf

Red. 2r/h µ = 0.25 µ = 0.510 0.585 1.0 1.20 0.699 1.041 1.130 0.854 1.083 1.240 1.08 1.125 1.350 1.41 1.17 1.460 1.98 1.21 1.570 3.04 1.25 1.6

Recall from Problem 6.63 that

Yf = (46, 000 psi)ε0.54

where the absolute value of the strain is

ε = ln(ho

h

)From this, the following forces are calculated(recall that F = paveA):

% F , kipRed. r, in. µ = 0.25 µ = 0.510 0.527 11.9 11.920 0.559 20.9 22.130 0.598 32.0 35.540 0.646 47.1 54.450 0.707 69.1 83.060 0.791 104. 129.70 0.913 166. 213.

The results are plotted below. For comparisonpurposes, the results from Problem 6.63 are alsoincluded as dashed lines. As can be seen, theresults are fairly close, even with the rough in-terpolation done in this solution.

Forg

ing

forc

e, k

ip

0

50

100

150

200

% Reduction0 20 40 60 7010 30 50

=0.25=0.5

6.65 Calculate the work done for each case in Prob-lem 6.63.

The work done can best be calculated by ob-taining the area under the curve F vs. ∆h.From the solution to Problem 6.63, the force isgiven by

F = Yf

(1 +

2µr3h

)(πr2)

where

r =

√r2o

(ho

h

)and

Yf = (46, 000 psi)[ln(ho

h

)]0.54

38






Numerous mathematical software packages canperform this calculation. The results are as fol-lows for the data in Problem 6.63, at 70% re-duction in height (or ∆h = 1.4):

µ Work (in.-lb)0 62,4450.25 71,0650.5 79,685

6.66 Determine the temperature rise in the specimenfor each case in Problem 6.63, assuming thatthe process is adiabatic and the temperature isuniform throughout the specimen.

To determine the temperature rise at 70% re-duction in height, we obtain the work done fromProblem 6.65 above. Assuming there is negli-gible stored energy, this work is converted intoheat. Thus, we can calculate the temperaturerise using Eq. (2.65) on p. 73:

∆T =utotal

ρc

where u is the specific energy, or the energy pervolume. The volume of the specimen is

V = πr2h = π(0.5)2(2) = 1.57 in3

The specific heat of copper is given in Table3.3 on p. 106 as 385 J/kgK or 0.092 BTU/lbF.Since 1 BTU= 780 ft-lb, the specific heat ofcopper is 861 in-lb/lbF. The density of copperis, from the same table, 8970 kg/m3 or 0.324lb/in3. Thus, using the work values obtainedin Problem 6.65, the temperature rise is as fol-lows:

µ ∆T , F0 1420.25 1620.5 182

6.67 To determine its forgeability, a hot-twist testis performed on a round bar 25 mm in diame-ter and 200 mm long. It is found that the barunderwent 200 turns before it fractured. Cal-culate the shear strain at the outer surface ofthe bar at fracture.

The shear strain can be calculated fromEq. (2.22) on p. 49 where r = 25/2 = 12.5 mm,

l = 200 mm, and φ = 2π(200) = 1257 radians.Therefore, the shear strain is

γ =(12.5)(1257)

200= 78.6

6.68 Derive an expression for the average pressure inplane-strain compression under the condition ofsticking friction.

Sticking friction refers to the condition wherea Tresca friction model is used with m = 1[see Eq. (4.5) on p. 140]. Therefore, the fol-lowing figure represents the applied stresses toan element of forging, which can be comparedto Fig. 6.4b on p. 269. The approach in Sec-tion 6.2.2 starting on p. 269 is followed closelyin this derivation.

dx

x

a

y

x + dxh

y

mk

mk

x

From equilibrium in the x-direction,

(σx + dσx)h+ 2mkdx− σxh = 0

Solving for dσx,

dσx = −2mkh

dx

Integrating,

σx = −2mkh

x+ C

where C is a constant. The boundary conditionis that at x = a, σx = 0, so that

0 = −2mkh

(a) + C

Therefore,

C =2mkh

a

andσx =

2mkh

(a− x)

39






The die pressure is obtained by applyingEq. (2.36) on p. 64:

σy − σx = Y ′

Note that in plane strain, Y ′ = 2√3Y , and

k = Y/√

3 (see Section 2.11.3 on p. 66), so thatk = Y ′/2. Therefore

σy −mY ′

h(a− x) = Y ′

orσy = Y ′

[1 +

m

h(a− x)

]Note that this relationship is consistent withFig. 6.10 on p. 273 for 0 ≤ x ≤ a. Since therelationship is linear, then we can note that

σy =Y ′

2

[(1 +

ma

h

)+ (1)

]or

σy = Y ′(1 +

ma

2h

)For sticking, m = 1 and

σy = Y ′(1 +

a

2h

)6.69 What is the magnitude of µ when, for plane-

strain compression, the forging load with slid-ing friction is equal to the load with stickingfriction? Use average-pressure formulas.

The average pressure with sliding friction is ob-tained from Eq. (6.15) on p. 271, and for stick-ing friction it is obtained from the answer toProblem 6.68 using m = 1. Equating these twoaverage pressures, we obtain

Y ′(1 +

µa

h

)= Y ′

(1 +

a

2h

)Therefore, µ = 0.5.

6.70 Note that in cylindrical upsetting, the frictionalstress cannot be greater than the shear yieldstress, k, of the material. Thus, there may bea distance x in Fig. 6.8 where a transition oc-curs from sliding to sticking friction. Derive anexpression for x in terms of r, h, and µ only.

The pressure curve for the solid cylindrical caseis similar to Fig. 6.5 on p. 270 and is given by

Eq. (6.17) on p. 272. Following the same proce-dure as in Example 6.2, the shear stress at theinterface due to friction can be expressed as

τ = µp

However, we know that the shear stress cannotexceed the yield shear stress, k, of the materialwhich, for the cylindrical state of stress, is Y

2 .Thus, in the limit, we have the condition

µY e2µ(r−x)/h =Y

2

or2µ(r − x)

h= ln

(12µ

)Hence,

x = r −(h

2µ

)ln(

12µ

)Note that this answer is the same as in the ex-ample problem for plane strain.

6.71 Assume that the workpiece shown in the accom-panying figure is being pushed to the right by alateral force F while being compressed betweenflat dies. (a) Make a sketch of the die-pressuredistribution for the condition for which F is notlarge enough to slide the workpiece to the right.(b) Make a similar sketch, except that F is nowlarge enough so that the workpiece slides to theright while being compressed.

F

Applying a compressive force to the left bound-ary of the workpiece in Fig. 6.5 on p. 270 raisesthe pressure at that boundary. The higher theforce F , the higher the pressure. Eventually theworkpiece will slide completely to the right, in-dicating that the neutral point has now movedall the way to the left boundary. These are de-picted in the figure below.

40






Y'

F

p

6.72 For the sticking example given in Fig. 6.10, de-rive an expression for the lateral force F re-quired to slide the workpiece to the right whilethe workpiece is being compressed between flatdies.

Because we have sticking on both die-workpieceinterfaces and a plane-strain case, the frictionalstress will simply be Y ′/2. Hence, the lateralforce required must overcome this resistance onboth (top and bottom) surfaces. Thus,

F = 2Y ′(2a)(w) = 4Y ′aw

where w is the width of the workpiece, i.e., thedimension perpendicular to the page in the fig-ure.

6.73 Two solid cylindrical specimens, A and B, bothmade of a perfectly-plastic material, are beingforged with friction and isothermally at roomtemperature to a reduction in height of 25%.Originally, specimen A has a height of 2 in. anda cross-sectional area of 1 in2, and specimen Bhas a height of is 1 in. and a cross-sectional areaof 2 in2. Will the work done be the same forthe two specimens? Explain.

We can readily see that specimen B will re-quire higher work because it has a larger die-workpiece surface area, hence a higher fric-tional resistance as compared to specimen A.We can prove this analytically by the followingapproach. The work done is the integral of theforce and distance:

W =∫F dh

where F = (pave)(A), and pave for a cylindri-cal body is given by Eq. (6.18) on p. 272. Be-cause the volume V of the workpiece is con-stant, we have a relationship between its radius

and height as

r =

√V

πh

Hence,

pave = Y

[1 +

(2µ3

)(V

π

)(1

h3/2

)]Because it consists of constants, let c =(2µ/3)(V/π), which results in the following ex-pression:

F = Y(1 +

c

h3/2

)(Vh

)and hence work is

W = Y V

∫ ho/2

ho

[(1h

)+( c

h5/2

)]dh

= Y V

[ln 0.75− 3c

21

h3/2o

(0.540)]

= Y V

[−0.288− c

h3/2o

(0.809)]

Thus, for this problem, we have

WA = Y V

[−0.288− c

(1

23/2

)(0.809)

]= Y V (−0.288− 0.286c)

WB = Y V

[−0.288− c

(1

13/2

)(0.809)

]= Y V (−0.288− 0.809c)

Comparing the two shows that WB > WA.

6.74 In Fig. 6.6, does the pressure distribution alongthe four edges of the workpiece depend on theparticular yield criterion used? Explain.

The answer is yes. This is a plane-stress prob-lem, but an element at the center of the edgesis subjected not only to a pressure p (due tothe dies) but also frictional constraint since thebody is expanding in all directions. Thus, anelement at the center of the edges is subjectedto biaxial compressive stresses. Because the lat-eral stress, σx, due to frictional forces is smallerthan the normal stress (pressure), we note thefollowing:

41






(a) According to the maximum-shear-stresscriterion, the pressure distribution alongthe edges should be constant because theminimum stress is zero. Hence,

p = Y

(b) According to the distortion-energy crite-rion for plane stress, the pressure distri-bution along the edges should be as givenin Fig. 6.6 on p. 271 and can be shown toobey the following relationship:

p2 + σ2 − pσ = Y 2

Note that at the corners p = Y , and thatp is highest at the center along the edgesbecause that is where the frictional stressis a maximum.

6.75 Under what conditions would you have a nor-mal pressure distribution in forging a solidcylindrical workpiece as shown in the accom-panying figure? Explain.

Y'

p

x

The pressure distribution is similar to the fric-tion hill shown in Fig. 6.5 on p. 270, with theexception that there are two symmetric regionswhere the pressure is constant. These regionssustain pressure but do not contribute to thefrictional stress. A trapped layer of incompress-ible lubricant in grooves machined on the sur-face of the workpiece, for example, would rep-resent such a condition. The grooves would befilled with the lubricant, which sustains pres-sure but would not contribute to shear at theinterface because of its low viscosity.

6.76 Derive the average die-pressure formula givenby Eq. (6.15). (Hint: Obtain the volume underthe friction hill over the surface by integration,and divide it by the cross-sectional area of theworkpiece.)

The area, A, under the pressure curve (from thecenterline to the right boundary a) in Fig. 6.5on p. 270 is given by

A =∫p dx

and the average pressure is

pave =1a

∫p dx

wherep = Y ′e2µ(a−x)/h

Integrating this equation between the limitsx = 0 and x = a, we obtain

pave =Y ′

2µa/h

(e2µa/h − 1

)Letting 2µa/h = m, and using a Taylor seriesexpansion of the exponent term,

em = 1 +m+m2

2!+m3

3!+ . . .

Ignoring third-order terms and higher as beingtoo small compared to other terms, we obtain

pave =Y ′

m

(1 +m+

m2

2− 1)

= Y ′(1 +

m

2

)= Y ′

(1 +

µa

h

)6.77 Take two solid cylindrical specimens of equal

diameter but different heights, and compressthem (frictionless) to the same percent reduc-tion in height. Show that the final diameterswill be the same.

Let’s identify the shorter cylindrical specimenwith the subscript s and the taller as t, andtheir original diameter as D. Subscripts f ando indicate final and original, respectively. Be-cause both specimens undergo the same percentreduction in height, we can write

htf

hto=hsf

hso

and from volume constancy,

htf

hto=(Dto

Dtf

)2

42






andhsf

hso=(Dso

Dsf

)2

Because Dto = Dso, we note from these rela-tionships that Dtf = Dsf .

6.78 A rectangular workpiece has the following orig-inal dimensions: 2a = 100 mm, h = 30 mm andwidth = 20 mm (see Fig. 6.5). The metal hasa strength coefficient of 300 MPa and a strain-hardening exponent of 0.3. It is being forged inplane strain with µ = 0.2. Calculate the forcerequired at a reduction of 20%. Do not useaverage-pressure formulas.

In this plane-strain problem note that the widthdimension remains at 20 mm. Thus, when thereduction in height is 20%, the final height ofthe workpiece is

h = (1− 0.2)(30) = 24 mm = 0.024 m

Since volume constancy has to be maintainedand we have a plane-strain situation, we canfind the new (final) dimension a from

(100)(30)(20) = (2a)(24)(20)

Thus, a = 62.5 mm = 0.0625 m. The absolutevalue of the true strain is

ε = ln(

3024

)= 0.223

and hence the uniaxial flow stress at the finalheight is

Yf = Kεn = (400)(0.223)0.3 = 255 MPa

and the flow stress in plane strain is Y ′f =

(1.15)(255) = 293 MPa. Thus, from Eq. (6.13)on p. 270 the pressure as a function of distancex is

p = Y ′e2µ(a−x)/h

= (293 MPa)e2(0.2)(0.0625−x)/0.024

= (293 MPa)e1.042−16.7x

To obtain the force required for one-half ofthe workpiece per unit width, we integrate theabove expression between the limits x = 0 andx = 0.0625, which gives the force per unit widthand one-half of the length as F = 32.2 MN/m.

The total force is the product of this force andthe specimen width times two, or

Ftotal = 2(32.2)(0.02) = 1.288 MN

Note that if we use the average pressure formulagiven by Eq. (6.16) on p. 271, the answer willbe

Ftotal = Y ′(1 +

µa

h

)(2a)(w)

= (293)[1 +

(0.2)(0.0625)0.024

]×(2)(0.0625)(0.02)

= 1.11 MN

The discrepancy is due to the fact that in de-riving the average pressure, a low value of µa/hhave been assumed for mathematical simplicity.

6.79 Assume that in upsetting a solid cylindricalspecimen between two flat dies with friction,the dies are rotated at opposite directions toeach other. How, if at all, will the forging forcechange from that for nonrotating dies? (Hint:Note that each die will now require a torque butin opposite directions.)

From the top view of the round specimen inFig. 6.8b on p. 272, we first note that the fric-tional stresses at the die-specimen interfaceswill essentially be tangential. (We say essen-tially because the rotational speed is assumedto be much higher than the vertical speed ofthe dies.) Consequently, the direction of µσz

will be tangential and, because there will nowbe no frictional stress in the radial direction,balancing forces in the radial direction will notinclude friction. Thus, the situation will be ba-sically similar to upsetting without friction, andthe forging force will be a minimum. However,additional work has to be done in supplyingtorque to the two dies that are rotating in oppo-site directions. Note also that we are assumingµ to be small, so that it will not cause plastictwisting of the specimen due to die rotation.

6.80 A solid cylindrical specimen, made of a per-fectly plastic material, is being upset betweenflat dies with no friction. The process is be-ing carried out by a falling weight, as in adrop hammer. The downward velocity of thehammer is at a maximum when it first con-tacts the workpiece and becomes zero when the

43






hammer stops at a certain height of the spec-imen. Establish quantitative relationships be-tween workpiece height and velocity, and makea qualitative sketch of the velocity profile of thehammer. (Hint: The loss in the kinetic energyof the hammer is the plastic work of deforma-tion; thus, there is a direct relationship betweenworkpiece height and velocity.)

For this problem we will use the energy method,in which case the kinetic energy of the fallingweight is being dissipated by the work of plas-tic deformation of the specimen. We know thatthe work, W , done on the specimen of volumeV is

W = uV

or in terms of specific heights,

W = Y ln(ho

h

)(V )

where h is the instantaneous height of the spec-imen. The kinetic energy, KE, of the fallingweight can be expressed in terms of the initialand instantaneous heights of the specimen:

KE =m(v2

o − v2)

2

where m is the mass of the falling body andvo is the velocity when the falling weight firstcontacts the specimen. Equating the two ener-gies, noting that Y , ho, V , m, and vo are con-stant, and simplifying, we find the relationshipbetween v and h as

v ∝√

lnh+ C

where C is a constant. Inspection of this equa-tion indicates that, qualitatively, the velocityprofile of the falling weight will be as shown inthe following figure:

h0

hf

v0

v

0

This behavior is to be expected because thespecimen cross-sectional area will be increas-ing rapidly with time, hence the upward resist-ing force on the falling weight will also increaserapidly and, thus, decelerate the weight rapidly.

6.81 Describe how would you go about estimatingthe force acting on each die in a swaging oper-ation.

First, an estimate has to be made of the con-tact area between the die and the workpiece;this can be done by studying the contact ge-ometry. Then, a flow stress, Yf , has to be de-termined, which will depend on the workpiecematerial, strain hardening exponent, n, and theamount of strain the material is undergoing.Also, as a first approximation, the hardness ofthe workpiece material can be used (with appro-priate units) since the deformation zone duringswaging is quite contrained, as in a hardnesstest. Note also that the swaging process can beassumed to be similar to hubbing, and conse-quently, Eq. (6.23) on p. 281 may be used toestimate the die force.

6.82 A mechanical press is powered by a 30-hp mo-tor and operates at 40 strokes per minute. Ituses a flywheel, so that the rotational speed of

44






the crankshaft does not vary appreciably duringthe stroke. If the stroke length is 6 in., what isthe maximum contact force that can be exertedover the entire stroke length? To what heightcan a 5052-O aluminum cylinder with a diam-eter of 0.5 in. and a height of 2 in. be forgedbefore the press stalls?

Note that the power is 30 hp = 198,000 in-lb/s.Assume that the press stroke has a constant ve-locity. Although this is a poor approximation,it does not affect the problem since a constantforce is assumed later; actually, both the forceand velocity will vary. At 40 strokes per minuteand with a 6 in. stroke, we would require a ve-locity of

V = (40 rpm)(12 in./rev)(60 min/s) = 8 in./s

The power exerted is the product of the forceand the velocity; thus,

198, 000 in.-lb/s = FV = F (8 in./s)

or F = 24.75 kip. For 5052-O, the yieldstrength is 90 MPa=13 ksi (from Table 3.7 onp. 116). Therefore, using Eqs. (6.18) and (6.19)on p. 272,

F = 24, 750 lb

= Yfπr2

(1 +

2µr3h

)= (13 ksi)π(r2)

(1 +

2(0.2)r3h

)Also, from volume constancy we have r2h =r2oho. Substituting this into the above equationand solving, yields r = 0.675 in., or a diameterof around 1.35 in. The height in this case isthen h = 0.274 in.

6.83 Estimate the force required to upset a 0.125-in-diameter C74500 brass rivet in order to forma 0.25-in-diameter head. Assume that the co-efficient of friction between the brass and thetool-steel die is 0.2 and that the rivet head is0.125 in. in thickness.

Since we are asked for the force required to per-form the forging operation, we use Eq. (6.18) onp. 272 to obtain the average pressure as evalu-ated at the end of the stroke where r = 0.125in and h = 0.125 in. From Table 3.11 on p. 119

and recognizing that the bronze is annealed, sothat its strength should be on the low end of theranges given, the yield stress of C74500 bronzecan be taken as Y = 170 MPa = 24.6 ksi. FromEq. (6.18) on p. 272, the pressure at the end ofthe press stroke is

pave = Y

(1 +

2µr3h

)= (24.6)

(1 +

2(0.20)(0.125)3(0.125)

)= 27.88 ksi

The force is the product of pressure and area,given in Eq. (6.19) as

F = pave

(πr2)

= (27.88)π(

0.1252

)2

or F = 0.342 kip = 342 lb.

6.84 Using the slab method of analysis, deriveEq. (6.17).

We use an element and the stresses acting onit as shown in Fig. 6.8. Balancing forces in theradial direction,

0 = σrxhdθ + 2σθhdxdθ

2− 2µσzxdθdx

−(σr + dσr)(x+ dx)dθh

Simplifying this equation, noting that the prod-uct dr dθ is very small and hence can be ne-glected, and dividing by xh dx, we obtain

dσr

dx+σr − σθ

x= −2µσz

h

Note that the circumferential and radial incre-mental strains are equal to each other by virtueof the fact that

dεθ =2π dx2πx

=dx

dand dεr =

dx

x

From Eq. (2.43), we can state that

dεrσr

=dεθσθ

=dεzσz

Consequently, we have

σr = σθ

45






and thusdσr

dx= −2µσz

h

Also, using the distortion-energy criterion,

(σr − σθ)2 + (σr − σz)

2 + (σz − σθ)2 = 2Y 2

we haveσrσz = Y

and since the yield stress, Y , is a constant, wenote that

dσr = dσz

Thus,dσz

dx= −2µσz

h

Note that this equation is similar to Eq. (6.12)for plane-strain forging. Following the sameprocedure as in the text, we can then obtainthe pressure p at any radius x as

p = Y e2µ(r−x)/h

Rolling

6.85 In Example 6.4, calculate the velocity of thestrip leaving the rolls.

Because mass continuity has to be maintained,we can write

Vf (0.80) = Vrhneutral

where hneutral is the thickness of the strip at theneutral point and

Vf = ωR = (2π)(100)(12) = 7540 in./min

The thickness at the neutral point can be cal-culated from Eqs. (6.32) and (6.33) on p. 294.In this problem, we can approximate a cer-tain thickness, based on observations regardingFigs. 6.33 on p. 293 and 6.34 on p. 294. Sincethe original and final thicknesses are 1.0 and0.8 in., respectively, let’s assume that hneutral =0.85. Thus

Vf =(7540)(0.85)

0.80= 8010 in./min

or Vf = 668 ft/min.

6.86 With appropriate sketches, explain the changesthat occur in the roll-pressure distribution ifone of the rolls is idling, i.e., power is shut offto that roll.

Note in this case that the idling roll cannot sup-ply power to the strip; hence, there cannot bea net torque acting on it (assuming no bearingfriction). Consequently, the roll-pressure dis-tribution will be such that the frictional forcesacting along the entry and exit zones, respec-tively, are equal. In the absence of strain hard-ening, the pressure distribution in the roll gapwill thus be symmetrical and the neutral axisshifts a little towards the entry. However, theother roll is still supplying power and its neutralaxis is more towards the exit. There is, there-fore, a zone in the roll gap on which the fric-tional stresses on the top and bottom surfacesare opposite in sign; this condition is known ascross shear.

6.87 It can be shown that it is possible to determineµ in flat rolling without measuring torque orforces. By inspecting the equations for rolling,describe an experimental procedure to do so.Note that you are allowed to measure any quan-tity other than torque or forces.

In this problem, we first measure the follow-ing quantities: vo, vf , vr, ho and hf . Fromthe available information and knowing R, wecan calculate the magnitude of the angle of ac-ceptance, α. From the velocity distribution, asin Fig. (6.32), we can now determine φn fromwhich we obtain Hn, using Eq. (6.32) on p. 294.To determine the coefficient of friction, we canrewrite Eq. (6.32) as

µ =ln(ho

hf

)Ho − 2Hn

in which Ho is obtained from Eq. (6.29) onp. 292 where φ is now the angle α.

6.88 Derive a relationship between back tension, σb,and front tension, σf , in rolling such that whenboth tensions are increased, the neutral pointremains in the same position.

We note that at the neutral point, the roll pres-sure, p, obtained from Eqs. (6.34) and (6.35) on

46






p. 294 must be equal. Therefore, we can write[Y ′

f − σb

] hn

hoeµ(Ho−Hn) =

[Y ′

f − σf

] hh

hfeµHn

thus

σb = Y ′f −

ho

hf

(e2µHn−µHo

) (Y ′

f − σf

)It should be noted that this equation can berearranged into different forms.

6.89 Take an element at the center of the deforma-tion zone in flat rolling. Assuming that allthe stresses acting on this element are princi-pal stresses, indicate the stresses qualitatively,and state whether they are tensile or compres-sive. Explain your reasoning. Is it possible forall three principal stresses to be equal to eachother in magnitude? Explain.

All stresses on the element will be compressivefor the following reasons:

(a) The vertical stress (pressure) is the com-pressive stress applied by the rolls.

(b) The longitudinal stress is compressive be-cause of the frictional forces in the roll gap.This can be shown by the free-body dia-gram given below, and the stress is alwayscompressive throughout the gap length (inthe absence of front or back tensions).

(c) The stress perpendicular to the page isalso compressive because we have a plane-strain state of stress. The material is notfree to expand laterally because it is con-strained both by frictional forces (alongthe length of the rolls) as well as the rigidregions of the strip ahead and behind theroll gap.

h + dhh

p

p

p

p

x + dx x

6.90 It was stated that in flat rolling a strip, the rollforce is reduced about twice as effectively byback tension as it is by front tension. Explainthe reason for this difference, using appropriatesketches. (Hint: Note the shift in the positionof the neutral point when tensions are applied.)

Referring to Figs. 6.33 on p. 293 or 6.34 onp. 294, we note that the neutral point is towardsthe exit, hence the area under the entry-sidecurve is larger than that for the exit-side curve.(This is in order to supply energy through a netfrictional force during ordinary rolling.) Conse-quently, a reduction in the height of the curveby back tension (σb) has a greater effect thanthat for the exit side by front tension.

6.91 It can be seen that in rolling a strip, the rollswill begin to slip if the back tension, σb, is toohigh. Derive an analytical expression for themagnitude of the back tension in order to makethe powered rolls begin to slip. Use the sameterminology as applied in the text.

Slipping of the rolls means that the neutralpoint has moved to the exit of the roll gap.Thus, the whole contact area becomes the entryzone and Eq. (6.34) on p. 294 is applicable. Weknow that when φ = 0, H = 0, and thus thepressure at the exit is

pφ=0 =(Y ′

f − σb

)(hf

ho

)eµHo

We also know that at the exit the pressure isequal to Y ′. Therefore, we obtain

Y ′f =

(Y ′

f − σb

)(hf

ho

)eµHo

Solving for σb,

σb = Y ′f

[1−

(ho

hf

)(e−µHo

)]where Ho is obtained from Eq. (6.29) withφ = α.

6.92 Derive Eq. (6.46).

Refer to the figure below.

47






Roll

/2x/2

h0 hf

R

z

Note that x = ho − hf , and is also called thedraft. For small α, z = R sinα. Also, note thatfor small angles,

x

2=z sinα

2

Therefore,

x = z sinα = R tan2 α

At small angles, the sine and tangent functionsare approximately equal; hence,

x = ho − hf = R tan2 α

Recall that the inclined-plane principle for fric-tion states that α = tan−1µ, or µ = tanα.Substituting, we have

ho − hf = Rµ2


6.93 In Steckel rolling, the rolls are idling, and thusthere is no net torque, assuming frictionlessbearings. Where, then, is the energy comingfrom to supply the necessary work of defor-mation in rolling? Explain with appropriatesketches, and state the conditions that have tobe satisfied.

The energy for work of deformation is suppliedby the tension required to pull the strip throughthe roll gap. Since the rolls are idling, the roll-pressure distribution will be such that the fric-tional forces in the entry and exit zones, respec-tively, are equal. The neutral point will shifttoward the entry zone, as if applying a fronttension in Fig. 6.35.

6.94 Derive an expression for the tension required inSteckel rolling of a flat sheet, without friction,for a workpiece with a true-stress-true-straincurve given by σ = a+ bε.

In this process, the work done in rolling is sup-plied by the front tension. Assuming a certainreduction in thickness per pass, we first deter-mine the absolute value of the true strain,

ε1 = ln(ho

hf

)Since we know the behavior of the material asσ = a + bε, we can determine the energy ofplastic deformation per unit volume, u, usingEq. (2.59) on p. 71. We also know the cross-sectional dimensions of the strip and the ve-locities vo and vf . The power dissipated isthe product of u and the volume rate of flowthrough the roll gap, which is given by thequantity wohovo. This product is equal to thepower supplied by the front tension that actson the exiting cross-sectional area of the rolledstrip. Hence, assuming a plane-strain condi-tion (that is, w = constant), we can write theexpression

uwhovo = σfwhfvf

from which the magnitude of the front tensioncan be determined.

6.95 (a) Make a neat sketch of the roll-pressure dis-tribution in flat rolling with powered rolls. (b)Assume now that the power to both rolls isshut off and that rolling is taking place by fronttension only, i.e., Steckel rolling. Superimposeon your diagram the new roll-pressure distri-bution, explaining your reasoning clearly. (c)After completing part (b), further assume thatthe roll bearings are becoming rusty and de-prived of lubrication although rolling is still tak-ing place by front tension only. Superimpose athird roll-pressure distribution diagram for thiscondition, explaining your reasoning.

The relevant pressure diagram for Steckelrolling can be obtained simply from Fig. 6.35on p. 295. Note that, with frictionless bearings,the front tension supplies the work of deforma-tion; thus, σf must be high enough such thatthe entry and exit zones have equal areas underthe pressure curve and the neutral point shiftsto the left. In the case of roll bearings with fric-tion, the front tension must increase in order tosupply the additional work required to rotatethe idling rolls with bearing friction. Thus, theneutral point will shift further to the left. We

48






can also assume that the bearing friction is sohigh that they freeze. In that case, the rollswill not rotate, which means that the neutralpoint must shift all the way to the entry andthus frictional forces are all in one direction.

6.96 Derive Eq. (6.28), based on the equation pre-ceding it. Comment on how different the h val-ues are as the angle φ increases.

Note that the procedure is identical to the an-swer to Problem 6.92 above. For small drafts,and hence small angles φ (which is typicallythe case in rolling practice), the expressionµ = tanα can be replaced by µ = α, or forthe more general case, µ = φ. Consequently,the expression

ho − hf = Rµ2

becomesh− hf = Rφ2

so thath = hf +Rφ2

which is Eq. (6.28) on p. 292.

6.97 In Fig. 6.34, assume that L = 2L2. Is the rollforce, F , for L now twice or more than twicethe roll force for L2? Explain.

An inspection of Fig. 6.34 on p. 294 clearly in-dicates that the roll force, F , will be more thantwice as high. This is due to the fact thatthe roll-pressure distribution has the shape ofa friction hill. Only when the pressure is con-stant through the roll gap (as in “frictionless”rolling), will the force be twice as high.

6.98 A flat-rolling operation is being carried outwhere h0 = 0.2 in., hf = 0.15 in., w0 = 10 in.,R = 8 in., µ = 0.25, and the average flow stressof the material is 40,000 psi. Estimate the rollforce and the torque. Include the effects of rollflattening.

The roll force can be estimated from Eq. (6.40)on p. 296, where the quantity L is obtained fromEq. (6.38). Therefore,

L =√R∆h =

√(8)(0.20− 0.15) = 0.632 in.

andhave =

0.20 + 0.152

= 0.175 in.

Therefore,

F = LwY ′[1 +

µL

2have

]= (0.632)(10)(40, 000)

[1 +

(0.25)(0.632)2(0.175)

]= 367, 000 lb

We check for roll flattening by using Eq. (6.48)on p. 299, where C = 1.6× 10−7 in2/lb, assum-ing steel rolls, and

F ′ =F

w=

367, 00010

= 36, 700 lb/in.

Thus,

R′ = R

(1 +

CF ′

ho − hf

)= (8)

[1 +

(1.6× 10−7)(36, 700)0.20− 0.15

]= 8.94 in.

Using this value in the force expression, we haveL = 0.668 in. and F = 395, 000 lb. This forcepredicts a flattened radius of R′ = 9.0 in. (Notethat the expression is converging.) This radiuspredicts L = 0.671 and F = 397, 000 lb, whichsuggests a radius of R′ = 9.02 in. Therefore,the roll force is around 397,000 lb, with an ef-fective roll radius of 9.0 in.

6.99 A rolling operation takes place under the condi-tions shown in the accompanying figure. Whatis the position xn of the neutral point? Notethat there are a front and back tension thathave not been specified. Additional data areas follows: Material is 5052-O aluminum; hard-ened steel rolls; surface roughness of the rolls =0.02 µm; rolling temperature = 210C.

R = 75 mm

x

V = 1.5 m/s3 mm

5 mmV = 2 m/s

49






Note that more data is given than is needed tosolve this problem. Assuming the material isincompressible, the velocity at the inlet is cal-culated as:

(2.0)(3 mm)w = Vi(5 mm)w

Therefore, Vi = 1.20 m/s. At the neutral point,the velocity is the roll velocity (or 1.5 m/s). As-suming incompressibility, we can compare theoutlet and the neutral point:

(1.5)(h) = (2.0)(3) → h = 4.0 mm

Consider the sketch of the roll bite geometrygiven below.

R

x

Rcos

5/2=2.5 mm3/2=1.5 mm

θ can be calculated from:

75− (75) cos θ =4− 3

2

or θ = 6.62. Therefore,

xn = R sin θ = (75) sin 6.62 = 8.64 mm

6.100 Estimate the roll force and power for annealedlow-carbon steel strip 200 mm wide and 10 mmthick, rolled to a thickness of 6 mm. The rollradius is 200 mm, and the roll rotates at 200rpm. Let µ = 0.1.

The roll force can be estimated from Eq. (6.40)on p. 296, where the quantity L is obtained fromEq. (6.38). Therefore,

L =√R∆h =

√(200)(4) = 28.3 mm

andhave =

10 + 62

= 8 mm

From Table 2.3 on p. 37, K = 530 MPa andn = 0.26. The strain is

ε = ln106

= 0.5108

The average yield stress can be obtained fromEq. (2.60) on p. 71 as

Y =Kεn+1

n+ 1=

(530)(0.5108)1.26

1.26= 180 MPa

andY ′ = (1.15)Y = 207 MPa

Therefore,

F = LwY ′(

1 +µL

2have

)= (0.0283)(0.2)(207)

[1 +

(0.1)(28.3)2(8)

]= 1.38 MN

The power per roll is given by Eq. (6.43) as

P =πFLN

60, 000=π(1.38× 106)(0.0283)(200)

60, 000

or P = 409 kW.

6.101 Calculate the individual drafts in each of thestands in the tandem-rolling operation shown.

50






0.26

30Stand 1 3 4 52

0.34

17.7 10.7 6.6 4.1 m/s

0.56 0.90 1.45 2.25 mm

Pay-offreel

Take-upreel

2.6 m/s

Stand 5

2.25 mm1.45 mm0.90 mm

Stand 4

In Section 6.3.1 starting on p. 290, the draft wasdefined as ∆h for a rolling operation. Therefore, theanswers are:

• Stand 5: 2.25 - 1.45 = 0.80 mm, or 36%.

• Stand 4: 1.45 - 0.90 = 0.55 mm, or 38%.

• Stand 3: 0.90 - 0.56 = 0.34 mm, or 38%.

• Stand 2: 0.56 - 0.34 = 0.22 mm, or 39%.

• Stand 1: 0.34 - 0.26 = 0.08 mm, or 24%.

6.102 Calculate the required roll velocities for eachroll in Problem 6.101 in order to maintain aforward slip of (a) zero and (b) 10%.

The forward slip is defined by Eq. (6.24) onp. 291 as:

Forward slip (FS) =Vf − Vr

Vr

For the forward slip to be zero, the roll velocityneeds to be the final velocity in the rolling op-eration. From the data given in the figure, thefollowing quantities can be determined:

Roll velocityFS=0 FS=10%

Stand (m/s) (m/s)1 30 27.32 17.7 16.13 10.7 9.734 6.6 6.05 2.6 2.36

Extrusion

6.103 Calculate the force required in direct extrusionof 1100-O aluminum from a diameter of 6 in. to2 in. Assume that the redundant work is 30% ofthe ideal work of deformation, and the frictionwork is 25% of the total work of deformation.

The extrusion ratio is R = 62/22 = 9, andthus the true strain is ε = ln(9) = 2.20. For1100-O aluminum, we have from Table 2.3 onp. 37, K = 180 MPa = 26,000 psi and n = 0.20.Therefore, from Eq. (2.60) on p. 71, the averageflow stress is

Y =Kεn

n+ 1=

(26, 000)(2.20)0.20

1.20= 25, 360 psi

The ideal extrusion pressure is, from Eq. (6.54)on p. 310,

p = Y lnR = (25, 360) ln 9 = 55, 700 psi

51






The ideal extrusion force is then

F = pA =πpd2

4=π(55, 700)(6)2

4

or F = 1.57×106 lb. The total force is the sumof the forces for ideal, friction, and redundantdeformation. In this case, we can write

Ftotal = Fideal + 0.30Fideal + 0.25Ftotal

Therefore,

Ftotal = 1.73Fideal = 2.72× 106 lb = 1360 tons

6.104 Prove Eq. (6.58).

Consider the sketch below for an extrusion op-eration.

rorx

ro-r

x

Note that the coordinate x has been measuredfrom the die entry, consistent with Example 6.5.Also, note that the sketch shows one-half of theextrusion operation, as the bottom boundaryis a centerline. The initial billet radius is ro.From the triangle indicated,

tanα =ro − r

x


6.105 Calculate the theoretical temperature rise inthe extruded material in Example 6.6, assum-ing that there is no heat loss. (See Section 3.9for information on the physical properties of thematerial.)

The temperature rise can be calculated fromthe work done in the process. In Example 6.6we note that the extrusion force is F = 3.2×106

lb. Thus, the work done in one inch of travel isW = 3.2 × 106 in-lb, and the extruded volumeis

V =πd2l

4=π(5)2(1)

4= 19.6 in3

For a density of 0.324 lb/in3 for copper, itsweight is (19.6)(0.324) = 6.36 lb. The spe-cific heat for copper is 385 J/kgC = 0.092BTU/lbF. Since 1 BTU = 778 ft-lb = 9336in.-lb, the work done is equivalent to 3.2 ×106/9336 = 343 BTU, or 343/6.36 = 54BTU/lb. Thus, the temperature rise will be54/0.092 = 590F, assuming that the process isadiabatic. Note that the final temperature willbe 1500 + 590 = 2090F, which is much abovethe melting temperature of copper. In practice,extrusion is carried out relatively slowly, so thatsignificant heat can be lost to the environment;also, a 1500F preheat is very unusual for cop-per.

6.106 Using the same approach as that shown in Sec-tion 6.5 for wire drawing, show that the extru-sion pressure is given by the expression

p = Y

(1 +

tanαµ

)[1−

(Ao

Af

)µ cot α],

where Ao and Af are the original and finalworkpiece diameters, respectively.

Refer to Fig. 6.61 on p. 321 for a stress ele-ment, from which we apply equilibrium in thex-direction as

0 = (σx + dσx)π

4(D + dD)2

−σxπ

4D2 + p

πDdx

cosα+ µp

πDdx

cosαSimplifying and ignoring second-order terms,

0 = Ddσx + 2σxdD + 2p(1 +

µ

tanα

)dD

From Eq. (2.36) on p. 64, and recognizing thatpositive pressure indicates negative stress,

σmax − σmin = σx + p = Y

Letting µ/ tanα = B, and using this relation-ship yields

dD

D=

dσx

2Bσx − 2Y (1 +B)

Integrating this equation between the limitsDf and Do and by noting that at D = Do,σx = −σd (because the stress is negative, al-though the pressure is positive) and at D = Df ,σx = 0,

σd = Y1 +B

B

[1−

(Df

Do

)2B]

52






or, noting that p = σd,

p = Y1 +B

B

[1−

(Af

Ao

)B]

= Y

(1 +

tanαµ

)[1−

(Af

Ao

)µ cot α]

6.107 Derive Eq. (6.56).

An estimate of the pressure p can be obtained asfollows, referring to the figure below for nomen-clature.

p A0

45°Af

The total power input, P , with a ram velocityof vo is

Pinput = pvo

(πD2

o

4

)The power due to plastic work of deformationis the product of the volume rate of flow andthe energy per unit volume. Thus,

Pplastic = vo

(πD2

o

4

)(Y )

[ln(Do

Df

)2]

As in the problem statement, let’s take the deadzone to imply a 45 die angle as shown, and thatthe frictional stress is equal to the shear yieldstress, k = Y/2, of the material. The power dis-sipated due to friction along the die angle canthen be calculated as

Pfriction = vo

(πD2

o√2

)(Y

2

)ln(Do

Df

)Equating the power input to the sum of theplastic deformation and friction powers, we ob-tain

pvo

(πD2

o

4

)= vo

(πD2

o

4

)(Y )

[ln(Do

Df

)2]

+vo

(πD2

o√2

)(Y

2

)ln(Do

Df

)

or

p = 3.41Y ln(Do

Df

)= 1.7Y ln

(Do

Df

)2

= 1.7Y lnR

In this analysis, we have neglected the forcerequired to overcome friction at the billet-container interface. Assume that the frictionalstress is equal to the shear yield stress of thematerial, k, we can obtain the additional rampressure required due to friction, pf , as

pf

(πD2

o

4

)= πDokL

or

pf = k4LDo

= Y2LDo

6.108 A planned extrusion operation involves steel at800C, with an initial diameter of 100 mm anda final diameter of 20 mm. Two presses, onewith a capacity of 20 MN and the other of 10MN, are available for this operation. Obviously,the larger press requires greater care and moreexpensive tooling. Is the smaller press sufficientfor this operation? If not, what recommenda-tions would you make to allow the use of thesmaller press?

For steel at 800C, k = 425 MPa (FromFig. 6.53 on p. 313). The initial and final areasare 0.00785 m2 and 3.14×10−4 m2, respectively.From Eq. (6.62) on p. 313, the extrusion forcerequired is

F = Aok ln(Ao

Af

)= (0.0078)(425) ln

(0.00785

3.14× 10−4

)= 10.6 MN

Thus, the smaller and easier to use press is notsuitable for this operation, but it almost hassufficient capacity. If the extrusion tempera-ture can be increased or if friction can be re-duced sufficiently, it may then be possible touse this machine.

53






6.109 Estimate the force required in extruding 70-30brass at 700C, if the billet diameter is 125 mmand the extrusion ratio is 20.

From Fig. 6.53 on p. 313, k for copper at 700 Cis approximately 180 MPa. Noting that R is 20and do = 125 mm = 0.125 m; using Eq. (6.62)on p. 313, we find that

F = (π/4)(0.125)2(180)(ln 20) = 6.62 MN

Drawing

6.110 Calculate the power required in Example 6.7 ifthe workpiece material is annealed 70-30 brass.

For annealed 70-30 brass we have, from Table2.3 on p. 37, K = 895 MPa and n = 0.49. Thus,the average flow stress is

Y =(895)(0.466)0.49

1.49= 413 MPa

Since all other quantities are the same as in theexample, the power will be

P = 12.25(

413785

)= 6.4 kW

6.111 Using Eq. (6.63), make a plot similar toFig. 6.63 for the following conditions: K = 100MPa, n = 0.3, and µ = 0.04.

Using Eq. (2.60) on p. 71, we can rewriteEq. (6.66) on p. 321 as

σd =[Kεn+1

n+ 1

] [1 +

tanαµ

][1−

(Af

Ao

)µ cot α]

where ε is the final strain. From Eqs. (2.7) onp. 34 and (2.10) on p. 35 it can be shown thatAf/Ao = e−ε and that ε = − ln(1 − R), whereR is the reduction. Therefore, the expressionbecomes:

σd =[Kεn+1

n+ 1

] [1 +

tanαµ

] [1− (eε)µ cot α

]=

[Kεn+1

n+ 1

] [1 +

tanαµ

] [1− e−µ cot α

]This allows construction of the curve given be-low.

Dra

win

g st

ress

(MP

a)

0

10

20

30

40

50

0 4 8 12 16Die angle (°)

Reduction = 40%

30%

20%10%

6.112 Using the same approach as that described inSection 6.5 for wire drawing, show that thedrawing stress, σd, in plane-strain drawing ofa flat sheet or plate is given by the expression

σd = Y ′(

1 +tanαµ

)[1−

(h− f

ho

)µ cot α],

where ho and hf are the original and final thick-ness, respectively, of the workpiece.

For a plane-strain element, we can apply equi-librium in the x direction as

0 = (σx + dσx) (h+ dh)w − σxhw

+pwdx

cosα+ µp

dx

cosα

Simplifying and ignoring second order terms,

hdσx + σxdh+ p(1 +

µ

tanα

)dx = 0

From Eq. (2.36), and recognizing that positivepressure indicates negative stress,

σmax − σmin = σx + p = Y

Letting µ/ tanα = B,

dh

h=

dσx

2Bσx − 2Y (1 +B)

Integrating this equation between the limits hf

and ho and by noting that at h = ho, σx = −σd

(because the stress is negative, although thepressure is positive) and at h = hf , σx = 0,

σd = Y1 +B

B

[1−

(hf

ho

)B]

or

σd = Y

(1 +

tanαµ

)[1−

(hf

ho

)µ cot α]

54






6.113 Derive an analytical expression for the die pres-sure in wire drawing, without friction or redun-dant work, as a function of the instantaneousdiameter in the deformation zone.

From Eq. (6.71) on p. 322 we know that thefollowing condition must be satisfied in the de-formation zone:

σx + p = Y

where the tensile stress σx is defined as

σx = Y ln(Do

Dx

)2

Consequently,

p = Y

[1− ln

(Do

Dx

)2]

Thus, at the die entry, for example, whereσx = 0, we have p = Y . As we approach the dieexit, σx increases and hence p decreases. (SeeFig. 6.62.)

6.114 A linearly strain-hardening material with atrue-stress-true-strain curve σ = 5, 000 +25, 000ε psi is being drawn into a wire. If theoriginal diameter of the wire is 0.25 in., whatis the minimum possible diameter at the exitof the die? Assume that there is no redundantwork and that the frictional work is 15% of theideal work of deformation. (Hint: The yieldstress of the exiting wire is the point on thetrue-stress-true-strain curve that correspondsto the total strain that the material has un-dergone.)

The drawing stress can be expressed in termsof the average flow stress, Y , as follows:

σd = 1.25Y ε1

For this linearly strain-hardening material, theaverage flow stress is

Y =5, 000 + (5, 000 + 25, 000ε1)

2= 5, 000 + 12, 500ε1

As the problem states, the yield stress of theexiting wire is

Y1 = 5, 000 + 25, 000ε1

Equating both expressions, as was done inEq. (6.71) on p. 322, we obtain

5, 000+25, 000ε1 = σd = 1.15(5, 000+12, 500ε1)ε1

Note that the coefficient of 1.15 takes the fric-tional work into account. This is a quadraticequation with one negative root and one posi-tive root, which is ε1 = 1.56. Thus,

ε1 = ln(

0.25Df

)2

= 1.56

This is solved as Df = 0.11 in.

6.115 In Fig. 6.65, assume that the longitudinal resid-ual stress at the center of the rod is -80,000 psi.Using the distortion-energy criterion, calculatethe minimum yield stress that this particularsteel must have in order to sustain these levelsof residual stresses.

This problem requires that (a) elements betaken at different radial positions, (b) the re-spective residual stresses are determined fromthe figure, and (c) substituted into the effec-tive stress for the distortion-energy criterion,given by Eq. (2.56) on p. 70. Four locationsare checked below.

(a) At the center, we have σL = −80, 000 psi,σR = −60, 000 psi, and σT = −45, 000 psi,thus the effective stress is

σ =1√2

[(σL − σR)2 + (σR − σT )2

+(σL − σT )2]1/2

= 30, 500 psi

(b) At R = 0.375 in., we have σL = −5, 000psi, σR = −40, 000 psi and σT = −15, 000psi, and thus the effective stress is 31,000psi.

(c) At R = 0.5 in, we have σL = 5, 000 psi,σR = −20, 000 psi, and σT = −35, 000 psi,and the effective stress is 35,000 psi.

(d) At the surface of the bar, σL = 50, 000psi, σR = 0 and σT = 50, 000 psi, and theeffective stress is 50,000 psi.

Since the effective stress is equal to the uniaxialstress in a tension test, it can be concluded thatthis part must have a minimum yield stress ofY = 50, 000 psi in order to sustain these resid-ual stresses.

55






6.116 Derive an expression for the die-separatingforce in frictionless wire drawing of a perfectlyplastic material. Use the same terminology asin the text.

For a straight conical die with an angle α andoriginal and final cross-sectional areas Ao andAf , respectively, it can be shown that the con-tact area, A, (in the shape of a truncated cone)is given by

A =Ao −Af

π sinαFrom the die-pressure curve such as that shownin Fig. 6.62 on p. 322, which can be obtainedanalytically, we determine an average pressurepave. Assuming a small die angle, as is generallythe case in wire drawing, the radial componentof this pressure (i.e., perpendicular to the longaxis), is the same as pave. Further assumingthat the die is split in half into two half circles,the die-separating force will act on one-half ofthe contact area, and thus the force will be

F =(Ao −Af ) pave

2π sinα

6.117 A material with a true-stress-true-strain curveσ = 10, 000ε0.3 is used in wire drawing. As-suming that the friction and redundant workcompose a total of 50% of the ideal work of de-formation, calculate the maximum reduction incross-sectional area per pass that is possible.

The maximum reduction per pass for a strain-hardening material was derived in Example 6.8.Using a similar approach, the following relation-ship can be written:

Kεn1 = (1.5)(Kεn1n+ 1

)ε1

orε1 =

n+ 11.5

Since n = 0.3 for this problem, we have ε1 =0.867. Note that the magnitude of K is notrelevant in this problem.

6.118 Derive an expression for the maximum reduc-tion per pass for a material of σ = Kεn as-suming that the friction and redundant workcontribute a total of 25% to the ideal work ofdeformation.

This problem requires the same approach as inProblem 6.86 above. Thus, referring to Exam-ple 6.8,

1.25ε1 = n+ 1

and hence

Max reduction per pass = 1− e−(n+1)/1.25

This means that, as expected, the maximum re-duction is lower than that obtained for the idealcase in Example 6.8.

6.119 Prove that the true-strain rate, ε, in drawing orextrusion in plane strain with a wedge-shapeddie is given by the expression

ε = − 2 tanαVoto

(to − 2x tanα)2,

where α is the die angle, to is the original thick-ness, and x is the distance from die entry (Hint:Note that dε = dA/A.)

This problem is very similar to Example 6.5.To avoid confusion between time and thicknessvariables, lets use h to denote thickness. Fromgeometry in the die gap,

tanα =(ho − h)/2

x

orh = ho − 2x tanα.

The incremental true strain can be defined as

dε =dA

A

where A = wh, and w is the (constant) width.Therefore, dA = wdh, and hence

dε =wdh

wh=dh

h

where dh = − tanαdx. We also know that

ε =dε

dt= −2 tanα

h

dx

dt

However, dx/dt = V , which is the velocity ofthe material at any location x in the die. Hence,

ε =dε

dt= −2V tanα

h

From constancy of flow rate, we can write

V = hovo/h,

56






and hence,

ε = −2Voho tanαh2

=2Voho tanα

(ho − 2x tanα)2

which is the desired relation. The negative signis due to the fact that true strain is definedin terms of the cross-sectional area, which de-creases as x increases.

6.120 In drawing a strain-hardening material withn = 0.25 what should be the percentage offriction plus redundant work, in terms of idealwork, so that the maximum reduction per passis 63%?

Referring to Example 6.8, we can write thefollowing expression to represent this situationwhere x is a multiplying factor that includesfriction and redundant work in terms of theideal work of deformation:

Kεn = x

(Kεn

n+ 1

)ε1

from which we obtain

x =n+ 1ε1

=1.25ε1

A reduction of 63% indicates a true strain ofε1 = 1. Hence, x = 1.25, and thus the sum offriction and redundant work is 25% of the idealwork.

6.121 A round wire made of a perfectly plastic ma-terial with a yield stress of 30,000 psi is beingdrawn from a diameter of 0.1 to 0.07 in. in adraw die of 15. Let the coefficient of frictionbe 0.1. Using both Eqs. (6.61) and (6.66), esti-mate the drawing force required. Comment onany differences in your answers.

In this problem, do = 0.1 in, so that the initialcross-sectional area is

Ao =π

4d2

o =π

4(0.1 in.)2 = 0.00785 in2

Similarly, since df = 0.07 in., Af = 0.00385 in2.From Eq. (6.61) on p. 313, the force required fordrawing is

F = YavgAf lnAo

Af

= (30, 000)(0.00385) ln(

0.007850.00385

)= 82.3 lb

For µ = 0.1 and α = 15 = 0.262 radians,Eq. (6.66) on p. 321 yields

F = YavgAf

[(1 +

µ

α

)ln(Ao

Af

)+

23α

]= (30, 000)(0.00385)

×[[

1 +0.1

0.262

]ln[0.007850.00385

]+

23(0.262)

]or F = 134 lb. Note that Eq. (6.61) does notinclude friction or redundant work effects. Bothof these factors will increase the forging force,and this is reflected by these results.



Design

6.123 Forging is one method of producing turbineblades for jet engines. Study the design of suchblades and, referring to the relevant technicalliterature, prepare a step-by-step procedure formaking these blades. Comment on the difficul-

ties that may be encountered in this operation.

By the student. A typical sequence wouldinclude cutting off a blank from bar stock,block forging, rough forging, finish forging, flash

57






removal, inspection, finishing, and cleaning.There may be quenching operations involved,depending on the material and desired proper-ties.

6.124 In comparing some forged parts with cast parts,it will be noted that the same part may be madeby either process. Comment on the pros andcons of each process, considering factors suchas part size, shape complexity, and design flex-ibility in the event that a particular design hasto be modified.

By the student. Typical answers may addresscost issues (forging will be expensive for shortproduction runs), performance (castings maylack ductility), fatigue performance, and mi-crostructure and grain flow.

6.125 Referring to Fig. 6.25, sketch the intermediatesteps you would recommend in the forging of awrench.

This is an open-ended problem, and there wouldbe several acceptable answers. Students shouldbe encouraged to describe the benefits of theirdie layouts, including their limitations, if any.If bar stock is the input material, an edging op-eration is useful to distribute material to theends where the sockets will require extra ma-terial. The blocking, finishing, and trimmingoperations, as sketched in Fig. 6.25 on p. 285,are conceptually the same as those required tomake a wrench.

6.126 Review the technical literature, and make a de-tailed list of the manufacturing steps involvedin the manufacture of hypodermic needles.

There are several manufacturers of hypodermicneedles, and while each one uses a somewhatdifferent process for production, the basic stepsremain the same, including shaping of the nee-dle, plastic-components molding, piece assem-bly, packaging, and labeling. This is a goodtopic for a paper by the student.

6.127 Figure 6.48a shows examples of products thatcan be obtained by slicing long extruded sec-tions into discrete parts. Name several otherproducts that can be made in a similar man-ner.

By the student. Examples include cookies,pasta, blanks for bearing races, and supportbrackets of all types. The case study for Chap-ter 6 shows a support bracket for an automobileaxle that was made in this manner. Using theInternet, the students should be able to givenumerous other examples.

6.128 Make an extensive list of products that eitherare made of or have one or more componentsof (a) wire, (b) very thin wire, and (c) rods ofvarious cross-sections.

By the student. This is an open-ended problemand students should be encouraged to developtheir lists based on their experiences and re-search. Some answers are:

• Wire is commonly found as electrical con-ductors, wire rope and cable, coat hangers,and nails.

• Very thin wire as integrated circuit pack-ages, communication cable (such as coax-ial cable) shielding, and steel wool.

• Rods as axles, bolts and other fasteners,reinforcing bars for concrete.

6.129 Although extruded products are typicallystraight, it is possible to design dies wherebythe extrusion is curved, with a constant radiusof curvature. (a) What applications could youthink of for such products? (b) Describe yourthoughts as to the shape such a die should havein order to produce curved extrusions.

The applications are limited for curved extru-sions. Students should be encouraged to de-velop their own solutions to this problem; someanswers are:

(a) For escalators, there are handrails thathave a large and constant radius of cur-vature.

(b) Many architectural shapes are curved.

(c) Bicycle frames and aircraft panels are of-ten constructed by bending an extrudedworkpiece; if the extruded section is pro-duced pre-bent, then the bending opera-tions would no longer be necessary.

The die shape is difficult to design. If the diecross section is not symmetric, then a curvature

58






will naturally develop. However, this would bedifficult to control, and it is probably better toincorporate forming rolls that develop a con-trolled radius of curvature, similar to the guiderolls in ring rolling (see Fig. 6.43 on p. 305).

6.130 Survey the technical literature, and describe thedesign features of the various roll arrangementsshown in Fig. 6.41.

By the student. This is an open-ended problem,and a wide variety of answers are possible basedon the student’s interpretation of “design fea-tures”. Students can develop descriptions basedon cost, process capability, stiffness of the rollarrangement or the materials rolled.

6.131 The beneficial effects of using ultrasonic vibra-tion to reduce friction in some of the processeswere described in this chapter. Survey the tech-nical literature and offer design concepts to ap-ply such vibrations.

By the student. This is a good topic for litera-ture search for a student paper.

6.132 In the Case Study at the end of this chapter,it was stated that there was a significant costimprovement using forgings when compared tothe extrusion-based design. List and explainthe reasons why you think these cost savingswere possible.

There are several acceptable answers to thisquestion. Students should, for example, con-sider:

(a) By reducing the total number of parts, thetooling and assembly cost is significantlyreduced.

(b) Because it is a near net-shape process, ma-chining and finishing costs are significantlyreduced.

(c) Material costs may be very different; theextruded alloy, for example, may be moreexpensive than the forged alloy.

6.133 In the extrusion and drawing of brass tubesfor ornamental architectural applications, it isimportant to produce very smooth surface fin-ishes. List the relevant process parameters andmake manufacturing recommendations to pro-duce such tubes.

By the student. This is an open-ended problem,and the students should consider, at a mini-mum, the following factors:

• A thick lubricant film will generally leadto orange peel.

• A thin lubricant film can lead to wear andmaterial transfer to the tooling and a grad-ual degradation in the surface produced.

• A polished die surface can produce asmooth workpiece surface; a rough surfacecannot.

• The workpiece material used may play arole in the shininess that can be achieved.

• The lubricant can stain the workpiece un-less properly formulated.

59






60






Chapter 7

Sheet-Metal Forming Processes

Questions

7.1 Select any three topics from Chapter 2, and,with specific examples for each, show their rel-evance to the topics described in this chapter.

This is an open-ended problem, and studentscan develop a wide range of acceptable answers.Some examples are:

• Yield stress and elastic modulus, describedin Section 2.2 starting on p. 30, have,for example, applicability to prediction ofspringback.

• Ultimate tensile strength is important fordetermining the force required in blanking;see Eq. (7.4) on p. 353.

• Strain-hardening exponent has been re-ferred to throughout this chapter, espe-cially as it relates to the formability ofsheet metals.

• Strain is used extensively, most directly inthe development of a forming limit dia-gram, such as that shown in Fig. 7.63a onp. 399.

7.2 Do the same as for Question 7.1, but for Chap-ter 3.

This is an open-ended problem, and studentscan develop a wide range of acceptable answers.Consider, for examples:

• Grain size and its effects on strength (Sec-tion 3.4 starting on p. 91), as well as theeffect of cold working on grain size, (see

Section 3.3.4) have a major influence onformability (Section 7.7 on p. 397).

• The material properties of the differentmaterials, described in Section 3.11, indi-cating materials that can be cold rollinginto sheets.

7.3 Describe (a) the similarities and (b) the differ-ences between the bulk-deformation processesdescribed in Chapter 6 and the sheet-metalforming processes described in this chapter.

By the student. The most obvious differencebetween sheet-metal parts and those made bybulk-deformation processes, described in Chap-ter 6, is the difference in cross section or thick-ness of the workpiece. Sheet-metal parts typi-cally have less net volume and are usually mucheasier to deform or flex. Sheet-metal parts arerarely structural unless they are loaded in ten-sion (because otherwise their small thicknesscauses them to buckle at relatively low loads)or they are fabricated to produce high sectionmodulus. They can be very large by assemblingindividual pieces, as in the fuselage of an air-craft. Structural parts that are made by forgingand extrusion are commonly loaded in variousconfigurations.

7.4 Discuss the material and process variables thatinfluence the shape of the curve for punch forcevs. stroke for shearing, such as that shownin Fig. 7.7 on p. 354, including its height andwidth.

61






The factors that contribute to the punch forceand how they affect this force are:

(a) the shear strength of the material and itsstrain-hardening exponent; they increasethe force,

(b) the area being sheared and the sheetthickness; they increase the force and thestroke,

(c) the area that is being burnished by rub-bing against the punch and die walls; itincreases the force, and

(d) parameters such as punch and die radii,clearance, punch speed, and lubrication.

7.5 Describe your observations concerning Figs. 7.5and 7.6.

The student should comment on the magnitudeof the deformation zone in the sheared region,as influenced by clearance and speed of opera-tion, and its influence on edge quality and hard-ness distribution throughout the edge. Note thehigher temperatures observed in higher-speedshearing. Other features depicted in Fig. 7.5on p. 352 should also be commented upon.

7.6 Inspect a common paper punch and commenton the shape of the tip of the punch as com-pared with those shown in Fig. 7.12.

By the student. Note that most punches areunlike those shown in Fig. 7.12 on p. 346; theyhave a convex curved shape.

7.7 Explain how you would estimate the tempera-ture rise in the shear zone in a shearing opera-tion.

Refer to Fig. 7.6 on p. 353 and note that wecan estimate the shear strain γ to which theshearing zone is subjected. This is done by con-sidering the definition of simple shear, given byEq. (2.2) on p. 30, and comparing this defor-mation with the deformation of grid patterns inthe figure. Then refer to the shear stress-shearstrain curve of the particular material beingsheared, and obtain the area under the curveup to that particular shear strain, just as wehave done in various other problems in the text.This will give the shearing energy per unit vol-ume. We then refer to Eq. (2.65) on p. 73 and

knowing the physical properties of the material,calculate the theoretical temperature rise.

7.8 As a practicing engineer in manufacturing, whywould you be interested in the shape of thecurve shown in Fig. 7.7? Explain.

The shape of the curve in Fig. 7.7 on p. 354 willgive us the following information:

(a) height of the curve: the maximum punchforce,

(b) area under the curve: the energy requiredfor this operation,

(c) horizontal magnitude of the curve: thepunch travel required to complete theshearing operation.

It is apparent that all this information shouldbe useful to a practicing engineer in regard tothe machine tool and the energy level required.

7.9 Do you think the presence of burrs can be ben-eficial in certain applications? Give specific ex-amples.

The best example generally given for this ques-tion is mechanical watch components, such assmall gears whose punched holes have a verysmall cross-sectional area to be supported bythe spindle or shaft on which it is mounted. Thepresence of a burr enlarges this contact areaand, thus, the component is better supported.As an example, note how the burr in Fig. 7.5on p. 352 effectively increases the thickness ofthe sheet.

7.10 Explain why there are so many different typesof tool and die materials used for the processesdescribed in this chapter.

By the student. Among several reasons are thelevel of stresses and type of loading involved(such as static or dynamic), relative sliding be-tween components, temperature rise, thermalcycling, dimensional requirements and size ofworkpiece, frictional considerations, wear, andeconomic considerations.

7.11 Describe the differences between compound,progressive, and transfer dies.

This topic is explained in Section 7.3.2 startingon p. 356. Basically, a compound die performs

62






several operations in one stroke at one die sta-tion. A progressive die performs several opera-tions, one per stroke, at one die station (morethan one stroke is necessary). A transfer dieperforms one operation at one die station.

7.12 It has been stated that the quality of thesheared edges can influence the formability ofsheet metals. Explain why.

In many cases, sheared edges are subjected tosubsequent forming operations, such as bend-ing, stretching, and stretch flanging. As statedin Section 7.3 starting on p. 351, rough edgeswill act as stress raisers and cold-worked edges(see Fig. 7.6b on p. 353) may not have suffi-cient ductility to undergo severe tensile strainsdeveloped during these subsequent operations.

7.13 Explain why and how various factors influencespringback in bending of sheet metals.

Plastic deformation (such as in bending pro-cesses) is unavoidably followed by elastic re-covery, since the material has a finite elasticmodulus (see Fig. 2.3 on p. 33). For a givenelastic modulus, a higher yield stress results ina greater springback because the elastic recov-ery strain is greater. A higher elastic modu-lus with a given yield stress will result in lesselastic strain, thus less springback. Equation(7.10) on p. 364 gives the relation between ra-dius and thickness. Thus, increasing bend ra-dius increases springback, and increasing thesheet thickness reduces the springback.

7.14 Does the hardness of a sheet metal have an ef-fect on its springback in bending? Explain.

Recall from Section 2.6.8 on p. 54 that hard-ness is related to strength, such as yield stressas shown in Fig. 2.24 on p. 55. Referring toEq. (7.10) on p. 364 , also note that the yieldstress, Y , has a significant effect on springback.Consequently, hardness is related to spring-back. Note that hardness does not affect theelastic modulus, E, given in the equation.

7.15 As noted in Fig. 7.16, the state of stress shiftsfrom plane stress to plane strain as the ratioof length-of-bend to sheet thickness increases.Explain why.

This situation is somewhat similar to rollingof sheet metal where the wider the sheet, thecloser it becomes to the plane-strain condition.In bending, a short length in the bend areahas very little constraint from the unbent re-gions, hence the situation is one of basicallyplane stress. On the other hand, the greaterthe length, the more the constraint, thus even-tually approaching the state of plane strain.

7.16 Describe the material properties that have aneffect on the relative position of the curvesshown in Fig. 7.19.

Observing curves (a) and (c) in Fig. 7.19 onp. 364, note that the former is annealed andthe latter is heat treated. Since these are allaluminum alloys and, thus, have the same elas-tic modulus, the difference in their springbackis directly attributable to the difference in theiryield stress. Likewise, comparing curves (b),(d), and (e), note that they are all stainlesssteels and, thus, have basically the same elas-tic modulus. However, as the amount of coldwork increases (from annealed to half-hard con-dition), the yield stress increases significantlybecause austenitic stainless steels have a high nvalue (see Table 2.3 on p. 37). Note that thesecomparisons are based on the same R/T ratio.

7.17 In Table 7.2, we note that hard materials havehigher R/t ratios than soft ones. Explain why.

This is a matter of the ductility of the material,particularly the reduction in area, as depictedby Eqs. (7.6) on p. 361 and (7.7) on p. 362.Thus, hard material conditions mean lower ten-sile reduction and, therefore, higher R/T ra-tios. In other words, for a constant sheet thick-ness, T , the bend radius, R, has to be larger forhigher bendability.

7.18 Why do tubes have a tendency to buckle whenbent? Experiment with a straight soda straw,and describe your observations.

Recall that, in bending of any section, one-halfof the cross section is under tensile stresses andthe other half under compressive stresses. Also,compressing a column tends to buckle it, de-pending on its slenderness. Bending of a tubesubjects it to the same state of stress, and since

63






most tubes have a rather small thickness com-pared to their diameter, there is a tendencyfor the compression side of the tube to buckle.Thus, the higher the diameter-to-thickness ra-tio, the greater the tendency to buckle duringbending.

7.19 Based on Fig. 7.22, sketch and explain theshape of a U-die used to produce channel-shaped bends.

The design would be a mirror image of thesketches given in Fig. 7.22b on p. 356 alonga vertical axis. For example, the image be-low was obtained from S. Kalpakjian, Manu-facturing Processes for Engineering Materials,1st ed., 1984, p. 415.

7.20 Explain why negative springback does not oc-cur in air bending of sheet metals.

The reason is that in air bending (shown inFig. 7.24a on p. 368), the situation depicted inFig. 7.20 on p. 365 cannot develop. Bendingin the opposite direction, as depicted betweenstages (b) and (c), cannot occur because of theabsence of a lower “die” in air bending.

7.21 Give examples of products in which the pres-ence of beads is beneficial or even necessary.

The student is encouraged to observe vari-ous household products and automotive com-ponents to answer this question. For example,along the rim of many sheet-metal cooking pots,a bead is formed to confine the burr and preventcuts from handling the pot. Also, the bead in-creases the section odulus, making th pot stifferin the diametral direction.

7.22 Assume that you are carrying out a sheet-forming operation and you find that the mate-rial is not sufficiently ductile. Make suggestionsto improve its ductility.

By the student. This question can be answeredin a general way by describing the effects oftemperature, state of stress, surface finish, de-formation rate, etc., on the ductility of metals.

7.23 In deep drawing of a cylindrical cup, is it alwaysnecessary that there to be tensile circumferen-tial stresses on the element in the cup wall, ashown in Fig. 7.50b? Explain.

The reason why there may be tensile hoopstresses in the already formed cup in Fig. 7.50bon p. 388 is due to the fact that the cup can betight on the punch during drawing. That is whythey often have to be stripped from the punchwith a stripper ring, as shown in Fig. 7.49a onp. 387. There are situations, however, whereby,depending on material and process parameters,the cup is sufficiently loose on the punch so thatthere are no tensile hoop stresses developed.

7.24 When comparing hydroforming with the deep-drawing process, it has been stated that deeperdraws are possible in the former method. Withappropriate sketches, explain why.

The reason why deeper draws can be obtainedby the hydroform process is that the cup beingformed is pushed against the punch by the hy-drostatic pressure in the dome of the machine(see Fig. 7.34 on p. 375). This means that thecup is traveling with the punch in such a waythat the longitudinal tensile stresses in the cupwall are reduced, by virtue of the frictional re-sistance at the interface. With lower tensilestresses, deeper draws can be made, i.e., theblank diameter to punch diameter ratio can begreater. A similar situation exists in drawingof tubes through dies with moving or station-ary mandrels, as discussed in O. Hoffman andG. Sachs, Introduction to the Theory of Plastic-ity for Engineers, McGraw-Hill, 1953, Chapter17.

7.25 We note in Fig. 7.50a that element A in theflange is subjected to compressive circumferen-tial (hoop) stresses. Using a simple free-bodydiagram, explain why.

This is shown simply by a free-body diagram,as illustrated below. Note that friction betweenthe blank and die and the blankholder also con-tribute to the magnitude of the tensile stress.

64






+

7.26 From the topics covered in this chapter, list andexplain specifically several examples where fric-tion is (a) desirable and (b) not desirable.

By the student. This is an open-ended problem.For example, friction is desirable in rolling, butit is generally undesirable for most forming op-erations.

7.27 Explain why increasing the normal anisotropy,R, improves the deep drawability of sheet met-als.

The answer is given at the beginning of Section7.6.1. The student is encouraged to elaboratefurther on this topic.

7.28 What is the reason for the negative sign in thenumerator of Eq. (7.21)?

The negative sign in Eq. (7.21) on p. 392 is sim-ply for the purpose of indicating the degree ofplanar anisotropy of the sheet. Note that if theR values in the numerator are all equal, then∆R = 0, thus indicating no planar anisotropy,as expected.

7.29 If you could control the state of strain in asheet-forming operation, would you rather workon the left or the right side of the forming-limitdiagram? Explain.

By inspecting Fig. 7.63a on p. 399, it is appar-ent that the left side has a larger safe zone thanthe right side, under each curve. Consequently,it is more desirable to work in a state of strainon the left side.

7.30 Comment on the effect of lubrication of thepunch surfaces on the limiting drawing ratio indeep drawing.

Referring to Fig. 7.49 on p. 387, note that lu-bricating the punch is going to increase the lon-gitudinal tensile stress in the cup being formed

(Fig. 7.50b on p. 388). Thus, deep drawabil-ity will decrease, hence the limited drawing ra-tio will also decrease. Conversely, not lubricat-ing the punch will allow the cup to travel withthe punch, thus reducing the longitudinal ten-sile stress.

7.31 Comment on the role of the size of the circlesplaced on the surfaces of sheet metals in deter-mining their formability. Are square grid pat-terns, as shown in Fig. 7.65, useful? Explain.

We note in Fig. 7.65 on p. 400 that, obviously,the smaller the inscribed circles, the more ac-curately we can determine the magnitude andlocation of strains on the surface of the sheetbeing formed. These are important consider-ations. Note in the figure, for example, howlarge the circles are as compared with the sizeof the crack that has developed. As for squaregrid patters, their distortion will not give a clearand obvious indication of the major and minorstrains. Although they can be determined fromgeometric relationships, it is tedious work to doso.

7.32 Make a list of the independent variables thatinfluence the punch force in deep drawing of acylindrical cup, and explain why and how thesevariables influence the force.

The independent variables are listed at the be-ginning of Section 7.6.2. The student should beable to explain why each variable influences thepunch force, based upon a careful reading of thematerials presented. The following are sampleanswers, but should not be considered the onlyacceptable ones.

(a) The blank diameter affects the forcebecause the larger the diameter, thegreater the circumference, and thereforethe greater the volume of material to bedeformed.

(b) The clearance, c, between the punch anddie directly affects the force; the smallerthe clearance the greater the thickness re-duction and hence the work involved.

(c) The workpiece properties of yield strengthand strain-hardening exponent affect theforce because as these increase, greaterforces will be required to cause deforma-tion beyond yielding.

65






(d) Blank thickness also increases the vol-ume deformed, and therefore increases theforce.

(e) The blankholder force and friction affectthe punch force because they restrict theflow of the material into the die, hence ad-ditional energy has to be supplied to over-come these forces.

7.33 Explain why the simple tension line in theforming-limit diagram in Fig. 7.63a states thatit is forR = 1, where R is the normal anisotropyof the sheet.

Note in Fig. 7.63a on p. 399 that the slope forsimple tension is 2, which is a reflection of thePoisson’s ratio in the plastic range. In otherwords, the ratio of minor strain to major strainis -0.5. Recall that this value is for a mate-rial that is homogeneous and isotropic. Isotropymeans that the R value must be unity.

7.34 What are the reasons for developing forming-limit diagrams? Do you have any specific criti-cisms of such diagrams? Explain.

The reasons for developing the FLD diagramsare self-evident by reviewing Section 7.7.1.Criticisms pertain to the fact that:

(a) the specimens are still somewhat idealized,

(b) frictional conditions are not necessarilyrepresentative of actual operations, and

(c) the effects of bending and unbending dur-ing actual forming operations, the pres-ence of beads, die surface conditions, etc.,are not fully taken into account.

7.35 Explain the reasoning behind Eq. (7.20) fornormal anisotropy, and Eq. (7.21) for planaranisotropy, respectively.

Equation (7.20) on p. 391 represents an averageR value by virtue of the fact that all directions(at 45circ intervals) are taken into account.

7.36 Describe why earing occurs. How would youavoid it? Would ears serve any useful purposes?Explain.

Earing, described in Section 7.6.1 on p. 394, isdue to the planar anisotropy of the sheet metal.Consider a round blank and a round die cavity;

if there is planar anisotropy, then the blank willhave less resistance to deformation in some di-rections compared to others, and will thin morein directions of greater resistance, thus develop-ing ears.

7.37 It was stated in Section 7.7.1 that the thickerthe sheet metal, the higher is its curve in theforming-limit diagram. Explain why.

In forming-limit diagrams, increasing thicknesstends to raise the curves. This is because thematerial is capable of greater elongations sincethere is more material to contribute to length.

7.38 Inspect the earing shown in Fig. 7.57, and esti-mate the direction in which the blank was cut.

The rolled sheet is stronger in the directionof rolling. Consequently, that direction resistsflow into the die cavity during deep drawing andthe ear is at its highest position. In Fig. 7.57on p. 394, the directions are at about ±45 onthe photograph.

7.39 Describe the factors that influence the size andlength of beads in sheet-metal forming opera-tions.

The size and length of the beads depends on theparticular blank shape, die shape, part depth,and sheet thickness. Complex shapes requirecareful placing of the beads because of the im-portance of sheet flow control into the desiredareas in the die.

7.40 It is known that the strength of metals dependson their grain size. Would you then expectstrength to influence the R value of sheet met-als? Explain.

It seen from the Hall-Petch Eq. (3.8) on p. 92that the smaller the grain size, the higher theyield strength of the metal. Since grain size alsoinfluences the R values, we should expect thatthere is a relationship between strength and Rvalues.

7.41 Equation (7.23) gives a general rule for dimen-sional relationships for successful drawing with-out a blankholder. Explain what would happenif this limit is exceeded.

66






If this limit is exceeded, the blank will beginto wrinkle and we will produce a cup that haswrinkled walls.

7.42 Explain why the three broken lines (simple ten-sion, plane strain, and equal biaxial stretching)in Fig. 7.63a have those particular slopes.

Recall that the major and minor strains shownin Fig. 7.63 on p. 399 are both in the plane ofthe sheet. Thus, the simple tension curve has anegative slope of 2:1, reflecting the Poisson’s ra-tio effect in plastic deformation. In other words,the minor strain is one-half the major strain insimple tension, but is opposite in sign. Theplane-strain line is vertical because the minorstrain is zero in plane-strain stretching. Theequal (balanced) biaxial curve has to have a45 slope because the tensile strains are equalto each other. The curve at the farthest left isfor pure shear because, in this state of strain,the tensile and compressive strains are equal inmagnitude (see also Fig. 2.20 on p. 49).

7.43 Identify specific parts on a typical automobile,and explain which of the processes described inChapters 6 and 7 can be used to make thosepart. Explain your reasoning.

By the student. Some examples would be:

(a) Body panels are obtained through sheet-metal forming and shearing.

(b) Frame members (only visible when lookedat from underneath) are made by rollforming.

(c) Ash trays are made from stamping, com-bined with shearing.

(d) Oil pans are classic examples of deep-drawn parts.

7.44 It was stated that bendability and spinnabilityhave a common aspect as far as properties ofthe workpiece material are concerned. Describethis common aspect.

By comparing Fig. 7.15b on p. 360 on bend-ability and Fig. 7.39 on p. 379 on spinnabil-ity, we note that maximum bendability andspinnability are obtained in materials with ap-proximately 50% tensile reduction of area. Anyfurther increase in ductility does not improvethese forming characteristics.

7.45 Explain the reasons that such a wide varietyof sheet-forming processes has been developedand used over the years.

By the student, based on the type of productsthat are made by the processes described inthis chapter. This is a demanding question;ultimately, the reasons that sheet-forming pro-cesses have been developed are due to demandand economic considerations.

7.46 Make a summary of the types of defects foundin sheet-metal forming processes, and includebrief comments on the reason(s) for each de-fect.

By the student. Examples of defects include(a) fracture, which results from a number ofreasons including material defects, poor lubri-cation, etc; (b) poor surface finish, either fromscratching attributed to rough tooling or to ma-terial transfer to the tooling or orange peel; and(c) wrinkles, attributed to in-plane compressivestresses during forming.

7.47 Which of the processes described in this chap-ter use only one die? What are the advantagesof using only one die?

The simple answer is to restrict the discussionto rubber forming (Fig. 7.33 on p. 375) andhydroforming (Fig. 7.34 on p. 375), althoughexplosive forming or even spinning could alsobe discussed. The main advantage is that onlyone tool needs to be made or purchased, asopposed to two matching dies for conventionalpressworking and forming operations.

7.48 It has been suggested that deep drawability canbe increased by (a) heating the flange and/or(b) chilling the punch by some suitable means.Comment on how these methods could improvedrawability.

Refering to Fig. 7.50, we note that:

(a) heating the flange will lower the strengthof the flange and it will take less energyto deform element A in the figure, thus itwill require less punch force. This will re-duce the tendency for cup failure and thusimprove deep drawability.

67






(b) chilling the punch will increase thestrength of the cup wall, hence the ten-dency for cup failure by the longitudinaltensile stress on element B will be less, anddeep drawability will be improved.

7.49 Offer designs whereby the suggestions given inQuestion 7.48 can be implemented. Would pro-duction rate affect your designs? Explain.

This is an open-ended problem that requiressignificant creativity on the part of the stu-dent. For example, designs that heat the flangemay involve electric heating elements in theblankholder and/or the die, or a laser as heatsource. Chillers could be incorporated in thedie and the blankholder, whereby cooled wateris circulated through passages in the tooling.

7.50 In the manufacture of automotive-body pan-els from carbon-steel sheet, stretcher strains(Lueder’s bands) are observed, which detrimen-tally affect surface finish. How can stretcherstrains be eliminated?

The basic solution is to perform a temperrolling pass shortly before the forming opera-tion, as described in Section 6.3.4 starting onp. 301. Another solution is to modify the de-sign so that Lueders bands can be moved toregions where they are not objectionable.

7.51 In order to improve its ductility, a coil of sheetmetal is placed in a furnace and annealed. How-ever, it is observed that the sheet has a lowerlimiting drawing ratio than it had before beingannealed. Explain the reasons for this behavior.

When a sheet is annealed, it becomes lessanisotropic; the discussion of LDR in Section7.6.1 would actually predict this behavior. Themain reason is that, when annealed, the mate-rial has a high strain-hardening exponent. Asthe flange becomes subjected to increasing plas-tic deformation (as the cup becomes deeper),the drawing force increases. If the material isnot annealed, then the flange does not strainharden as much, and a deeper container can bedrawn.

7.52 What effects does friction have on a forming-limit diagram? Explain.

By the student. Friction can have a strong ef-fect on formability. High friction will cause lo-calized strains, so that formability is decreased.Low friction allows the sheet to slide more eas-ily over the die surfaces and thus distribute thestrains more evenly.

7.53 Why are lubricants generally used in sheet-metal forming? Explain, giving examples.

Lubricants are used for a number of reasons.Mainly, they reduce friction, and this improvesformability as discussed in the answer to Prob-lem 7.52. As an example of this, lightweightoils are commonly applied in stretch formingfor automotive body panels. Another reason isto protect the tooling from the workpiece mate-rial; an example is the lubricant in can ironingwhere aluminum pickup can foul tooling andlead to poor workpiece surfaces. The student isencouraged to pursue other reasons. (See alsoSection 4.4 starting on p. 138.)

7.54 Through changes in clamping, a sheet-metalforming operation can allow the material to un-dergo a negative minor strain in the FLD. Ex-plain how this effect can be advantageous.

As can be seen from Fig. 7.63a on p. 399, ifa negative minor strain can be induced, thena larger major strain can be achieved. If theclamping change is less restrictive in the mi-nor strain direction, then the sheet can contractmore in this direction and thus allow larger ma-jor strains to be achieved without failure.

7.55 How would you produce the parts shown inFig. 7.35b other than by tube hydroforming?

By the student. The part could be producedby welding sections of tubing together, or by asuitable casting operation. Note that in eithercase production costs are likely to be high andproduction rates low.

7.56 Give three examples each of sheet metal partsthat (a) can and (b) cannot be produced byincremental forming operations.

By the student. This is an open-ended problemthat requires some consideration and creativityon the part of the student. Consider, for exam-ple:

68






(a) Parts that can be formed are light fixtures,automotive body panels, kitchen utensils,and hoppers.

(b) Incremental forming is a low force oper-ation with limited size capability (limitedto the workspace of the CNC machine per-forming the operation). Examples of partsthat cannot be incrementally formed arespun parts where the thickness of the sheetis reduced, or very large parts such as theaircraft wing panels in Fig. 7.30 on p. 372.Also, continuous parts such as roll-formedsections and parts with reentrant cornerssuch as those with hems or seams are notsuitable for incremental forming.

7.57 Due to preferred orientation (see Section 3.5),materials such as iron can have higher mag-netism after cold rolling. Recognizing this fea-ture, plot your estimate of LDR vs. degree ofmagnetism.

By the student. There should be a realizationthat there is a maximum magnetism with fullyaligned grains, and zero magnetism with fullyrandom orientations. The shape of the curvebetween these extremes is not intuitively obvi-ous, but a linear relationship can be expected.

7.58 Explain why a metal with a fine-grain mi-crostructure is better suited for fine blankingthan a coarse-grained metal.

A fine-blanking operation can be demanding;the clearances are very low, the tooling is elab-orate (including stingers and a lower pressurecushion), and as a result the sheared surfacequality is high. The sheared region (see Fig. 7.6on p. 353) is well defined and constrained toa small volume. It is beneficial to have manygrain boundaries (in the volume that is frac-turing) in order to have a more uniform andcontrolled crack.

7.59 What are the similarities and differences be-tween roll forming described in this chapter andshape rolling in Chapter 6?

By the student. Consider, for example:

(a) Similarities include the use of rollers tocontrol the material flow, the productionof parts with constant cross section, andsimilar production rates.

(b) Differences include the mode of deforma-tion (bulk strain vs. bending and stretch-ing of sheet metal), and the magnitude ofthe associated forces and torques.

7.60 Explain how stringers can adversely affectbendability. Do they have a similar effects onformability?

Stingers, as shown in Fig. 7.17, have an adverseaffect on bendability when they are orientedtransverse to the bend direction. The basic rea-son is that stringers are hard and brittle inclu-sions in the sheet metal and thus serve as stressconcentrations. If they are transverse to thisdirection, then there is no stress concentration.

7.61 In Fig. 7.56, the caption explains that zinc hasa high c/a ratio, whereas titanium has a lowratio. Why does this have relevance to limitingdrawing ratio?

This question can be best answered by refer-ring to Fig. 3.4 and reviewing the discussion ofslip in Section 3.3. For titanium, the c/a ra-tio in its hcp structure is low, hence there areonly a few slip systems. Thus, as grains becomeoriented, there will be a marked anisotropy be-cause of the highly anisotropic grain structure.On the other hand, with magnesium, with ahigh c/a ratio, there are more slip systems (out-side of the close-packed direction) active andthus anisotropy will be less pronounced.

7.62 Review Eqs. (7.12) through (7.14) and explainwhich of these expressions can be applied to in-cremental forming.

By the student. These equations are applicablebecause the deformation in incremental form-ing is highly localized. Note that the strain re-lationships apply to a shape as if a mandrel waspresent.

69






Problems

7.63 Referring to Eq. (7.5), it is stated that actualvalues of eo are significantly higher than val-ues of ei, due to the shifting of the neutral axisduring bending. With an appropriate sketch,explain this phenomenon.

The shifting of the neutral axis in bending isdescribed in mechanics of solids texts. Briefly,the outer fibers in tension shrink laterally dueto the Poisson’ effect (see Fig. 7.17c), and theinner fibers expand. Thus, the cross sectionis no longer rectangular but has the shape of atrapezoid, as shown below. The neutral axis hasto shift in order to satisfy the equilibrium equa-tions regarding forces and internal moments inbending.

Before After

Change in neutral axis

location

7.64 Note in Eq. (7.11) that the bending force is afunction of t2. Why? (Hint: Consider bending-moment equations in mechanics of solids.)

This question is best answered by referring toformulas for bending of beams in the study ofmechanics of solids. Consider the well-knownequation

σ =Mc

I

where c is directly proportional to the thick-ness, and I is directly proportional to the thirdpower of thickness. For a cantilever beam, theforce can be taken as F = M/L, where L is themoment arm. For plastic deformation, σ is thematerial flow stress. Therefore:

σ =Mc

I∝ FLt

t3

and thus,

F ∝ σt2

L

7.65 Calculate the minimum tensile true fracturestrain that a sheet metal should have in orderto be bent to the following R/t ratios: (a) 0.5,(b) 2, and (c) 4. (See Table 7.2.)

To determine the true strains, we first refer toEq. (7.7) to obtain the tensile reduction of areaas a function of R/T as

R

T=

60r− 1

orr =

60(R/T + 1)

The strain at fracture can be calculated fromEq. (2.10) as

εf = ln(Ao

Af

)= ln

(100

100− r

)

= ln

100

100−(

60(R/T + 1)

)

This equation gives for R/T = 0.5, and εf isfound to be 0.51. For R/T = 2, we have εf =0.22, and for R/T = 4, εf = 0.13.

7.66 Estimate the maximum bending force requiredfor a 1

8 -in. thick and 12-in. wide Ti-5Al-2.5Sntitanium alloy in a V -die with a width of 6 in.

The bending force is calculated from Eq. (7.11).Note that Section 7.4.3 states that k takes arange from 1.2 to 1.33 for a V-die, so an aver-age value of k = 1.265 will be used. From Table3.14, we find that UTS=860 MPa = 125,000 psi.Also, the problem statement gives us L = 12in., T = 1

8 in = 0.125 in, and W = 6 in. There-fore, Eq. (7.11) gives

Fmax = k(UTS)LT 2

W

= (1.265)(125, 000)(12)(0.125)2

6= 4940 lb

7.67 In Example 7.4, calculate the work done by theforce-distance method, i.e., work is the integral

70






product of the vertical force, F , and the dis-tance it moves.

Let the angle opposite to α be designated as βas shown.

10 in. 5 in.

F

a b

α β

Since the tension in the bar is constant, theforce F can be expressed as

F = T (sinα+ sinβ)

where T is the tension and is given by

T = σA =(100, 000ε0.3

)A

The area is the actual cross section of the barat any position of the force F , obtained fromvolume constancy. We also know that the truestrain in the bar, as it is being stretched, isgiven by

ε = ln(a+ b

15

)Using these relationships, we can plot F vs. d.Some of the points on the curve are:

α () d (in.) ε T (kip) F (kip)5 0.87 0.008 11.5 2.9810 1.76 0.03 16.9 8.5815 2.68 0.066 20.7 15.120 3.64 0.115 23.3 21.7

The curve is plotted as follows and the integralis evaluated (from a graphing software package)as 34,600 in-lb.

20,000

15,000

10,000

5,000

0

F, lb

0 1 2 3d, in

7.68 What would be the answer to Example 7.4 ifthe tip of the force, F , were fixed to the stripby some means, thus maintaining the lateralposition of the force? (Hint: Note that the leftportion of the strip will now be strained morethan the right portion.)

In this problem, the work done must be calcu-lated for each of the two members. Thus, forthe left side, we have

a =10 in.cos 20

= 10.64 in.

where the true strain is

εa = ln(

10.6410

)= 0.062

It can easily be shown that the angle β corre-sponding to α = 20 is 36. Hence, for the leftportion,

b =(5in.)cos 36

= 6.18 in.

and the true strain is

εb = ln(

6.185

)= 0.21

Thus, the total work done is

W = (10)(0.5)(100, 000)∫ 0.062

0

ε0.3 dε

+(5)(0.5)(100, 000)∫ 0.21

0

ε0.3 dε

= 35, 700 in.-lb

7.69 Calculate the magnitude of the force F in Ex-ample 7.4 for α = 30.

See the solution to Problem 7.67 for the rele-vant equations. For α = 30,

d = (10 in.) tanα = 5.77 in.

also, T = 25.7 kip and F = 32.2 kip.

7.70 How would the force in Example 7.4 vary if theworkpiece were made of a perfectly-plastic ma-terial?

We refer to the solution to Problem 7.67 andcombine the equations for T and F ,

F = σA (sinα+ sinβ)

71






Whereas Problem 7.67 pertained to a strain-hardening material, in this problem the truestress σ is a constant at Y regardless of themagnitude of strain. Inspecting the table in theanswer, we note that as the downward travel,d, increases, F must increase as well becausethe rate of increase in the term (sinα+sinβ) ishigher than the rate of decrease of the cross-sectional area. However, F will not rise asrapidly as it does for a strain-hardening ma-terial because σ is constant.

Note that an equation such as Eq. (2.60) onp. 71 can give an effective yield stress for astrain-hardening material. If such a value isused, F would have a large value for zero de-flection. The effect is that the curve is shiftedupwards and flattened. The integral under thecurve would be the same.

7.71 Calculate the press force required in punch-ing 0.5-mm-thick 5052-O aluminum foil in theshape of a square hole 30 mm on each side.

The approach is the same as in Example 7.1.The press force is given by Eq. (7.4) on p. 353:

Fmax = 0.7(UTS)(t)(L)

For this problem, UTS=190 MPa (see Table 3.7on p. 116). The distance L is 4(30 mm) = 120mm, and the thickness is given as t=0.5 mm.Therefore,

Fmax = 0.7(190)(0.5)(120) = 7980 N

7.72 A straight bead is being formed on a 1-mm-thick aluminum sheet in a 20-mm-diameter diecavity, as shown in the accompanying figure.(See also Fig. 7.25a.) Let Y = 150 MPa. Con-sidering springback, calculate the outside diam-eter of the bead after it is formed and unloadedfrom the die.

20

R

For this aluminum sheet, we have Y = 150 MPaand E = 70 GPa (see Table 2.1 on p. 32). UsingEq. (7.10) on p. 364 for springback, and notingthat the die has a diameter of 20 mm and thesheet thickness is T = 1 mm, the initial bendradius is

Ri =20 mm

2− 1 mm = 9 mm

Note thatRiY

ET=

(0.009)(150)(70, 000)(0.001)

= 0.0193

Therefore, Eq. (7.10) on p. 364 yields

Ri

Rf= 4

(RiY

ET

)3

− 3(RiY

ET

)+ 1

= 4(0.0193)3 − 3(0.0193) + 1= 0.942

and,

Rf =Ri

0.942=

9 mm0.942

= 9.55 mm

Hence, the final outside diameter will be

OD = 2Rf + 2T= 2(9.55 mm) + 2(1 mm)= 21.1 mm

7.73 Inspect Eq. (7.10) and substituting in some nu-merical values, show whether the first term inthe equation can be neglected without signifi-cant error in calculating springback.

As an example, consider the situation in Prob-lem 7.72 where it was shown that

RiY

ET=

(0.009)(150)(70, 000)(0.001)

= 0.0193

72






Consider now the right side of Eq. (7.10) onp. 364 :

4(RiY

ET

)3

− 3(RiY

ET

)+ 1

Substituting the value from Problem 7.72,

4(0.0193)3 − 3(0.0193) + 1

which is

2.88× 10−5 − 0.058 + 1

Clearly, the first term is small enough to ignore,which is the typical case.

7.74 In Example 7.5, calculate the amount of TNTrequired to develop a pressure of 10,000 psi onthe surface of the workpiece. Use a standoff ofone foot.

Using Eq. (7.17) on p. 381 we can write

p = K

(3√W

R

)a

Solving for W ,

W =( pK

)3/a

R3

=(

1000021600

)3/1.15

(1)3 = 0.134 lb

7.75 Estimate the limiting drawing ratio (LDR) forthe materials listed in Table 7.3.

Referring to Fig. 7.58 on p. 395, we constructthe following table:

Average Limitednormal drawing

Material anisotropy ratioZinc alloys 0.4-0.6 1.8Hot-rolled steel 0.8-1.0 2.3-2.4Cold-rolled rimmed 1.0-1.4 2.3-2.5

steelCold-rolled Al-killed 1.4-1.8 2.5-2.6

steelAluminum alloys 0.6-.8 2.2-2.3Copper and brass 0.6-0.9 2.3-2.4Ti alloys (α) 3.0-5.0 2.9-3.0

7.76 For the same material and thickness as in Prob-lem 7.66, estimate the force required for deepdrawing with a blank of diameter 10 in. and apunch of diameter 9 in.

Note that Dp = 9 in., Do = 10 in., t0 =0.125 in., and UTS = 125,000 psi. Therefore,Eq. (7.22) on p. 395 yields

Fmax = πDpto(UTS)(Do

Dp− 0.7

)= π(9)(0.125)(125, 000)

(109− 0.7

)= 181, 000 lb

or Fmax = 90 tons.

7.77 A cup is being drawn from a sheet metal thathas a normal anisotropy of 3. Estimate themaximum ratio of cup height to cup diameterthat can successfully be drawn in a single draw.Assume that the thickness of the sheet through-out the cup remains the same as the originalblank thickness.

For an average normal anisotropy of 3, Fig. 7.56on p. 392 gives a limited drawing ratio of 2.68.Assuming incompressibility, one can equate thevolume of the sheet metal in a cup to the vol-ume in the blank. Therefore,(π

4D2

o

)T = πDphT +

(π4D2

p

)T

This equation can be simplified as

π

4(D2

o −D2p

)= πDph

where h is the can wall height. Note that theright side of the equation includes a volume forthe wall as well as the bottom of the can. Thus,since Do/Dp = 2.68,

π

4

[(2.68Dp)

2 −D2p

]= πDph

orh

Dp=

2.682 − 14

= 1.55

7.78 Obtain an expression for the curve shown inFig. 7.56 in terms of the LDR and the averagenormal anisotropy, R (Hint: See Fig. 2.5b).

Referring to Fig. 7.56 on p. 392, note that thisis a log-log plot with a slope that is measured

73






to be 8. Therefore the exponent of the powercurve is tan 8 = 0.14. Furthermore, it canbe seen that, for R = 1.0, we have LDR=2.3.Therefore, the expression for the LDR as a func-tion of the average strain ratio R is given by

LDR = 2.3R0.14

7.79 A steel sheet has R values of 1.0, 1.5, and 2.0 forthe 0, 45 and 90 directions to rolling, respec-tively. If a round blank is 150 mm in diameter,estimate the smallest cup diameter to which itcan be drawn in one draw.

Substituting these values into Eq. (7.20) onp. 391 , we have

R =1.0 + 2(1.5) + 2.0

4= 1.5

The limiting-drawing ratio can be obtainedfrom Fig. 7.56 on p. 392, or it can be obtainedfrom the expression given in the solution toProblem 7.78 as

LDR = 2.3R0.14 = 2.43

Thus, the smallest diameter to which this ma-terial can be drawn is 150/2.43 = 61.7 mm.

7.80 In Problem 7.79, explain whether ears will formand, if so, why.

Equation (7.21) on p. 392 yields

∆R =R0 − 2R45 +R90

2

=1.0− 2(1.5) + 2.0

2= 0

Since ∆R = 0, no ears will form.

7.81 A 1-mm-thick isotropic sheet metal is inscribedwith a circle 4 mm in diameter. The sheet isthen stretched uniaxially by 25%. Calculate (a)the final dimensions of the circle and (b) thethickness of the sheet at this location.

Referring to Fig. 7.63b on p. 399 and notingthat this is a case of uniaxial stretching, thecircle will acquire the shape of an ellipse with apositive major strain and negative minor strain(due to the Poisson effect). The major axis ofthe ellipse will have undergone an engineeringstrain of (1.25-1)/1=0.25, and will thus have

the dimension (4)(1+0.25)=5 mm. Because wehave plastic deformation and hence the Pois-son’s ratio is ν = 0.5, the minor engineeringstrain is -0.25/2=-0.125; see also the simple-tension line with a negative slope in Fig. 7.63aon p. 399. Thus, the minor axis will have thedimension

x− 4 mm4 mm

= −0.125

or x = 3.5 mm. Since the metal is isotropic, itsfinal thickness will be

t− 1 mm1 mm

= 0− 0.125

or t = 0.875 mm. The area of the ellipse willbe

A = π

(5 mm

2

)(3.5 mm

2

)= 13.7 mm2

The volume of the original circle is

V =π

4(4 mm)2 (1 mm) = 12.6 mm3

7.82 Conduct a literature search and obtain theequation for a tractrix curve, as used inFig. 7.61.

The coordinate system is shown in the accom-panying figure.

y

x

The equation for the tractrix curve is

x = a ln

(a+

√a2 − y2

y

)−√a2 − y2

= a cosh−1

(a

y

)−√a2 − y2

where x is the position along the direction ofpunch travel, and y is the radial distance of thesurface from the centerline.

74






7.83 In Example 7.4, assume that the stretching isdone by two equal forces F , each at 6 in. fromthe ends of the workpiece. (a) Calculate themagnitude of this force for α = 10. (b)If we want the stretching to be done up toαmax = 50 without necking, what should bethe minimum value of n of the material?

(1) Refer to Fig. 7.31 on p. 373 and note thefollowing: (a) For two forces F at 6 in. fromeach end, the dimensions of the edge portionsat α = 10 will be 6/ cos 10 = 6.09 in. Thetotal deformed length will thus be

Lf = 6.09 + 3.00 + 6.09 = 15.18 in.

With a the true strain of

ε = ln(

15.1815

)= 0.0119

and true stress of

σ = Kεn = (100, 000)(0.0119)0.3 = 26, 460 psi

From volume constancy we can determine thestretched cross-sectional area,

Af =AoLo

Lf=

(0.05 in2)(15 in.)15.18

= 0.0494 in2

Consequently, the tensile force, which is uni-form throughout the stretched part, is

Ft = (26, 460 psi)(0.0495 in2) = 1310 lb

The force F will be the vertical component ofthe tensile force in the stretched member (not-ing that the middle horizontal 3-in. portiondoes not have a vertical component). There-fore

F =1310 lbtan 10

= 7430 lb

(2) For α = 50, we have the total length of thestretched part as

Lf = 2(

6 in.cos 50

)+ 3.00 in. = 21.67 in.

Hence the true strain will be

ε = ln(

21.6715

)= 0.368

The necking strain should be equal to thestrain-hardening exponent, or n = 0.368. Typ-ical values of n are given in Table 2.3 on p. 37.

Thus, 304 annealed stainless steel, phosphorbronze, or 70-30 annealed brass would be suit-able metals for this application, as n > 0.368for these materials.

7.84 Derive Eq. (7.5).

Referring to Fig. 7.15 on p. 360 and letting thebend-allowance length (i.e., length of the neu-tral axis) be lo, we note that

lo =(R+

T

2

)α

and the length of the outer fiber is

lf = (R+ T )α

where the angle α is in radians. The engineer-ing strain for the outer fiber is

eo =lf − lolo

=lflo− 1

Substituting the values of lf and lo, we obtain

eo =1(

2RT

)+ 1

7.85 Estimate the maximum power in shear spinninga 0.5-in. thick annealed 304 stainless-steel platethat has a diameter of 12 in. on a conical man-drel of α = 30. The mandrel rotates at 100rpm and the feed is f = 0.1 in./rev.

Referring to Fig. 7.36b on p. 377 we note that,in this problem, to = 0.5 in., α = 30, N = 100rpm, f = 0.1 in./rev., and, from Table 2.3 onp. 37, for this material K = (1275)(145) =185,000 psi and n = 0.45. The power requiredin the operation is a function of the tangentialforce Ft, given by Eq. (7.13) as

Ft = utof sinα

In order to determine u, we need to knowthe strain involved. This is calculated fromEq. (7.14) for the distortion-energy criterion as

ε =cotα√

3=

cot 30√3

= 1.0

and thus, from Eq. (2.60),

u =Kεn+1

n+ 1=

(185, 000)(1)1.45

1.45

75






or u = 127, 000 in-lb/in3. Therefore,

Ft = (127, 000)(0.5)(0.1)(sin 30) = 3190 lb

and the maximum torque required is at the 15in. diameter, hence

T = (3190 lb)(

12 in.2

)= 19, 140 in-lb

or T = 1590 ft-lb. Thus the maximum powerrequired is

Pmax = Tω

= (19, 140 in.-lb)(100 rev/min)×(2π rad/rev)

= 12.03× 106 in-lb/min

or 30.3 hp. As stated in the text, because ofredundant work and friction, the actual powermay be as much as 50% higher, or up to 45 hp.

7.86 Obtain an aluminum beverage can and cut it inhalf lengthwise with a pair of tin snips. Using amicrometer, measure the thickness of the bot-tom of the can and of the wall. Estimate (a)the thickness reductions in ironing of the walland (b) the original blank diameter.

Note that results will vary depending on thespecific can design. In one example, results fora can diameter of 2.6 in. and a height of 5in., the sidewall is 0.003 in. and the bottom is0.0120 in. thick. The wall thickness reductionin ironing is then

%red =to − tfto

× 100%

=0.0120− 0.003

0.012× 100%

= 75%

The initial blank diameter can be obtained byvolume constancy. The volume of the can ma-terial after deep drawing and ironing is

Vf =πd2

c

4to + πdtwh

=π(2.5)2

4(0.012) + π(2.5)(0.003)(5)

= 0.1767 in3

Since the initial blank has a thickness equal tothe final can bottom (i.e., 0.0120 in.) and adiameter d, the volume is

0.1767 in3 =πd2

4to =

πd2

4(0.012 in)

or d = 4.33 in.

7.87 What is the force required to punch a squarehole, 150 mm on each side, from a 1-mm-thick5052-O aluminum sheet, using flat dies? Whatwould be your answer if beveled dies were usedinstead?

This problem is very similar to Problem 7.71.The punch force is given by Eq. (7.4) on p. 353.Table 3.7 on p. 116 gives the UTS of 5052-O aluminum as UTS=190 MPa. The sheetthickness is t = 1.0 mm = 0.001 m, and L =(4)(150mm) = 600 mm = 0.60 m. Therefore,from Eq. (7.4) on p. 353,

Fmax = 0.7(UTS)(t)(L)= 0.7(190 MPa)(0.001 m)(0.60 m)= 79, 800 N = 79.8 kN

If the dies are beveled, the punch force couldbe much lower than calculated here. For a sin-gle bevel with contact along one face, the forcewould be calculated as 19,950 N, but for double-beveled shears, the force could be essentiallyzero.

7.88 Estimate the percent scrap in producing roundblanks if the clearance between blanks is onetenth of the radius of the blank. Consider sin-gle and multiple-row blanking, as shown in theaccompanying figure.

(a) A repeating unit cell for the part the upperillustration is shown below.

76






2R

0.1R

2.1R

The area of the unit cell is A =(2.2R)(2.1R) = 4.62R2. The area of thecircle is 3.14R2. Therefore, the scrap is

scrap =4.62R2 − 3.14R2

4.62R2× 100 = 32%

(b) Using the same approach, it can be shownthat for the lower illustration the scrap is26%.

7.89 Plot the final bend radius as a function of ini-tial bend radius in bending for (a) 5052-O alu-minum; (b) 5052-H34 Aluminum; (c) C24000brass and (d) AISI 304 stainless steel sheet.

The final bend radius can be determined fromEq. (7.10) on p. 364 . Solving this equation forRf gives:

Rf =Ri

4(RiY

Et

)3

− 3(RiY

Et

)+ 1

Using Tables 2.1 on p. 32, 3.4, 3.7, and 3.10,the following data is compiled:

Material Y (MPa) E (GPa)5052-O Al 90 735052-H34 210 73C24000 Brass 265 127AISI 304 SS 265 195

where mean values of Y and E have been as-signed. From this data, the following plot isobtained. Note that the axes have been definedso that the value of t is not required.

Rf/R

i

Ri/t

0 5 10 15 20

5052-H34C24000 Brass

304 SS5052-O

1.0

1.05

1.10

1.15

1.2

1.25

7.90 The accompanying figure shows a parabolicprofile that will define the mandrel shape in aspinning operation. Determine the equation ofthe parabolic surface. If a spun part is to beproduced from a 10-mm thick blank, determinethe minimum blank diameter required. Assumethat the diameter of the profile is 6 in. at a dis-tance of 3 in. from the open end.

12 in.

4 in.

Since the shape is parabolic, it is given by

y = ax2 + bx+ c

where the following boundary conditions can beused to evaluate constant coefficients a, b, andc:

(a) at x = 0, dydx = 0.

(b) at x = 3 in., y = 1 in.(c) at x = 6 in., y = 4 in.

The first boundary condition gives:

dy

dx= 2ax+ b

Therefore,0 = 2a(0) + b

or b = 0. Similarly, the second and third bound-ary conditions result in two simultaneous alge-braic equations:

36a+ c = 4

77






and9a+ c = 1

Thus, a = 19 and c = 0, so that the equation for

the mandrel surface is

y =x2

9

If the part is conventionally spun, the surfacearea of the mandrel has to be calculated. Thesurface area is given by

A =∫ 6

0

2πRds

where R = x and

ds =

√1 +

(dy

dx

)2

dx =

√1 +

(29x

)2

dx

Therefore, the area is given by

A =∫ 6

0

2πx

√1 +

(29x

)2

dx

=∫ 6

0

2πx

√1 +

481x2 dx

To solve this integral, substitute a new variable,u = 1 + 4

81x2, so that

du =881x dx

and so that the new integration limits are fromu = 1 to u = 225

81 . Therefore, the integral be-comes

A =∫ 225/81

1

2π818√u du

=81π4

(23u3/2

)225/81

1

= 154 in2

For a disk of the same surface area and thick-ness,

Ablank =π

4d2 = 154 in2

or d = 14 in.

7.91 For the mandrel needed in Problem 7.90, plotthe sheet-metal thickness as a function of radiusif the part is to be produced by shear spinning.Is this process feasible? Explain.

As was determined in Problem 7.90, the equa-tion of the surface is

y =x2

9

The sheet-metal thickness in shear spinning isgiven by Eq. (7.12) on p. 377 as

t = to sinα

where α is given by (see Fig. 7.36 on p. 377)

α = 90 − tan−1

(dy

dx

)= 90 − tan−1

(29x

)This results in the following plot of sheet thick-ness:

t/t o

or

0

0.2

0.4

0.6

0.8

1.0

x0 2 4 6

t/to

Note that at the edge of the shape, t/to = 0.6,corresponding to a strain of ε = ln 0.6 = −0.51.This strain is achievable for many materials, sothat the process is feasible.

7.92 Assume that you are asked to give a quiz to stu-dents on the contents of this chapter. Preparefive quantitative problems and five qualitativequestions, and supply the answers.


78






Design

7.93 Consider several shapes (such as oval, triangle,L-shape, etc.) to be blanked from a large flatsheet by laser-beam cutting, and sketch a nest-ing layout to minimize scrap.

Several answers are possible for this open-endedproblem. The following examples were obtainedfrom Altan, T., ed., Metal Forming Handbook,Springer, 1998:

7.94 Give several structural applications in whichdiffusion bonding and superplastic forming areused jointly.

By the student. The applications for super-plastic forming are mainly in the aerospace in-dustry. Some structural-frame members, whichnormally are placed behind aluminum sheet andare not visible, are made by superplastic form-ing. Two examples below are from Hosford andCadell, Metal Forming, 2nd ed., pp. 85-86.

Aircraft wing panel, produced through internalpressurization. See also Fig. 7.46 on p. 384.

Sheet-metal parts.

7.95 On the basis of experiments, it has beensuggested that concrete, either plain or rein-forced, can be a suitable material for dies insheet-metal forming operations. Describe yourthoughts regarding this suggestion, consideringdie geometry and any other factors that may berelevant.

By the student. Concrete has been used in ex-plosive forming for large dome-shaped parts in-tended, for example, as nose cones for intercon-tinental ballistic missiles. However, the use ofconcrete as a die material is rare. The moreserious limitations are in the ability of consis-tently producing smooth surfaces and accept-able tolerances, and the tendency of concreteto fracture at stress risers.

7.96 Metal cans are of either the two-piece variety(in which the bottom and sides are integral) orthe three-piece variety (in which the sides, thebottom, and the top are each separate pieces).For a three-piece can, should the seam be (a) inthe rolling direction, (b) normal to the rollingdirection, or (c) oblique to the rolling directionof the sheet? Explain your answer, using equa-tions from solid mechanics.

The main concern for a beverage containeris that the can wall should not fail understresses due to internal pressurization. (Inter-nal pressurization routinely occurs with car-bonated beverages because of jarring, dropping,and rough handling and can also be caused bytemperature changes.) The hoop stress and theaxial stress are given, respectively, by

σh =pr

t

σa =12σh =

pr

2twhere p is the internal pressure, r is the canradius, and t is the sheet thickness. These areprincipal stresses; the third principal stress is inthe radial direction and is so small that it canbe neglected. Note that the maximum stressis in the hoop direction, so the seam should beperpendicular to the rolling direction.

79






7.97 Investigate methods for determining optimumshapes of blanks for deep-drawing operations.Sketch the optimally shaped blanks for draw-ing rectangular cups, and optimize their layouton a large sheet of metal.

This is a topic that continues to receive consid-erable attention. Finite-element simulations, aswell as other techniques such as slip-line fieldtheory, have been used. An example of an opti-mum blank for a typical oil-pan cup is sketchedbelow.

Die cavity profile

Optimum blank shape

7.98 The design shown in the accompanying illustra-tion is proposed for a metal tray, the main bodyof which is made from cold-rolled sheet steel.Noting its features and that the sheet is bent intwo different directions, comment on relevantmanufacturing considerations. Include factorssuch as anisotropy of the cold-rolled sheet, itssurface texture, the bend directions, the natureof the sheared edges, and the method by whichthe handle is snapped in for assembly.

By the student. Several observations can bemade. Note that a relief notch design, as shownin Fig. 7.68 on p. 405 has been used. It is avaluable experiment to have the students cutthe blank from paper and verify that the tray is

produced by bending only because of this notch.As such, the important factors are bendabil-ity, and scoring such as shown in Fig. 7.71 onp. 406, and avoiding wrinkling such as discussedin Fig. 7.69 on p. 405.

7.99 Design a box that will contain a 4 in. × 6 in. × 3in. volume. The box should be produced fromtwo pieces of sheet metal and require no toolsor fasteners for assembly.

This is an open-ended problem with a widevariety of answers. Students should considerthe blank shape, whether the box will be deep-drawn or produced by bending operations (seeFig. 7.68), the method of attaching the parts(integral snap-fasteners, folded flaps or loose-fit), and the dimensions of the two halves areall variables. It can be beneficial to have thestudents make prototypes of their designs fromcardboard.

7.100 Repeat Problem 7.99, but the box is to be madefrom a single piece of sheet metal.

This is an open-ended problem; see the sugges-tions in Problem 7.99. Also, it is sometimeshelpful to assign both of these problems, or toassign each to one-half of a class.

7.101 In opening a can using an electric can opener,you will note that the lid often develops a scal-loped periphery. (a) Explain why scallopingoccurs. (b) What design changes for the canopener would you recommend in order to min-imize or eliminate, if possible, this scallopingeffect? (c) Since lids typically are recycled ordiscarded, do you think it is necessary or worth-while to make such design changes? Explain.

By the student. The scalloped periphery isdue to the fracture surface moving ahead ofthe shears periodically, combined with the load-ing applied by the two cutting wheels. Thereare several potential design changes, includingchanging the plane of shearing, increasing thespeed of shearing, increasing the stiffness of thesupport structure, or using more wheels. Scal-lops on the cans are not normally objectionable,so there has not been a real need to make open-ers that avoid this feature.

7.102 A recent trend in sheet-metal forming is to pro-vide a specially-textured surface finish that de-

80






velops small pockets to aid lubricant entrain-ment. Perform a literature search on this tech-nology, and prepare a brief technical paper onthis topic.

This is a valuable assignment, as it encour-ages the student to conduct a literature review.This is a topic where significant research hasbeen done, and a number of surface texturesare available. A good starting point is to ob-tain the following paper:

Hector, L.G., and Sheu, S., “Focused energybeam work roll surface texturing science andtechnology,” J. Mat. Proc. & Mfg. Sci., v. 2,1993, pp. 63-117.

7.103 Lay out a roll-forming line to produce any threecross sections from Fig. 7.27b.

By the student. An example is the followinglayout for the structural member in a steel doorframe:

Stage 1 Stage 2

Stage 3 Stage 4

Stage 5 Stage 6

Stage 7

A B

7.104 Obtain a few pieces of cardboard and carefullycut the profiles to produce bends as shown inFig. 7.68. Demonstrate that the designs labeledas “best” are actually the best designs. Com-ment on the difference in strain states betweenthe designs.

By the student. This is a good project thatdemonstrates how the designs in Fig. 7.68 onp. 405 significantly affect the magnitude andtype of strains that are applied. It clearly showsthat the best design involves no stretching, butonly bending, of the sheet metal.

81






82






Chapter 8

Material-Removal Processes: Cutting

Questions

8.1 Explain why the cutting force, Fc, increaseswith increasing depth of cut and decreasingrake angle.

(a) Increasing the depth of cut means morematerial being removed per unit time.Thus, all other parameters remaining con-stant, the cutting force has to increase lin-early because the energy requirement in-creases linearly.

(b) As the rake angle decreases, the shear an-gle decreases and hence the shear strainincreases. Therefore, the energy perunit volume of material removed increases,thus the cutting force has to increase.Note that the rake angle also has an ef-fect on the frictional energy (see Table 8.1on p. 430).

8.2 What are the effects of performing a cuttingoperation with a dull tool tip? A very sharptip?

There are several effects of a dull tool. Notethat a dull tool is one having an increased tipradius (see Fig. 8.28 on p. 449). As the tip ra-dius increases (i.e., as the tool dulls), the cut-ting force increases due to the fact that the ef-fective rake angle is now decreased. In fact,shallow depths of cut may not be possible. An-other effect is the possibility for surface residualstresses, tearing, and cracking of the machinedsurface, due to severe surface deformation and

the heat generated by the dull tool tip rubbingagainst this surface. Dull tools also increasethe tendency for BUE formation, which leadsto poor surface finish.

8.3 Describe the trends that you observe in Tables8.1 and 8.2.

By the student. A review of Tables 8.1 and 8.2on pp. 430-431 indicates certain trends that areto be expected, including:

(a) As the rake angle decreases, the shearstrain and hence the specific energy in-crease.

(b) Cutting force also increases with decreas-ing rake angle;

(c) Shear plane angle decreases with increas-ing rake angle.

8.4 To what factors would you attribute the largedifference in the specific energies within eachgroup of materials shown in Table 8.3?

The differences in specific energies seen in Ta-ble 8.3 on p. 435, whether among different ma-terials or within types of materials, can basi-cally be attributed to differences in the mechan-ical and physical properties of these materials,which affect the cutting operation. For exam-ple, as strength increases, so does the total spe-cific energy. Differences in tool-chip interfacefriction characteristics would also play a signif-icant role. Physical properties, such as thermal

83






conductivity and specific heat, both of whichincrease cutting temperatures as they decrease,could be responsible for such differences. Thesepoints are supported when one closely exam-ines this table and observes that the ranges formaterials such as steels, refractory alloys, andhigh-temperature alloys are large, in agreementwith our knowledge of the great variety of theseclasses of materials.

8.5 Describe the effects of cutting fluids on chip for-mation. Explain why and how they influencethe cutting operation.

By the student. In addition to the effects dis-cussed in Section 8.7 starting on p. 464, cuttingfluids influence friction at the tool-chip inter-face, thus affecting the shear angle and chipthickness. These, in turn, can influence thetype of chip produced. Also, note that witheffective cutting fluids the built-up edge can bereduced or eliminated.

8.6 Under what conditions would you discouragethe use of cutting fluids? Explain.

By the student. The use of cutting fluids couldbe discouraged under the following conditions:

(a) If the cutting fluid has any adverse ef-fects on the workpiece and/or machine-tool components, or on the overall cuttingoperation.

(b) In interrupted cutting operations, such asmilling, the cutting fluid will, by its cool-ing action, subject the tool to large fluc-tuations in temperature, possibly causingthermal fatigue of the tool, particularly inceramics.

8.7 Give reasons that pure aluminum and copperare generally rated as easy to machine.

There are several reasons that aluminum andcopper are easy to machine. First, they are rel-atively soft, hence cutting forces and energy arelow compared to many other materials. Fur-thermore, they are good thermal conductors.Also, they are ductile and can withstand thestrains in cutting and still develop continuouschips. These materials do not generally forma built-up edge, depending on cutting parame-ters.

8.8 Can you offer an explanation as to why themaximum temperature in cutting is located atabout the middle of the tool-chip interface?(Hint: Note that there are two principal sourcesof heat: the shear plane and the tool-chip in-terface.)

It is reasonable that the maximum tempera-ture in orthogonal cutting is located at aboutthe middle of the tool-chip interface. The chipreaches high temperatures in the primary shearzone; the temperature would decrease fromthen on as the chip climbs up the rake face ofthe tool. If no frictional heat was involved, wewould thus expect the highest temperature tooccur at the shear plane. However, recall thatfriction at the tool-chip interface also increasesthe temperature. After the chip is formed itslides up the rake face and temperature beginsto build up. Consequently, the temperature dueonly to frictional heating would be highest atthe end of the tool-chip contact. These two op-posing effects are additive, and as a result thetemperature is highest somewhere in betweenthe tip of the tool and the end of contact zone.

8.9 State whether or not the following statementsare true for orthogonal cutting, explaining yourreasons: (a) For the same shear angle, thereare two rake angles that give the same cuttingratio. (b) For the same depth of cut and rakeangle, the type of cutting fluid used has no influ-ence on chip thickness. (c) If the cutting speed,shear angle, and rake angle are known, the chipvelocity can be calculated. (d) The chip be-comes thinner as the rake angle increases. (e)The function of a chip breaker is to decrease thecurvature of the chip.

(a) To show that for the same shear anglethere are two rake angles and given thesame cutting ratio, recall the definition ofthe cutting ratio as given by Eq. (8.1) onp. 420. Note that the numerator is con-stant and that the cosine of a positive andnegative angle for the denominator has thesame value. Thus, there are two rake an-gles that give the same r, namely a rakeangle, α, greater than the shear angle, φ,and a rake angle smaller than the shearangle by the same amount.

84






(b) Incorrect, because the cutting fluid will in-fluence friction, hence the shear angle and,consequently, the chip thickness.

(c) Correct, because if the cutting speed, V ,shear angle, φ, and rake angle, α, are allknown, the velocity of the chip up the faceof the tool (Vo) can be calculated. This isdone simply by using Eq. (8.5).

(d) Correct, as can be seen in Table 8.1 onp. 430.

(e) Incorrect; its function is to decrease the ra-dius of curvature, that is, to increase cur-vature.

8.10 It has been stated that it is generally undesir-able to allow temperatures to rise excessively inmachining operations. Explain why.

By the student. This is an open-ended prob-lem with a large number of acceptable answers.The consequences of allowing temperatures torise to high levels in cutting include:

(a) Tool wear will be accelerated due to hightemperatures.

(b) High temperatures will cause dimensionalchanges in the workpiece, thus reducing di-mensional accuracy.

(c) Excessively high temperatures in thecutting zone may induce metallurgicalchanges and cause thermal damage to themachined surface, thus affecting surfaceintegrity.

8.11 Explain the reasons that the same tool life maybe obtained at two different cutting speeds.

Tool life in this case refers to flank wear. Atlow cutting speeds, the asperities at the tool-workpiece interface have more time to form astronger junction, thus wear is likely to increase(see Section 4.4.2 starting on p. 144). Further-more, at low speeds some microchipping of cut-ting tools have been observed (due possibly tothe same reasons), thus contributing to toolwear. At high cutting speeds, on the otherhand, temperature increases, thus increasingtool wear.

8.12 Inspect Table 8.6 and identify tool materialsthat would not be particularly suitable for in-terrupted cutting operations, such as milling.Explain your choices.

By the student. In interrupted cutting opera-tions, it is desirable to have tools with high im-pact strength and toughness. From Table 8.6on p. 454 the tool materials that have the bestimpact strength are high-speed steels, and, toa lesser extent, cast alloys and carbides. Notealso that carbon steels and alloy steels also havehigh toughness. In addition, with interruptedcutting operations, the tool is constantly beingsubjected to thermal cycling. It is thus desir-able to utilize materials with low coefficients ofthermal expansion and high thermal conductiv-ity to minimize thermal stresses in the tool (seepp. 107-108).

8.13 Explain the possible disadvantages of a machin-ing operation if a discontinuous chip is pro-duced.

By the student. The answer is given in Section8.2.1. Note that:

(a) The forces will continuously vary, possiblyleading to chatter and all of its drawbacks.

(b) Tool life will be reduced.

(c) Surface finish may be poor surface.

(d) Tolerances may not be acceptable.

8.14 It has been noted that tool life can be almostinfinite at low cutting speeds. Would you thenrecommend that all machining be done at lowspeeds? Explain.

As can be seen in Fig. 8.21 on p. 441, toollife can be almost infinite at very low cuttingspeeds, but this reason alone would not al-ways justify using low cutting speeds. Lowcutting speeds will remove less material in agiven time which could be economically unde-sirable. Lower cutting speeds often also leadto the formation of built-up edge and discon-tinuous chips. Also, as cutting speed decreases,friction increases and the shear angle decreases,thus generally causing the cutting force to in-crease.

8.15 Referring to Fig. 8.31, how would you explainthe effect of cobalt content on the properties ofcarbides?

Recall that tungsten-carbide tools consist oftungsten-carbide particles bonded together in

85






a cobalt matrix using powder-metallurgy tech-niques. Increasing the amount of cobalt willmake the material behave in a more ductilemanner, thus adversely affecting the strength,hardness, and wear resistance of the tungsten-carbide tools. The property which cobaltimproves is toughness and transverse-rupturestrength. The accompanying figure was takenfrom p. 502 of S. Kalpakjian, ManufacturingProcesses for Engineering Materials, 3d ed.,1997.

Wea

r (m

g), c

omp

ress

ive

and

tra

ns-

vers

e-ru

ptu

re s

tren

gth

(kg/

mm

2 )

600

0

200

400

Vic

kers

har

dne

ss (H

V)

Cobalt (% by weight)0 10 20 30

500

1000

1500

750

1250

1750HRA92.4

90.5

88.5

85.7

Com

pressive strengthHardness

Tr

ansv

erse-rupture strength

Wear

8.16 Explain why studying the types of chips pro-duced is important in understanding machiningoperations.

By the student. The study the types of chipsproduced is important because the type of chipsignificantly influences the surface finish pro-duced as well as the overall cutting operation.For example, continuous chips are generally as-sociated with good surface finish. Built-up-edgechips usually result in poor surface finish. Ser-rated chips and discontinuous chips may resultin poor surface finish and dimensional accuracy,and possibly lead to chatter.

8.17 How would you expect the cutting force to varyfor the case of serrated-chip formation? Ex-plain.

By the student. One would expect the cuttingforce to vary under cutting conditions produc-ing serrated chips. During the continuous-chipformation period, the cutting force would berelatively constant. As this continuous region

becomes segmented, the cutting force wouldrapidly drop to some lower value, and then be-gin rising again, starting a new region of contin-uous chip. The whole process is repeated overand over again.

8.18 Wood is a highly anisotropic material; that is,it is orthotropic. Explain the effects of orthog-onal cutting of wood at different angles to thegrain direction on the types of chips produced.

When cutting a highly anisotropic materialsuch as wood (orthotropic), the chip formationwould depend on the direction of the cut withrespect to the wood grain direction and the rakeangle of the tool. The shear strength of woodis low (and tensile strength is high) in the graindirection, and high when perpendicular to thegrain direction. Cutting wood along the graindirection would produce long continuous chipsby virtue of a splitting action ahead of the tool.Thus, the chip is more like a shaving or veneer(and can become a polygonal in shape at largedepths of cut, like cracking a toothpick at con-stant intervals along its length). Cutting acrossthe grain would produce discontinuous chips;cutting along a direction where the shear planeis in the same direction as the grain of the woodcan produce continuous chips, similar to thoseobserved in metal cutting. These phenomenacan be demonstrated with a wood plane andpiece of pine (see, for example, Kalpakjian, Me-chanical Processing of Materials, 1963, p. 315).These observations are also relevant to cut-ting single-crystal materials, which exhibit highanisotropy.

8.19 Describe the advantages of oblique cutting.Which machining proceses involve oblique cut-ting? Explain.

A major advantage of oblique cutting is that thechip moves off to the side of the cutting zone,thus out of the way of the working area (seeFig. 8.9 on p. 426). Thus it is better suited forcutting operations involving a cross feed as inturning. Note also that the effective rake angleis increased and the chip is thinner.

8.20 Explain why it is possible to remove more ma-terial between tool resharpenings by loweringthe cutting speed.

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This situation can be visualized by referring toFig. 8.21a on p. 441. Note that at any loca-tion on a particular curve, the product of cut-ting speed (ft/min) and tool life (min) is thedistance (ft) the tool travels before it reachedthe end of its life (a specified wear land). Thedistance traveled is directly proportional to thevolume of material removed. Note also in thefigure that at very high speeds, tool life is vir-tually zero, so is the material removed. Con-versely, at very low speeds, tool life is virtuallyinfinite, thus the volume removed is almost in-finite.

It is therefore apparent that more materialcan be removed by lowering the cutting speed.However, there are two important considera-tions:

(a) The economics of the machining processwill be adversely affected if cutting speedsare low, as described in Section 8.15 andshown in Fig. 8.75 on p. 509.

(b) As stated in Section 8.3.1, tool-life curvescan curve downward at low cutting speeds.Consequently, there would be a specificcutting speed where material removal be-tween tool changes is a maximum.

8.21 Explain the significance of Eq. (8.8).

The main significance of Eq. (8.8) on p. 427is that it determines an effective rake angle foroblique cutting (a process of more practical sig-nificance than orthogonal cutting), which canbe related back to the simpler orthogonal cut-ting models for purposes of analysis.

8.22 How would you go about measuring the hothardness of cutting tools? Explain any diffi-culties that may be involved.

Hot hardness refers to the hardness of the ma-terial at the elevated temperatures typical ofthe particular cutting operation (see Fig. 8.30on p. 453). Once the temperature is known(which can be measured with thermocouples orcan be estimated), the hardness of the materialcan be evaluated at this temperature. A simplemethod of doing so is by heating the tool mate-rial, then subjecting it to a hardness test whileit is still hot.

8.23 Describe the reasons for making cutting toolswith multiphase coatings of different materials.Describe the properties that the substrate formultiphase cutting tools should have for effec-tive machining.

By the student; see Section 8.6.5. One can com-bine benefits from different materials. For ex-ample, the outermost layer can be the coatingwhich is best from hardness or low frictionalcharacteristics to minimize tool wear. The nextlayer can have the benefit of being thermally in-sulating, and a third layer may be of a materialwhich bonds well to the tool. Using these multi-ple layers allows a synergistic result in that thelimitations of one coating can be compensatedfor with another layer.

8.24 Explain the advantages and any limitations ofinserts. Why were they developed?

With inserts, a number of new cutting edges areavailable on each tool, so that the insert merelyneeds to be indexed. Also, since inserts areclamped relatively easily, they allow for quicksetups and tool changes. There are no signif-icant limitations to inserts other than the factthat they require special toolholders, and thatthey should be clamped properly. Their recy-cling and proper disposal is also an importantconsideration.

8.25 Make a list of alloying elements in high-speed-steel cutting tools. Explain why they are used.

Typical alloying elements for high-speed steelare chromium, vanadium, tungsten, and cobalt(see Section 8.6.2). These elements serve to pro-duce a material with higher strength, hardness,and wear resistance at elevated temperatures.(See also Section 3.10.3.)

8.26 What are the purposes of chamfers on cuttingtools? Explain.

Chamfers serve to increase the strength of in-serts by effectively increasing the included angleof the insert. This trend is shown in Fig. 8.34on p. 458. The tendency of edge chipping isthus reduced.

8.27 Why does temperature have such an importanteffect on cutting-tool performance?

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Temperature has a large effect on the life ofa cutting tool. (a) Materials become weakerand softer as they become hotter (see Fig. 8.30on p. 453), hence their wear resistance is re-duced. (b) Chemical reactivity generally in-creases with increasing temperature, thus in-creasing the wear rate. (c) The effectiveness ofcutting fluids can be compromised at excessivetemperatures. (d) Because of thermal expan-sion, workpiece tolerances will be adversely af-fected.

8.28 Ceramic and cermet cutting tools have certainadvantages over carbide tools. Why, then, arecarbide tools not replaced to a greater extent?

Ceramics are preferable to carbides in that theyhave a lower tendency to adhere to metals be-ing cut, and have very high abrasion resistanceand hot hardness. However, ceramics are sen-sitive to defects and are generally brittle, andthus can fail prematurely. Carbides are muchtougher than ceramics, and are therefore muchmore likely to perform as expected even whenconditions such as chatter occur. (See also Sec-tion 11.8.)

8.29 Why are chemical stability and inertness im-portant in cutting tools?

Chemical stability and inertness are importantfor cutting tools to maintain low friction andwear (see also Section 4.4). A major cause offriction is the shear stress required to break themicrowelds in the contact area between the twomaterials. If the tool material is inert, the mi-crowelds are less likely to occur with the work-piece material, and friction and wear will thusbe reduced.

8.30 What precautions would you take in machiningwith brittle tool materials, especially ceramics?Explain.

With brittle tool materials, we first want to pre-vent chipping, such as by using negative rakeangles and reduce vibration and chatter. Also,brittleness of ceramic tools applies to thermalgradients, as well as to strains. To prevent toolfailures due to thermal gradients, a steady sup-ply of cutting fluid should be applied, as wellas selecting tougher tool materials.

8.31 Why do cutting fluids have different effectsat different cutting speeds? Is the control ofcutting-fluid temperature important? Explain.

A cutting fluid has been shown to be drawninto the asperities between the tool and chipthrough capillary action. At low cutting speeds,the fluid has longer time to penetrate more ofthe interface and will thus be effective in re-ducing friction acting as a lubricant. At highercutting speeds, the fluid will have less time topenetrate the asperities; therefore, it will be lesseffective at higher speeds. Furthermore, cut-ting fluids whose effectiveness depends on theirchemical reactivity with surfaces, will have lesstime to react and to develop low-shear-strengthfilms. At higher cutting speeds, temperaturesincrease significantly and hence cutting fluidsshould have a cooling capacity as a major at-tribute.

8.32 Which of the two materials, diamond or cubicboron nitride, is more suitable for machiningsteels? Why?

Of the two choices, cubic boron nitride is moresuitable for cutting steel than diamond tools.This is because cBN, unlike diamond, is chem-ically inert to iron at high temperatures, thustool life is better.

8.33 List and explain the considerations involved indetermining whether a cutting tool should bereconditioned, recycled, or discarded after use.

By the student. This is largely a matter ofeconomics. Reconditioning requires skilled la-bor, grinders, and possibly recoating equip-ment. Other considerations are the cost ofnew tools and possible recycling of tool mate-rials, since many contain expensive materialsof strategic importance such as tungsten andcobalt.

8.34 List the parameters that influence the temper-ature in machining, and explain why and howthey do so.

By the student. An inspection of Eq. (8.29)on p. 438 indicates that temperature increaseswith strength, cutting speed, and depth of cut.This is to be expected because:

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(a) strength indicates energy dissipation, thushigher heat content,

(b) the higher the cutting speed, the less timefor heat to be dissipated, and

(c) the greater the depth of cut, the smallerthe surface area-to-thickness ratio of thechip, thus less heat dissipation. In thedenominator of this equation are specificheat and thermal conductivity, both ofwhich influence heat conduction and dis-sipation.

8.35 List and explain the factors that contribute topoor surface finish in machining operations.

By the student. Recall, for example, in turningor milling, as the feed per tooth increases oras the tool radius decreases, the roughness in-creases. Other factors that contribute to poorsurface finish are built-up edge, tool chipping orfracture, and chatter. Each of these factors canadversely affect any of the processes describedin the chapter. See also Section 8.4.

8.36 Explain the functions of the different angles ona single-point lathe cutting tool. How does thechip thickness vary as the side cutting-edge an-gle is increased? Explain.

These are described in Section 8.8.1 and canalso be found in various handbooks on machin-ing. As the side cutting-edge angle is increased,the chip becomes thinner because it becomeswider (see Fig. 8.41 on p. 470).

8.37 It will be noted that the helix angle for drills isdifferent for different groups of workpiece ma-terials. Why?

The reasons are to control chip flow through theflutes and to avoid excessive temperature rise,which would adversely affect the drilling oper-ation. These considerations are especially im-portant in drilling thermoplastics, which tendto become gummy. The student is encouragedto survey the literature and give a comprehen-sive answer.

8.38 A turning operation is being carried out on along, round bar at a constant depth of cut. Ex-plain what differences, if any, there may be inthe machined diameter from one end of the bar

to the other. Give reasons for any changes thatmay occur.

The workpiece diameter can vary from one endof the bar to the other because the cutting toolis expected to wear, depending on workpiecematerials, processing parameters, and the effec-tiveness of the cutting fluid. It can be seen thatwith excessive flank wear, the diameter of thebar will increase towards the end of the cut.Temperature variations will also affect work-piece diameter.

8.39 Describe the relative characteristics of climbmilling and up milling and their importance inmachining operations.

By the student. The answer can be found inSection 8.10.1. Basically, in up (conventional)milling, the maximum chip thickness is at theexit of tooth engagement and, thus, contami-nation and scale on the workpiece surface doesnot have a significant effect on tool life. Climbmilling has been found to have a lower ten-dency to chatter, and the downward compo-nent of the cutting force holds the workpiecein place. Note, however, that workpiece surfaceconditions can affect tool wear.

8.40 In Fig. 8.64a, high-speed-steel cutting teeth arewelded to a steel blade. Would you recommendthat the whole blade be made of high-speedsteel? Explain your reasons.

It is desirable to have a hard, abrasion-resistanttool material (such as HSS or carbide) on thecutting surface and a tough, thermally conduc-tive material in the bulk of the blade. This is aneconomical method of producing high-qualitysteel saw blades. To make the whole blade fromHSS would be expensive and unnecessary.

8.41 Describe the adverse effects of vibrations andchatter in machining.

By the student. The adverse effects of chatterare discussed in Section 8.11 and are summa-rized briefly below:

• Poor surface finish, as shown in the rightcentral region of Fig. 8.72 on p. 501.

• Loss of dimensional accuracy of the work-piece.

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• Premature tool wear, chipping, and fail-ure, a critical consideration with brittletool materials, such as ceramics, some car-bides, and diamond.

• Possible damage to the machine-tool com-ponents from excessive vibration and chat-ter.

• Objectionable noise, particularly if it is ofhigh frequency, such as the squeal heardwhen turning brass on a lathe with a lessrigid setup.

8.42 Make a list of components of machine tools thatcould be made of ceramics, and explain why ce-ramics would be a suitable material for thesecomponents.

By the student. Typical components wouldbe members that reciprocate at high speedsor members that move at high speeds and arebrought to rest in a short time (inertia effects).Bearing components are also suitable applica-tions by virtue of the hardness, resistance, andlow inertial forces with ceramics (due to theirlower density).

8.43 In Fig. 8.12, why do the thrust forces start at afinite value when the feed is zero? Explain.

The reason is likely due to the fact that the toolhas a finite tip radius (see Fig. 8.28 on p. 449),and that some rubbing along the machined sur-face takes place regardless of the magnitude offeed.

8.44 Is the temperature rise in cutting related to thehardness of the workpiece material? Explain.

Because hardness and strength are related (seeSection 2.6.8), the hardness of the workpiecematerial would influence the temperature risein cutting by requiring higher energy.

8.45 Describe the effects of tool wear on the work-piece and on the overall machining operation.

By the student. Tool wear can adversely affecttemperature rise of the workpiece, cause exces-sive rubbing of the machined surface resultingin burnishing, and induce residual stresses, sur-face damage, and cracking. Also, the machiningoperation is influenced by increased forces andtemperatures, loss of dimensional control, andpossibly causing vibration and chatter as well.

8.46 Explain whether or not it is desirable to have ahigh or low (a) n value and (b) C value in theTaylor tool-life equation.

As we can see in Fig. 8.22a on p. 442, high nvalues are desirable because for the same toollife, we can cut at higher speeds, thus increas-ing productivity. Conversely, it can also be seenthat for the same cutting speed, high n valuesgive longer tool life. Note that as n approacheszero, tool life becomes extremely sensitive tocutting speed, with rapidly decreasing tool life.

8.47 Are there any machining operations that can-not be performed on (a) machining centers and(b) turning centers? Explain.

By the student. By the student; see Section8.11. In theory, every cutting operation can beperformed on a machining center, if we considerthe term in its broadest sense, but in practice,there are many that are not reasonable to per-form. For example, turning would not be per-formed on a machining center, nor would bor-ing; for these, turning centers are available.

8.48 What is the significance of the cutting ratio inmachining?

Note that the cutting ratio is easily calculatedby measuring the chip thickness, while the un-deformed chip thickness is a machine setting.Once calculated, the shear angle can be directlyobtained through Eq. (8.1) on p. 420, and thusmore knowledge is obtained on cutting mechan-ics, as described in detail in Section 8.2.

8.49 Emulsion cutting fluids typically consist of 95%water and 5% soluble oil and chemical addi-tives. Why is the ratio so unbalanced? Is theoil needed at all? Explain.

The makeup of emulsions reflects the fact thatmachining fluids have, as their primary pur-pose, the cooling of the cutting zone (water be-ing an excellent coolant). However, the oil isstill necessary; it can attach itself to surfacesand provide boundary lubrication, especially ifthe cutting process is interrupted, as in milling.See also Section 8.7.

8.50 It was stated that it is possible for the n valuein the Taylor tool-life equation to be negative.Explain.

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In machining steel with carbides, for example,it has been noted that at low speeds wear ishigh, while at intermediate speeds it is muchlower. Thus, at low speeds, the Taylor tool-lifeequation may have a negative value of n. Aprobable reason is that low cutting speeds al-low for greater interaction between the tool andthe workpiece, thus causing higher wear. Thistopic can be a good term paper for students.

8.51 Assume that you are asked to estimate the cut-ting force in slab milling with a straight-toothcutter. Describe the procedure that you wouldfollow.

By the student. The student should first make alarge, neat sketch of the cutter tooth-workpieceinteraction, based on Fig. 8.53a on p. 483; thenconsider factors such as rake angle, shear an-gle, varying chip thickness, finite length of chip,etc., remembering that the depth of cut is verysmall compared to the cutter diameter. See alsoSection 8.2.

8.52 Explain the possible reasons that a knife cutsbetter when it is moved back and forth. Con-sider factors such as the material being cut, in-terfacial friction, and the shape and dimensionsof the knife.

By the student. One obvious effect is that thelongitudinal movement of the knife reduces thevertical component of the friction force vec-tor, thus the material being cut is not draggeddownward. (Consider, for example, cutting ablock of relatively cheese with a wide knife andthe considerable force required to do so.) An-other factor is the roughness of the cutting edgeof the knife. No matter how well it is sharpenedand how smooth it appears to be, it still hassome finite roughness which acts like the cut-ting teeth of a very fine saw (as can be observedunder high magnification). The students is en-couraged to inspect the cutting edge of knives,especially sharp ones, under a microscope andrun some simple cutting experiments and de-scribe their observations.

8.53 What are the effects of lowering the frictionat the tool-chip interface (say with an effectivecutting fluid) on the mechanics of cutting oper-ations? Explain, giving several examples.

The most obvious effect of lowering fric-tion through application of a more effectivecoolant/lubricant is that the cutting and nor-mal forces will be reduced. Also, the shear an-gle will be affected [see Eq. (8.20) on p. 433], sothat the cutting ratio will be significantly differ-ent. This also implies that the chip will undergoa different shear strain, and that chip morphol-ogy is likely to be different. The student shouldelaborate further on this topic.

8.54 Why is it not always advisable to increase cut-ting speed in order to increase production rate?Explain.

From the Taylor tool-life equation, V Tn =C, it can be seen that tool wear increasesrapidly with increasing speed. When a toolwears excessively, it causes poor surface fin-ish and higher temperatures. With continualtool replacement, more time is spent indexingor changing tools than is gained through fastercutting. Thus, higher speeds can lead to lowerproduction rates.

8.55 It has been observed that the shear-strain ratein metal cutting is high even though the cuttingspeed may be relatively low. Why?

By the student. The reason is explained in Sec-tion 8.2, and is associated with Eqs. (8.6) and(8.7) on p. 421.

8.56 We note from the exponents in Eq. (8.30) thatthe cutting speed has a greater influence ontemperature than does the feed. Why?

The difference is not too large; it is likely due tothe fact that as cutting speed increases, there islittle time for the energy dissipated to be con-ducted or dissipated from the tool. The feedhas a lower effect because its speed is so muchlower than the cutting speed.

8.57 What are the consequences of exceeding the al-lowable wear land (see Table 8.5) for cuttingtools? Explain.

The major consequences would be:

(a) As the wear land increases, the wear flatwill rub against the machined surface andthus temperature will increase due to fric-tion.

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(b) Surface damage may result and dimen-sional control will become difficult.

(c) Some burnishing may also take place onthe machined surface, leading to residualstresses and temperature rise.

(d) Cutting forces will increase because ofthe increased wear land, requiring greaterpower for the same machining operation.

8.58 Comment on and explain your observations re-garding Figs. 8.34, 8.38, and 8.43.

By the student. For example, from Fig. 8.34on p. 458 it is clear that edge strength can beobtained from tool geometry; from Fig. 8.38 onp. 461, it is clear that strength is also obtainedthrough the tool material used. Figure 8.43 onp. 472 shows the allowable speeds and feeds fordifferent materials; the materials generally cor-respond to the strengths given in Fig. 8.38. Therange in feeds and speeds can be explained bythe range of strengths for different tool geome-tries in Fig. 8.34.

8.59 It will noted that the tool-life curve for ceramiccutting tools in Fig. 8.22a is to the right of thosefor other tools. Why?

Ceramic tools are harder and have higher resis-tance to temperature; consequently, they resistwear better than other tool materials shown inthe figure. Ceramics are also chemically inert,even at the elevated temperatures of machin-ing. The high hardness leads to abrasive wearresistance, and the chemical inertness leads toadhesive wear resistance.

8.60 In Fig. 8.18, it can be seen that the percentageof the energy carried away by the chip increaseswith cutting speed. Why?

Heat is removed from the cutting zone mainlyby conduction through the workpiece, chip, andtool. Also note the temperature distributionshown in Fig. 8.16 on p. 437 and how high thetemperatures are. Consequently, as the cuttingspeed increases, the chip will act more and moreas a heat sink and carry away much of the heatgenerated in the cutting zone, and less and lessof the heat will be conducted away to the toolor the workpiece.

8.61 How would you go about measuring the effec-tiveness of cutting fluids? Explain.

By the student. The most effective and obviousmethod is to test different cutting fluids in ac-tual machining operations. Other methods areto heat the fluids to the temperatures typicallyencountered in machining, and measure theirviscosity and other relevant properties such aslubricity, specific heat, and chemical reactions(see Chapter 4 for details). The students areencouraged to develop their own ideas for suchtests.

8.62 Describe the conditions that are critical in ben-efiting from the capabilities of diamond andcubic-boron-nitride cutting tools.

Because diamond and cBN are brittle, impactdue to factors such as cutting-force fluctuationsand poor quality of the machine tools usedare important. Thus, interrupted cutting (suchas milling or turning spline shafts) should beavoided as much as possible. Machine toolsshould have sufficient stiffness to avoid chatterand vibrations (see Section 8.12). Tool geom-etry and setting is also important to minimizestresses and possible chipping. The workpiecematerial must be suitable for diamond or cBN;for example, carbon is soluble in iron and steelsat elevated temperatures as encountered in cut-ting, and therefore diamond would not be suit-able for these materials.

8.63 The last two properties listed in Table 8.6 canbe important to the life of the cutting tool. Ex-plain why. Which of the properties listed arethe least important in machining operations?Explain.

Thermal conductivity is important becausewith increasing thermal conductivity, heat isconducted away from the cutting zone morequickly through the tool, leading to lower tem-peratures and hence lower wear. Coefficientof thermal expansion is especially significantfor thermal fatigue and for coated tools, wherethe coating and the substrate must have simi-lar thermal expansion coefficients to avoid largethermal stresses. Of the material propertieslisted, density, elastic modulus, and meltingtemperature are the least important.

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8.64 It will be noted in Fig. 8.30 that the tool mate-rials, especially carbides, have a wide range ofhardness at a particular temperature. Why?

By the student. There are various reasons forthe range of hardness, including the following:

• All of the materials can have variations intheir microstructure, thus significantly af-fecting hardness. For example, comparethe following two micrographs of tung-sten carbide, showing a fine-grained (left)and coarse-grained (right) tungsten car-bide. (Source: Trent, E.M., and Wright,P.K., Metal Cutting 4th ed., ButterworthHeinemann, 2000, pp. 178-185).

• There can be a wide range in the concen-tration of the carbide as compared to thecobalt binder.

• For materials such as carbon tool steels,the carbon content can be different, as canthe level of case hardening of the tool.

• High-speed steels and ceramics are genericterms, with a wide range of individualchemistries and compositions.

8.65 Describe your thoughts on how would you goabout recycling used cutting tools. Commenton any difficulties involved, as well as on eco-nomic considerations.

By the student. Recycling is a complicatedsubject and involves economic as well as envi-ronmental considerations (see also pp. 12-15).

Fortunately, cutting-tool materials are gener-ally non-toxic (with the exception of cobalt incarbide tools), and thus they can be disposedof safely. The main consideration is economics:Is recycling of the tool material cost effective?Considerations include energy costs in recyclingthe tool and processing costs in refurbishment,compared to the material costs savings. This isan appropriate topic for a student term paper.

8.66 As you can see, there is a wide range of toolmaterials available and used successfully today,yet much research and development continuesto be carried out on these materials. Why?

By the student. The reasons for the avail-ability of a large variety of cutting-tool ma-terials is best appreciated by reviewing Ta-ble 8.6 on p. 454. Among various factors,the type of workpiece material machined, thetype of machining operation, and the surfacefinish and dimensional accuracy required allaffect the choice of a cutting-tool material.For example, for interrupted cutting operationssuch as milling, we need toughness and im-pact strength. For operations where much heatis generated due, for example, to high cuttingspeeds, hot hardness is important. If very finesurface finish is desired, then ceramics and dia-mond would be highly desirable. Tool materialscontinue to be investigated further because, asin all other materials, there is much progress tobe made for reasons such as to improve consis-tency of properties, extend their applications,develop new tool geometries, and reduce costs.The students are encouraged to comment fur-ther on this topic.

8.67 Drilling, boring, and reaming of large holes isgenerally more accurate than just drilling andreaming. Why?

The boring process has generally better controlof dimensional accuracy than drilling becauseof the overall stiffness of the setup. However,a boring tool requires an initial hole, so thedrilling step cannot be eliminated. Reamingis a generally slow process and produces goodsurface finish on a precisely produced hole.

8.68 A highly oxidized and uneven round bar is be-ing turned on a lathe. Would you recommend a

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relatively small or large depth of cut? Explainyour reasons.

Because oxides are generally hard and abrasive(see p. 146), light cuts will cause the tool towear rapidly, and thus it is highly desirable tocut right through the oxide layer during the firstpass. Note that an uneven round bar indicatessignificant variations in the depth of cut beingtaken; thus, depending on the degree of eccen-tricity, it may not always be possible to do sosince this can lead to self-excited vibration andchatter.

8.69 Does the force or torque in drilling change asthe hole depth increases? Explain.

Both the torque and the thrust force generallyincrease as the hole depth increases, althoughthe change is more pronounced on the torque.Because of elastic recovery along the cylindri-cal surface of the hole, there is a normal stressexerted on the surface of the drill while in thehole. Consequently, the deeper the hole, thelarger the surface area and thus the larger theforce acting on the periphery of the drill, lead-ing to a significant increase in torque.

8.70 Explain the advantages and limitations of pro-ducing threads by forming and cutting, respec-tively.

By the student. Thread rolling is described inSection 6.3.5. The main advantages of threadrolling over thread cutting are the speeds in-volved (thread rolling is a very high-production-rate operation). Also, the fact that the threadsundergo extensive cold working will lead tostronger work-hardened threads. Cutting con-tinues to be used for making threads becauseit is a very versatile operation and much moreeconomical for low production runs (since ex-pensive dies are not required). Note that inter-nal threads also can be rolled, but this is notnearly as common as machining and can be adifficult operation to perform.

8.71 Describe your observations regarding the con-tents of Tables 8.8, 8.10, and 8.11.

By the student. Note, for example, that theside rake angle is low for the ductile materialssuch as thermoplastics, but is high for materials

more difficult to machine, such as refractory al-loys and some cast irons with limited ductility.Similar observations can be made for the drillgeometries and the point angle.

8.72 The footnote to Table 8.10 states that as thedepth of the hole increases, speeds and feedsshould be reduced. Why?

As hole depth increases, elastic recovery in theworkpiece causes normal stresses on the surfaceof the drill, thus the stresses experienced by thedrill are higher than they are in shallow holes.These stresses, in turn, cause the torque on thedrill to increase and may even lead to its failure.Reduction in feeds and speeds can compensatefor these increases. (See also answer to Ques-tion 8.69.)

8.73 List and explain the factors that contribute topoor surface finish in machining operations.

By the student. As an example, one factor is ex-plained by Eq. (8.35) on p. 449, which gives theroughness in a process such as turning. Clearly,as the feed increases or as the tool nose radiusdecreases, roughness will increase. Other fac-tors that affect surface finish are built-up edge(see, for example, Figs. 8.4 and 8.6), dull toolsor tool-edge chipping (see Fig. 8.28), or vibra-tion and chatter (Section 8.11.1).

8.74 Make a list of the machining operations de-scribed in this chapter, according to the diffi-culty of the operation and the desired effective-ness of cutting fluids. (Example: Tapping ofholes is a more difficult operation than turningstraight shafts.)

By the student. Tapping is high in operationalseverity because the tool produces chips thatare difficult to dispose of. Tapping has a veryconfined geometry, making effective lubricationand cooling difficult. Turning, on the otherhand, is relatively easy.

8.75 Are the feed marks left on the workpiece bya face-milling cutter segments of a true circle?Explain with appropriate sketches.

By the student. Note that because there is al-ways movement of the workpiece in the feed di-rection, the feed marks will not be segments oftrue circles.

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8.76 What determines the selection of the numberof teeth on a milling cutter? (See, for example,Figs. 8.53 and 8.55.)

The number of teeth will affect the surface fin-ish produced, as well as vibrations and chatter,depending on the machine-tool structural char-acteristics. The number is generally chosen toachieve the desired surface finish at a given setof machining parameters. Note also that thefiner the teeth, the greater the tendency for chipto clog. At many facilities, the choice of a cut-ter may simply be what tooling is available inthe stock room.

8.77 Explain the technical requirements that led tothe development of machining and turning cen-ters. Why do their spindle speeds vary over awide range?

By the student. See Section 8,11. Briefly, ma-chining centers, as a manufacturing concept,serve two basic purposes:

(a) save time by rapid tool changes,

(b) eliminating part handling and mounting inbetween operations, and

(c) rapid changeover for machining differentparts in small lots.

Normally, much time would be spent transfer-ring and handling the workpiece between dif-ferent machine tools. Machining centers elim-inate or greatly reduce the need for part han-dling and, consequently, reduce manufacturingtime and costs.

8.78 In addition to the number of components, asshown in Fig. 8.74, what other factors influencethe rate at which damping increases in a ma-chine tool? Explain.

By the student. The most obvious factors arethe damping characteristics of the machine-toolstructure and its foundation; vibration isolat-ing pads are commonly installed under machinetools. The type and quality of joints, as wellas the quality of the sliding surfaces and theirlubrication, and the manner in which the indi-vidual components are assembled also have asignificant effect. (See Section 8.11.1.)

8.79 Why is thermal expansion of machine-tool com-ponents important? Explain, with examples.

When high precision is required, thermal distor-tion is very important and must be eliminatedor minimized. This is a serious concern, as evena few degrees of temperature rise can be signifi-cant and can compromise dimensional accuracy.The student should elaborate further.

8.80 Would using the machining processes describedin this chapter be difficult on nonmetallic orrubber like materials? Explain your thoughts,commenting on the influence of various physicaland mechanical properties of workpiece materi-als, the cutting forces involved, the parts ge-ometries, and the fixturing required.

By the student. Rubber like materials are dif-ficult to machine mainly because of their lowelastic modulus and very large elastic strainsthat they can undergo under external forces.Care must be taken to properly support theworkpiece and minimize the cutting forces.Note also that these materials become stifferwith lower temperatures, which suggests an ef-fective cutting strategy and chilling of the work-piece.

8.81 The accompanying illustration shows a partthat is to be machined from a rectangularblank. Suggest the type of operations requiredand their sequence, and specify the machinetools that are needed.

Drilled andtapped holes

Steppedcavity

By the student. The main challenge with thepart shown is in designing a fixture that allowsall of the operations to be performed withoutinterference. Clearly, a milling machine will berequired for milling the stepped cavity and the

95






slots; the holes could be produced in the millingmachine as well, although a drill press may beused instead. Note that one hole is drilled ona milled surface, so drilling and tapping haveto follow milling. If the surface finish on theexterior is not critical, a chuck or vise can beused to grip the surface at the corners, which isplausible if the part has sufficient height. Thegrips usually have rough surfaces, so they willleave marks which will be more pronounced inaluminum than in stainless steels.

8.82 Select a specific cutting-tool material and esti-mate the machining time for the parts shownin the accompanying three figures: (a) pumpshaft, stainless steel; (b) ductile (nodular) ironcrankshaft; (c) 304 stainless-steel tube with in-ternal rope thread.

(b)

(a)

(c)

30 mm

160 mm

4 mm

5 mm 24 mmLead 100 mm

250 mm

Pitch: 12.7 mm

75 mm

50 mm

By the student. Students should address themethods and machinery required to producethese components, recognizing the economicimplications of their selection of materials.

8.83 Why is the machinability of alloys generally dif-ficult to assess?

The machinability of alloys is difficult to as-sess because of the wide range of chemical,mechanical, and physical properties that canbe achieved in alloys, as well as their varyingamounts of alloying elements. Some mildly al-loyed materials may be machined very easily,whereas a highly alloyed material may be brit-tle, abrasive, and thus difficult to machine.

8.84 What are the advantages and disadvantages ofdry machining?

By the student. See Section 8.7.2. The advan-tages of dry machining include:

(a) no lubricant cost;(b) no need for lubricant disposal;(c) no environmental concerns associated with

lubricant disposal;(d) no need to clean the workpiece, or at least

the cleaning is far less difficult.


(a) possibly higher tool wear;(b) oxidation and discoloration of the work-

piece surface since no lubricant is presentto protect surfaces;

(c) possibly higher thermal distortion of theworkpiece, and

(d) washing away chips may become difficult.

8.85 Can high-speed machining be performed with-out the use of cutting fluids? Explain.

This can be done, using appropriate tool mate-rials and processing parameters. Recall that inhigh speed machining, most of the heat is con-veyed from the cutting zone through the chip,so the need for a cutting fluid is less.

8.86 If the rake angle is 0, then the frictional forceis perpendicular to the cutting direction and,therefore, does not contribute to machiningpower requirements. Why, then, is there an in-crease in the power dissipated when machiningwith a rake angle of, say, 20?

Lets first note that although the frictional force,because of its vertical position, does not directlyaffect the cutting power at a rake angle of zero,

96






it does affect it indirectly by influencing theshear angle. Recall that the higher the friction,the lower the shear angle and the higher theenergy required. As the rake angle increases,say to 20, the friction force (see Fig. 8.11 onp. 428) will now affect the position of the re-sultant force, R, and thus have a componentcontributing to the cutting force. These com-plex interactions result in the kind of force vari-ations, as a function of rake angle, shown inTables 8.1 and 8.2 on pp. 430-431.

8.87 Would you recommend broaching a keyway ona gear blank before or after the teeth are ma-chined? Explain.

By the student. The keyway should be ma-chined before the teeth is machined. The rea-son is that in hobbing or related processes (seeSection 8.10.7), the gear blank is indexed. Thekeyway thus serves as a natural guide for index-ing the blank.

8.88 Given your understanding of the basic metal-cutting process, describe the important physi-cal and chemical properties of a cutting tool.

By the student. Generally, the important prop-erties are hardness (especially hot hardness),toughness, thermal conductivity, and thermalexpansion coefficient. Chemically, the toolmust be inert to the workpiece material at thecutting temperatures developed. See also Sec-tion 8.6 and Table 8.6 on p. 454.

8.89 Negative rake angles are generally preferredfor ceramic, diamond, and cubic boron nitridetools. Why?

By the student. Although hard and strong incompression, these materials are brittle and rel-atively weak in tension. Consequently, negativerake angles, which indicate larger included an-gle of the tool tip (see, for example, Fig. 8.2on p. 419) are preferred mainly because of thelower tendency to cause tensile stresses andchipping of the tools.

8.90 If a drill bit is intended for woodworking ap-plications, what material is it most likely to bemade from? (Hint: Temperatures rarely rise to400C in woodworking.) Are there any reasonswhy such a drill bit cannot be used to drill afew holes in a piece of metal? Explain.

Because of the lower forces and temperaturesinvolved, as well as economic considerations,woodworking tools are typically made of car-bon steels, with some degree of hardening byheat treatment. Note from Fig. 8.30 on p. 453that carbon steels maintain a reasonably highhardness for temperatures less than 400oF. Fordrilling metals, however, the temperatures arehigh enough to significantly soften the carbonsteel (unless drilling at low rotational speeds),thus quickly dulling the drill bit.

8.91 What are the consequences of a coating ona cutting tool that has a different coefficientof thermal expansion than does the substrate?Explain.

Consider the situation where a cutting tool andthe coating are stress-free at room temperaturewhen the tool is inserted; then consider the sit-uation when the tool is used in cutting and thetemperatures are very high. A mismatch inthermal expansion coefficients will cause highthermal strains at the temperatures developedduring machining. This can result in a sepa-ration (delamination) of the coating from thesubstrate. (See also pp. 107-108.)

8.92 Discuss the relative advantages and limitationsof near-dry machining. Consider all relevanttechnical and economic aspects.

The advantages are mostly environmental asthere is no cutting fluid involved, which wouldadd to the manufacturing cost, or to dispose ofor treat before its disposal. This has other im-plications in that the workpiece doesn’t have tobe cleaned, so no cleaning fluids, such as sol-vents, have to be used. Also, lubricants areexpensive and difficult to control. However,cutting-fluid residues provide a protective oilfilm on the machined surfaces, especially withfreshly machined metals that begin to rapidlyoxidize, as described in Section 4.2. (See alsoanswer to Question 8.84.)

8.93 In modern manufacturing with computer-controlled machine tools, which types of metalchips are undesirable and why?

By the student. Continuous chips are not desir-able because (a) the machines are now mostlyuntended and operate at high speeds, thus chip

97






generation is at a high rate (see also chip collec-tion systems, p. 700) and (b) continuous chipswould entangle on spindles and machine com-ponents, and thus severely interfere with themachining operation. Conversely and for thatreason, discontinuous chips or segmented chipswould be desirable, and indeed are typicallyproduced using chip-breaker features on tools,Note, however, that such chips can lead to vi-bration and chatter, depending on the work-piece material, processing parameters, and thecharacteristics of the machine tool (see p. 487).

8.94 Explain why hacksaws are not as productive asband saws.

A band saw has continuous motion, whereas ahacksaw reciprocates. About half of the time,the hacksaw is not producing any chips, andthus it is not as productive.

8.95 Describe workpieces and conditions underwhich broaching would be the preferred methodof machining.

By the student. Broaching is very attractivefor producing various external and internal ge-ometric features; it is a high-rate productionprocess and can be highly automated. Al-though the broach width is generally limited(see p. 491), typically a number of passes aretaken to remove a volume of material, such ason the top surface of engine blocks. Producingnotches, slots, or keyways are common applica-tions where broaching is very useful.

8.96 With appropriate sketches, explain the differ-ences between and similarities among the fol-lowing processes: (a) shaving, (b) broaching,and (c) turn broaching.

By the student. Note, for example, that thesimilarities are generally in the mechanics ofcutting, involving a finite-width chip and usu-ally orthogonal. The differences include partic-ulars of tooling design, the machinery used, andworkpiece shapes.

8.97 Why is it difficult to use friction sawing on non-ferrous metals? Explain.

As stated in Section 8.10.5, nonferrous metalshave a tendency to adhere to the blade, caused

by adhesion at the high temperatures and at-tributable to the softness of these materials.Note also that these materials typically havehigh thermal conductivity, so if the metal hasmelted, it will quickly solidify and make the op-eration more difficult.

8.98 Review Fig. 8.68 on modular machining cen-ters, and explain workpieces and operationsthat would be suitable on such machines.

By the student. The main advantages to thedifferent modular setups shown in Fig. 8.68 onp. 498 are that various workpiece shapes andsizes can be accommodated and the tool sup-port can be made stiffer by minimizing the over-hang. (See Section 8.11.3 for the benefits ofreconfigurable machines.)

8.99 Describe types of workpieces that would not besuitable for machining on a machining center.Give specific examples.

By the student. There are some workpieces thatcannot be produced on machining centers, as bytheir nature they are very flexible. Consider, forexample:

• Workpieces that are required in muchhigher quantities than can be performedeconomically on machining centers.

• Parts that are too large for the machining-center workspace, such as large forgings orcastings.

• Parts that require specialized machines,such as rifling of gun barrels.

8.100 Give examples of (a) forced vibration and (b)self-excited vibration in general engineeringpractice.

By the student. See Section 8.12. Simple exam-ples of forced vibration are a punching bag, apogo stick, vibrating pages and cell phones, andtiming clocks in computers. Examples for self-excited vibration include musical instrumentsand human speech. The collapse of the TacomaNarrows Bridge in Washington State in 1940 isa major example of self-excited vibration. (Seealso engineering texts on vibration.)

8.101 Tool temperatures are low at low cutting speedsand high at high cutting speeds, but low againat even higher cutting speeds. Explain why.

98






At low cutting speeds, energy is dissipated inthe shear plane and at the chip-tool interface,and conducted through the workpiece and/ortool and eventually to the environment (see alsoFig. 8.18 on p. 439). At higher speeds, conduc-tion cannot take place rapidly enough. At evenhigher speeds, the heat will be carried awayby the chip, hence the workpiece will remaincooler. This is one of the major advantages ofhigh speed machining, described in Section 8.8.

8.102 Explain the technical innovations that havemade high-speed machining advances possible,and the economic motivations for high-speed

machining.

This topic is described in Section 8.8. Thetechnical advances that have made high-speedmachining possible include the availability ofadvanced cutting-tool materials, design of ma-chine tools, stiff and lightweight spindles, andadvanced methods of chip disposal. The eco-nomic motivations for high-speed machining arethat dimensional tolerances can be improved,mainly because of the absence or reduction ofthermal distortion, and the labor cost per partcan be greatly reduced.

Problems

8.103 Assume that in orthogonal cutting the rake an-gle is 15 and the coefficient of friction is 0.2.Using Eq. (8.20), determine the percentage in-crease in chip thickness when friction is dou-bled.

We begin with Eq. (8.1) on p. 420 which showsthe relationship between chip thickness anddepth of cut. Assuming that the depth of cutand the rake angle are constant, we can rewritethis equation as

totc

=cos (φ2 − α) sinφ2

cos (φ1 − α) sinφ2

Now, using Eq. (8.20) on p. 433 we can estimatethe two shear angles. For Case 1, we have fromEq. (8.12) on p. 429 that µ = 0.2 = tanβ, orβ = 11.3, and hence

φ1 = 45 +15

2− 11.3

2= 46.85

and for Case 2, where µ = 0.4, we have β =tan−1 0.4 = 21.8 and hence φ2 = 41.6. Sub-stituting these values in the above equation forchip thickness ratio, we obtain

totc

=cos (φ2 − α) sinφ1

cos (φ1 − α) sinφ2

=cos (41.6 − 15) sin 46.85

cos (46.85 − 15) sin 41.6

or to/tc = 1.16. Therefore, the chip thicknessincreased by 16%.

8.104 Prove Eq. (8.1).

Refer to the shear-plane length as l and notefrom Fig. 8.2a on p. 419 that the depth of cut,to, is

to = l sinφ

Similarly, from Fig. 8.3, the chip thickness is

tc = l cos(φ− α)

Substituting these relationships into the defini-tion of cutting ratio gives

r =totc

=l sinφ

l cos(φ− α)=

sinφcos(φ− α)

8.105 With a simple analytical expression prove thevalidity of the statement in the last paragraphin Example 8.2.

The work involved in tension and machining,respectively, can be expressed as

Wtension ∝ D2o

[ln(Do

Df

)n+1]

and

Wmachining ∝(D2

o −D2f

)umachining

Since umachining is basically a constant, the ra-tio of Wt/Wm is a function of the original and

99






final diameters of the part. Either by inspectionof these equations, or by substituting numbers(such as letting Do = 0.100 in and Df = 0.080in.) and comparing the results, we find that asDo decreases, the ratio of Wt/Wm increases.

8.106 Using Eq. (8.3), make a plot of the shear strain,γ, vs. the shear angle, φ, with the rake angle,α, as a parameter. Describe your observations.

The plot is as follows:

Shear plane angle, φ (°)0 30 60 90

She

ar s

trai

n, γ

1

2

3

4

5

6α

= 0°

α =

10°

α =

10°

α = 20°

At high shear angles, the effect of α is more pro-nounced. At low shear angles, the rake angle αhas a much lower effect. This can be visualizedfrom the geometry of the cutting zone.

8.107 Assume that in orthogonal cutting, the rake an-gle is 10. Plot the shear plane angle and cut-ting ratio as a function of the friction coefficient.

Note from Eq. (8.12) that β = tan−1 µ. Theshear angle can be estimated, either fromEq. (8.20) or (8.21), as

φ = 45 +α

2− β

2

orφ = 45 + α− β

Substituting for α and β gives

φ = 50 − 12

tan−1 µ

orφ = 55 − tan−1 µ

These are plotted as follows:

60

40

20

0She

ar p

lane

ang

le, φ

Friction coefficient, µ

0 0.2 0.4 0.6 0.8 1.0

Eq. (8.21)

Eq. (8.20)

The cutting ratio is given by Eq. (8.1) on p. 420as

r =sinφ

cos(φ− α)

The two expressions for φ can be used to ob-tain the cutting ratio as a function of µ, whichis plotted below. This can be compared to theresults for Problem 8.103.

Cu

ttin

g ra

tio,

r

Friction coefficient, µ

1.2

0.8

0.4

00 0.2 0.4 0.6 0.8 1.0

Eq. (8.21)

Eq. (8.20)

8.108 Derive Eq. (8.12).

From the force diagram in Fig. 8.11a on p. 428,we express the following:

F = (Ft + Fc tanα) cosα

andN = (Fc − Ft tanα) cosα

Therefore, by definition,

µ =F

N=

(Ft + Fc tanα)(Fc − Ft tanα)

8.109 Determine the shear angle in Example 8.1. Isthis calculation exact or an estimate? Explain.

For the cutting ratio of r = 0.555, obtained inExample 8.1 on p. 435, and using Eq. (8.1) onp. 420 , with α = 10, we find that φ = 31.17.Assuming that shear takes place along a plane,

100






this is an exact calculation. If shear takes placein a zone (Fig. 8.2b), this is an approximation.Note that we can estimate φ theoretically usingEq. (8.20).

8.110 The following data are available from orthogo-nal cutting experiments. In both cases, depth ofcut (feed) to = 0.13 mm, width of cut w = 2.5mm, rake angle α = −5, and cutting speedV = 2 m/s.

Workpiece materialAluminum Steel

Chip thickness, tc, mm 0.23 0.58Cutting force, Fc, N 430 890Thrust force, Ft, N 280 800

Determine the shear angle φ [do not useEq. (8.20)], friction coefficient µ, shear stressτ and shear strain γ on the shear plane, chipvelocity Vc and shear velocity Vs, as well as en-ergies uf , us and ut.

First, consider the aluminum workpiece, wheretc = 0.23 mm, Fc = 430 N, Ft = 280 N,to = 0.13 mm, w = 2.5 mm, α = −5 andV = 2 m/s. From Eq. (8.1) on p. 420 ,

r =totc

=0.130.23

= 0.565

Also from Eq. (8.1),

sinφcos(φ− α)

= r

orsinφ

cos(φ+ 5)= 0.565

This equation is solved numerically as φ =28.2. From Eq. (8.12), the coefficient of fric-tion is given by

µ =Ft + Fc tanαFc − Ft tanα

=280 + 430 tan(−5)430− 280 tan(−5)

or µ = 0.533. Therefore, β = tan−1 µ = 28.0.To obtain the shear stress on the shear plane,we solve Eq. (8.11) for τ :

Fc =wtoτ cos(β − α)

sinφ cos(φ+ β − α)

or

τ =Fc sinφ cos(φ+ β − α)

wto cos(β − α)

=(430) sin 28.2 cos(28.2 + 28.0 + 5)

(0.0025)(0.00013) cos(28.0 + 5)= 359 MPa

From Eq. (8.3) the shear strain is given by

γ = cotφ+ tan(φ− α)= cot 28.2 + tan(28.2 + 5) = 2.52

The chip velocity is obtained from Eq. (8.5):

Vc = Vsinφ

cos(φ− α)

= (2)sin(28.2)

cos(28.2 + 5)= 1.13 m/s

The shear velocity, Vs, is obtained fromEq. (8.6):

Vs = Vccosαsinφ

= (1.13)cos(−5)sin(28.2)

= 2.38 m/s

The energies are given by Eqs. (8.24)-(8.25) and(8.27) as:

ut =Fc

wto=

430(2.5)(0.13)

= 1323 MN-m/m3

uf =(Fc sinα+ Ft cosα)r

wto= 420 MN-m/m3

us = ut − uf = 1323− 419 = 903 MN-m/m3

The same approach is used for the steel work-piece, with the following results:

rc = 0.224 φ = 12.3

µ = 0.752 τ = 458 MPaγ = 4.90 Vc = 0.448 m/s

Vs = 2.08 m/s ut = 2738 MN-m/m3

us = 2244MN-m/m3 uf = 494 MN-m/m3

8.111 Estimate the temperatures for the conditions ofProblem 8.110 for the following workpiece prop-erties:

Workpiece materialAluminum Steel

Flow strengthYf , MPa 120 325

Thermal diffusivity,K, mm2/s 97 14

Volumetric specific heat,ρc, N/mm2C 2.6 3.3

101






From Problem 8.110, we note that V = 2 m/s =2000 mm/s and to = 0.13 mm. Equation (8.29) canbe used to calculate the temperature rise, but theequation requires English units. It can be shownthat the equivalent form of Eq. (8.29) for SI unitsis

T =3.8Yf

ρc3

rV to

K

Therefore, the temperature for the aluminum isgiven as:

Tal =3.8Yf

ρc3

rV to

K=

3.8(120)

2.6

3

r(2000)(0.13)

97

or Tal = 244C. For steel,

Ts =3.8Yf

ρc3

rV to

K=

3.8(325)

3.3

3

r(2000)(0.13)

14

or Ts = 990C

8.112 In a dry cutting operation using a −5 rake an-gle, the measured forces were Fc = 1330 N andFt = 740 N. When a cutting fluid was used, theseforces were Fc = 1200 N and Ft = 710 N. What isthe change in the friction angle resulting from theuse of a cutting fluid?

Equation (8.12) allows calculation of the frictionangle, β, as:

tan β =Ft + Fc tan α

Fc − Ft tan α

For the initial case,

tan β =740 + (1330) tan−5

1330− 740 tan−5= 0.447

Therefore, β = 24.1. With a cutting fluid,Eq. (8.12) gives:

tan β =710 + (1200) tan−5

1200− 710 tan−5= 0.479

or β = 25.6. Thus, the cutting fluid has caused achange in β of 25.6-24.1 = 1.5.

8.113 In the dry machining of aluminum with a 10 rakeangle tool, it is found that the shear angle is 25.Determine the new shear angle if a cutting fluid isapplied which decreases the friction coefficient by15%.

From Eq. (8.20) and solving for β,

β = 90 + α− 2φ = 90 + 10 − 2(25) = 50

Therefore, from Eq. (8.12), µ = tan β = 1.19 Ifthe friction coefficient is reduced by 15%, then

µ = 1.01, so that β = tan µ = 45.4. Therefore,from Eq. (8.20),

φ = 45 +α

2− β

2= 27.3

8.114 Taking carbide as an example and using Eq. (8.30),determine how much the feed should be changed inorder to keep the mean temperature constant whenthe cutting speed is tripled.

We begin with Eq. (8.32) which, for this case, canbe rewritten as

V a1 fb

1 = (3V1)afb

2

Rearranging and simplifying this equation, we ob-tain

f2

f1= 3−a/b

For carbide tools, approximate values are given onin Section 8.2.6 as a = 0.2 and b = 0.125. Substi-tuting these values, we obtain

f2

f1= 3−(0.2/0.125) = 0.17

Therefore, the feed should be reduced by (1-0.17)= 0.83, or 83%.

8.115 With appropriate diagrams, show how the use of acutting fluid can affect the magnitude of the thrustforce, Ft, in orthogonal cutting.

Note in Fig. 8.11 on p. 428 that the use of a cut-ting fluid will reduce the friction force, F , at thetool-chip interface. This, in turn, will change theforce diagram, hence the magnitude of the thrustforce, Ft. Consider the sketch given below. Theleft sketch shows cutting without an effective cut-ting fluid, so that the friction force, F is large com-pared to the normal force, N . The sketch on theright shows the effect if the friction force is a smallerfraction of the normal force because of the cuttingfluid. As can be seen, the cutting force is reducedwhen using the fluid. The largest effect is on thethrust force, but there is also a noticeable effect onthe cutting force, which becomes larger as the rakeangle increases.

102






Chip

Tool

Workpiece

FcFt

R FN

α

Chip

Tool

Workpiece

Fc

Fs

Ft RF

N

α

φ

β

β−α

8.116 An 8-in-diameter stainless-steel bar is being turnedon a lathe at 600 rpm and at a depth of cut, d = 0.1in. If the power of the motor is 5 hp and has a me-chanical efficiency of 80%, what is the maximumfeed that you can have at a spindle speed of 500rpm before the motor stalls?

From Table 8.3 on p. 435, we estimate the powerrequirement for this material as 1.5 hp-min/in3 (amean value for stainless steel). Since the motor hasa capacity of 5 hp, the maximum volume of ma-terial that can be removed per unit time is 5/1.5= 3.33 in3/min. Because the depth of cut is muchsmaller than the workpiece diameter and referringto Fig. 8.42, we note that the material removal ratein this operation is

MRR = πDdfN

Thus, the maximum feed can now be calculated as

f =MRR

πDdN=

3.33

π(8)(0.1)(600)

or f = 0.0022 in./rev.

8.117 Using the Taylor equation for tool wear and lettingn = 0.3, calculate the percentage increase in toollife if the cutting speed is reduced by (a) 30% and(b) 60%.

The Taylor equation for tool wear is given byEq. (8.31), which can be rewritten as

C = V T n

We can compare two cases as

V1Tn1 = V2T

n2

or

V2

V1=

„T1

T2

«n

solving for T1/T2,

T1

T2=

„V2

V1

«1/n

(a) For the case where the speed is reduced by30%, then V2 = 0.7V1, and thus

T1

T2=

„0.7V1

V1

«1/0.3

= 0.30

or the new life T2 is 3.3 times the original life.

(b) For a speed reduction of 60%, the new toollife is T2 = 21.2T1, or a 2120% increase.

8.118 The following flank wear data were collected in aseries of machining tests using C6 carbide toolson 1045 steel (HB=192). The feed rate was 0.015in./rev and the width of cut was 0.030 in. (a) Plotflank wear as a function of cutting time. Usinga 0.015 in. wear land as the criterion of tool fail-ure, determine the lives for the four cutting speedsshown. (b) Plot the results on log-log plot and de-termine the values of n and C in the Taylor toollife equation. (Assume a straight line relationship.)(c) Using these results, calculate the tool life for acutting speed of 300 ft/min.

103






Cutting speed Cutting time Flank wearV , ft/min min in.

400 0.5 0.00142.0 0.00234.0 0.00308.0 0.005516.0 0.008224.0 0.011254.0 0.0150

600 0.5 0.00182.0 0.00354.0 0.00608.0 0.010013.0 0.014514 0.0160

800 0.5 0.00502.0 0.01004.0 0.01405.0 0.0160

1000 0.5 0.01001.0 0.01301.8 0.01502.0 0.0160

The plot of flank wear as a function of cuttingtime is as follows:

Cutting time, min

0 20 40 60

Flan

k w

ear,

in.

0.020

0

0.010 V=400V=600V=800V=000

The 0.015 in. threshold for flank wear is indi-cated by the dashed line. From this, the follow-ing are the estimated tool life:

Speed (ft/min) Life (min)400 54600 13.5800 4.51000 1.8

The log-log plot of cutting speed vs. tool life isas follows:

Tool

life

(min

)

100

10

1100 500 1000

Cutting speed (ft/min)

5

50

From which a curve fit suggests n = 0.262 andC = 1190. Therefore, the Taylor equation forthis material is

V T 0.262 = 1190

If V = 300, then T = 192 min.

8.119 Determine the n and C values for the four toolmaterials shown in Fig. 8.22a.

From Eq. (8.31) on p. 441 note that the valueof C corresponds to the cutting speed for a toollife of 1 minute. From Fig. 8.22a, and by ex-trapolating the tool-life curves to a tool life of1 min, the C values can be estimated as (rang-ing from ceramic to HSS): 11,000, 3,000, 400,and 200. Likewise the n values are obtainedfrom the negative inverse slopes, and are esti-mated as 0.73 (36), 0.47 (25), 0.14 (8), and0.11 (6), respectively. Note that these n valuescompare well with those given in Table 8.4 onp. 442.

8.120 Using Eq. (8.30) and referring to Fig. 8.18a, es-timate the magnitude of the coefficient a.

For this problem, assume (although it is notstrictly correct) that the mean temperature, T ,is equal to the flank surface temperature, asgiven in Fig. 8.18a. We can then determinethe values of temperature as a function of thecutting speed, V , and obtain a curve fit. Theparticular answers obtained by the students willvary, depending on the distance from the tooltip taken to obtain the estimate. However, asan example, note that at a value of 0.24 in. fromthe tool tip, we have

Speed (ft/min) 200 300 550Flank temperature (F) 900 1030 1270

104






The resulting curve fit of the form of Eq. (8.30)on p. 439 gives the value of a as 0.34. Note thatthis is within a reasonable range of the valuegiven on p. 439.

8.121 (a) Estimate the machining time requiredin rough turning a 1.5-m-long, annealedaluminum-alloy round bar 75-mm in diameter,using a high-speed-steel tool. (b) Estimate thetime for a carbide tool. Let feed = 2 mm/rev.

Let’s assume that annealed aluminum alloyscan be machined at a maximum cutting speedof 4 m/s using high-speed steel tools and 7 m/sfor carbide tools (see Table 8.9 on p. 472). Themaximum cutting speed is at the outer diame-ter, and for high-speed steel it is

V = NπD

or

N =V

πD=

4π(0.075)

= 16.97 rev/s

or N = 1018 rpm. For carbide, the speed is1782 rpm. For a feed of 2 mm/rev, the time toperform one pass is given for high-speed steelby

t =L

fN=

1.5(0.002)(1018)

= 0.74 min = 44 s

Similarly, the machining time per pass for car-bide is 0.42 min or 25 s.

8.122 A 150-mm-long, 75-mm-diameter titanium-alloy rod is being reduced in diameter to 65 mmby turning on a lathe in one pass. The spindlerotates at 400 rpm and the tool is traveling atan axial velocity of 200 mm/min Calculate thecutting speed, material removal rate, time ofcut, power required, and the cutting force.

First note that the spindle speed is 400 rpm =41.89 rad/s. The depth of cut can be calculatedfrom the information given as

d =75− 65

2= 5 mm

and the feed is

f =200 mm/min400 rev/min

= 0.50 mm/rev

Therefore, the material removal rate can be cal-culated from Eq. (8.38) on p. 470 as

MRR = πDavedfN

= π(70)(5)(0.50)(400)= 2.2× 105 mm3/min

or MRR=3660 mm3/s. The actual time to cutis given by Eq. (8.39) as

t =l

fN=

150 mm200 mm/min

= 0.75 min

or t = 45 s. From Table 8.3, the unit energy re-quired is between 3.0 and 4.1 W-s/mm3, so letsuse an average value of 3.5 W-s/mm3. Thus,the power required is

P = u(MRR) = (3.5)(3660) = 12, 810 W

or 12.8 kW. The cutting force, Fc, is the tan-gential force exerted by the tool. Since poweris the product of torque and rotational speed,ω, we have

T =P

ω=

12, 810 W41.89 rad/s

= 306 Nm

Dividing the torque by the average workpieceradius, we have

Fc =T

Dave/2=

306 Nm0.035 m

= 8740 N

8.123 Calculate the same quantities as in Example 8.4but for high-strength cast iron and at N = 500rpm. .

The maximum cutting speed is at the outer di-ameter, Do, and is obtained from the expression

V = πDN = π(0.500)(300) = 471 in./min

The cutting speed at the machined diameter is

V = πDN = π(0.480)(300) = 452 in./min

The depth of cut is unaffected and is d = 0.010in. The feed is

f =v

N=

8 in./min300 rpm

= 0.0267 in/rev

Thus, according to Eq. (8.38), the material re-moval rate is

MRR = πDavedfN

= π(0.490)(0.010)(0.02)(300)= 0.0924 in3/min

105






The actual time to cut, according to Eq. (8.39),is

t =l

fN=

60.0267

300 = 0.75 min = 45 s.

The power required can be calculated by re-ferring to Table 8.3. Taking a value for highstrength cast iron as 2.0 hp-min/in3, the powerdissipated is

P = (2.0)(0.0924) = 0.1848 hp

and since 1 hp = 396,000 in.-lb/min, the poweris 73,180 in.-lb/min. The cutting force, Fc, isthe tangential force exerted by the tool. Sincepower is the product of torque, T , and rota-tional speed in radians per unit time, we have

T =73, 180(300)2π

= 38.8 in.-lb

Since T = (Fc)(Davg/2),

Fc =T

Dave/2=

38.80.490/2

= 158 lb

8.124 A 0.75-in-diameter drill is being used on adrill press operating at 300 rpm. If the feedis 0.005 in./rev, what is the material removalrate? What is the MRR if the drill diameter istripled?

The metal removal rate for drilling is given byEq. (8.40) on p. 480 as

MRR =πD2

4fN

=π(0.75)2

4(0.005)(300)

= 0.66 in3/min

If the drill diameter is tripled (that is, it is now2.25 in.), then the metal removal rate is

MRR =πD2

4fN

=π(2.25)2

4(0.005)(300)

= 5.96 in3/min

It can be seen that this is a ninefold increase inmetal removal rate.

8.125 A hole is being drilled in a block of magne-sium alloy with a 15-mm drill at a feed of 0.1mm/rev. The spindle is running at 500 rpm.Calculate the material removal rate, and esti-mate the torque on the drill.

The material removal rate can be calculatedfrom Eq. (8.40) as

MRR =πD2

4fN

=π(15 mm)2

4(0.1 mm/rev)(500 rpm)

= 8840 mm3/min

or 147 mm3/s. Referring to Table 8.3, lets takean average specific energy of 0.5 W-s/mm3 formagnesium alloys. Therefore

P =(0.5 W-s/mm3

) (147 mm3/s

)= 73.5 W

Power is the product of the torque on the drilland the rotational speed in radians per second,which, in this, case is (500 rpm)(2π)/60=52.36rad/s. Therefore, the torque is

T =P

ω=

73.5 W52.36 rad/s

= 1.40 Nm

8.126 Show that the distance lc in slab milling is ap-proximately equal to

√Dd for situations where

D d.

R=D/2

dx

lc

Referring to the figure given above, the hy-potenuse of the right triangle is assigned thevalue of x. From the triangle sketched insidethe tool,

sinθ

2=x/2R

=x

2RFrom the lower triangle,

sinθ

2=d

x

106






Thus, eliminating sin θ2 ,

x

2R=d

x

or, solving for x,

x =√

2Rd =√Dd

From the lower triangle,

cosθ

2=lcx

If θ is small, then cos θ2 can be taken as 1.

Therefore, lc ≈ x, and

lc =√Dd

8.127 Calculate the chip depth of cut in Example 8.6.

The chip depth of cut, tc, is given by Eq. (8.42)as

tc = 2f

√d

D= 2(0.1)

√0.125

2= 0.05 in.

8.128 In Example 8.6, which of the quantities will beaffected when the spindle speed is increased to200 rpm?

By the student. The quantities affected will beworkpiece speed, v, torque, T , cutting time, t,material removal rate, and power.

8.129 A slab-milling operation is being carried outon a 20-in.-long, 6-in.-wide high-strength-steelblock at a feed of 0.01 in./tooth and a depth ofcut of 0.15 in. The cutter has a diameter of 2.5in, has six straight cutting teeth, and rotates at150 rpm. Calculate the material removal rateand the cutting time, and estimate the powerrequired.

From the data given we can calculate the work-piece speed, v, from Eq. (8.43) as

v = fNn = (0.01)(150)(6) = 9 in./min

Using Eq. (8.45) on p. 484, the material removalrate is

MRR = wdv = (6)(0.15)(9) = 8.1 in3/min

Since the workpiece is high-strength steel, thespecific energy can be estimated from Table 8.3

as 3.4 hp-min/in3, as this is the largest value inthe range given. Therefore,

P =(3.4 hp-min/in3

) (8.1in3/min

)= 27.5 hp

The cutting time is given by Eq. (8.44) in whichthe quantity lc can be shown to be (see answerto Problem 8.126)

lc =√Dd =

√(2.5)(0.15) = 0.61 in.

Therefore the cutting time is

t =l + lcv

=20 in. + 0.61 in.

9 in./min= 2.29 min

8.130 Referring to Fig. 8.54, assume that D = 200mm, w = 30 mm, l = 600 mm, d = 2 mm,v = 1 mm/s, and N = 200 rpm. The cutterhas 10 inserts, and the workpiece material is304 stainless steel. Calculate the material re-moval rate, cutting time, and feed per tooth,and estimate the power required.

The cross section of the cut is

wd = (30)(2) = 60 mm2

Noting that the workpiece speed is v = 1 mm/s,the material removal rate can be calculated as

MRR = (60 mm2)(1 mm/s) = 60 mm3/s

The cutting time is given by Eq. (8.44) in whichthe quantity lc can be shown to be (see answerto Problem 8.126)

lc =√Dd =

√(200)(2) = 20 mm

Therefore, the cutting time is

t =l + lcv

=600 mm + 20 mm

1 mm/s= 620 s

The feed per tooth is obtained from Eq. (8.43).Noting that N = 200 rpm = 3.33 rev/s and thenumber of inserts is 10, we have

f =v

Nn=

1 mm/s(3.33 rev/s)(10)

= 0.030 mm/tooth

For 304 stainless steel, the unit power can be es-timated from Table 8.3 as 4 W-s/mm3. There-fore,

P = (4 W-s/mm3)(60 mm3/s) = 240 W

107






8.131 Estimate the time required for face milling an8-in.-long, 3-in.-wide brass block using a 8-in-diameter cutter with 12 HSS teeth.

Using the high-speed-steel tool, let’s take a rec-ommended cutting speed for brass (a copperalloy) at 90 m/min = 1.5 m/s, or 59 in./s (seeTable 8.12 on p. 489), and the maximum feedper tooth as 0.5 mm, or 0.02 in., The rotationalspeed of the cutter is then calculated from

V = πDN

or, solving for N ,

N =V

πD=

59 in./sπ(8 in.)

= 2.34 rev/s = 131 rpm

The workpiece speed can be obtained fromEq. (8.43) as

v = fNn = (0.02 in.)(141 rpm)(12)

or v = 0.56 in./s. The cutting time is givenby Eq. (8.44) in which the quantity lc can beshown to be (see answer to Problem 8.126)

lc =√Dd =

√(8)(3) = 4.90 in.

Therefore the cutting time is

t =l + lcv

=8 in. + 4.9 in.

0.56 in./s= 23.0 s

8.132 A 12-in-long, 2-in-thick plate is being cut on aband saw at 150 ft/min The saw has 12 teethper in. If the feed per tooth is 0.003 in., howlong will it take to saw the plate along itslength?

The workpiece speed, v, is the product of thenumber of teeth (12 per in.), the feed pertooth (0.003 in.), and the band saw speed (150ft/min). The speed is thus

v = (12)(0.003)(150) = 5.4 ft/min = 1.08 in./s

For a 12-in. long plate, the cutting time is then(12)/(1.08)=11.1 s. Note that plate thicknesshas no effect on the answer.

8.133 A single-thread hob is used to cut 40 teeth on aspur gear. The cutting speed is 200 ft/min andthe hob has a diameter of 4 in. Calculate therotational speed of the spur gear.

If a single-threaded hob is used to cut fortyteeth, the hob and the blank must be geared sothat the hob makes forty revolutions while theblank makes one revolution. The expression forthe cutting speed of the hob is

V = πDN or N =V

πD

Since the cutting speed is given as 200 ft/min= 2400 in./min, we have

N =V

πD=

2400π(4)

= 190 rad/min = 30.2 rpm

Therefore, the rotational speed of the blank is30.2/40 = 0.75 rpm.

8.134 In deriving Eq. (8.20) it was assumed that thefriction angle, β, was independent of the shearangle, φ. Is this assumption valid? Explain.

We observe from Table 8.1 that the friction an-gle, β, and the shear angle, φ, are interrelated;thus, β is not independent of φ. Note how-ever, that β varies at a much lower rate thanφ does. Therefore, while it is not strictly true,the assumption can be regarded as a valid ap-proximation.

8.135 An orthogonal cutting operation is being car-ried out under the following conditions: depthof cut = 0.10 mm, width of cut = 5 mm, chipthickness = 0.2 mm, cutting speed = 2 m/s,rake angle = 15, cutting force = 500 N, andthrust force = 200 N. Calculate the percent-age of the total energy that is dissipated in theshear plane during cutting.

The total power is

Ptot = FcV = (500 N)(2 m/s) = 1000 Nm/s

The power dissipated in the shear zone is

Pshear = FsVs

whereFs = R cos(φ+ β − α)

and

R =√F 2

c + F 2t =

√5002 + 2002 = 538 N

Also, note that the cutting ratio is given byEq. (8.1) on p. 420 as

r =totc

=0.100.20

= 0.5

108






From Fig. 8.2, it can be shown that

φ = tan−1

(r cosα

1− r sinα

)= tan−1

((0.5) cos 15

1− (0.5) sin 15

)= 29.0

Note that because all necessary data is given,we should not use the approximate shear-anglerelationships in Section 8.2.4 to estimate thefriction angle. Instead, to find β, we useEq. (8.11):

Fc = R cos(β − α)

solving for β,

β = cos−1

(Fc

R

)+ α = cos−1

(500538

)+ 15

or β = 36.7 Also, Fs is calculated as

Fs = R cos (φ+ β − α)= (538 N) cos (29.0 + 36.7 − 15)= 340 N

Also, from Eq. (8.6),

Vs =V cosα

cos (φ− α)=

(2) cos 15

cos(29.0 − 15)

or Vs = 1.99 m/s. Therefore,

Pshear = FsVs = (340 N)(1.99 m/s)

or Pshear = 677 N-m/s. Hence the percentage is677/1000=0.678 or 67.7%. Note that this valuecompares well with the data in Table 8.1 onp. 430.

8.136 An orthogonal cutting operation is being car-ried out under the following conditions: depthof cut = 0.020 in., width of cut = 0.1 in., cuttingratio = 0.3, cutting speed = 300 ft/min, rakeangle = 0, cutting force = 200 lb, thrust force= 150 lb, workpiece density = 0.26 lb/in3, andworkpiece specific heat = 0.12 BTU/lbF. As-sume that (a) the sources of heat are the shearplane and the tool-chip interface; (b) the ther-mal conductivity of the tool is zero, and thereis no heat loss to the environment; (c) the tem-perature of the chip is uniform throughout. If

the temperature rise in the chip is 155F, cal-culate the percentage of the energy dissipatedin the shear plane that goes into the workpiece.

The power dissipated in the shear zone is givenas

Pshear = FsVs

whereFs = R cos (φ+ β − α)

and

R =√F 2

c + F 2t =

√2002 + 1502 = 250 lb

Therefore, from Problem 8.135 above,

φ = tan−1

(r cosα

1− r sinα

)= tan−1

(0.3 cos 0

1− 0.3 sin 0

)= 16.7

Note that because all the necessary data isgiven, we should not use the shear-angle rela-tionships in Section 8.2.4 to estimate the fric-tion angle. Instead, to find β, we use Eq. (8.11)to obtain

Fc = R cos (β − α)

or, solving for β,

β = cos−1

(Fc

R

)+ α

= cos−1

(200250

)+ 0

= 36.9

Therefore,

Fs = R cos (φ+ β − α)= (250) cos (16.7 + 36.9 − 0)= 148 lb

Also, from Eq. (8.6),

Vs =V cosα

cos(φ− α)=

(300) cos 0

cos(16.7 − 0)

or Vs = 313 ft/min. Therefore,

Pshear = FsVs = (148)(313) = 46, 350 ft-lb/min

or Pshear = 59.6 BTU/min. Thevolume rate of material removal is

109






(300)(0.020)(0.10)(12)=14.4 in3/min. Thus,the heat content, Q, of the chip is

Qchip = cρV∆T

= (0.12 BTU/lbF)(0.26 lb/in3

)×(14.4 in3/min

)(155F)

= 70 BTU/min

The total power dissipated is

Ptotal = (200)(300)(1/778) = 77.1 BTU/min.

Hence, the ratio of heat dissipated into theworkpiece is (77.1-70)=7.1 BTU/min. In termsof the shear energy, this represents a percentageof 7.1/59.6=0.12, or 12%.

8.137 It can be shown that the angle ψ between theshear plane and the direction of maximum grainelongation (see Fig. 8.4a) is given by the expres-sion

ψ = 0.5 cot−1(γ

2

),

where γ is the shear strain, as given byEq. (8.3). Assume that you are given a pieceof the chip obtained from orthogonal cutting ofan annealed metal. The rake angle and cuttingspeed are also given, but you have not seen thesetup on which the chip was produced. Outlinethe procedure that you would follow to estimatethe power required in producing this chip. As-sume that you have access to a fully equippedlaboratory and a technical library.

Remembering that we only have a piece of thechip and we do not know its relationship to theworkpiece, the procedure will consist of the fol-lowing steps:

(a) Referring to Fig. 8.4a on p. 422, let theangle between the direction of maximumgrain elongation (grain-flow lines) and avertical line be denoted it as η. Sincewe know the rake angle, we can positionthe chip in its proper orientation and thenwrite

φ+ γ + η = 90

Note that we can now measure the angleη, but we still have two unknowns.

(b) From the formula given in the statement ofthe problem, we have a direct relationship

between the angles ψ and γ. Also fromEq. (8.3) we have a relationship betweenφ and γ. Therefore, we can determine thevalue of γ.

(c) From an analysis of the material andits hardness, its shear stress-shear straincurve can be estimated.

(d) We can then determine the value of us.(e) Since φ and α are known, we are now

able to determine the depth of cut, to,and consequently, the volume rate of re-moval, since V and the width of cut arealso known. The product of us and vol-ume removal rate is the power dissipatedin the shear plane.

(f) We must add to this the energy dissi-pated in friction, uf , at the tool-chip inter-face. Based on observations such as thosegiven in Table 8.1, we may estimate thisquantity, noting that as the rake angle in-creases, the percentage of the friction en-ergy to total energy increases. A conser-vative estimate is 50%.

8.138 A lathe is set up to machine a taper on a barstock 120 mm in diameter; the taper is 1 mmper 10 mm. A cut is made with an initial depthof cut of 4 mm at a feed rate of 0.250 mm/revand at a spindle speed of 150 rpm. Calculatethe average metal removal rate.

For an initial depth of cut of 4 mm and a taperof 1 mm/10 mm, there will be a 40 mm lengthwhich is tapered. If the depth of cut were a con-stant at 4 mm, the metal removal rate would begiven by Eq. (8.38) as

MRR = πDavedfN

= π

(120 + 116

2

)(4)(0.250)(150)

= 55, 600 mm3/min

Since the bar has a taper, the average metalremoval rate is one-half this value, or 27,800mm3/min.

8.139 Develop an expression for optimum feed ratethat minimizes the cost per piece if the tool lifeis as described by Eq. (8.34).

There can be several solutions for this problem,depending on the type of the machine tool. For

110






example, Section 8.15 considers the case wherean insert is used. The insert has a number offaces that can be used before the tool is re-placed. Other tools may be used only once; oth-ers (such as drills) can be reground and reused.Since inserts are used in the textbook, the fol-lowing solution considers a tool that can be pe-riodically reground. From Eq. (8.46) on p. 507the total cost per piece can be written as

Cp = Cm + Cs + Cl + Ct

Note that Cl and Cs will not be dependenton the feed rate. However, in turning, themachining cost can be obtained by combiningEqs. (8.47) and (8.51) to obtain

Cm = Tm (Lm +Bm)

=πLD

fV(Lm +Bm)

=[πLD

V(Lm +Bm)

]1f

The number of parts per tool grind is given as

Np =T

Tm=C7V −7d−1f−4

πLD/fV=(

C7

πLDdV 6

)1f3

so that the tooling cost, equivalent toEq. (8.49), is

Ct =1Np

[Tc (Lm +Bm) + Tg (Lg +Bg) +Dc]

where Lg and Bg are the labor and overheadrate associated with the tool grinding opera-tion, respectively. We can define a function Ψas:

Ψ = [Tc (Lm +Bm) + Tg (Lg +Bg) +Dc]

which is a function of labor, overhead, and toolreplacement costs, and is independent of feed.Therefore, the tooling cost is:

Ct =1Np

Ψ =(πLDdV 6

C7

)Ψf3

The total cost per piece can be expressed as afunction of feed, f :

Cp = Cm + Cs + Cl + Ct + Cl + Cs

=[πLD

V(Lm +Bm)

]1f

+(πLDdV 6

C7

)Ψf3

Taking the derivative with respect to the con-stant feed (f) and setting it equal to zero gives:

dCp

df= 0

= −[πLD

V(Lm +Bm)

]1f2

+3Ψ(πLDdV 6

C7

)f2

Solving for f then gives

f =(C7 (Lm +Bm)

3dV 7Ψ

)1/4

8.140 Assuming that the coefficient of friction is 0.25,calculate the maximum depth of cut for turninga hard aluminum alloy on a 20-hp lathe (witha mechanical efficiency of 80%) at a width ofcut of 0.25 in., rake angle of 0, and a cuttingspeed of 300 ft/min. What is your estimate ofthe material’s shear strength?

The maximum allowable cutting force that willstall the lathe is given as:

P = (0.8)(20 hp) = 528, 000 ft-lb/min

Solving for Fc,

Fc =528, 000 ft-lb/min

V=

528, 000 ft-lb/min300 ft/min

or Fc = 1760 lb. From Eq. (8.11),

Fc =wtoτ cos(β − α)

sinφ cos(φ+ β − α)

or

to =Fc sinφ cos(φ+ β − α)

wτ cos(β − α)

It is known that α = 0 and w = 0.25 in. FromEq. (8.12),

β = tan−1 µ = tan−1 0.25 = 14.0

Using Eq. (8.20), the shear angle, φ, is found as

φ = 45 +α

2− β

2= 45 + 0 − 14

2= 38

The strength of an aluminum alloy varieswidely, as can be seen from Table 3.7 on p. 116.

111






Lets use Y = 300 MPa = 43.5 ksi as typicalfor a very hard aluminum alloy, thus the shearstrength is τ = Y/2 = 21.7 ksi. Hence the max-imum depth of cut is

to =Fc sinφ cos(φ+ β − α)

wτ cos(β − α)

=(1760) sin 31 cos(31 + 14 − 0)

(0.25)(21, 700) cos(14 − 0)= 0.121 in.

The maximum depth of cut is just under 18 in.

8.141 Assume that, using a carbide cutting tool, youmeasure the temperature in a cutting opera-tion at a speed of 250 ft/min and feed of 0.0025in./rev as 1200F. What would be the approx-imate temperature if the cutting speed is in-creased by 50%? What should the speed be tolower the maximum temperature to 800F?

From Eq. (8.30) we know that

T ∝ V af b

orT = kV af b

where k is a constant. From Section 8.2.6, fora carbide tool, a = 0.2 and b = 0.125. Forthe first problem, where the cutting speed isincreased by 50%, we can write

T1

T2=kV a

1 fb1

kV a2 f

b2

=kV a

1 fb1

k(2V1)af b1

=1

1.5a=

11.50.2

or T1T2

= 0.92. Therefore, the temperature in-crease is 15% over the first case. Note that thisequation is problematic if either of the tem-peratures T1 or T2 is zero or negative; there-fore, an absolute temperature scale is required.The problem states that T1 = 1200F, thus, onan absolute scale, T1 = 1660 R, and therefore,T2 = 1908 R, or T2 = 1448F.

For the second problem, where T2 =800F=1260 R, the temperature ratio is T1

T2=

1.317. Therefore(V1

V2

)a

= 1.317

orV1

V2= (1.317)1/a = 1.3175 = 3.97

So that the speed has to be 250/3.97 = 63ft/min.

8.142 A 3-in-diameter gray cast-iron cylindrical partis to be turned on a lathe at 500 rpm. Thedepth of cut is 0.25 in. and a feed is 0.02 in./rev.What should be the minimum horsepower ofthe lathe?

The metal removal rate is given as

MRR = πDavedfN

= π(3.875)(0.25)(0.02)(600)= 36.5 in3/min

The energy requirement for cast irons is, atmost, 2.0 hp.min/in3 (see Table 8.3). There-fore, the horsepower needed in the lathe motor,corrected for 80% efficiency, is

P =2.0 hp-min/in3

36.5 in3/min= 0.05 hp

This is a small number and suitable for afractional-power lathe.

8.143 (a) A 6-in.-diameter aluminum bar with alength of 12 in. is to have its diameter reducedto 5 in. by turning. Estimate the machiningtime if an uncoated carbide tool is used. (b)What is the time for a TiN-coated tool?

(a) From Table 8.9 on p. 472, the range ofparameters for machining aluminum with un-coated carbide tools is estimated as:

d = 0.01− 0.35 in.

f = 0.003− 0.025 in.

V = 650− 2000 ft/min.

This table gives a wide range of recommenda-tions and states that coated and ceramic toolsare on the high end of the recommended values.There is some variability in the actual speedsthat can be selected by the student for analy-sis; the following solution will use these valuesfor the uncoated carbide.

It is not advisable to produce this part in a sin-gle machining operation, since the depth of cutwould exceed the recommendations given in Ta-ble 8.9. Also, as described in Section 8.9, usu-ally one or more roughing cuts are followed by afinishing cut to meet surface finish and dimen-sional tolerance requirements. Since the total

112






depth of cut is to be 0.5 in., it would be ap-propriate to perform two equal roughing cuts,each with d = 0.24 in., and a finishing cut atd = 0.02 in. For the roughing cuts, the maxi-mum allowable feed and speed can be used, sothat f = 0.025 in./rev and V = 2000 ft/min.For the finishing cuts, the feed is determinedby surface finish requirements, but is assignedthe minimum value of 0.003 in./rev; the speedis similarly set at a value of V = 1000 ft/min.The average diameter for the first roughing cutis 5.76 in., and 5.28 in. for the second cut. Therotational speeds for first and second roughingand finishing cuts are (from V = πDaveN) 110rpm, 120 rpm, and 60 rpm, respectively. Thetotal machining time is thus

t =∑ l

fN=

12 in.(0.025 in./rev)(110 rpm)

+12 in.

(0.025 in./rev)(120 rpm)

+12 in.

(0.02 in./rev)(60 rpm)= 18.36 min

(b) For a coated tool, such as TiN, the cut-ting speed can be higher than the values usedabove. Consequently, the cutting time will belower than that for uncoated tools.

8.144 Calculate the power required for the cases givenin Problem 8.143.

Note that Problem 8.143 was an open-endedproblem, and thus the specific feeds, speeds,and depths of cut depend on the number andcharacteristics of the roughing and finishingcuts selected. This answer will be based thesolution to Problem 8.143.

For aluminum, Table 8.3 gives a specific en-ergy of between 0.15 and 0.4 hp-min/in3, thus amean value of u = 0.275 hp-min/in3 is chosen.Consider the first roughing cut, where d = 0.24in, f = 0.025 in., N = 110 rpm, and Davg isgiven as

Davg =6 in. + 5.52 in.

2= 5.76 in.

Therefore, the metal removal rate is given by

Eq. (8.38) as:

MRR = πDavgdfN

= π(5.76)(0.24)(0.025)(110)= 11.94 in3/min

Therefore, the power required is:

P = u(MRR) = (0.275)(11.94)

or P = 3.28 hp. Similarly, for the secondroughing cut, d = 0.24 in, f = 0.025 in./rev,N = 120 rpm, and Davg = 5.28 in. Therefore,MRR=11.94 in3/min and P = 3.28 hp. Forthe finishing cut, d = 0.01 in., f = 0.02 in/rev,N = 60 rpm and Davg = 5.01 in. Therefore,MRR=0.19 in3/min and P = 0.052 hp.

8.145 Using trigonometric relationships, derive an ex-pression for the ratio of shear energy to fric-tional energy in orthogonal cutting, in terms ofangles α, β, and φ only.

We begin with the following expressions for us

and uf , respectively, (see Section 8.2.5):

us =FsVs

wtoVand uf =

FVc

wtoV

Thus, their ratio becomes

us

uf=FsVs

FVc

The terms involved above can be defined as

F = R sinβ

and from Fig. 8.11,

Fs = R cos(φ+ β − α)

However, this expression can be simplified fur-ther by noting in the table for Problem 8.107that the magnitudes of φ and α are close toeach other. This expression can thus be ap-proximated as

Fs = R cosβ

Also,

Vs =V cosα

cos(φ− α)

Vc =V sinα

cos(φ− α)Combining these expressions and simplifying,we obtain

us

uf= cotβ cotα

113






8.146 For a turning operation using a ceramic cuttingtool, if the cutting speed is increased by 50%,by what factor must the feed rate be modifiedto obtain a constant tool life? Let n = 0.5 andy = 0.6.

Equation (8.33) will be used for this problem.Since the tool life is constant, we can write thefollowing:

C1/nd−x/n1 f

−y/n1

V1/n1

=C1/nd

−x/n2 f

−y/n2

V1/n2

Note that the depth of cut is constant, henced1 = d2, and also it is given that V2 = 1.5V1.Substituting the known values into this equa-tion yields:

V −21 f

−0.6/0.51 = (1.5V1)

−2f−0.6/0.52

or

1.52 =(f1f2

)−1.2

so thatf1f2

=(1.52

)1/1.2= 1.96

or the feed has to be reduced by about 50%.

8.147 Using Eq. (8.35), select an appropriate feed forR = 1 mm and a desired roughness of 1 µm.How would you adjust this feed to allow for nosewear of the tool during extended cuts? Explainyour reasoning.

If Ra = 1 µm, and R = 1 mm, then

f2 = (1 µm)(8)(1 mm) = 8× 10−9 m2

Therefore,

f = 0.089 mm/rev

If nose wear occurs, the radius will increase.The feed will similarly have to increase, per theequation above.

8.148 In a drilling operation, a 0.5-in. drill bit is be-ing used in a low-carbon steel workpiece. Thehole is a blind hole which will then be tappedto a depth of 1 in. The drilling operation takesplace with a feed of 0.010 in./rev and a spindlespeed of 700 rpm. Estimate the time requiredto drill the hole prior to tapping.

The velocity of the drill into the workpieceis v = fN = (0.010 in./rev)(700 rpm) = 7in./min. Since the hole is to be tapped to adepth of 1 in., it should be drilled deeper thanthis distance. Note from Section 8.9.4 that thepoint angle for steels ranges from 118 to 135,so that (using 118 to get a larger number andconservative answer) the drill actually has topenetrate at least a distance of

l = 1 +d

2sin(90 − 118/2)

= 1 +(

0.5 in.2

)(sin 31)

= 1.13 in.

In order to ensure that the tap doesn’t strike thebottom of the hole, let’s specify that the drillshould penetrate 1.25 in., which is the nearest1/4 in. over the minimum depth of the hole.Therefore, the time required for this drilling op-eration is 1.25 in./(7 in./min) = 0.18 min = 11s.

8.149 Assume that in the face-milling operationshown in Fig. 8.54, the workpiece dimensionsare 5 in. by 10 in. The cutter is 6 in. in di-ameter, has 8 teeth, and rotates at 300 rpm.The depth of cut is 0.125 in. and the feed is0.005 in./tooth. Assume that the specific en-ergy required for this material is 2 hp-min/in3

and that only 75% of the cutter diameter is en-gaged during cutting. Calculate (a) the powerrequired and (b) the material removal rate.

From the information given, the material re-moval rate is

MRR = (0.005 in./tooth)(8 teeth/rev)×(300 rev/min)(0.125 in.)×(0.75)(6 in.)

or MRR = 6.75 in3/min. Since the specificenergy of material removal is given as 2 hp-min/in3,

Power = (6.75)(2) = 13.5 hp

8.150 Calculate the ranges of typical machining timesfor face milling a 10-in.-long, 2-in.-wide cutterand at a depth of cut of 0.1 in. for the followingworkpiece materials: (a) low-carbon steel, (b)titanium alloys, (c) aluminum alloys, and (d)thermoplastics.

114






The cutting time, t, in face milling is byEq. (8.44) as

t =l + lcv

We know that l = 10 in., hence, as calculated inExample 24.1 (and proven in Problem 24.36), lcis obtained as

lc =√Dd =

√(2 in.)(0.1 in.) = 0.45 in.

The remaining main variable is the feed, a rangeof which can be seen in Table 8.12 for the ma-terials listed in the problem. For example, withlow-carbon steel, the feed per tooth is 0.003-0.015 in/tooth. The cutting time, as obtainedfor 10 teeth in the cutter. is given below:

Maximum MinimumMaterial time (s) time (s)Low-carbon steel 348 70Titanium alloys 348 70Aluminum alloys 348 58Thermoplastics 348 58

8.151 A machining-center spindle and tool extend 12in. from its machine-tool frame. What temper-ature change can be tolerated to maintain a tol-erance of 0.0001 in. in machining? A toleranceof 0.001 in.? Assume that the spindle is madeof steel.

The extension due to a change in temperatureis given by

∆L = α∆TL

where α is the coefficient of thermal expansion,which, for carbon steels, is α = 6.5× 10−6/F.If ∆L = 0.0001 in. and L = 12 in., then ∆Tcan easily be calculated to be 1.28F. Also, for∆L = 0.001 in., we have ∆T = 12.8F. Notingthat the temperatures involved are quite small,this example clearly illustrates the importanceof environmental control in precision manufac-turing operations, where dimensional tolerancesare extremely small.

8.152 In the production of a machined valve, the laborrate is $19.00 per hour and the general overheadrate is $15.00 per hour. The tool is a square ce-ramic insert and costs $25.00; it takes five min-utes to change and one minute to index. Es-timate the optimum cutting speed from a costperspective. Let C = 100 for Vo in m/min.

The optimum cutting speed is given byEq. (8.57) as:

Vo =C (Lm +Bm)n(

1n − 1

)n Ψn

where

Ψ =1m

[Tc (Lm +Bm) +Di] + Ti(Lm +Bm)

Note that for a ceramic tool, n is estimatedfrom Table 8.4 as 0.50. For Lm = $19.00,Bm = $15.00, m = 4, Di = $25.00, Tc = 5min, and Ti = 1 min,

Ψ =14

[560

(19 + 15) + 25]

+ 1(19 + 15)

or Ψ = 40.95. Therefore, the optimum cuttingspeed is

Vo =C (Lm +Bm)n(

1n − 1

)n Ψn

=(100) (19 + 15)0.5(1

0.5 − 1)0.5 7.5250.5

= 91 m/min.

8.153 Estimate the optimum cutting speed in Prob-lem 8.152 for maximum production.

Using the same values as in Problem 8.152, theoptimum cutting speed for maximum produc-tion is determined from Eq. (8.60) as:

Vo =C[(

1n − 1

) (Tc

m + Ti

)]nSubstituting appropriate values, this gives acutting speed of 66.67 m/min for the ceramicinserts.

8.154 Develop an equation for optimum cutting speedin a face milling operation using a milling cutterwith inserts.

The analysis is similar to that for a turning op-eration presented in Section 8.15. Note thatsome minor deviations from this analysis shouldbe acceptable, depending on the specific as-sumptions made. The general approach should,however, be consistent with the following solu-tion.

The main differences between this problem andthat in Section 8.15 are

115






(a) The tool cost is given by:

Ct =1Np

[Tc (Lm +Bm)

+Dc + Ti (Lm +Bm)]

where Np = number of parts produced pertool change, Ti=time required to changeinserts, Dc is the cost of the inserts, andremaining terminology is consistent withthat in the textbook.

(b) If the approach distance, lc, can be ig-nored, the machining time is obtainedfrom Eqs. (8.43) and (8.44) as

Tm =lD

fV m

where l is the cutting length, D is the cut-ter diameter, f is the feed per tooth, mis the number of teeth on the cutter pe-riphery and C is the cutting speed. Notethat m is the variable used to representthe number of inserts, whereas n is used inEq. (8.43). This substitution of variableshas been made to avoid confusion with theexponent in the Taylor tool life equation.Note that this equation for cutting time isonly slightly different than Eq. (8.51).

Also note that the Taylor tool life equation re-sults in:

T =(C

V

)1/n

so that the number of parts per tool change is:

Np =T

Tm=C1/nfV (n−1)/nm

lD

Substituting into Eq. (8.46),

Cp = V −1 lD

fm(Lm +Bm) + Cs + Cl

+lD

C1/nfmV n/(n−1)Ψ

where

Ψ = Tc (Lm +Bm) +Dc + Ti (Lm +Bm)

Taking the derivative with respect to V :

dCp

dV= 0 = −V −2 lD(Lm +Bm)

fm

+n

n− 1lD

C1/nfmΨV n/(n−1)−1

Solving for V ,

V −2

V 1/(n−1)=

nn−1

lDC1/nfm

ΨlD(Lm+Bm)

fm

simplifying,

V =

(n

n−1C−1/nΨ

Lm +Bm

) n−12n−3

8.155 Develop an equation for optimum cutting speedin turning where the tool is a high speed steeltool that can be reground periodically.

Compared to Section 8.15, the main differenceis in the equation for Ct. Thus, Eq. (8.49) be-comes

Ct =1Np

[Tc (Lm +Bm) + Tg (Lg +Bg) +Dc]

or, in order to simplify,

Ct =ΨNp

Note that Ψ is independent of cutting and thuscan be assumed to be constant in this deriva-tion. Just as done in Section 8.15, these re-lations are substituted into the cost per piecegiven by Eq. (8.46), the derivative with respectto V is taken and set equal to zero. The resultis

Vo =C(Lm +Bm)n[(

1n − 1

)Ψ]n

If we substitute for Ψ, this is an expression verysimilar to Eq. (8.57).

8.156 Assume that you are an instructor covering thetopics in this chapter, and you are giving a quizon the quantitative aspects to test the under-standing of the students. Prepare several nu-merical problems, and supply the answers tothem.


116






.

Design

8.157 Tool life could be greatly increased if an effec-tive means of cooling and lubrication were de-veloped. Design methods of delivering a cuttingfluid to the cutting zone and discuss the advan-tages and shortcomings of your design.

By the student. This is an open-ended problemand students are encouraged to pursue creativesolutions. Methods of delivering fluid to thecutting zone include (see also Section 8.7.1):

(a) Flooding or mist cooling of the cuttingzone, which has been the traditional ap-proach.

(b) High-pressure coolant application.

(c) Using a tool with a central hole or otherpassageway to allow for the fluid to bepumped into the cutting zone; an exam-ple is the end mill shown below.

8.158 Devise an experimental setup whereby you canperform an orthogonal cutting operation on alathe using a short round tubular workpiece.

By the student. This can be done simply byplacing a thin-walled tube in the headstock ofa lathe (see Fig. 8.19, where the solid bar is nowreplaced with a tube) and machining the end ofthe tube with a simple, straight tool (as if toshorten the length of the tube). Note that thefeed on the lathe will become the depth of cut,to, in orthogonal cutting, and the chip widthwill be the same as the wall thickness of thetube.

8.159 Cutting tools are sometimes designed so thatthe chip-tool contact length is controlled byrecessing the rake face some distance awayfrom the tool tip (see the leftmost design inFig. 8.7c). Explain the possible advantages ofsuch a tool.

By the student. The principal reason is thatby reducing the tool-chip contact, the frictionforce, F , is reduced, thus friction and cuttingforces are reduced. Chip morphology may alsochange. The student is encouraged to searchthe technical literature regarding this topic.

8.160 The accompanying illustration shows drawingsfor a cast-steel valve body before (left) and af-ter (right) machining. Identify the surfaces thatare to be machined (noting that not all sur-faces are to be machined). . What type ofmachine tool would be suitable to machine thispart? What type of machining operations areinvolved, and what should be the sequence ofthese operations?

Casting After machining

100 mm

100 mm

By the student. Note that the dimensions of thepart suggest that most of these surfaces are producedin a drill press, although a milling machine could alsobe used. However, the sharp radius in the enlargedhole on the right side cannot be produced with a drill;this hole was bored on a lathe.

8.161 Make a comprehensive table of the process ca-pabilities of the machining operations describedin this chapter. Use several columns describethe (a) machines involved, (b) type of toolsand tool materials used, (c) shapes of blanksand parts produced, (d) typical maximum andminimum sizes produced, (e) surface finish pro-duced, (f) dimensional tolerances produced,and (g) production rates achieved.

By the student. This is a challenging and com-prehensive problem with many possible solu-tions. Some examples of acceptable answerswould be:

117






Pro

cess

Mach

ine

Cutt

ing-t

ool

Shapes

Typic

al

tools

mate

rials

size

sTurn

ing

Lath

eA

ssort

ed;se

eA

xis

ym

met

ric

1-1

2in

.Table

23.4

dia

met

er,4-4

8in

.le

ngth

Dri

llin

gLath

e,m

ill

Ass

ort

ed,

Cir

cula

rhole

s1-1

00

mm

(50

dri

llpre

ssusu

ally

HSS

µm

poss

ible

)K

nurl

ing

Lath

e,m

ill

Ass

ort

ed,

Rough

surf

ace

son

Sam

eas

inusu

ally

HSS

axis

ym

met

ric

turn

ing

part

s

8.162 A large bolt is to be produced from hexagonalbar stock by placing the hex stock into a chuckand machining the cylindrical shank of the boltby turning on a lathe. List and explain the dif-ficulties that may be involved in this operation.

By the student. There could several diffi-culties with this operation. Obviously theprocess involves interrupted cutting, with re-peated impact between the cutting-tool andthe workpiece surface, and the associated dy-namic stresses which, in turn, could lead to toolchipping and breakage. Even if the tool sur-vives, chatter may be unavoidable in the earlystages (depending on the characteristics of themachine-tool and of the fixtures used) when thedepth of cut variations are at their maximum.Note that the ratio of length-to-cross-sectionalarea of the bolt also will have an influence onpossible vibration and chatter.

8.163 Design appropriate fixtures and describe themachining operations required to produce thepiston shown in Fig. 12.62.

By the student. Note that the piston has to be

turned on a lathe to establish the exterior sur-face and the grooves for the piston rings, andcan be fixtured on an internal surface for theseoperations. The seat for the main piston bear-ing requires end milling and boring, and canbe fixtured on its external surface. The face ofthe piston needs contour milling because of theclose tolerances with the cylinder head.

8.164 In Figs. 8.16 and 8.17b, we note that the maxi-mum temperature is about halfway up the faceof the tool. We have also described the adverseeffects of temperature on various tool materi-als. Considering the mechanics of cutting oper-ations, describe your thoughts on the technicaland economic merits of embedding a small in-sert, made of materials such as ceramic or car-bide, halfway up the rake face of a tool made ofa material with lower resistance to temperaturethan ceramic or carbide.

By the student. This is an interesting problemthat has served well as a topic of classroom dis-cussion. The merits of this suggestion include:

(a) If performed properly, the tool life couldbe greatly improved, and thus the eco-nomics of the cutting operation could begreatly affected in a favorable way.

(b) Brazing or welding an insert is probablyeasier than applying a coating at an ap-propriate location.

The drawbacks of this approach include:

(a) The strength of the joint between insertand tool material must be high in order towithstand machining operation.

(b) It is likely that the tool will wear beyondwhere the insert is placed.

(c) Thermal stresses can develop, especially atthe interface where coefficients of thermalexpansion may be significantly different.

8.165 Describe your thought on whether chips pro-duced during machining can be used to makeuseful products. Give some examples of possi-ble products and comment on their character-istics and differences as compared to the sameproducts made by other manufacturing pro-cesses. Which types of chips would be desirablefor this purpose? Explain.

118






By the student. This is an interesting designproject and represents an example of cradle-to-cradle life-cycle design (see Section 1.4). Someexamples of possible applications include:

(a) If the chips are discontinuous, they canhave a high aspect ratio transverse to thecutting direction; these chips can thenserve as metal reinforcement in compositematerials.

(b) Filters can be made by compacting metalchips into suitable shapes, such as cylin-drical or tubular.

(c) The chips can be used as a vibration-isolating elastic support.

(d) The chips can be further conditioned (suchas in a ball mill) to produce different formsor powders.

(e) The chips can be used as a precursor inchemical vapor deposition.

(f) Numerous artwork can be developed forunique chips.

8.166 Experiments have shown that it is possible toproduce thin, wide chips, such as 0.08 mm(0.003 in.) thick and 10 mm (4 in.) wide, whichwould be similar to rolled sheet. Materials usedhave been aluminum, magnesium, and stainlesssteel. A typical setup would be similar to or-thogonal cutting, by machining the periphery ofa solid round bar with a straight tool moving ra-dially inward (plunge). Describe your thoughtson producing thin metal sheet by this method,its surface characteristics, and its properties.

By the student. This is an interesting problemthat has served well as a topic of classroom con-versation. This process does not appear to be inany way advantageous to metal rolling. How-ever, many aerospace alloys are too brittle orhard to be rolled economically, and this methodoffers a possible manufacturing approach. Thistechnique has also been used to develop materi-als that are highly oriented, which, for example,can, for example, positively influence magnetic.Note from Figs. 8.2 and 8.5 that the sheet wouldhave a smooth surface on one side (where it hasrubbed against the tool face) and a rough sur-face on the opposite side.

8.167 One of the principal concerns with coolants isdegradation due to biological attack by bacte-ria. To prolong their life, chemical biocides areoften added, but these biocides greatly compli-cate the disposal of coolants. Conduct a liter-ature search regarding the latest developmentsin the use of environmentally-benign biocides incutting fluids.

By the student. This is an interesting topic fora research paper. New and environmentally be-nign biocides are continuously being developed,with some surprising requirements. For exam-ple, the economic and safety and ecological con-cerns are straightforward. However, there isalso the need to consider factors such as thetaste of the biocide. That is, if a food containeris produced, trace amounts of lubricant and bio-cide will remain on the surface and can influencethe taste of the contents. Note that these tracesare not considered hazardous. Also, the re-peatability of the biocide is an issue; it must becontrollable to fulfill TQM considerations (seeSection 4.9).

8.168 If expanded honeycomb panels (see Section7.5.5) were to be machined in a form milling op-eration (see Fig.8.58b), what precautions wouldyou take to keep the sheet metal from bucklingdue to cutting forces? Think of as many solu-tions as you can.

By the student. This is an open-ended problemcan be interpreted in two ways: That the hon-eycomb itself is being pocket machined, or thata fabricated honeycomb is being contoured. Ei-ther problem is a good opportunity to challengestudents to develop creative solutions. Accept-able approaches include:

(a) high-speed machining, with properly cho-sen processing variables,

(b) using alternative processes, such as chem-ical machining,

(c) filling the cavities of the honeycomb struc-ture with a low-melting-point metal (toprovide strength to the thin layers of mate-rial being machined) which is then meltedaway after the machining operation hasbeen completed, and

(d) filling the cavities with wax, or with water

119






(which is then frozen), and melted afterthe machining operation is completed.

8.169 The part shown in the accompanying figure is a

power-transmitting shaft; it is to be producedon a lathe. List the operations that are ap-propriate to make this part and estimate themachining time.

7.2824.625

4.093

0.439

0.625

0.5910.591

1.7411.156

0.125

0.75

0.460 0.5000.375

1.2070.813

0.500

0.500

0.062 R

9030

13030

6030

0.38-24 UNF

Key seat width 0.096 x depth 0.151

Dimensions in inches

By the student. Note that the operations should be designed to incorporate, as appropriate, roughingand finishing cuts and should minimize the need for tool changes or refixturing.

120






Chapter 9

Material-Removal Processes:Abrasive, Chemical, Electrical, andHigh-Energy Beams

Questions

9.1 Why are grinding operations necessary for partsthat have been machined by other processes?

The grinding operations are necessary for sev-eral reasons, as stated in Section 9.1. For ex-ample, the hardness and strength of the work-piece may be too high to be machined to fi-nal dimensions economically; a better surfacefinish and dimensional tolerance is needed; orthe workpiece is too slender to support machin-ing forces. Students are encouraged to expandon these statements, giving specific examplesbased on the contents of Chapters 8 and 9.

9.2 Explain why there are so many different typesand sizes of grinding wheels.

There numerous types and sizes of grindingwheels because of the different types of opera-tions performed on a variety of materials. Thegeometry of a grinding wheel and the mate-rial and structural considerations for a grindingwheel depend upon the workpiece shape andcharacteristics, surface finish desired, produc-tion rate, heat generation during the process,economics of wheel wear, and type of grindingfluids used.

9.3 Why are there large differences between thespecific energies involved in grinding (Table 9.3)

and in machining (Table 8.3)? Explain.

Specific energies in grinding, as compared tomachining, are much higher (see Table 9.3 onp. 534) due to:

(a) The presence of wear flats, causing highfriction.

(b) The large negative rake angles of the abra-sive grains, whereby the chips formed dur-ing grinding undergo higher deformation,and thus require more energy.

(c) Size effect, due to very small chips pro-duced (see Example 9.1 on p. 532), hasalso been discussed as a contributing fac-tor.

9.4 Describe the advantages of superabrasives overconventional abrasives.

By the student. See also Sections 8.6.7 and8.6.9. Superabrasives are extremely hard (dia-mond and cubic boron nitride are the two hard-est materials known), thus they are able to re-move material even from the hardest workpiece.Their higher costs are an important economicconsideration.

9.5 Give examples of applications for the grindingwheels shown in Fig. 9.2.

121






By the student; see also the Bibliography atthe end of the chapter. As an example, theflaring cup wheel shown in Fig. 9.2d is com-monly used for surface grinding with hand-heldgrinders, and the mounted wheel shown in Fig9.2g is a common wheel for manual rework ofdies.

9.6 Explain why the same grinding wheel may actsoft or hard.

An individual grinding wheel can act soft orhard depending on the particular grinding con-ditions. The greater the force on the grindingwheel grains, the softer the wheel acts; thus, agrinding wheel will act softer as the workpiecematerial strength, work speed, and depth of cutincrease. It will act harder as the wheel speedand wheel diameter increase. Equation (9.6)gives the relationship between grain force andthe process parameters. See also Section 9.5.2.

9.7 Describe your understanding of the role of fri-ability of abrasive grains on the performance ofgrinding wheels.

By the student. High friability means that thegrains will fracture with relative ease duringgrinding. In effect, this allows for sharp cut-ting points to be developed, leading to moreeffective grinding. If, on the other hand, thegrains do not fracture easily, the cutting pointswill become dull and grinding will become in-efficient; this situation will then lead to unac-ceptable temperature rise and adversely affect-ing surface integrity.

9.8 Explain the factors involved in selecting the ap-propriate type of abrasive for a particular grind-ing operation.

By the student. Consider, for example, the fol-lowing: Abrasives should be inert to the work-piece material so that the material does notbond to the abrasive grain during the grind-ing operation, as this will reduce the effective-ness of the abrasive. The abrasives should beof appropriate size for the particular applica-tion. Applications that require better surfacefinish require smaller grains, while those wheresurface finish is a secondary consideration to re-moval rate should use larger grains. The grind-ing wheel should provide for heat removal from

the cutting zone, either through the chips gen-erated or the use of grinding fluids.

9.9 What are the effects of wear flat on the grind-ing operation? Are there similarities with theeffects of flank wear in metal cutting? Explain.

A wear flat causes dissipation of energy andincreases the temperature of the operationthrough friction. Wear flats are undesirable be-cause they provide no useful work (they playno obvious role in producing the chip) but theysignificantly increase the frictional forces andcan cause severe temperature rise of the work-piece. Recall that in orthogonal cutting, flankwear is equivalent to wear flats in grinding (see,for example, Fig. 8.20a on p. 440).

9.10 It was stated that the grinding ratio, G, de-pends on the following factors: (1) type ofgrinding wheel, (2) workpiece hardness, (3)wheel depth of cut, (4) wheel and workpiecespeeds, and (5) type of grinding fluid. Explainwhy.

The grinding ratio, G, decreases as the grainforce increases and is associated with high at-tritious wear of the wheel. Consider also:

(a) The type of wheel will have an effect onwheel wear; vitrified wheels generally wearslower than resinoid bonded.

(b) Depth of cut has a similar effect.(c) Workpiece hardness will lower G because

of increased wear, if all other process pa-rameters are kept constant.

(d) Wheel and workpiece speed affect wearin opposite ways; higher wheel speed re-duces the force on the grains, which re-duces wheel wear.

(e) Type of grinding fluid, as it reduces wearand thus improves the efficiency of grind-ing.

9.11 List and explain the precautions you would takewhen grinding with high precision. Commenton the role of the machine, process parameters,the grinding wheel, and grinding fluids.

By the student. When grinding for high pre-cision (see also p. 477), it is essential that theforces involved remain low so that workpieceand machine deflections are minimal. As can

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be seen from Eq. (9.6) on p. 532, to mini-mize grinding forces, hence minimize deflec-tions, the wheel speed should preferably behigh, the workpiece speed should be low, andthe depth of cut should be small. The ma-chine used should have high stiffness with goodbearings. The temperature rise, as given byEq. (9.9) on p. 535, should be minimized.

The grinding wheel should have fine grains andthe abrasive should be inert to the workpiecematerial to avoid any adverse reactions. Agrinding fluid should be selected to provide lowwheel loading and wear, and also to provide foreffective cooling. Automatic dressing capabili-ties should be included and the wheel should bedressed often.

9.12 Describe the methods you would use to deter-mine the number of active cutting points perunit surface area on the periphery of a straight(Type 1; see Fig. 9.2a) grinding wheel. Whatis the significance of this number?

By the student. One method is to examine thewheel periphery under a microscope, and countthe points that are in sharp focus. Anothermethod is to measure the chip thickness andother variables in a known grinding operationand use Eq. (9.5) on p. 532 to determine C.Another method involves rolling the grindingwheel over a flat glass coated with soot; eachpoint on the periphery of the wheel contact-ing the soot removes a small amount of soot.The glass is then placed under a microscopeand with back lighting, the points are countedand expressed as a number per unit area. Note,however, soot thickness will affect the results.

9.13 Describe and explain the difficulties involved ingrinding parts made of (a) thermoplastics, (b)thermosets, and (c) ceramics.

By the student. Some of the difficulties encoun-tered would be:

(a) Thermoplastics have a low melting pointand have a tendency to soften and becomegummy; thus, they tend to bond to grind-ing wheels by mechanical locking. An ef-fective coolant, including cool air jet, canbe used to keep temperatures low.

(b) The low elastic modulus of thermoplasticscan make it difficult to hold dimensionaltolerances during grinding.

(c) Thermosets are harder and do not softenwith temperature (although they decom-pose and crumble at high temperatures).Consequently, grinding by using appropri-ate wheels and processing parameters isrelatively easy.

(d) Grinding of ceramics is relatively easyby using diamond wheels, appropriateprocessing parameters, and implementingductile-regime grinding. Note also theavailability of machinable ceramics (seep. 702).

9.14 Explain why ultrasonic machining is not suit-able for soft and ductile metals.

In ultrasonic machining, the stresses developedfrom particle impact should be sufficiently highto cause spalling of the workpiece. This in-volves surface fracture on a very small scale.If the workpiece is soft and ductile, the impactforce will simply deform the workpiece locally(as does the indenter in a hardness test), in-stead of causing fracture.

9.15 It is generally recommended that a soft-gradewheel be used for grinding hardened steels. Ex-plain why.

Note that grinding hardened steels involveshigher forces and the use of hard-grade wheels(meaning higher bond strength; see Section9.3.2) will tend to cause wear and dulling ofthe abrasive grains. As a result, temperaturewill increase, possibly causing surface damageand loss of dimensional accuracy. The use of asoft-grade wheel (see Figs. 9.4 and 9.5) meansthat under the high grinding forces present, dullgrains will be dislodged more easily, exposingsharp new cutting edges and thus leading tomore efficient grinding.

9.16 Explain the reasons that the processes de-scribed in this chapter may adversely affect thefatigue strength of materials.

Fatigue (see Section 2.7) is a complex phe-nomenon which accounts for a vast majorityof component failures. It is known that cracks

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generally start at or just below the workpiecesurface, and grow with repeated cyclic load-ings. Since different material-removal processesresult in different surface finishes (see, for ex-ample, Fig. 8.26 on p. 448), the size and shapeof cracks, and also similar stress raisers, varywith the particular process employed. This isthe basic reason why smooth polished surfacesare best suited for fatigue applications. Recallalso the role of residual stresses, particularly thebeneficial effects of compressive surface resid-ual stresses, in improving the fatigue strengthof materials.

9.17 Describe the factors that may cause chatter ingrinding operations and give the reasons whythey cause chatter.

Grinding chatter (see Section 9.6.8) is similarto chatter in machining, hence many of thefactors discussed in Section 8.12 apply here aswell. Basically, chatter is caused by any peri-odic variation in grinding forces. Factors thatcontribute to chatter are: stiffness of the ma-chine and damping of vibration, irregular grind-ing wheels, improper dressing techniques, un-even wheel wear, high material-removal rates,eccentric support or mounting of wheels, gearsand shafts, vibrations from nearby machinesthrough foundations, and inadequate clampingof the workpiece. Sources of regenerative chat-ter, such as workpiece material inhomogeneityand surface irregularities (such as from a previ-ous machining operation), also can cause chat-ter.

9.18 Outline the methods that are generally avail-able for deburring manufactured parts. Discussthe advantages and limitations of each method.

By the student. See Section 9.8. Some exam-ples of methods for deburring include grind-ing using bench or hand grinders, using wirebrushes, filing, scraping, chemical machining,and tumbling in a ball mill.

9.19 In which of the processes described in this chap-ter are the physical properties of the workpiecematerial important? Explain.

By the student. Recall that advances machin-ing processes generally depend on the electrical

and chemical properties of the workpiece ma-terial. Thus, for example, hardness, which isan important factor in conventional machiningprocesses, is not significant in chemical machin-ing because it does not adversely affect the abil-ity of the chemical to react with the workpiece.The student should elaborate further based onthe contents of this chapter.

9.20 Give all possible technical and economic reasonsthat the material removal processes described inthis chapter may be preferred, or even required,over those described in Chapter 8.

By the student. Note that the main reasons arelisted in Section 9.1. Students are encouragedto give specific examples after studying each ofthe individual processes.

9.21 What processes would you recommend for diesinking in a die block, such as that used forforging? Explain. (See also Section 6.7.)

By the student. Review the die manufacturingmethods described in Section 6.7, and note thatthe most commonly used die-sinking methodsare:

(a) milling, using rounded-tip end mills, fol-lowed by finishing processes, includinggrinding and polishing, and

(b) electrical-discharge machining.

1. smaller dies may be made by processes such aselectrochemical machining and hubbing.

9.22 The proper grinding surfaces for each type ofwheel are shown in Fig. 9.2. Explain why grind-ing on other surfaces of the wheel is improperand/or unsafe.

Because the wheels are designed to resist grind-ing forces, the proper grinding faces indicatedin Fig. 9.2 on p. 507 should be utilized. Note,for example, that if grinding forces act normalto the plane of a thin straight wheel (Type 1),the wheel will flex and may eventually fracture.Thus, from a functional standpoint, grindingwheels are made stiffer in the directions inwhich they are intended to be used. There areserious safety and functional considerations in-volved. For example, an operator grinding onthe side surface of a flared-cup wheel causes

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wear to take place. The flange thickness is thensignificantly reduced and the wheel may even-tually fracture, exploding with violent force andpotentially causing serious injury or death.

9.23 Note that wheel (b) in Fig. 9.3 has serrationsalong its periphery. Explain the reason for sucha design.

The basic advantages of this design are the fol-lowing:

(a) The stresses developed (rotational as wellas thermal) along the periphery of thewheel are lower.

(b) The serrations allow increased flow ofgrinding fluid, thus lowering temperaturesand reducing wheel wear,

(c) The grinding chips can be ejected eas-ier from the grinding zone through thesegrooves.

9.24 In Fig. 9.10, it will be noted that wheel speedand grinding fluids can have a major effect onthe type and magnitude of residual stresses de-veloped in grinding. Explain the possible rea-sons for these phenomena.

Grinding wheel speed affects temperature inthe same way that cutting-tool speed affectstemperature (see Section 8.2.6), but the effectis more complex and even greater in grindingsince a significant portion of the energy is dis-sipated in plowing and sliding abrasive grainsover the workpiece surface without chip gener-ation (see Figs. 9.7 and 9.9). The three grindingfluids indicated in the figure have different effec-tiveness on grinding mechanics and thus in re-ducing the temperature, leading to lower resid-ual stresses. This is a good topic for a studentpaper.

9.25 Explain the consequences of allowing the work-piece temperature to rise excessively in grindingoperations.

Recall the discussion of residual stresses in theanswer to Question 9.24. Temperature rise canhave additional major effects in grinding, in-cluding:

(a) If excessive, it can cause metallurgicalburn and heat checking.

(b) The workpiece may distort due to thermalgradients.

(c) With increasing temperature, the part willexpand, and thus the actual depth of cutwill be greater. Upon cooling, the partwill contract and the dimensional toler-ances may not be within the desired range.

9.26 Comment on any observations you have regard-ing the contents of Table 9.4.

By the student. Students should be encouragedto make comparisons and list advantages anddisadvantages of the processes listed in the ta-ble. An instructor may ask students to answerthis question for a particular workpiece mate-rial, such as carbon steel, Ti-6Al-4V, or a hardceramic, or to calculate the grinding time toproduce a simple part.

9.27 Why has creep-feed grinding become an impor-tant manufacturing process? Explain.

Recall that the advantages of creep-feed grind-ing (see Section 9.6.6) is the ability for highmaterial-removal rates while still maintainingthe advantages in high dimensional toleranceand surface finish of grinding operations, andthus significant economic advantages. Notethat these advantages are most pronouncedwith highly-alloyed materials which are difficultto machine, and where an abrasive process isrequired.

9.28 There has been a trend in manufacturing indus-tries to increase the spindle speed of grindingwheels. Explain the possible advantages andlimitations of such an increase in speed.

Increasing spindle speed has the benefit of in-creasing the material-removal rate, thus in-creasing productivity and reducing costs; seealso high-speed machining, Section 8.8. Thedrawbacks could include increases in tempera-tures [see Eq. (9.9) on p. 535] and associatedproblems, and more importantly the need formore stiff machine tools and better bearings toavoid chatter. Also, grinding wheels, if improp-erly designed, manufactured, selected, used, orhandled, can explode at high spindle speeds.

9.29 Why is preshaping or premachining of partsgenerally desirable in the advanced machiningprocesses described in this chapter? Explain.

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By the student. This is basically a matter ofeconomics, since large amounts of material mayfirst be removed by other means in less time andat lower cost. Surface finish and dimensionalaccuracy is not important in these preshapingoperations, unless they cause serious substratedamage that cannot be removed by subsequentmaterial removal and finishing processes.

9.30 Why are finishing operations sometimes neces-sary? How could they be minimized to reduceproduct costs? Explain, with examples.

By the student. Finishing operations are nec-essary when the dimensional tolerances or sur-face finish required cannot be obtained fromprimary processing. For example, sand cast-ing cannot produce a very smooth surface fin-ish whereas grinding can. However, if the partcould be roll forged instead of cast, smooth sur-faces can be obtained. The trend towards near-net-shape manufacturing (see p. 18) is drivenby a desire to avoid time-consuming and costlyfinishing operations.

9.31 Why has the wire-EDM process become sowidely used in industry, especially in tool anddie manufacturing? Explain.

Wire EDM has become widely accepted for sev-eral reasons (see also Section 9.13.2). The pro-cess is relatively easy to automate, and numer-ical control can be applied to machine tapers,inclines, or complex contours. Wire EDM is aprocess that can be used on any electrically con-ducting workpiece, regardless of its mechanicalproperties, so it can be preferred over processessuch as band sawing where wear and dullingof the blade would otherwise be an importantconcern. With increasing strength and tough-ness and various other properties of advancedengineering materials, there was a need to de-velop processes that were not sensitive to theseproperties. As in all other processes, it hasits advantages as well as limitations, regardingparticularly the material-removal rate and pos-sible surface damage, which could significantlyreduce fatigue life.

9.32 Make a list of the material removal processesdescribed in this chapter that may be suitablefor the following workpiece materials: (1) ce-ramics, (2) cast iron, (3) thermoplastics, (4)

thermosets, (5) diamond, and (6) annealed cop-per. Explain.

By the student. It will be noted that, as de-scribed in Chapter 8, most of these materialscan be machined through conventional means.Consider the following processes:

(a) Ceramics: water-jet machining, abrasive-jet machining, chemical machining.

(b) Cast iron: chemical machining, elec-trochemical machining, electrochemicalgrinding, EDM, laser-beam and electron-beam machining, and water- and abrasive-jet machining.

(c) Thermoplastics: water-jet and abrasive-jet machining; electrically-conductingpolymers may be candidates for EDMprocessing.

(d) Thermosets: similar consideration as forthermoplastics.

(e) Diamond: None, because diamond wouldnot be responsive to any of the methodsdescribed in this chapter.

(f) Annealed copper: Chemical and electro-chemical processes, EDM, and laser-beammachining.

9.33 Explain why producing sharp corners and pro-files using some of the processes described inthis chapter can be difficult.

By the student. Some of the processes are func-tionally constrained and cannot easily providevery small radii. Consider water-jet machining:the minimum radius which can be cut will de-pend on the ability to precisely focus the waterjet. With wire EDM, the minimum radius de-pends on the wire diameter. Small radii arepossible with small wires, but small wires havelow current-carrying capacity, thus compromis-ing the speed of the process. With laser-beamcutting, radii are adversely affected by mate-rial melting away from the cutting zone, as wellas beam diameter. Similar problems exist inchemical machining as the chemical tends to re-move a wider area than that required for sharpprofiles.

9.34 How do you think specific energy, u, varies withrespect to wheel depth of cut and hardness ofthe workpiece material? Explain.

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The specific energy, u, will decrease with in-creasing depth of cut, d, according to the sizeeffect, discussed on p. 533. It will increasewith workpiece hardness because of the higherstrength and hence the more energy required.An increase in the wheel depth of cut will re-sult in higher forces on the grains, as seen inEq. (9.6) on p. 532. Increasing workpiece hard-ness also means higher forces.

9.35 It is stated in Example 9.2 that the thrust forcein grinding is about 30% higher than the cut-ting force. Why is it higher?

We note in Fig. 9.7 that abrasive grains typi-cally have very high negative rake angles. Let’snow compare the force differences in grindingwith that for machining. Referring to Fig. 8.12we note that as the rake angle decreases, thethrust force increases rapidly. Inspecting thedata in Tables 8.1 and 8.2 on pp. 430-431, wenote the same phenomenon, and particularlythe fact that the difference between the twoforces becomes smaller as the rake angle be-comes negative. Based on these observations, itis to be expected that the thrust force in grind-ing will be higher than the cutting force.

9.36 Why should we be interested in the magnitudeof the thrust force in grinding? Explain.

By the student. Major considerations includethe fact that the thrust force determines thestrength required in supporting the grindingwheel on the machine. Note also the load thatis exerted onto the workpiece, which influencesthe elastic recovery in the workpiece and thusaffect the dimensional accuracy.

9.37 Why is the material removal rate in electrical-discharge machining a function of the meltingpoint of the workpiece material? Explain.

By the student. As described in Section 9.13,material removal in EDM is accomplished bymelting small amounts of material throughsparks supplied by electrical energy. Con-sequently, the higher the melting point, thehigher the energy required.

9.38 Inspect Table 9.4 and, for each process, list anddescribe the role of various mechanical, physi-cal, and chemical properties of the workpiecematerial on performance.

By the student. This problem is a good topicfor classroom discussion. Students may, for ex-ample, be asked which of the processes are af-fected by hardness, melting temperature, andelectrical and thermal conductivity.

9.39 Which of the processes listed in Table 9.4 wouldnot be applicable to nonmetallic materials? Ex-plain.

By the student. The following are generallynot applicable to nonmetallic materials: elec-trochemical machining, electrochemical grind-ing, EDM, and wire EDM.

9.40 Why does the machining cost increase rapidlyas surface finish requirements become finer?

By the student. As surface finish requirementsbecome finer, the depth of cut must be de-creased, and the grit size must also be de-creased. The operation must be carried outcarefully using rigid machines, proper control ofprocessing variables, and effective metalwork-ing fluids. These generally lead to longer ma-chining times and thus higher costs.

9.41 Which of the processes described in this chapterare particularly suitable for workpieces made of(a) ceramics, (b) thermoplastics, and (c) ther-mosets? Explain.

By the student. Note that many processes havelimited suitability for difficult-to-process work-pieces. However, an example of an acceptableanswer is:

(a) Ceramics: grinding, ultrasonic machining,chemical machining;

(b) thermoplastics: chemical machining, high-energy-beam machining, water-jet andabrasive-jet machining;

(c) thermosets: grinding, ultrasonic machin-ing, chemical machining, and water-jetand abrasive-jet machining.

9.42 Other than cost, is there a reason that a grind-ing wheel intended for a hard workpiece cannotbe used for a softer workpiece? Explain.

By the student. Recall that a soft workpiecemay load a grinding wheel unless it is specif-ically intended for use on that material. This

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would mean that the grinding wheel would needto be dressed and trued more often for efficientgrinding.

9.43 How would you grind the facets on a diamond,such as for a ring, since diamond is the hardestmaterial known?

Diamond grinding is typically done using finediamond powder. It should be realized thatjust because diamond is the hardest materialknown does not mean that it does not wear.Hardness not only arises from material proper-ties but also local geometry (see Section 2.6), soat the asperity scale it is possible for abrasionto occur on diamond.

9.44 Define dressing and truing, and describe the dif-ference between them.

These two terms are sometimes confused or dif-ficult to differentiate, since they usually are per-formed at the same time. As discussed in Sec-tion 9.5.1, dressing is the process of condition-ing worn grains to expose new and aggressivegrains. Truing involves reshaping an out-of-round wheel.

9.45 What is heat checking in grinding? What is itssignificance? Does heat checking occur in othermanufacturing processes? Explain.

Heat checking refers to small surface cracks ona workpiece, and in grinding this is caused byhigh stresses and excessively high temperatures(see also section 9.4.3). Heat checking is oftenassociated with development of tensile residualstresses on a surface. This is significant becauseit compromises both the fatigue properties ofthe workpiece as well as its appearance. Heatchecking also occurs in casting, especially in diecasting.

9.46 Explain why parts with irregular shapes, sharpcorners, deep recesses, and sharp projectionscan be difficult to polish.

By the student. Students are likely to have hadsome experience relevant to this question. Thebasic reason why these shapes may be difficultto polish is that it is difficult to have a polishingmedium to can properly follow an intricate sur-face, penetrate corners or depths, and be able

to apply equal pressure on all surfaces for uni-form polishing.

9.47 Explain the reasons why so many different de-burring operations have been developed overthe years.

By the student. There are several deburringoperations because of the wide variety of work-piece materials, their characteristics, shapes,surface features, and textures involved. Thereis also the requirement for different levels of au-tomation in deburring operations.

9.48 Note from Eq. (9.9) that the grinding tem-perature decreases with increasing work speed.Does this mean that for a work speed of zero,the temperature is infinite? Explain.

Consider the heat flow in grinding: The heatsource is at the wheel/workpiece interface andis caused by the work of plastic deformation inproducing chips and by friction (as it is in metalcutting). The heat is removed through the fol-lowing mechanisms:

(a) Chips leaving the ground surface.

(b) By conduction to the workpiece

(c) By convection in the workpiece, with theheat being physically moved with the ma-terial.

(d) By the grinding fluid, if used.

(e) Radiation, although this is usually muchsmaller than the other forms of heat trans-fer in the system.

According to the equation (which is an approx-imation), decreasing the work speed will in-crease the temperature, but the temperaturecannot be infinite because there are still theother means of heat transfer listed above.

9.49 Describe the similarities and differences in theaction of metalworking fluids in machining vs.grinding operations.

Compare the contents of Sections 8.7 and 9.6.9.From a functional standpoint, the purposes ofthese fluids are primarily cooling and lubricat-ing to reduce friction, temperature, wear, andpower requirements. There are many similari-ties between the two groups, including chemi-cal, rheological, and tribological properties. As

128






for differences, note that the dimensions in-volved in grinding are much smaller than thosein machining, and consequently the fluids mustbe able to penetrate the small interfaces. Thus,properties such as viscosity, wetting, surface-tension characteristics, and method of applica-tion would be more important in grinding. (Seealso Section 4.4.3.)

9.50 Are there any similarities among grinding, hon-ing, polishing, and buffing? Explain.

By the student. All of these processes use abra-sive particles of various types, sizes, and shapes,as well as various equipment to remove materialin very small amounts. Based on the details ofeach process described in this chapter, the stu-dent should elaborate further on this topic.

9.51 Is the grinding ratio an important factor in eval-uating the economics of a grinding operation?Explain.

A high grinding ratio, G, is high, means thatmuch material is removed with relatively lit-tle wear of the grinding wheel. Note, however,that this is not always desirable because it couldindicate that abrasive grains may be dulling,raising the workpiece temperature and possiblycausing surface damage. Low grinding ratios,on the other hand, indicate high wheel-wearrate, leading to the need to dress wheels morefrequently and eventually replacing the wholewheel. These considerations also involve thecost of the wheel, as well as the costs incurredin replacing the wheel and the economic im-pact of having to interrupt the production run.Consequently, as in all aspects of manufactur-ing, an optimum set of parameters have to beestablished to minimize any adverse economicimpact.

9.52 Although grinding can produce a very fine sur-face finish on a workpiece, is this necessarily anindication of the quality of a part? Explain.

The answer is not necessarily so because surfaceintegrity includes factors in addition to surfacefinish (which is basically a geometric feature).As stated on p. 133, surface integrity includesseveral mechanical and metallurgical parame-ters which, in turn, can have adverse effects onthe performance of a ground part, such as its

strength, hardness, and fatigue life. The stu-dents are encouraged to explore this topic fur-ther.

9.53 If not performed properly, honing can pro-duce holes that are bellmouthed, wavy, barrel-shaped, or tapered. Explain how this is possi-ble.

If the honing tool is mounted properly on itscenter and the axis of the tool is aligned withthe axis of the hole, the hole will be cylindri-cal. However, if this is not the case, the pathfollowed by the hone will not be circular. Itsshape will depend on the geometric relation-ships of the axes involved. This topic could beinteresting exercise in solid and descriptive ge-ometry, referring also to the literature on hon-ing practices.

9.54 Which of the advanced machining processes de-scribed in this chapter causes thermal damageto workpieces? List and explain the possibleconsequences of such damage.

The advanced machining processes which causethermal damage are obviously those that in-volve high levels of heat, that is, EDM, laser-beam, and electron-beam machining. The ther-mal effect can cause the material to developa heat-affected zone (see Fig. 12.15), thus ad-versely affecting hardness and ductility. For thevarious effects of temperature in machining andgrinding, see Sections 8.2.6 and 9.4.3, respec-tively.

9.55 Describe your thoughts regarding laser-beammachining of nonmetallic materials. Give sev-eral possible applications and include their ad-vantages as compared to other processes.

By the student. Most nonmetallic materials,including polymers and ceramics, can be laser-beam machined using different types of lasers.The presence of a concentrated heat source andits various adverse effects on a particular mate-rial and workpiece must of course be considered.Some materials can involve additional concerns;wood, for example, is flammable and may re-quire an oxygen-free environment.

9.56 It was stated that graphite is the generallypreferred material for EDM tooling. Would

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graphite also be appropriate for wire EDM? Ex-plain.

It is presently impossible to produce graphitewires, although significant effort has been di-rected towards impregnating tungsten wirewith graphite to improve its performance inEDM. An important consideration is their lackof ductility, which is essential in wire EDM(note the spools and wire guides in Fig. 9.35).Such hybrid wires have considerable promise,

but to date they have not produced sufficientutility, especially when compared to their cost.

9.57 What is the purpose of the abrasives in electro-chemical grinding? Explain.

By the student. The purpose of the abrasives inelectrochemical grinding are described in Sec-tion 9.12. Namely, they act as insulators and,in the finishing stages, help produce a surfacewith good surface finish and dimensional accu-racy.

Problems

9.58 In a surface-grinding operation, calculate thechip dimensions for the following process vari-ables: D = 8 in., d = 0.001 in., v = 30 ft/min,V = 5000 ft/min, C = 500 per in2, and r = 20.

The approximate chip length, l, is given byEq. (9.1) on p. 530 as

l =√Dd =

√(8)(0.001) = 0.0894 in.

The undeformed chip thickness, t, is given byEq. (9.5) on p. 532 as

t =

√4vV Cr

√d

D

=

√4(30)

(5000) (500) (20)

√0.001

8

= 1.64× 10−4 in.

9.59 If the workpiece strength in grinding is in-creased by 50%, what should be the percentagedecreases in the wheel depth of cut, d, in or-der to maintain the same grain force, all othervariables being the same?

From Section 9.4.1, it is apparent that if theworkpiece-material strength is doubled, thegrain force will be doubled. Since the grainforce is dependent on the square root of thedepth of cut, the new depth of cut would beone-fourth the original depth of cut. Thus, thereduction in the wheel depth of cut would be75%.

9.60 Taking a thin, Type 1 grinding wheel, as anexample, and referring to texts on stresses inrotating bodies, plot the tangential stress, σt,and radial stress, σr, as a function of radial dis-tance (from the hole to the periphery of thewheel). Note that because the wheel is thin,this situation can be regarded as a plane-stressproblem. How would you determine the max-imum combined stress and its location in thewheel? Explain.

The tangential and radial stresses in a rotatingcylinder are given, respectively, by (see Ham-rock, Jacobson, and Schmid, Fundamentals ofMachine Elements, McGraw-Hill, 1999, p. 401)

σθ =3 + ν

9ρω2

[r2i + r2o +

r2i r2o

r2− 1 + 3ν

3 + νr2]

and

σr =3 + ν

8ρω2

[r2i + r2o −

r2i r2o

r2− r2

]

where ρ is the material density, ω is the angu-lar velocity, ri and ro are the inner and outerradii, respectively, and ν is Poisson’s ratio forthe material. These are plotted as follows:

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ro

ri

Normalstress,

Radius, r

θ

r

The combined stresses can be calculated foreach radial position by referring to Section 2.11.

9.61 Derive a formula for the material removal ratein surface grinding in terms of process parame-ters. Use the same terminology as in the text.

The material removal rate is defined as

MRR =Volume of material removed

TimeIn surface grinding, the situation is similar tothe metal removal rate in slab milling (see Sec-tion 8.10.1). Therefore,

MRR =lwd

t= vwd

where w is the width of the grinding wheel.

9.62 Assume that a surface-grinding operation is be-ing carried out under the following conditions:D = 250 mm, d = 0.1 mm, v = 0.5 m/s, andV = 50 m/s. These conditions are then changedto the following: D = 150 mm, d = 0.1 mm,v = 0.3 m/s, and V=25 m/s. What is the dif-ference in the temperature rise from the initialcondition?

The temperature rise is given by Eq. (9.9) onp. 535. Note that the value of C is not known,but we can assume that it does not change be-tween the two cases, so it can be ignored in thisanalysis. For the initial case, we have

∆T ∝ D1/4d3/4

(V

v

)0.5

∝ (250)1/4(0.1)3/4

(500.5

)0.5

= 7.071

For the second case, we have

∆T ∝ D1/4d3/4

(V

v

)0.5

∝ (150)1/4(0.1)3/4

(250.3

)0.5

= 5.681

Therefore, we expect the temperature to de-crease, with the temperature rise to be about20% lower in the second case.

9.63 For a surface-grinding operation, derive an ex-pression for the power dissipated in impartingkinetic energy to the chips. Comment on themagnitude of this energy. Use the same termi-nology as in the text.

The power, P , in terms of kinetic energy perunit time, can be expressed as

P =12

(Volume of chips

Time

)ρV 2

or

P =12w

(rt2

4

)(V Cρ)

(V 2)

Since (rt2

4

)(V C) = vd

we have

P =dwρV 2v

2

The same expression can be derived by notingthat the volume of the chips removed is dwL,where L is the length ground. The work donein imparting velocity V to the chips is

Work =mV 2

2=dwLρV 2

2

Since power is the time rate of work,

P =dW

dt=dwρV 2

2dL

dt=dwρV 2v

2

which is the same expression as before.

9.64 The shaft of a Type 1 grinding wheel is attachedto a flywheel only, which is rotating at a certaininitial rpm. With this setup, a surface-grinding

131






operation is being carried out on a long work-piece and at a constant workpiece speed, v. Ob-tain an expression for estimating the linear dis-tance ground on the workpiece before the wheelcomes to a stop. Ignore wheel wear.

By the student. Note that, because of the vari-ables involved, there will be many possible an-swers. The specific energy (energy per unit vol-ume) is given by Eq. (9.7) as

u = uchip + uplowing + usliding

The magnitude of uchip can be found using anegative rake angle and the material propertiesas described in Chapter 8. uplowing can be foundthrough various means, including upper-boundanalysis. Equating the energy in the flywheelto the work done per unit length plowed, onecan then calculate the total length.

9.65 Calculate the average impact force on a steelplate by a 1-mm spherical aluminum-oxideabrasive grain, dropped from heights of (a) 1m, (b) 2 m, and (c) 10 m. Plot the results andcomment on your observations.

(a) The velocity of the particle as it strikes thesurface from an initial height of one meteris given by

v =√

2gh =√

2(9.81 m/s2)(1 m) = 4.43 m/s

The solid wave velocity in the workpiece isgiven by

c =

√E

ρ=

√200× 109 Pa7680 kg/m3 = 5103 m/s

and hence the contact time is calculatedfrom Eq. (9.11) as

to =5rco

(cov

)1/5

=5(0.0005)

5000

(50004.43

)1/5

or to = 2.01 × 10−6 s. From Table 11.7,the density of aluminum oxide is, on aver-age, 4250 kg/m3. Therefore, the mass ofthe particle is

m =43πd3ρ =

43π(0.001)3(4250)

or m = 1.78 × 10−5 kg. Therefore, fromEq. (9.13),

Fave =2mvto

=2(1.78× 10−5)(4.43)

2.04× 10−6

or Fave = 77.3 N(b) Repeating this calculation for a height of

2 m gives a force of Fave = 119 N.(c) For a height of 10 m, Fave = 312.7 N.

Note that these calculations are for free fall anddo not include air resistance on the particle.

9.66 A 50-mm-deep hole, 25 mm in diameter, isbeing produced by electrochemical machining.Assuming that high production rate is more im-portant than the quality of the machined sur-face, estimate the maximum current and thetime required to perform this operation.

The maximum current density for electrochem-ical machining is 8 A/mm2 (see Table 9.4 onp. 554). The area of the hole is

A =πD2

4=π(25)2

4= 491 mm2

The current is the product of the current den-sity and the cathode area, which is assumed tobe the same as the cross-sectional area of thehole. Thus,

i =(8 A/mm2

) (491 mm2

)= 3927 A

Note also that the maximum material removalrate in Table 9.4 (given in terms of penetrationrate) is 12 mm/min. Since the depth of the holeis 50 mm, the time required is

t =50 mm

12 mm/min= 4.17 min

9.67 If the operation in Problem 9.66 were performedon an electrical-discharge machine, what wouldbe the estimated machining time?

For electrical discharge machining, Table 9.4gives the material removal rate as typically 300mm3/min. The volume to be removed is

V = Ah = (491)(50) = 24, 550 mm3

hence the time required is

t =24, 550 mm3

300 mm3/min= 81.83 min

132






The required time could, however, be much less.If, for example, a through hole is being ma-chined, then the entire hole volume does nothave to be machined, only a volume associatedwith the hole periphery, the depth of the hole,and the kerf (known as trepanning). For largeblind holes or for deep cavities, a more commonapproach is to rough machine by end milling(see Fig. 8.1d), then follow EDM.

9.68 A cutting-off operation is being performed witha laser beam. The workpiece being cut is 1

4in. thick and 4 in. long. If the kerf is 1

6 in. wide,estimate the time required to perform this op-eration.

From Table 9.4, a typical traverse rate is 0.5-7m/min. For a 4 in. (0.10 m) length, the rangeof machining time is 12-0.85 s. The 1

4 -in. work-piece thickness is a moderate workpiece thick-ness, so an average traverse rate is a reasonableapproximation. Therefore, an estimate for ma-chining time is around 6 s. It should be rec-ognized, however, that the time required willdepend greatly on the power available in themachinery.

9.69 Referring to Table 3.3, identify two metals ormetal alloys that, when used as workpiece andelectrode, respectively, in EDM would give the(1) lowest and (2) highest wear ratios, R. Cal-culate these quantities.

(1) For lowest wear ratio (workpiece to elec-trode): tungsten/lead alloys (R = 0.00266), al-though the use of lead would be unrealistic forsuch an application. (2) For highest wear ratio:lead alloys/tungsten (R = 1902). An exam-ple of a more realistic value of the highest wearratio is for tungsten electrode/tantalum work-piece (R = 3.03).

9.70 It was stated in Section 9.5.2 that, in practice,grinding ratios typically range from 2 to 200.Based on the information given in Section 9.13,estimate the range of wear ratios in electrical-discharge machining and then compare themwith grinding ratios.

For grinding ratios we refer to Section 9.5.2,where we note that this ratio ranges between 2and 200, and even higher. Thus, the values arevery comparable.

9.71 It is known that heat checking occurs whengrinding under the following conditions: Spin-dle speed of 4000 rpm, wheel diameter of 10in., and depth of cut of 0.0015 in., and a feedrate of 50 ft/min. For this reason, the spindlespeed is to be kept at 3500 rpm. If a new, 8-in-diameter wheel is now used, what should bethe spindle speed before heat checking occurs?What spindle speed should be used to main-tain the same grinding temperatures as thoseencountered with the existing operating condi-tions?

Heat checking is associated with high surfacetemperatures, so the temperature rise given byEq. (9.9) on p. 535 will be used to solve thisproblem. For the known case where heat check-ing occurs, the velocity is calculated to be

V = rω = (5)(4000)(

112

)= 1667 ft/min

Eq. (9.9) gives

∆T ∝ D1/4d3/4

(V

v

)1/2

or

∆T = AD1/4d3/4

(V

v

)1/2

= A(10)1/4(0.0015)3/4

(166750

)1/2

= 0.0783A

where A is a constant. The known safe operat-ing condition has a speed of 1460 ft/min. Thisleads to a temperature rise of

∆T = AD1/4d3/4

(V

v

)1/2

= A(10)1/4(0.0015)3/4

(146050

)1/2

= 0.0732A

If an 8-in. wheel is used, the speed at whichheat checking occurs is:

0.0783A = AD1/4d3/4

(V

v

)= A(8)1/4(0.0015)3/4

(V

50

)0.5

133






or V = 1865 ft/min. This indicates a spindlespeed of 5600 rpm. For the known safe con-dition, we perform the same calculation using∆T = 0.0732A to obtain:

0.0732A = AD1/4d3/4

(V

v

)= A(8)1/4(0.0015)3/4

(V

50

)0.5

or V = 1630 ft/min. This corresponds to aspindle speed of 4900 rpm.

9.72 A hard aerospace aluminum alloy is to beground. A depth of 0.003 in. is to be removedfrom a cylindrical section 8 in. long and with a3-in. diameter. If each part is to be ground innot more than one minute, what is the approxi-mate power requirement for the grinder? Whatif the material is changed to a hard titaniumalloy?

The volume to be removed is

Volume = πDavgdl = π(3− 0.003)(0.003)(8)

or 0.226 in3. Therefore, the minimum metal re-moval rate is 0.226 in3/min. Taking the specificenergy requirement as 10 hp-min/in3 (see Table9.3), the power requirement is

P = (10 hp-min/in3)(0.226 in3/min) = 2.26 hp

For the hard titanium, let u = 20 hp-min/in3;thus, the new power would be 4.52 hp.

9.73 A grinding operation is taking place with a 10-in. grinding wheel at a spindle rotational speedof 4000 rpm. The workpiece feed rate is 50ft/min, and the depth of cut is 0.002 in. Con-tact thermometers record an approximate max-imum temperature of 1800F. If the workpieceis steel, what is the temperature if the spindlespeed is increased to 5000 rpm? What if it isincreased to 10,000 rpm?

At a rotational speed of 4000 rpm, the surfacespeed is

V = rω = (5)(4000)(

112

)= 1667 ft/min

The temperature rise for d = 0.002 in., v = 50ft/min, and D = 10 in. is 1800F; therefore,

Eq. (9.9) gives

∆T ∝ D1/4d3/4

(V

v

)1/2

or

∆T = AD1/4d3/4

(V

v

)1/2

so that

1800F = A(10)1/4(0.002)3/4

(166750

)1/2

hence A = 18, 500. If the spindle speed is now5000 rpm, or a surface speed of 2080 ft/min,the temperature rise will be

∆T = AD1/4d3/4

(V

v

)1/2

= (18, 500)(10)1/4(0.002)3/4

(208050

)1/2

= 2010F

At a spindle speed of 10,000 rpm, the surfacespeed is 4167 ft/min, and the temperature riseis

∆T = AD1/4d3/4

(V

v

)1/2

= (18, 500)(10)1/4(0.002)3/4

(416750

)1/2

= 2840F

Note that this temperature is above the melt-ing point of steel (see Table 3.3 on p. 106).Clearly, the temperature cannot increase abovethe melting point of the workpiece material.This indicates that the 10,000 rpm speed, com-bined with the other process parameters, wouldnot be a realistic process parameter.

9.74 The regulating wheel of a centerless grinder isrotating at a surface speed of 25 ft/min and isinclined at an angle of 5. Calculate the feedrate of material past the grinding wheel.

The feed rate is merely the component of thevelocity in the feed direction, given by

f = V sinα = (25 ft/min)(sin 5)

or f = 2.18 ft/min or 0.44 in./s.

134






9.75 Using some typical values, explain whatchanges, if any, take place in the magnitudeof the impact force of a particle in ultrasonicmachining of a hardened-steel workpiece as itstemperature is increased?

Inspecting Eqs. (9.11) and (9.13) we can seethat the force of a particle on a surface is givenby

Fave =2mv[

5rco

(cov

)1/5] =

2mv6/5c4/5o

5r

=2mv6/5E2/5

5rρ2/5

From Fig. 2.9, the stiffness of carbon steel overa temperature increase of 700C changes from27 to 20 × 106 psi. For the same temperaturerange, there is a thermal strain of

ε = 1 + α∆T = 1 +(11.7× 10−6

)(700)

or ε = 1.00819. Comparing the two states, oneat room temperature and the other at elevatedtemperature, and noting that the density is af-fected by thermal strain, we can write

Fave,1

Fave,2=

(2mv6/5E0.4

1

5rρ0.41

)(

2mv6/5E0.42

5rρ0.42

)=

(E1

E2

)0.4(ρ2

ρ1

)0.4

=(E1

E2

)0.4(ε31ε32

)0.4

Since E1/E2 = 20/27 and ε1/ε2 = (0.00819)3,we calculate the right-hand side of this equationas 0.89. Therefore, it can be concluded that theforce decreases with an increase in temperature.It should be noted, however, that the change issmall. For example, a 700C temperature riseis required for a force reduction of around 10%.

9.76 Estimate the percent increase in the cost of thegrinding operation if the specification for thesurface finish of a part is changed from 63 to 16µin.

Referring to Fig. 9.41, note that changing thesurface finish from 63 µin. to 16 µin. would

involve an increased cost of about 400%. Thisis a very significant increase in cost, and is agood example of the importance of the state-ment made throughout the book that, in or-der to minimize manufacturing costs (see alsoFig. 16.6), dimensional accuracy and surfacefinish should be specified as broadly as is per-missible.

9.77 Assume that the energy cost for grinding analuminum part is $0.90 per piece. Letting thespecific energy requirement for this material be8 Ws/mm3, what would be the energy cost ifthe workpiece material is changed to T15 toolsteel?

From Table 9.3 on p. 534, note that the powerrequirement for T15 tool steel ranges from 17.7to 82 W-s/mm3. Consequently, the costs wouldrange from 2.5 to 11.7 times that for the alu-minum. This means an energy cost between $2and $9.36 per part.

9.78 Derive an expression for the angular velocity ofthe wafer as a function of the radius and angu-lar velocity of the pad in chemical mechanicalpolishing.

x

y

ωt

ωw

r*

rw

r

+

Wafer

Table

By the student. Refer to the figure above andconsider the case where a wafer is placed on thex-axis, as shown. Along this axis there is no ve-locity in the x-direction. The y-component ofthe velocity has two sources: the rotation of

135






the table and the rotation of the carrier. Con-sidering the table movement only, the velocitydistribution can be expressed as

Vy = rωt

and for the carrier

Vy = r∗ωw

where r∗ can be positive or negative, and isshown positive in the figure. Note that r =rw + r∗, so that we can substitute this equationinto Vy and combine the velocities to obtain thetotal velocity as

Vy,tot = (rw + r∗)ωt + r∗ωw

If ωw = −ωt, then Vy,tot = rwωt. Since the lo-cation of the wafer and the angular velocity ofthe carrier are fixed, the y-component of veloc-ity is constant across the wafer.

9.79 A 25-mm-thick copper plate is being machinedby wire EDM. The wire moves at a speed of 1.5m/min and the kerf width is 1.5 mm. Calculatethe power required. (Assume that it takes 1550J to melt one gram of copper.)

Note from Table 3.3 on p. 106 that the den-sity of copper is ρ = 8970 kg/m3. The metalremoval rate is given by Eq. (9.22) on p. 565 as

MRR = Vfhb = (1500)(25)(1.5)

or MRR=56,250 mm3/min = 56.25 × 10−6

m3/min. Therefore, we can calculate the rateof mass removal as:

Mass MRR = ρ(MRR) = (8970)(56.25× 10−6)

or 505 g/min. Therefore, the required power iscalculated as

P = (505)(1550)(

160

)= 13, 046 Nm/s

or P = 13 kW.

9.80 An 8-in. diameter grinding wheel, 1 in. wide,is used in a surface grinding operation per-formed on a flat piece of heat-treated 4340steel. The wheel is rotating with a surface speedV = 5, 000 fpm, depth of cut d = 0.002 in./pass,and cross feed w = 0.15 in. The reciprocating

speed of the work is v = 20 ft/min. and the op-eration is performed dry. (a) What is the lengthof contact between the wheel and the work? (b)What is the volume rate of metal removed? (c)Letting C = 300, estimate the number of chipsproduced per unit time. (d) What is the av-erage volume per chip? (e) If the tangentialcutting force on the workpiece is Fc = 10 lb,what is the specific energy for the operation?

(a) The length of contact between the wheeland the workpiece is given by Eq. (9.1) as

l =√Dd =

√(8)(0.002) = 0.1265 in.

(b) The metal removal rate is given by (SeeExample 9.2 on p. 533):

MRR = dwv = (0.002)(20)(12)(0.15)

or MRR = 0.072 in3/min.(c) The rate of chip production is given by

n = V wC = (5000)(12)(0.15)(300)

or n = 2.7× 106 chips/min.(d) The average volume is:

Vol =MRR

Chips/min=

0.0722.7× 106

or Vol= 2.67× 10−8 in3/chip.(e) Note that the power required is P = FcV .

The specific energy is the ratio of power tomaterial removal rate, or

u =FcV

MRR=

(10)(5000)(12)0.072

or u = 8.33 × 106 in-lb/in3. Since 1hp=396,000 in-lb/min,

u =8.33× 106

396, 000= 21 hp-min/in3

As can be seen from Table 9.3, this is areasonable specific energy for grinding ahard (i.e., heat-treated) steel.

9.81 A 150-mm-diameter tool steel (u = 60 W-s/mm3) work roll for a metal rolling operationis being ground using a 250-mm-diameter, 75-mm-wide, Type 1 grinding wheel. The workroll rotates at 10 rpm. Estimate the chip di-mensions and grinding force if d = 0.04 mm,

136






N = 3000 rpm, r = 12, C = 5 grains per mm2,and the wheel rotates at N = 3000 rpm.

This solution is similar to that given in Exam-ple 9.1 on p. 532, except the undeformed chiplength, l, is given by Eq. (9.2) since the work-piece is cylindrical. Therefore, the chip lengthis:

l =

√Dd

1 + (D/Dw)=

√(0.25)(0.00004)

1 + (0.25)/(0.15)

which is solved as l = 0.00194 m = 1.94 mm.Note that the velocities are

v = 2πrN = 2π(0.075)(10) = 4.7 m/min

or v = 0.0785 m/s, and

V = rω = 2πrN = 2π(0.075)(3000)

or V = 1413.7 m/min = 23.6 m/s. Therefore,the undeformed chip thickness is given by (seeSection 9.4):

V Crt2l

4= vd or t =

√4vdV Crl

Substituting into this expression, t is found tobe

t =

√4(0.0785)(0.00004)(23.6)(5)(0.00194)

= 0.0074 mm

The material removal rate is

MRR = dwv

= (0.00004)(0.075)(4.7)= 1.41× 10−5 m3/min = 235 mm3/s

Since u is given as 60 W-s/mm3, the power con-sumed will be:

P = u(MRR) = (60)(235) = 14.1 kW

Also, we know

P = FcV or Fc =P

V

so thatFc =

14, 10023.6

= 597 N

9.82 Estimate the contact time and average forcefor the following particles striking a steel work-piece at 1 m/s. Use Eqs. (9.11) and (9.13) andcomment on your findings. (a) 5-mm-diametersteel shot; (b) 0.1-mm-diameter cubic boronnitride particles; (c) 3-mm-diameter tungstensphere; (d) 75-mm-diameter rubber ball; (e) 3-mm-diameter glass beads. (Hint: See Tables2.1, 3.3 and 8.6.)

The time of contact depends on the elasticwave velocity in the workpiece; for steel, whereE = 195 GPa (from Table 2.1) and ρ = 8025kg/m3 (from Table 3.3), the wave velocity iscalculated as:

co =

√E

ρ=

√195× 109

8025= 4930 m/s

Therefore, for the steel shot with a diameter of5 mm (r = 0.0025 m), Eq. (9.11) gives:

to =5rco

(cov

)1/5

=5(0.0025)(4930)

(4930

1

)1/5

or to = 13.9 µs. Therefore, the average forceexerted is

Fave =2πr3ρvto

=2π(0.0025)3(8025)(1)

13.9× 10−6

or Fave = 56.7 N. Using the same calculations,the following table can be constructed:

r ρ to Fave

Material (mm)

„kg

m3

«(µs) (N)

Steel 2.5 80251 13.9 56.7cBN 0.1 35002 2.77 0.00794Tungsten 3 19,2901 8.32 393.3Rubber 75 9001 208 11,470Glass 3 25501 8.32 52

Notes:1. From Table 2.1.2. From Table 8.6.

Note that these results are reasonable for thecBN particles in part (b), as the values are wellwithin the parameters suggested in Section 9.9.However, as particle size increases, the applica-bility of the equations is compromised and un-reasonable results are obtained. Consider theextreme case of a rubber ball, similar to a toy

137






ball that is gently tossed. These equations pre-dict a force of over 1400 N, an answer that isclearly unrealistic. Equations (9.11) and (9.13)are based upon stress waves; for compliant andlarge objects, the stress waves interact and thecontact is pseudostatic, so that these equationsno longer apply.

9.83 Assume that you are an instructor covering thetopics in this chapter, and you are giving a quiz

on the quantitative aspects to test the under-standing of the students. Prepare three quan-titative problems, and supply the answers.


Design

9.84 Would you consider designing a machine toolthat combines, in one machine, two or more ofthe processes described in this chapter? Ex-plain. For what types of parts could such amachine be useful? Make preliminary sketchesfor such machines.

By the student. This is a valuable though dif-ficult exercise. Note that, in some respects,processes such as chemical mechanical polish-ing and electrochemical machining satisfy thecriteria stated in this problem.

9.85 With appropriate sketches, describe the prin-ciples of various fixturing methods and devicesthat can be used for each of the processes de-scribed in this chapter.

By the student. This is an open-ended problemthat would also be suitable for a project. Thestudents are encouraged to conduct literaturesearch on the topic, as well as recall the typeof fixtures used and described throughout thechapters. See especially Section 14.9.

9.86 As also described in Section 4.3, surface finishcan be an important consideration in the de-sign of products. Describe as many parametersas you can that could affect the final surfacefinish in grinding, including the role of processparameters as well as the setup and the equip-ment used.

By the student. Note that among major param-eters are the grit size and shape of the abrasive,

dressing techniques, and the processing param-eters such as feed, speed, and depth of cut.

9.87 Size effect in grinding was described in Section9.4.1. Design a setup and suggest a series ofexperiments whereby size effect can be investi-gated.

By the student. Refer to various sources listedin the bibliography. The experiments can bemacroscaled by measuring power consumptionas a function of chip thickness (see Eq. (9.5) forthe important parameters affecting chip thick-ness). The experiments could also utilize aneffective system on a microscale, such as in-denters mounted on piezo-electric load cells anddragged across a surface.

9.88 Describe how the design and geometry of theworkpiece affects the selection of an appropri-ate shape and type of a grinding wheel.

By the student. The workpiece shape and sizehave a direct role on grinding wheel selection(see, for example, pp. 527, 539, and 543). Thepart geometry places restrictions on the grind-ing surfaces, such as with gear teeth where thewheel edge radius must be less than the geartooth notch radius in order to properly grindthe teeth. (See also Sections 8.10.7 and 9.6.)

9.89 Prepare a comprehensive table of the capabili-ties of abrasive machining processes, includingthe shapes of parts ground, types of machinesinvolved, typical maximum and minimum work-piece dimensions, and production rates.

138






By the student. This is a challenging assign-ment. The following should be considered as anexample of the kinds of information that can becontained in such a table.

Pro

cess

Abra

sives

Part

shapes

Maxim

um

size

Typic

al

use

dsu

rface

finis

h(µ

m)

Gri

ndin

gA

l 2O

3Fla

t,ro

und

or

Fla

t:no

lim

it.

0.2

SiC

,cB

Nci

rcula

rR

ound:

12

in.

Dia

mond

Cir

cula

r:12

in.

Barr

elfinis

hin

gA

l 2O

3,SiC

Lim

ited

asp

ect

6in

.0.2

rati

oC

hem

ical-

Al 2

O3,SiC

Fla

tsu

rface

s13

in.

0.0

5m

echanic

al

and

low

erpolish

ing

Shot

bla

stin

gSand,SiO

2A

llty

pes

No

lim

it1-1

0

9.90 How would you produce a thin circular diskwith a thickness that decreases linearly fromthe center outward?

By the student. If the part is sufficiently thick,one method is to machine it on a CNC millingmachine, but the stiffness of the workpiece isan important factor due to cutting forces thatwould deflect thin parts. A similar methodwould be grinding the part, using numericalcontrol equipment. A simpler method is to takea round disc with a constant thickness, insert itfully into a chemical-machining solution (Sec-tion 9.10), and withdraw it slowly while it isbeing rotated; such a part has been made suc-cessfully by this method. Note that there willbe a major difference in production rates.

9.91 Marking surfaces of manufactured parts withletters and numbers can be done not only with

labels and stickers but also by various mechan-ical and nonmechanical means (see also Section9.14.1). Make a list of some of these methods,explaining their advantages and limitations.

By the student. Processes include laser ma-chining, where the laser path is computer con-trolled, chemical etching, where a droplet ofsolution is placed in similar fashion to ink-jetprinters, and machining on a CNC milling ma-chine.

9.92 On the basis of the information given in Chap-ters 8 and 9, comment on the feasibility ofproducing a 10-mm diameter, 100-mm deepthrough hole in a copper alloy by (a) conven-tional drilling and (b) other methods.

By the student. Note that this is a generalquestion and it can be interpreted as either athrough hole or a blind hole (in which case itdoes not specify the shape of the bottom of thehole). Furthermore, the quality of the hole, itsdimensional accuracy, and the surface finish ofthe cylindrical surface are not specified. It isintended that the students be observant and re-sourceful to ask such questions so as to supplyappropriate answers.

Briefly, a through hole with the dimensionsspecified can easily be drilled; if the dimen-sional accuracy and surface finish are not ac-ceptable, the hole can subsequently be reamedand honed. Holes can be internally ground, de-pending on workpiece shape and accuracies re-quired; note, however, that there has to be ahole first in order to be ground internally.

For blind holes, the answers will depend on therequired shape of the hole bottom. Drills typi-cally will not produce flat bottom, and will re-quire an operation such as end milling. Internalgrinding is possible on an existing hole, notingalso the importance of internal corner features(relief) as stated in the design considerationsfor grinding in Section 9.16.

9.93 Conduct a literature search and explain howobserving the color, brightness, and shape ofsparks produced in grinding can be a usefulguide to identifying the type of material beingground and its condition.

By the student. Various charts, showing pho-

139






tographs or sketches of the type and color ofsparks produced, have been available for yearsas a useful but general guide for material iden-tification at the shop level, especially for steels.Some of these charts can be found in textbooks,such as in Fig. 24.15 on p. 458 of MachiningFundamentals, by J.R. Walker.

9.94 Visit a large hardware store and inspect the var-ious grinding wheels on display. Make a noteof the markings on the wheels and, based onthe marking system shown in Figs. 9.4 and 9.5,comment on your observations, including themost commonly found types and sizes of wheelsin the store.

By the student. This is a good opportunity toencourage students to gain some exposure togrinding wheels. The authors have observed anincreased reluctance on students to gain prac-tical experience and exposure to machinery byactually visiting vendors, and have found thisexercise to be very valuable. The markingson grinding wheels will have the type of in-formation given in Figs. 9.4 or 9.5. It willalso be noted that the most common grindingwheels are basically the same as those shown inFig. 9.2. Those in Fig. 9.3 are less common andalso more expensive. Students may also com-ment on sizes; many grinding wheel shapes areavailable for hobbyists but not on a larger scale.

9.95 Obtain a small grinding wheel and observe itssurfaces using a magnifier or a microscope, andcompare with Fig. 9.6. Rub the periphery ofthe wheel while pressing it hard against a va-riety of flat metallic and nonmetallic materi-als. Describe your observations regarding (a)the type of chips produced, (b) the surfacesdeveloped, and (c) the changes, if any, to thegrinding wheel surface.

By the student. This is a good project and canbecome a component of a laboratory course.

9.96 In reviewing the abrasive machining processesin this chapter it will noted that some processesuse bonded abrasives while others involve looseabrasives. Make two separate lists for these twotypes and comment on your observations.

By the student. This is an open-ended problemand the following table should be regarded as

only an illustration of an answer. The studentsshould give further details, based on a study ofeach of the processes covered in the chapter.

Process CommentsBonded abrasivesGrindingBelt grindingSandingHoningSuperfinishing

These processes are basicallysimilar to each other andhave a wide range abrasivesizes, the material removalrates, surface finish, and lay(see Fig. 33.2 on p. 1039).

Loose abrasivesUltrasonicmachining

A random surface lay is mostcommon for these processes

Chemical-mechanicalpolishing

Barrel finishingAbrasive-flow

machining

9.97 Based on the topics covered in Chapters 6through 9, make a comprehensive table of hole-making processes. (a) Describe the advantagesand limitations of each method, (b) commenton the quality and surface integrity of the holesproduced, and (c) give examples of specific ap-plications.

By the student. This is a challenging topic forstudents. The statement of the problem impliesthat holes are to generated on a sheet or a blockof solid material, and that it does not includefinishing processes for existing holes. It shouldbe recalled that holemaking processes include(a) piercing, (b) punching, (c) drilling and bor-ing, (d) chemical machining, (e) electrochemi-cal machining, (f) electrical-discharge machin-ing, (g) laser-beam and electron-beam machin-ing, and (h) water-jet and abrasive water-jetmachining. The students should prepare a com-prehensive answer, based on the study of theseprocesses in various chapters.

9.98 Precision engineering is a term used to describemanufacturing high-quality parts with close di-mensional tolerances and good surface finish.Based on their process capabilities, make a listof advanced machining processes (in decreasingorder of quality of parts produced). Include abrief commentary on each method.

By the student. This is a challenging task. Stu-dents should carefully review the contents ofFigs. 4.20, 8.26, 9.27, 16.4, and 16.5, as well

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as the relevant sections in the text. Note thatthe order in such a listing will depend on thesize of the parts to be produced, the quantityrequired, the workpiece materials, and the de-sired dimensional tolerances and surface finish.

9.99 It can be seen that several of the processes de-scribed in this chapter can be employed, ei-ther singly or in combination, to produce orfinish tools and dies for metalworking opera-tions. Prepare a brief technical paper on thesemethods, describing their advantages and limi-tations, and giving typical applications.

By the student. See also Section 6.7. Thisis a valuable exercise for students, and the re-sponses should include the latest technical inno-vations, including rapid prototyping and rapidtooling (described in Chapter 10). Tradition-ally, processes such as casting, die-sinking (suchas with an end mill), and plunge EDM was mostcommonly used for these applications, althoughpolishing and electrochemical grinding may alsobe used for near-net-shape parts to improvetheir surface finish. Laser-beam and electrical-discharge machining is sometimes performed toroughen tool and die surfaces for improved ma-terial formability (by virtue of its effects on tri-bological behavior at workpiece-die interfaces).

9.100 List the processes described in this chapter thatwould be difficult to apply to a variety of non-metallic or rubberlike materials. Explain yourthoughts, commenting on such topics as partgeometries and the influence of various physicaland mechanical properties of workpiece materi-als.

By the student. Some materials will be diffi-cult for some of the processes. For example, achemically inert material will obviously be dif-ficult to machine chemically. Grinding may bedifficult if the workpiece is nonmetallic, and thecompliance of rubber materials may limit thegrinding force (and thus material removal rate)that can be achieved. Furthermore, a rubber-like material may quickly load a grinding wheel,

requiring frequent redressing. Recall also thatan electrically-insulating material is impossiblefor EDM; a tough material can be difficult tocut with a water jet; and a shiny or transparentmaterial is difficult for laser machining. Notethat it is rare that a workpiece material has allof these properties simultaneously.

9.101 Make a list of the processes described in thischapter in which the following properties arerelevant or significant: (a) mechanical, (b)chemical, (c) thermal, and (d) electrical. Arethere processes in which two or more of theseproperties are important? Explain.

By the student. Because the term relevantmay be interpreted as subjective, the stu-dents should be encouraged to be responsive asbroadly as possible. Also, the question can beinterpreted as properties that are important inthe workpiece or the phenomenon that is thebasic principle of the advanced machining pro-cess. Some suggestions are:

Mechanical: Electrochemical grind-ing, water-jet machining,abrasive-jet machining.

Chemical: Chemical machining, electro-chemical machining, electro-chemical grinding.

Thermal: Chemical machining, elec-trochemical machining,electrochemical grinding,plunge EDM, wire EDM,laser-beam machining,electron-beam machining.

Electrical: Electrochemical machining,electrochemical grinding,plunge EDM, wire EDM,electron-beam machining.

Clearly, there are processes (such as chemicalmachining) where two properties are impor-tant: the chemical reactivity of workpiece andreagents, and the corrosion processes (the prin-ciple of chemical machining) which are temper-ature dependent.

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Chapter 10

Properties and Processing ofPolymers and Reinforced Plastics;Rapid Prototyping and Rapid Tooling

Questions

10.1 Summarize the most important mechanical andphysical properties of plastics in engineering ap-plications.

The most important mechanical and physicalproperties of plastics are described in Sections10.3 through 10.8. Students may create sum-maries of mechanical properties, make compar-isons with other material classes, or investigatenovel graphical methods of summarizing theproperties.

10.2 What are the major differences between theproperties of plastics and of metals?

There are several major differences that can beenumerated, as described throughout the chap-ter. Some examples are:

(a) Plastics are much less stiff than metals.

(b) They have lower strength than metals andare lighter.

(c) The thermal and electrical conductivitiesof metals are much higher than those forplastics.

(d) There are much wider color choices forplastics than for metals.

10.3 What properties are influenced by the degree ofpolymerization?

By the student. As described in Section 10.2.1,the degree of polymerization directly influencesviscosity. In addition, as can be understood byreviewing Figs. 10.3 and 10.5, a higher degree ofpolymerization will lead to higher strength andstrain hardening in thermoplastics, and will ac-centuate the rubber-like behavior of networkedstructures.

10.4 Give applications for which flammability ofplastics would be a major concern.

By the student. There are several applica-tions where flammability of plastics is a majorconcern. These include aircraft, home insula-tion (thermal and electrical), cookware, cloth-ing, and components for ovens and stoves (in-cluding components such as handles and dials).Students should be encouraged to describe ad-ditional applications.

10.5 What properties do elastomers have that ther-moplastics, in general, do not have?

By the student. By virtue of their chemicalstructures, elastomers have the capability ofreturning to their original shape after being

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stretched, while thermoplastics cannot. Elas-tomers can do so because they have a low elasticmodulus and can undergo large elastic deforma-tion without rupture.

10.6 Is it possible for a material to have a hystere-sis behavior that is the opposite of that shownin Fig. 10.14, whereby the arrows are counter-clockwise? Explain.

If the arrows were counterclockwise, the mate-rial would have a hysteresis gain. This wouldmean that the energy put into the material islower than the energy recovered during unload-ing, which, of course, is impossible.

10.7 Observe the behavior of the tension-test spec-imen shown in Fig. 10.13, and state whetherthe material has a high or low m value. (SeeSection 2.2.7.) Explain why.

Recall that the m value indicates the strainrate sensitivity of a material. The material inFig. 10.13 on p. 598 elongates extensively byorientation of the polymer molecules, thus itwould be expected to have high strain-rate sen-sitivity. This is related to diffuse necking, as op-posed to localized necking observed with metalsin tension tests (see Fig. 7.1d).

10.8 Why would we want to synthesize a polymerwith a high degree of crystallinity?

This is an open-ended question that can be an-swered in several ways. Students may rely uponparticular applications or changes in materialproperties. One can refer to Section 10.2.1 andFig. 10.4, and note that a high degree of crys-tallinity leads to increased stiffness, especiallyat higher temperatures.

10.9 Add more to the applications column in Table10.3.

By the student. Some additional examples are:

(a) Mechanical strength: rope, hangers.

(b) Functional and decorative: electrical out-lets, light switches.

(c) Housings, etc.: pens, electrical plugs.

(d) Functional and transparent: food and bev-erage containers, packaging, cassette hold-ers.

(e) Wear resistance: rope, seats.

10.10 Discuss the significance of the glass-transitiontemperature, Tg, in engineering applications.

By the student. The glass-transition tempera-ture is the temperature where a thermoplasticbehaves in a manner that is hard, brittle andglassy below this temperature, and rubbery orleathery above it (see Section 10.2.1). Sincethermoplastics begin to lose their load-carryingcapacity above this temperature, there is an up-per useful temperature range for the plastic. Inengineering applications where thermoplasticswould be expected to carry a load, the materialfor the part would have to have a glass tran-sition temperature higher than the maximumtemperature to which it would be subjected inservice.

10.11 Why does cross-linking improve the strength ofpolymers?

Cross-linked polymers have additional bondslinking adjacent chains together (see Fig. 10.3on p. 589). The strength is increased with ther-moplastics because these cross links give ad-ditional resistance to material flow since theymust be broken before the molecules can slidepast one another. With thermosets, they rep-resent additional bonds that must be brokenbefore fracture can occur.

10.12 Describe the methods by which optical proper-ties of polymers can be altered.

Optical properties can be altered by additiveswhich can alter the color or translucence ofthe plastic. Additives can either be dyes orpigments. Recall also that stress whiteningmakes the plastic appear lighter in color or moreopaque. As stated in Section 10.2.1, opticalproperties are also affected by the degree ofcrystallinity of the polymer.

10.13 Explain the reasons that elastomers were devel-oped. Based on the contents of this chapter, arethere any substitutes for elastomers? Explain.

By the student. Elastomers (Section 10.8) weredeveloped to provide a material that could un-dergo a large amount of deformation withoutfailure. They provide high friction and nonskid

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surfaces, abrasive-wear resistance, shock and vi-bration isolation, and protection against corro-sion. They are also used in rubber-pad formingoperations.

10.14 Give several examples of plastic products orcomponents for which creep and stress relax-ation are important considerations.

By the student. Recall that creep in polymersis particularly important in high-temperature,low-stress applications. In low temperature,high-stress circumstances stress relaxation isimportant. As an example of the importanceof creep, consider polymers as pot handles incookware. As an example of stress relaxation,seat cushions will deform to provide a uniformstress distribution and thus provide better com-fort for the occupant.

10.15 Describe your opinions regarding recycling ofplastics versus developing plastics that arebiodegradable.

By the student. Some arguments may be madeare that recycling actually has a cost associ-ated with it, such as costs involved in collect-ing the materials to be recycled and the en-ergy required in recycling methods. Note alsothat the properties of the recycled polymer willlikely be inferior as compared to the virgin poly-mer. Biodegradable plastics have drawbacks aswell; it is difficult to design them to degradein the intended time frame, and they may havemore failures in service. They can be signifi-cantly more expensive than polymers that arenot biodegradable.

10.16 Explain how you would go about determiningthe hardness of the plastics described in thischapter.

Many of the hardness tests described in Sec-tion 2.6 (see also Fig. 2.22 on p. 52) are notsuitable for polymers, for reasons such as in-elastic recovery of the surface indentation andtime-dependent stress relaxation. Recall thatdurometer testing is an appropriate approachfor such materials.

10.17 Distinguish between composites and alloys.Give several examples.

By the student. Consider the following:

• Alloys are mixtures of metals, whereascomposites are not necessarily metals.Metal-matrix composites require a rein-forcement in the form of fibers or whiskers.

• With composites, the reinforcement ismuch stronger than the matrix. Even intwo-phased alloys, where a matrix couldbe defined, such a difference in strength orstiffness is not necessarily significant.

10.18 Describe the functions of the matrix and thereinforcing fibers in reinforced plastics. Whatfundamental differences are there in the char-acteristics of the two materials?

By the student. As described in Section 10.9,consider, for example, the fact that reinforcingfibers are generally stronger and/or stiffer thanthe polymer matrix. The function of the rein-forcement is therefore to increase the mechan-ical properties of the composite. On the otherhand, the fibers are less ductile and generallyhave limited resistance to chemicals or mois-ture (graphite, for example, decomposes whenexposed to oxygen). The matrix is very resis-tant to chemical attack and thus protects thefibers.

10.19 What products have you personally seen thatare made of reinforced plastics? How can youtell that they are reinforced?

By the student. This is an open-ended ques-tion and students can develop a wide variety ofanswers. Some suggestions are tennis rackets,baseball bats, chairs, and boat hulls. Some-times, it is readily apparent that the part hasbeen produced through lay-up; other times thefiber reinforcements can be seen directly on thesurface of the part.

10.20 Referring to earlier chapters, identify metalsand alloys that have strengths comparable tothose of reinforced plastics.

By the student. See Table 16.1 on p. 956. Atypical comparison is given below:

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ReinforcedMetal plastics(MPa) (MPa)

Magnesium (165-195) Nylon (70-210)Polyester (110-160)

Al alloys (90-600) ABS (100)Acetal (135)

Nylon (70-210)Polycarbonate (110)Polyester (110-160)

Polypropylene (40-100)Cu alloys (140-1310) Nylon (70-210)

Polyester (40-100)Iron (185-285) Nylon (70-210)

10.21 Compare the relative advantages and limita-tions of metal-matrix composites, reinforcedplastics, and ceramic-matrix composites.

By the student. This is a challenging questionand the students are encouraged to develop acomprehensive table based on their understand-ing of the contents of this chapter.

10.22 This chapter has described the many advan-tages of composite materials. What limita-tions or disadvantages do these materials have?What suggestions would you make to overcomethese limitations?

By the student. Consider, for example, two dis-advantages as anisotropic properties and pos-sible environmental attack of the fibers (espe-cially water adsorption). Anisotropy, thoughnot always undesirable, can be reduced by hav-ing a random orientation of reinforcing ma-terials. Environmental attack of the fiberswould cause loss of fiber strength and possiblydebonding from the matrix.

10.23 A hybrid composite is defined as a material con-taining two or more different types of reinforc-ing fibers. What advantages would such a com-posite have over other composites?

By the student. The hybrid composite can havetailored properties. Thus, a certain strengthlevel could be obtained at a lower cost by usinga combination of fibers, rather than just onefiber. The anisotropic properties could also becontrolled in different ways, such as having, forexample, Kevlar fibers oriented along the ma-jor stress direction and other fibers dispersed

randomly in the composite. The student is en-couraged to elaborate further in greater detail.

10.24 Why are fibers capable of supporting a majorportion of the load in composite materials? Ex-plain.

By the student. Refer to Example 10.4 onp. 617. The reason that the fibers can carrysuch a large portion of the load is that they arestiffer than the matrix. Although both the ma-trix and the fibers undergo the same strain, thefibers will this support a larger portion of theload.

10.25 Assume that you are manufacturing a productin which all the gears are made of metal. Asalesperson visits you and asks you to considerreplacing some of the metal gears with plas-tic ones. Make a list of the questions that youwould raise before making such a decision. Ex-plain.

By the student. Consider, for example, the fol-lowing questions:

(a) Will the plastic gear retain its requiredstrength, stiffness, and tolerances if thetemperature changes during its use?

(b) How acceptable is the wear resistance andfatigue life of the plastic gears?

(c) Is it compatible with meshing metal gearsand other components in the gear train?

(d) Are there any backlash problems?(e) What are its frictional characteristics?(f) Is the lighter weight of the plastic gear sig-

nificant?(g) Is noise a problem?(h) Is the plastic gear affected adversely by lu-

bricants present?(i) Will the supplier be able to meet the quan-

tity demanded?(j) How much cost savings are involved? (See

also Section 16.9.)

10.26 Review the three curves in Fig. 10.8, and de-scribe some applications for each type of be-havior. Explain your choices.

By the student. See also Section 10.5. Severalexamples can be given; consider the followingsimple examples:

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• Rigid and brittle: Handles, because theyshould not flex significantly.

• Tough and ductile: Helmets, to dissipatethe energy from impact without fractur-ing.

• Soft and flexible: Beverage bottles, be-cause they can flex when dropped andregain their shape and not break, unlikeglass bottles.

10.27 Repeat Question 10.26 for the curves inFig. 10.10.

By the student. See also Section 10.5 and con-sider the following:

• Low-density polyethylene: nonbreakablefood containers with impact strength atlow temperatures, such as in freezers.

• High-impact polypropylene: productsthat that when dropped or in a collision,will not crack at a wide range of tempera-tures.

• Polyvinyl chloride (PVC): it can be eitherflexible or hard, and either type can beused for tubing; since it is not too strongor impact resistant, it must be limited tolow pressure tubing.

• Polymethylmethacrylate: has moderatestrength, good optical properties, and isweather resistant; these properties makethem useful for lighting fixtures that donot require high impact resistance.

10.28 Do you think that honeycomb structures couldbe used in passenger cars? If so, which compo-nents? Explain.

By the student. As an example, two sugges-tions concerning automobiles: (1) Radiatorswith copper honeycomb structure to improveheat conduction, and (2) passenger compart-ment walls consisting of honeycomb structureswith cavities filled with noise- and vibration-damping materials, making the compartmentmore sound proof.

10.29 Other than those described in this chapter,what materials can you think of that can beregarded as composite materials?

By the student. Some examples are:

• Wood: a composite consisting of blockcells and long fibrous cells.

• Particle board: a composite that is a com-bination of wood scraps and a binder.

• Winter coat: a layered type of compositeconsisting of an outer cloth material whichis weather resistant and an insulating in-ner material to prevent loss of body heat.

• Pencil: graphite rod core surrounded bywood covering.

• Walls: consists of a plaster matrix withwood stud or metal reinforcements.

The students are encouraged to cite severalother examples.

10.30 What applications for composite materials canyou think of in which high thermal conductivitywould be desirable? Explain.

By the student. See also Sections 3.9.4and.3.9.5. Composite materials with high ther-mal conductivity could be useful as heat ex-changers, food and beverage containers, andmedical equipment.

10.31 Conduct a survey of a variety of sports equip-ment, and identify the components that aremade of composite materials. Explain the rea-sons for and advantages of using composites forthese specific applications.

By the student. Examples include rackets fortennis, badminton, and racquetball; baseballand softball bats; golf clubs; fishing rods; andskis and ski poles. The main reason is the lightweight of these materials, combined with highstiffness and strength, thus resulting in superiorperformance.

10.32 We have described several material combina-tions and structures in this chapter. In relativeterms, identify those that would be suitable forapplications involving one of the following: (a)very low temperatures; (b) very high tempera-tures; (c) vibrations; and (d) high humidity.

By the student. This is a challenging topicand the students are encouraged to develop re-sponses with appropriate rationale. For exam-ple, very low temperature applications require(1) considerations of the polymer or matrix me-chanical properties with appropriate ductility

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toughness, (2) warpage that can occur whenlowered from room temperature, and (3) ther-mal stresses that may develop due to differencesin thermal expansion between the polymer partand other parts that are in contact. Tempera-ture variations are particularly important, asdescribed in Sections 3.9.4 and 3.9.5.

10.33 Explain how you would go about determin-ing the hardness of the reinforced plastics andcomposite materials described in this chapter.What type of tests would you use? Are hard-ness measurements for these types of materialsmeaningful? Does the size of the indentationmake a difference in your answer? Explain.

By the student. The important considerationhere is the fact that the smaller the indentation,the more localized the hardness measurementwill be (see Section 2.6). Consequently, onecan then distinguish the hardness of the ma-trix and the reinforcements separately by usingsmall indentations. A large indentation, suchas resulting from a Brinell test, will only givean overall hardness value. (Note that this is aconsideration similar to microhardness testingof individual components of an alloy or of theindividual grains.)

10.34 Describe the advantages of applying traditionalmetalworking techniques to the forming andshaping of plastics.

By the student. Review Section 10.10 andChapters 6 and 7. Note also that this topic isbriefly described in Section 10.10.9. Applyingtraditional metalworking techniques to shapingof plastics is advantageous for several reasons.Since the stock shapes are similar (sheet, rod,tubing, etc.), well-known and reliable processescan be applied efficiently. Being able to utilizesimilar machines and many years of research,development, and experience associated withmachine design and process optimization willhave major significance in plastics applicationsas well.

10.35 Describe the advantages of cold forming of plas-tics over other methods of processing.

By the student. See Section 10.10.9 where fourmain advantages are outlined.

10.36 Explain the reasons that some forming andshaping processes are more suitable for certainplastics than for others.

By the student. Consider, for example, the fol-lowing: It is difficult to extrude thermosets be-cause curing is not feasible during the contin-uous extrusion process. Injection molding ofcomposites is difficult because the fluidity ofthe material is essential to ensure proper fillingof the die, but characteristics and presence ofthe fibers interferes with this process. Plasticswhich are produced through reaction moldingare difficult to produce through other means,and other processes are not readily adaptableto allow sufficient mixing of the two ingredi-ents. These difficulties should, however, be re-garded as challenges and thus novel approachesare encouraged. The students are encouragedto develop additional answers.

10.37 Would you use thermosetting plastics in injec-tion molding? Explain.

Thermosetting plastics are suitable for injectionmolding (Section 10.10.2), although the processis often referred to as reaction injection mold-ing; see p. 629. The basic modification whichmust be made to the process is that the moldsmust be heated to allow polymerization andcross-linking of the material. The major draw-back associated with this change is that, be-cause of the longer cycle times, the process willnot have as high a production rate as for ther-moplastics.

10.38 By inspecting plastic containers, such as forbaby powder, you can see that the lettering onthem is raised and not sunk in. Offer an ex-planation as to why they are molded in thatway.

The reason is that in making molds and diesfor plastics processing, it is much easier to pro-duce letters and numbers by removing materialfrom mold surfaces, such as by grinding or endmilling, similar to carving of wood. As a re-sult, the molded plastic part will have raisedletters and numbers. On the other hand, if wewant depressed letters on the product itself, themarkings on the molds would have to protrude.This is possible to do but would be costly and

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time consuming to make such a mold, and itswear resistance will likely be lower.

10.39 Give examples of several parts that are suitablefor insert molding. How would you manufacturethese parts if insert molding were not available?

By the student. See also the parts shown inFig. 10.30 on p. 628. Some common parts arescrew drivers with polymer handles, electricaljunction boxes with fasteners that are insertmolded, screws and studs in polymer parts toaid assembly, and some writing instruments.Usually, these parts would have to be mechan-ically assembled or adhesively bonded if insertmolding was not an option.

10.40 What manufacturing considerations are in-volved in making a metal beverage containerversus a plastic one?

By the student. See also Fig. 16.31 on p. 452 ofKalpakjian and Schmid, Manufacturing Engi-neering and Technology, 3d ed. and the Bibliog-raphy at the end of Chapter 16. Since beveragecans are mass produced in the range of millionsper day, the processing must be simple and eco-nomical. Other important considerations arechilling characteristics, labeling, feel, aesthet-ics, and ease of opening. Students should com-ment on all these aspects. Note also that thebeverage can must have sufficient strength toprevent from rupturing under internal pressure(which is on the order of about 120 psi), orbeing dropped, or buckling under a compres-sive load during stacking in stores. The canshould maintain its properties from low temper-atures in the refrigerator to hot summer tem-peratures, especially under the sun in hot cli-mates. Particularly important is the gas per-meability of plastic containers which will sig-nificantly reduce their shelf life. Also note howsoft drinks begin to lose their carbonation inunopened plastic bottles after a certain periodof time. (See also Section 10.10.)

10.41 Inspect several electrical components, such aslight switches, outlets, and circuit breakers, anddescribe the process or processes used in mak-ing them.

By the student. The plastic components areusually injection molded and then mechanically

assembled. Several ingenious designs use insertmolding. Integrated circuits and many otherelectrical components may be potted.

10.42 Inspect several similar products that are madeof metals and plastics, such as a metal bucketand a plastic bucket of similar shape and size.Comment on their respective thicknesses, andexplain the reasons for their differences, if any.

By the student. Recall that the basic differ-ence between metals and plastics have been dis-cussed in detail in the text. Consider the fol-lowing examples:

(a) Metal buckets are thinner than plasticones, and are more rigid; plastic bucketsthus have to be thicker because of theirmuch lower elastic modulus, as well as in-volve designs with higher section modulus.

(b) Mechanical pencils vs. plastic pens; thepolymer pens are much thicker, becausethey must be rigid for its intended use.

(c) Plastic vs. metal forks and spoons; al-though no major difference in overall size,the plastic ones are more flexible but canbe made more rigid by increasing the sec-tion modulus (as can be observed by in-specting their designs).

10.43 Make a list of processing methods used for rein-forced plastics. Identify which of the followingfiber orientation and arrangement capabilitieseach has: (1) uniaxial, (2) cross-ply, (3) in-plane random, and (4) three-dimensional ran-dom.

By the student. An example of a partial answeris the following:

Process Uni

axia

l

Cro

ss-p

ly

In-p

lane

rand

om

3Dra

ndom

Sheet-molding compound XTape lay-up X XContact molding XInjection molding X XPultrusion XPulforming X

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10.44 As you may have observed, some plastic prod-ucts have lids with integral hinges; that is, noother material or part is used at the junctionof the two parts. Identify such products, anddescribe a method for making them.

Such parts with integral lids are produced withone shot in processes such as injection molding.The hinge is actually a much thinner (reduced)section, which can bend easily, thus acting likea hinge. It should be noted that there are signif-icant material requirements that must be metbefore such a design can be achieved, includingstiffness and fatigue strength. Many polymersare ideally suited for such applications.

10.45 Explain why operations such as blow moldingand film-bag making are done vertically andwhy buildings that house equipment for theseoperations have ceilings 10 m to 15 m (35 ft to50 ft) high.

They are done vertically so that the gravita-tional force does not interfere with the opera-tion. The height of ceilings is dictated by prod-uct requirements. The height is large enoughso that the blown tube can be cooled from asemi-molten state to a solid state suitable forcompression, cutting and recoiling.

10.46 Consider the case of a coffee mug being pro-duced by rapid prototyping. Describe how thetop of the handle can be manufactured, sincethere is no material directly beneath the archof the handle.

By the student. Depending on the process usedand the particular shape of the mug handle,this may or may not be a difficult problem;even if difficult, it can be overcome fairly easily.Some processes, such as stereolithography andfused deposition modeling, can allow buildingof gradual arches, but a coffee mug is proba-bly too severe, and a ceiling design as shown inFig. 10.49b on p. 649 would have to be used.Other processes such as selective laser sinteringand laminated object manufacturing have noneed for supports, and thus a coffee mug canbe produced easily.

10.47 Make a list of the advantages and disadvantagesof each of the rapid-prototyping operations.

By the student. As examples, the studentscould investigate (a) cost (where FDM, 3DP,STL have advantages over SLS of metals, for ex-ample), (b) material properties (see Table 10.8on p. 646) where selective laser sintering withbronze-infiltrated steel powder would be supe-rior, or (c) dimensional tolerances or surface fin-ish.

10.48 Explain why finishing operations generally areneeded for rapid-prototyping operations. If youare making a prototype of a toy car, list the fin-ishing operations you would want to perform.

By the student. The finishing operations re-quired vary for different rapid prototyping ap-plications. For example, in stereolithography,the part has to be cured in order to fully de-velop its mechanical properties (the laser doesnot fully cure the photopolymer), and then thepart may need to be sanded or finely ground toobtain a desired surface. Also, often decorationis needed for aesthetic purposes. On the otherhand, in fused deposition modeling, the finish-ing operations would involve removal of sup-port material, followed by sanding and paint-ing, whenever necessary. For a prototype of atoy automobile, the finishing processes wouldbe as discussed.

10.49 A current topic of research involves producingparts from rapid-prototyping operations andthen using them in experimental stress analysis,in order to infer the strength of the final partsproduced by conventional manufacturing oper-ations. List the concerns that you may havewith this approach, and outline means of ad-dressing these concerns.

In theory, this technique can be successful fordetermining the stresses acting on a part of acertain geometry, as long as the part remainsin the linear elastic range and the strains aresmall. However, it is difficult, although notimpossible, to infer performance of convention-ally manufactured parts, especially, for exam-ple, with respect to fatigue or wear. The rea-son is that the material microstructure and re-sponse to loading will be very different thanthat for a rapid prototyped model.

10.50 Because of relief of residual stresses during cur-ing, long unsupported overhangs in parts from

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stereolithography will tend to curl. Suggestmethods of controlling or eliminating this prob-lem.

These problems can be minimized in design byreducing overhangs or changing the support ofthe overhang. Otherwise, as shown in Fig. 10.49on p. 649, gussets or ceilings could be used tosupport the material and minimize curl duringcuring.

10.51 One of the major advantages of stereolithog-raphy and cyberjet is that semi- and fully-transparent polymers can be used, so that in-ternal details of parts can readily be discerned.List parts or products for which this feature isvaluable.

By the student. Some examples are (a) heat ex-changers, where the fluid flow can be observed;(b) drug delivery systems, so that any blockageor residual medicines can be observed; (c) anyship-in-the-bottle type of part; d (d) market-ing models to explain the internal features of aproduct.

10.52 Based on the processes used to make fibersas described in this chapter, explain how youwould produce carbon foam. How would youmake a metal foam?

By the student. This is a good topic for a lit-erature search. One approach is to produce apolymer foam followed by a carburization pro-cess, as would be performed to produce graphitepowder. The result is a foam produced fromcarbon, a product that has value as a filter ma-terial because of the very large surface area-to-volume ratio. A metal foam can be easilyproduced at this point by placing the carbonfoam in a CVD reactor, whereby the metal willcoat the carbon foam. Other alternatives areto blow hot air through molten metal; the frothsolidifies into a metal foam, or to use a blowingagent in a P/M process (see Chapter 11).

10.53 Die swell in extrusion is radially uniform for cir-cular cross-sections, but is not uniform for othercross-sections. Recognizing this fact, makea qualitative sketch of a die profile that willproduce (a) square and (b) triangular cross-sections of extruded polymer, respectively.

The sketches are given below. Note that thereis expected to be greater recovery at cornerswhere the strain on the extruded polymer ishighest.

10.54 What are the advantages of using whiskers asa reinforcing material? Are there any limita-tions?

By the student. Whiskers are much strongerthan other fibers because of their small sizeand lack of defects (see pp. 105 and 463).Whiskers will yield composite materials withhigher strength-to-weight ratios.

10.55 By incorporating small amounts of blowingagent, it is possible to produce polymer fiberswith gas cores. List some applications for suchfibers.

By the student. Examples include applicationswhere weight is a primary concern, such asaerospace structures. Also, such a structure isvery common in foams, and the typical applica-tions are for flotation devices (life savers, surf-boards, etc), or thermal applications where thegas cores act as an effective insulators (coffeecups, thermos, etc.). If woven into a fabric, itcan be an effective insulator for winter clothing.

10.56 With injection-molding operations, it is com-mon practice to remove the part from its run-ner and then to place the runner into a shredderand recycle into pellets. List the concerns youwould have in using such recycled pellets as op-posed to so-called virgin pellets.

151






By the student. Consider the following con-cerns:

(a) The polymer may become chemically con-taminated by tramp oils or parting agentson the die;

(b) Wear particles from the shredder may con-taminate the polymer;

(c) The polymer may be chemically degradedfrom the heating and cooling cycles en-countered in injection molding;

(d) The molecular weight of the shreddedpolymer may be much lower than that forthe original polymer, so that the mechani-cal properties of the recycled stock can beinferior.

10.57 What characteristics make polymers attractivefor applications such as gears? What character-istics would be drawbacks for such applications?

By the student. Students should be encouragedto develop answers that rely on their personalexperience. The advantages include (a) the lowfriction of polymers, even when not lubricated),(b) wear resistance, (c) good damping charac-teristics, so that sound and impact forces arenot as severe with plastic gears), and (d) manu-facturing characteristics that allow the produc-tion of tooth profiles with superior surface fin-ish (see Section 8.10.7). The main drawbacks topolymer gears are associated with low stiffness,especially at elevated temperature, and lowerstrength than metals (so the loads that can betransferred for an equivalent sized gear is muchlower), but they would be suitable for motiontranslation.

10.58 Can polymers be used to conduct electricity?Explain, giving several examples.

Recall that polymers can be made to con-duct electricity (see Section 10.7.2), such aspolyacetylene, polyaniline, and polythiophene.Other polymers can be made more conduc-tive by doping them with metal particles orwhiskers. If continuous wire reinforcement ispresent, the polymer can be directionally con-ductive. It can also be conductive in a planeif a mesh reinforcement is used. An electrolessnickel plating (p. 160) of a polymer part can

also make it conductive. The students shouldcomment further on this topic.

10.59 Why is there so much variation in the stiffnessof polymers? What is its engineering signifi-cance?

By the student. Table 10.1 on p. 585 shows awide range of stiffness; note, for example, thatfor polyethylene the change can be 1400%. Thisis mainly due to the widely varying degree ofpolymerization and crystallinity, and the num-ber of crosslinks, if any, present, as well as theimportant effects of the reinforcements. Stiff-ness will increase with any of these variables.

10.60 Explain why thermoplastics are easier to recy-cle than thermosets.

If a polymer’s chemistry can be identified, thena polymer product can be cut into small pieces(such as pellets or particles) and fabricatedas is done with so-called virgin thermoplas-tics. There is some degradation of mechanicalproperties and a measurable loss of molecularweight, but if properly sorted (see top of p. 607),these drawbacks can be minimized. It is diffi-cult to recycle thermosets because it is impos-sible to break down a thermosetting resin intoits mer components. Thus, the manufactur-ing strategies for the original polymer and forits recycled counterparts have to be different.Furthermore, thermosets cannot be melted, orchopped up as would thermoplastics.

10.61 Describe how shrink-wrap works.

Shrink wrap consists of branched thermo-plastics. When deformed above their glass-transition temperature, the branches attain apreferred orientation, similar to the effect ofcombing hair. The plastic is then quickly low-ered in temperature, preventing stress relax-ation. When the sheet is then wrapped aroundan object (including food products) and thenheated, the stresses are relieved and the plasticsheet or film shrinks around the object.

10.62 List the characteristics required of a polymerfor the following applications: (a) a total hip re-placement insert, (b) a golf ball, (c) an automo-tive dashboard, (d) clothing, and (e) a child’sdoll.

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By the student. Consider, for example, the fol-lowing:

(a) Some of the characteristics required for apolymer insert in a total hip replacementare that it be biocompatible; not dissolveor warp in the presence of bodily fluids;support the loads developed during nor-mal walking, sitting, and standing; notwear excessively; and provide low friction.Cost is not as imperative as other appli-cations, given the high cost of surgery forhip replacements.

(b) For a golf ball, abrasion resistance is im-portant, as well as impact strength andtoughness. The polymer needs to have astiffness consistent with typical golf balls,and it must be coatable, so that it can bemade into a bright color. Cost is also im-portant.

(c) An automobile-dashboard polymer needsto be formable into the desired (and quitedemanding) shapes. It also has to beavailable in a range of desired colors,and should have acceptable manufacturingcost.

(d) The polymer in clothing needs to beproduced into fibers and in continuouslengths. The fibers must be sufficientlyflexible so that they can be woven intocloth and withstand normal wear and tear.The polymer must have low elastic modu-lus but sufficient strength so that the clothfeels soft but doesn’t tear easily. It mustalso be inexpensive.

(e) A child’s doll must be non-toxic, andshould be soft but tough so that the childcannot break off a piece of the doll andthus becoming a choking hazard. Thepolymer should be easy to decorate andcleanable.

10.63 How can you tell whether a part is made of athermoplastic or a thermoset? Explain.

By the student. There are several nondestruc-tive and destructive tests that can be per-formed. For example, tension tests will demon-strate the difference: a pronounced plasticity isindicative of a thermoplastic. Exposure to hightemperatures is another test: the presence of

a glass-transition temperature is indicative of athermoplastic. The shape of the part is oftena clue; for example, thin films must be madeof thermoplastics because they are blown fromextruded tubing.

10.64 Describe the features of an extruder screw andcomment on their specific functions.

By the student. A typical extruder is shownin Figs. 10.22 and 10.23 on p. 620. The threeprincipal features of the screw shown are:

• Feed section: In this region, the screw isintended to entrain powder or pellets fromthe hopper; as a result, the flight spacingand depth is larger than elsewhere on thescrew.

• Melt section: In the melt section, the flightdepth is very low and the plastic is meltedagainst the hot barrel; also, gases that areentrained in the feed section are vented.

• Metering section: This region producedthe pressure and flow rate needed for theextrusion operation.

Note that screws are designed for particularpolymers, so the feed, melt, and metering sec-tions are polymer-specific. Also, some extrud-ers use two screws to increase the internal shear-ing and mixing of the polymer.

10.65 An injection-molded nylon gear is found to con-tain small pores. It is recommended that thematerial be dried before molding it. Explainwhy drying will solve this problem.

The probable reason is that the porosity is dueto entrapped moisture in the material. Recallalso that nylon absorbs water (hygroscopy; seetop of p. 600), thus drying will alleviate thisproblem.

10.66 What determines the cycle time for (a) injectionmolding, (b) thermoforming, and (c) compres-sion molding?

The cycle time for injection molding is deter-mined by several factors, including:

• Material: Thermoplastics require muchless time than thermosets, and certainthermoplastics will require less time tocool and solidify than will others (i.e., dif-ferent thermal properties).

153






• Part shape: If the part has a high surfacearea-to-volume, it will cool rapidly.

• Initial temperature: If a polymer is in-jected at a temperature much above itssolidification temperature, it will requiremore time to cool.

The considerations for thermoforming and com-pression molding are similar. The students areencouraged to analyze and elaborate further onthis subject.

10.67 Does the pull-in defect (sink marks) shown inFig. 10.57 also occur in metal forming and cast-ing processes? Explain.

The type of defect shown in Fig. 10.57 also oc-curs in metal forming (because of the flow ofthe material into the die cavity) and castingprocesses (because of excessive, localized sur-face shrinkage during solidification and coolingin the mold). This is described in various hand-books, but it should be noted that sink marks isa terminology restricted to polymer parts. Forexample, in Bralla, J.G., Design for Manufac-turability Handbook, 2nd. ed., pp. 5.51, the sinkmarks are referred to as dishing for investmentcasting, and on p. 5.64 the same features arereferred to as shrink marks.

10.68 List the differences between the barrel sectionof an extruder and that on an injection-moldingmachine.

By the student. Some of the basic differencesbetween an extruder and an injection-moldingmachine barrel are:

• Extruders involve more heating from theheating elements and less from friction,so there will be more (or larger capacity)heating elements and temperature sensorsin an extruder barrel.

• Extruders do not utilize torpedoes or re-ciprocating screws.

• Extruders may use multiple screws to im-prove mixing in the barrel.

10.69 Identify processes that are suitable for makingsmall production runs of plastic parts, such asquantities of 100 or fewer. Explain.

By the student. Refer to Table 10.9 on p. 658and note that low quantities involve processesin which tooling costs must be kept low. Thus,the most suitable processes would be castingand machining (because of the readily avail-able and versatile machine tools). However,rapid prototyping operations can also be useddirectly if the quantities are sufficiently smalland part characteristics are acceptable. Also,tooling can be produced using the methods de-scribed in Section 10.12.6 to render processessuch as injection molding viable for small pro-duction runs. Note, however, that these toolsare not suitable for large production runs.

10.70 Review the Case Study to this chapter and ex-plain why aligners cannot be made directly byrapid prototyping operations.

As described in the Case Study on p. 658, thepolymers in stereolithography have a yellowtint, which is objectionable for cosmetic rea-sons. However, there are some clear polymersnow available (see WaterShed 11120 in Table10.8 on p. 646), but it is difficult to fully curethe monomer, making the aligners develop anunpleasant taste.

10.71 Explain why rapid prototyping approaches arenot suitable for large production runs.

The main reasons are the long times requiredfor producing parts (2 hours or so for a smallpart is rapid when only one part is required.Two hours per part is unacceptably long for amillion parts. Recall also that rapid prototyp-ing operations can be very demanding and re-quire high-quality materials that have high costassociated with them.

10.72 List and explain methods for quickly manufac-turing tooling for injection molding.

This topic is discussed in Section 10.12.6. De-pending on the material, the following are op-tions:

• A mold can be directly produced with arapid prototyping operation if the polymerto be injection molded has a lower meltingtemperature than the mold material.

• An RTV molding/urethane casting opera-tion can be employed (see p. 653), using arapid prototyped pattern.

154






• ACES injection molding uses a rapid-prototyped tooling shell backed with alow-melting-point metal.

• Sprayed metal tooling can produce ashell from a rapid-prototyped pattern.The shell is backed with an epoxy oraluminum-filled epoxy, for strength and toremove heat from injection molding.

• The Keltool process (see p. 654) can beused, using an RTV mold, filled with pow-dered A6 tool steel infiltrated with copper.

10.73 Careful analysis of a rapid-prototyped part in-dicates that it is made up of layers with a whitefilament outline visible on each layer. Is the ma-terial a thermoset or a thermoplastic? Explain.

The presence of the filament outline sug-gests that the material was produced in fused-deposition modeling (Section 10.12.3). Thisprocess requires adjacent layers to fuse afterbeing extruded. Extrusion and bonding is ob-viously possible with thermoplastics but verydifficult for a thermoset.

10.74 List the advantages of using a room-temperature vulcanized (RTV) rubber moldin injection molding.

The advantages include the following:

• The tooling cost is low.• Very detailed part geometry can be incor-

porated into the RTV mold.• The mold is flexible, so that it can be

peeled off of parts; thus, draft angles and

other design considerations for metallictooling can be relaxed.

• The RTV mold can be produced with theaid of rapid prototyping operations so thatthe mold is quickly produced.

Note that there are also disadvantages to thismethod, mainly the limited tool life.

10.75 What are the similarities and differences be-tween stereolithography and cyberjet?

As seen in Table 10.7 on p. 646, note that (a)both processes rely on the same layer-creationtechnique, namely liquid-layer curing, (b) bothuse a principle of using a photopolymer to cre-ate a thermoset part, and (c) both have com-parable materials, with similar characteristicsof strength, cost, and appearance.

10.76 Explain how color can be incorporated intorapid-prototyped components.

The following methods are the most straight-forward:

• The ZCorp (see Fig. 10.52 on p. 651)versions of three-dimensional printing ma-chines incorporate colored binders, so thatfull-color prototypes can be produced di-rectly.

• FDM machines usually have two heads, sothat two colors can be extruded as desired.

• Otherwise, color is most easily incorpo-rated by painting the prototyped part.

Problems

10.77 Calculate the areas under the stress-straincurve (toughness) for the material in Fig. 10.9,plot them as a function of temperature, and de-scribe your observations.

The area under the curves is estimated byadding the area under the initial elastic regionto that in the flat regions. The results are asfollows:

Temperature Toughess(C) (MJ/m3)-25 1400 63525 76050 73065 52080 500

155






Temperature (°C)

-50 0 50 100

800

600

400

200

0

Toug

hnes

s (M

J/m

3 )

Note that there is an optimum temperature formaximum toughness.

10.78 Note in Fig. 10.9 that, as expected, the elasticmodulus of the polymer decreases as temper-ature increases. Using the stress-strain curvesgiven in the figure, make a plot of the modulusof elasticity versus temperature.

Note that all curves start at the origin, and un-dergo a transition from linear elastic behaviorto plastic behavior at a strain of approximately4%. We can therefore estimate the elastic mod-ulus from the slope of the curves up to 4%strain. The following table can be constructed:

Stress at ElasticTemperature 4% strain modulus

(C) (MPa) (GPa)-25 70 1.750 60 1.525 40 1.050 25 0.62565 20 0.5080 13 0.325

The resulting plot is as follows:

Temperature (°C)-25 0 25 50 75 100

Ela

stic

Mod

ulus

(GP

a) 2.0

1.5

1.0

0.5

0

10.79 Calculate the percentage increase in mechani-cal properties of reinforced nylon from the datashown in Fig. 10.19.

The following data is obtained from Fig. 10.19on p. 615, with the last column calculated fromthe data. Also, note that the flexural modulusof nylon has been obtained from Table 10.1 onp. 585 as 1.4 GPa.

TemperatureProperty 0% 40% % Increase

Tensile strength 100 200a 100

(MPa) 250b,c 150Impact energy 70 150a 114

(J/m) 295b 32180c 14

Flexural modulus 1.4 12a,b 757(GPa) 25c 1686

Flexural strength 150 300a 100

(MPa) 330b 120350c 133

Notes: 1. short glass; 2. long glass; 3. car-bon.

10.80 A rectangular cantilever beam 75-mm high, 25-mm wide, and 1-m long is subjected to a con-centrated force of 100 N at its end. Select twodifferent unreinforced and reinforced materialsfrom Table 10.1, and calculate the maximumdeflection of the beam. Then select aluminumand steel, and for the same beam dimensions,calculate the maximum deflection. Comparethe results.

This is a simple mechanics of solids problem inwhich the governing equation for the deflection,d, of a cantilever beam with a concentrated loadof P (=100 N) at the end is

d =PL3

3EI

where L is the beam length (1 m), E the elas-tic modulus of the material chosen from Table10.1, and I is the moment of inertia, i.e.,

I =bh3

12=

(25)(75)3

12= 8.79× 105 mm4

Substituting for moment of inertia, load, and

156






length gives the deflection as

d =PL3

3EI

=(100 N)(1 m)3

3 (8.79× 105 mm4)1E

=3.792× 107 N/m

E

Thus, the higher the elastic modulus, thesmaller the deflection of the beam. Examplesof the results are as follows:

Material E (GPa) d (mm)Aluminum 70a 0.542Steel 200a 0.190ABS, nylon 1.4 27.1Polyesters 2.0 19.0Polystyrene 2.7 14.0Note: a From Table 2.1.

10.81 In Sections 10.5 and 10.6, we listed several plas-tics and their applications. Rearrange this in-formation, respectively, by making a table ofproducts and the type of plastics that can beused to make the products.

The following is an example of an acceptableanswer to this problem. Note that there aremany approaches and part classifications thatcould be used, and the information in thetextbook could be supplemented with Internetsearches. Also, many more products could belisted if desired.

Product Polymers

Bearings Acetals, fluorocarbons, nylon,polyimides

Gears Acetals, nylon, polyester,polyimides

Cams Acetals, polyesterLenses Acrylics, cellulosics,

polycarbonatesFurniture Acrylics, polystyreneHousings Acetals, ABS, polypropylene,

polysulfone, aminosLow wear Nylon, polyethyleneGuards Cellulosics, polycarbonatesPipes ABS, cellulosics, polypropylene,

PVCToys Cellulosics, polyethylene,

polystyreneElectrical Fluorocarbons, nylon, poly-

insulation carbonate, polypropylene,polysulfone, PVC

Food Polycarbonate, polypropylene,contact polystyrenes, melamine

Gaskets Fluorocarbons, PVC, silicones

10.82 Determine the dimensions of a tubular steeldrive shaft for a typical automobile. If you nowreplace this shaft with shafts made of unrein-forced and reinforced plastic, respectively, whatshould be the shaft’s new dimensions to trans-mit the same torque for each case? Choose thematerials from Table 10.1, and assume a Pois-son’s ratio of 0.4.

Note that the answers will vary widely depend-ing on the shaft dimensions. However,

J =π

32(D4

o −D4i

)The shear stress under pure torsion for a tubu-lar shaft is given by

τ =T (Do/2)

J=

16TDo

π (D4o −D4

i )

Therefore, the torque that can be carried by theshaft at the shear yield stress of the material is

T =kπ(D4

o −D4i

)16Do

Since steel has a higher shear stress than rein-forced polymers, the tube dimensions will haveto be modified in order to accommodate the

157






same torque. Using an s subscript for steel andp for polymer, it can be seen that

ksπ(D4

os −D4is

)16Dos

=kpπ

(D4

op −D4ip

)16Dop

orks

kp=D−1

op

(D4

op −D4ip

)D−1

os (D4os −D4

is)

As an example, compare a low-carbon steel(UTS=395 MPa) to reinforced ABS, with aUTS of 100 MPa. For solid shafts, (Dis =Dip = 0), the required outer diameter of theABS shaft is

395100

=(1/Dop)D4

op

(1/Dop)D4op

orDop = (3.95)1/3Dos = 1.58Dos

10.83 Calculate the average increase in the propertiesof the plastics listed in Table 10.1 as a result oftheir reinforcement, and describe your observa-tions.

The results are given in the following table.

Unrein- Rein- AverageProp- forced forced increase

Material propertya (ave) (ave) %ABS UTS 41.5 100 59

E 2.1 7.5 54Acetal UTS 62.5 135 73

E 2.45 10.0 75.5Epoxy UTS 87.5 735 648

E 10.2 36.5 263Nylon UTS 69 140 71

E 2.1 6.0 39Polycarb- UTS 62.5 110 48

onate E 2.1 6.0 39Polyester UTS 55 135 80

E 2.0 10.2 82Polyprop- UTS 27.5 70 43ylene E 0.95 4.75 38

Note: (a) UTS in MPa, E in GPa

10.84 In Example 10.4, what would be the percent-age of the load supported by the fibers if theirstrength is 1250 MPa and the matrix strengthis 240 MPa? What if the strength is unaffected,but the elastic modulus of the fiber is 600 GPawhile the matrix is 50 GPa?

A review of the calculations in Example 10.4on p. 617 indicates that the strength of the

materials involved does not influence the re-sults. Since the problem refers only to changesin strength, it is assumed that the moduli ofelasticity are the same as in the original exam-ple.

If the strength is unaffected, but the elasticmoduli are changed, there will be an effect onthe load supported by the fibers. The percent-age of the load supported by the fibers can thenbe calculated as follows:

Ec = (0.2)(600) + (1− 0.2)(50) = 120 + 40

or Ec = 160 GPa. Also,

Ff

Fm=

(0.20)(600)(0.8)(50)

=12040

= 3

orFc = Ff + Ff/3 = 1.33Ff

andFf = 0.75Fc

Thus, the fibers support 75% of the load in thiscomposite material. As expected, this percent-age is higher than the 43% in the sample calcu-lations given in Example 10.4.

10.85 Estimate the die clamping force required forinjection molding 10 identical 1.5-in.-diameterdisks in one die. Include the runners of appro-priate length and diameter.

Note that this question can be answered in sev-eral ways, and that the layout is somewhat ar-bitrary. In fact, the force could conceivablybe based on the thickness of the disc, but thiswould be a much more difficult cavity to ma-chine into a die, and a far more difficult part toeject. Instead, we will use central sprues withrunners to feed two rows of five discs each. Us-ing 0.25-in. diameter runners, their contribu-tion to the area is

Arunners = 2(0.25 in.)(10 in) = 5 in2

Note that we have allowed some extra space tohave clearance between the disks. The totaldisk surface area is then

Adiscs = 10(πd2

4

)= 10

(π(1.5 in.)2

4

)158






or Adiscs = 17.7 in2. Therefore, the total sur-face area of the mold is about 23 in2. Asstated in Section 10.10.2, injection pressuresrange from 10,000 to 30,000 psi. Therefore, theclamping force will range from 230,000 (115) to690,000 lb (345 tons).

10.86 A two-liter plastic beverage bottle is madefrom a parison with the same diameter as thethreaded neck of the bottle and has a lengthof 5 in. Assuming uniform deformation duringblow molding, estimate the wall thickness of thetubular section.

This problem will use typical values for two-literbottle dimensions, but small deviations fromthese numbers, and hence the answer, are likely.If open-ended problems are not desirable, thestudent can be asked to use L = 9 in., D = 4.25in., and t = 0.015 in. for the finished bottle,with a 1.125 in. diameter neck. These are rea-sonable values that are used in this solution.

The volume of the plastic material is estimatedas

V = πDLt = π(4.25)(9)(0.015) = 1.8 in3

As stated in the problem, the parison is a tubu-lar piece 5 in. long, and its diameter is the sameas the threaded neck of a two-liter bottle, i.e.,about 1 1

8 in., as measured. Let’s assume that,as in metals, the volume of the material doesnot change during processing (although this isnot a good assumption because of significantdensity variations in polymers due to changesin the free space or free volume in their molec-ular structure). Assuming volume constancy asan approximation, the thickness tp of the pari-son is calculated as

1.8 = π(1.125)tp(5) → tp = 0.10 in.

10.87 Estimate the consistency index and power-lawindex for the polymers in Fig. 10.12.

The solution requires consideration of the datagiven in Fig. 10.12b on p. 597. As an exam-ple, consider rigid PVC at 190C. The followingdata is interpolated from the curve:

Strain rate Viscosityγ (s−1) (Ns/m2)10 11,00023 8000100 6000230 10001000 500

The plot is constructed from this data as fol-lows:

Vis

cosi

ty, η

(Ns/

m2 )

0

4,000

8,000

12,000

Strain rate, γ (s-1)0 400 800 1200

η=72,465γ-0.707

A curve fit of the form of η = Aγ1−n is fit tothe data, suggesting that the consistency indexis A = 72, 465 Ns/m2, and that the power lawindex is

1− n = −0.707 → n = 1.707

10.88 An extruder has a barrel diameter of 100 mm.The screw rotates at 100 rpm, has a channeldepth of 6 mm, and a flight angle of 17.5.What is the highest flow rate of polypropylenethat can be achieved?

The highest flow rate is if there is zero pressureat the end of the barrel, and then we have puredrag flow, given by Eq. (10.20) as

Qd =π2D2HN sin θ cos θ

2Using D = 100 mm, H = 6 mm, N = 100 rpmand θ = 17.5 gives

Qd =π2(100)2(6)(100) sin 17.5 cos 17.5

2or Qd = 8.49 × 106 mm3/min = 141,500mm3/sec.

10.89 The extruder in Problem 10.88 has a pumpingsection that is 2.5 m long and is used to ex-trude round polyethylene solid rod. The diehas a land of 1 mm and a diameter of 5 mm.If the polyethylene is at a mean temperature of

159






250C, what is the flow rate through the die?What if the die diameter is 10 mm?

The extruder characteristic is given byEq. (10.23) on p. 621 as

Q =π2D2HN sin θ cos θ

2− πDH3 sin2 θ

12ηlp

For polyethylene at 250C, the viscosity η isabout 80 Ns/m2, as obtained from Fig. 10.12.Also, from the statement of Problem 10.88, weknow that D = 100 mm, H = 6 mm, N = 100rpm, and θ = 17.5; l is given as 2.5 m. There-fore, the extruder characteristic is

Q =π2(0.100)2(0.006)(100) sin 17.5 cos 17.5

2

−π(0.100)(0.006)3 sin2 17.5

12 (80) (3)p

or

Q = 0.00849 m3/min−(2.13× 10−12 m5/N-min

)p

For this die, the die characteristic is given byEq. (10.25) on p. 621, where K is evaluated fromEq. (10.26) as

K =πD4

d

128ηld=

π(0.005)4

128(80)(0.001)

or K = 1.15× 10−10 m5/N-min. Therefore, thedie characteristic is given by

Q = Kp =(1.15× 10−10 m5/N-min

)p

We now have two equations and two unknowns;these are solved as p = 72.5 MPa and Q =0.00833 m3/min.

If the die has a diameter of 10 mm, then

K =πD4

d

128ηld=

π(0.010)4

128(80)(0.001)

or K = 3.07 × 10−9, and the simultaneousequations then yield Q = 0.00848 m3/min andp = 2.7 MPa.

10.90 An extruder has a barrel diameter of 4 in., achannel depth of 0.25 in., a flight angle of 18,and a pumping zone that is 6 ft long. It is usedto pump a plastic with a viscosity of 100×10−4

lb-s/in2. If the die characteristic is experimen-tally determined as Qx = (0.00210 in5/lb-s)p,what screw speed is required to achieve a flowrate of 7 in3/s from the extruder?

The pressure at the die can be determined fromthe die characteristic and the required flow rate,using Eq. (10.25):

Q = 7 in3/s = Kp = (0.00210 in5/lb-s)p

Solving for p,

p =7

0.00210= 3.333 ksi

From Eq. (10.23), the extruder characteristic is

Q =π2D2HN sin θ cos θ

2− πDH3 sin2 θ

12ηlp

Solving for N ,

N =[

2π2D2H sin θ cos θ

] [Q+

πDH3 sin2 θ

12ηlp

]or

N =[

2π2(4)2(0.25) sin 18 cos 18

]×[7 +

π(4)(0.25)3 sin2 18

12 (100× 10−4) (72)(3333)

]or N = 2.45 rev/s, or 147 rpm.

10.91 What flight angle should be used on a screw sothat a flight translates a distance equal to thebarrel diameter with every revolution?

Refer to the following figure:

D

θ

BarrelL

Barrel

160






The relationship between the screw angle, θ,the lead, L, and diameter, D, can be best seenby unwrapping a revolution of the screw. Thisgives

θ = tan−1

(L

πD

)For L = D, we have

θ = tan−1

(D

πD

)= tan−1

(1π

)Therefore, θ = 17.6.

10.92 For a laser providing 10 kJ of energy to aspot with diameter of 0.25 mm, determine thecure depth and the cured line width in stere-olithography. Use Ec = 6.36 × 1010 J/m2 andDp = 100 µm.

The exposure at the surface of the material isgiven by

Eo =10 kJ

π4 (0.25 mm)2

= 2.04× 1011 J/m2

For Dp = 100 µm=0.1 mm, Eq. (10.29) onp. 644 gives the cure depth as

Cd = Dp ln(Eo

Ec

)= (0.1 mm) ln

(2.04× 1011

6.36× 1010

)= 0.116 mm

Therefore, the linewidth is given by Eq. (10.30)as

Lw = B

√Cd

2Dp= (0.25 mm)

√0.1162(0.1)

or Lw = 0.19 mm. Note that this is smallerthan the laser diameter of 0.25 mm.

10.93 For the stereolithography system described inProblem 10.92, estimate the time required tocure a layer defined by a 40-mm circle if adja-cent lines overlap each other by 10% and thepower available is 10 MW.

The following solution uses the results from thesolution to Problem 10.92. If the linewidth is0.19 mm, the allowable linewidth to incorporatea 10% overlap is 0.171 mm. A 40-mm diameter

circle has an area of 1257 mm2; thus, a laserwould have to travel a total distance of 7.35 musing the 0.171 mm linewidth. Since the re-quired energy is 10 kJ for a length of 0.25 mmand the power available is 10 MW, the spot ve-locity can be obtained from

10 MW =10 kJ

0.25 mmv

Solving for the velocity yields v = 0.25 m/s.Therefore, the laser will take 7.35/0.25=29.4 sto cure the circle.

10.94 The extruder head in a fused-deposition-modeling setup has a diameter of 1 mm (0.04in.) and produces layers that are 0.25 mm (0.01in.) thick. If the velocities of the extruder headand polymer extrudate are both 50 mm/s, esti-mate the production time for generating a 50-mm (2-in.) solid cube. Assume that there is a15-s delay between layers as the extruder headis moved over a wire brush for cleaning.

Note that although the calculations are givenbelow, in practice, the rapid-prototyping soft-ware can easily make such a calculation. Sincethe thickness of the cube is 50 mm and the lay-ers are 0.25 mm thick, there are 200 layers, for atotal inactive’ time of (200)(15 s)=3000 s. Notealso that the cross section of the extruded fila-ment in this case is highly elliptical, and thus itsshape is not easily determined from the infor-mation given in the problem statement. How-ever, the polymer extrudate speed is given as 50mm/s and the orifice diameter is 1 mm, hencethe volume flow rate is

Q = vA = (50 mm/s)[π4

(1 mm)2]

= 39.27 mm3/s

The cube has a volume of (50)(50)(50)=125,000mm3 and the time required to extrude this vol-ume is 125,000/39.27=3180 s. Hence, the totalproduction time is 3180 s + 3000 s = 6180 s= 1.7 hrs. Note that this estimate does notinclude any porosity, and it assumes that ex-trusion is continuous. In practice, however, theextruder has to periodically pick up and moveto a new location in a layer.

10.95 Using the data for Problem 10.94 and assum-ing that the porosity of the support material is

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50%, calculate the production rate for makinga 100-mm (4-in.) high cup with an outside di-ameter of 88 mm (3.5 in.) and wall thickness of6 mm (0.25 in.). Consider both the case withthe closed-end (a) down and (b) up.

(a) (a) Closed-end down. No support mate-rial is needed. There are 400 layers, sothe inactive’ time is 6000 s. The cup wallvolume is

V =π

4d2t+ πdht

=π

4(88 mm)2(6 mm)

+π(88 mm)(100 mm)(6 mm)

or V = 202, 000 mm3. It takes202, 000/39.27 = 5140 s to extrude; thetotal time is 6000+5140 = 11, 140 s = 3.1hrs.

(b) (b) Closed-end up. In addition to the wall,the interior must now be filled with sup-port for the closed-end on top. The vol-ume of the cup is

V =π

4d2h

=π

4(88 mm)2(100 mm)

= 608, 000 mm3

Since the support material has a porosityof 50%, the time required to extrude thesupport material is t = 304, 000/39.27 =7740 s = 2.2 hrs. Therefore, the total timefor producing the part and the support is3.1 + 2.2 = 5.3 hrs.

10.96 What would the answer to Example 10.5 beif the nylon has a power law viscosity withn = 0.5? What if n = 0.2?

Since the nylon has a power law viscosity, thenEq. (10.24) on p. 621 has to be used for theextruder characteristic, instead of Eq. (10.23).The extruder characteristic is thus given by

Q =(

4 + n

10

)(π2HD2N sin θ cos θ

)−pπDH

3 sin2 θ

(1 + 2n)4η

Q =(

4 + n

10


)−pπDH

3 sin2 θ

(1 + 2n)4η

= π2

(4 + 0.5

10

)×[(0.007)(0.05)2(0.833) sin 20 cos 20

]−π(0.05)(0.007)3 sin2 20

[1 + 2(0.5)](4)(300)p

= 2.08× 10−5 −(2.62× 10−12

)p

If the same die characteristic can be used, thenthere are two equations and two unknowns.This results in p = 4.01 MPa and Q = 1.03 ×10−5 m3/s.Using the more realistic value of n = 0.2, theextruder characteristics becomes

Q =(

4 + n

10


)−pπDH

3 sin2 θ

(1− 2n)4η

= π2

(4 + 0.2

10

)×[(0.007)(0.05)2(0.833) sin 20 cos 20

]−π(0.05)(0.007)3 sin2 20

[1 + 2(0.2)](4)(300)p

= 1.94× 10−5 −(3.75× 10−12

)p

If the same die characteristic can be used, thereare two equations and two unknowns. This re-sults in p = 3.07 MPa and Q = 7.87 × 10−6

m3/s.

10.97 Referring to Fig. 10.7, plot the relaxation curves(i.e., the stress as a function of time) if a unitstrain is applied at time t = to.

Consider first the simple spring and dashpotmodels shown in parts (a) and (b) of the fig-ure. If a unit strain is applied to a spring, theforce developed is F = k, where k is the stiff-ness of the spring. This force will be maintainedand will not change as long as the deformationis maintained. For the dashpot model, a unitchange in strain causes an infinite force, but theforce quickly drops to zero as the strain is main-tained, because the strain rate is zero. Thus,the relaxation curve for a spring is a constant,

162






and for the dashpot, it is a multiple of the Diracdelta function.

For the Maxwell model, when the unit strain isapplied, the spring immediately stretches sincethe dashpot has a high resistance to deforma-tion. As the deformation is held, the load istransferred from the spring. The relaxationfunction is given by

σ(t) = ke−(k/η)t

where k is the spring stiffness and η is the co-efficient of viscosity for the dashpot.

For the Voigt model, the application of a unitstrain causes the dashpot to develop infiniteforce. After t = 0, the strain rate is zeroand the dashpot develops no force, so that theforce is that generated by the spring under aunit strain. The relaxation curve for the Voigtmodel is given by

σ(t) = ηδ(t) + k

where δ is the Dirac delta function. Thesecurves are plotted below:

Stress

Timet=0

Stress

Timet=0

10.98 Derive a general expression for the coefficientof thermal expansion for a continuous fiber-reinforced composite in the fiber direction.

Note that, in this case, a temperature rise leadsto a thermal expansion of the composite, so thatits deflection can be written as

δc = αc∆tl

where α is the coefficient of thermal expansionand a c subscript indicates a property of thecomposite. For the fiber and matrix, there willbe an internal stress developed, unless the co-efficients of thermal expansion are the same forthe fiber and the matrix. If not, then an inter-nal stress is developed in order to ensure thatthe fiber and matrix undergo the same deforma-tion as the composite. Assume that the fiber

has a higher coefficient of thermal expansionthan the matrix, so that the fibers are com-pressed by the internal stress and the matrix isloaded in tension. Therefore, the deformationof the fibers is given by

δf = αf∆tl − σf

Efl

and for the matrix:

δm = αm∆tl +σm

Eml

Since the deformations have to be equal, wehave

αf∆tl − σf

Efl = αm∆tl +

σm

Eml

or(αf − αm)∆t =

σf

Ef+σm

Em

Note that the internal forces must balance eachother, so that

σfAf = −σmAm

where the minus sign indicates that the fibersare loaded in compression and the matrix intension (or vice-versa). Thus,

σm =x

1− xσf

Substituting, we have

(αf − αm)∆t = σf

[1Ef

+x

(1− x)Em

]Solving for σf ,

σf =(αf − αm)∆t[1

Ef+ x

(1−x)Em

]Therefore, the fiber deformation becomes

δf = αf∆tl − (αf − αm)∆t

Ef

[1

Ef+ x

(1−x)Em

] l=

αf −(αf − αm)

Ef

[1

Ef+ x

(1−x)Em

]∆tl

Since this is the same as the deformation of thecomposite,

αc∆tl =

αf −(αf − αm)[

1 + x(1−x)

Ef

Em

]∆tl

163






or

αc = αf −(αf − αm)[

1 + x(1−x)

Ef

Em

]10.99 Estimate the number of molecules in a typical

automobile tire. Estimate the number of atoms.

An automobile tire is an example of a highlycross-linked network structure (see Fig. 10.3).In theory, a networked structure can continueindefinitely, so that an automobile tire couldbe considered as one giant molecule. In reality,there are probably a few thousand molecules.

Although tires come in a wide variety of sizes,consider a tire with 10 kg of rubber, producedfrom polybutadiene, (C4H6)n. (Note that thereis additional weight associated with the rein-forcement and pigment in the tire.) The atomicweight of carbon is 12.011 and that of hydro-gen is 1.0079, as obtained from a periodic ta-ble of elements. Thus, a polybutadiene merhas a molecular weight of 54.09. Therefore,a mole of such mers (or 10 moles of atoms)would weigh 54.09 grams. In a 20-kg tire,there are 20000/54.09 = 370 moles of butadi-ene mers, or 3700 moles of atoms. Since 1 mole= 6.023 × 1023, there are 2.23 × 1027 atoms ina tire.

10.100 Calculate the elastic modulus and percentageof load supported by fibers in a composite withan epoxy matrix (E = 10 GPa), with 20%fibers made of (a) high-modulus carbon and (b)Kevlar 29.

From Table 10.4 on p. 609, for high-moduluscarbon, E = 415 GPa and for Kevlar 29,E = 62 GPa. For x = 0.2, Eq. (10.16) on p. 617gives, for the high-modulus carbon,

Ec = xEf + (1− x)Em

= (0.2)(415 GPa) + (1− 0.2)(10 GPa)= 91 GPa

The same calculation for Kevlar 29 gives Ec =20.4 GPa. Using Eq. (10.15),

Ff

Fm=

AfEf

AmEm=

xAEf

(1− x)AEm=

xEf

(1− x)Em

So that Fm is:

Fm =[(1− x)Em

xEf

]Ff

Substituting into Eq. (10.12) yields

Fc = Ff + Fm = Ff +[(1− x)Em

xEf

]Ff

or

Fc =[1 +

(1− x)Em

xEf

]Ff

For the high-modulus carbon reinforced epoxycomposite,

Fc =[1 +

(1− 0.2)(10)(0.2)(415)

]Ff = 1.10Ff

or Ff = 0.91Fc. For the Kevlar fiber-reinforcedcomposite,

Fc =[1 +

(1− 0.2)(10)(0.2)(62)

]Ff = 1.65Ff

or Ff = 0.61Fc.

10.101 Calculate the stress in the fibers and in the ma-trix for Problem 10.100. Assume that the cross-sectional area is 50 mm2 and Fc = 2000 N.

Using the results for Problem 10.100, we note:

(a) For the high-modulus carbon fibers,

Af = 0.2Ac = 0.2(50 mm2) = 10 mm2

Ff = 0.91Fc = 0.91(2000 N) = 1820 N

Therefore,

σf =1820 N10 mm2

= 182MPa

Similarly, Am = 40 mm2, Fm = 180 N,and σm = 4.5 MPa.

(b) For the Kevlar 29 fibers,

Af = 0.2Ac = 0.2(50 mm2) = 10 mm2

Ff = 0.61Fc = 0.61(2000 N) = 1220 N

Therefore,

σf =1220 N10 mm2

= 122MPa

and for the matrix, Am = 40 mm2, Fm =780 N, and σm = 19.5 MPa.

164






10.102 Consider a composite consisting of reinforcingfibers with Ef = 300 GPa. If the allowable fiberstress is 200 MPa and the matrix strength is 50MPa, what should be the matrix stiffness sothat the fibers and matrix fail simultaneously?

From Eq. (10.15),

Ff

Fm=

x

1− x

Ef

Em=

x

1− x

300Em

Since F = σA,

Ff

Fm=

σfAf

σmAm=

200xA50(1− x)A

=x

1− x

300Em

or20050

=300Em

which is solved as Em = 75MPa.

10.103 Assume that you are asked to give a quiz to stu-dents on the contents of this chapter. Preparefive quantitative problems and five qualitativequestions, and supply the answers.

By the student. This is a challenging ques-tion that requires considerable focus and un-derstanding on the part of the students, andhas been found to be a very valuable homeworkproblem.

DESIGN

10.104 Make a survey of the recent technical litera-ture and present data indicating the effects offiber length on such mechanical properties asthe strength, elastic modulus, and impact en-ergy of reinforced plastics.

By the student.

10.105 Discuss the design considerations involved in re-placing a metal beverage container with a con-tainer made of plastic.

By the student. See also Question 10.40 whichpertains to manufacturing considerations ofbeverage cans. This is an open-ended prob-lem that can involve a wide variety of topics.Some of the major concerns are as follows. Notethat the beverage can must be non-toxic andshould have sufficient strength resist rupturingunder internal pressure (which typically is onthe order of about 120 psi) or buckling undera compressive load during stacking in stores.The can should maintain its properties, fromlow temperatures in the refrigerator to hot sum-mer temperatures outside, especially under thesun in hot climates. Particularly important isthe gas permeability of plastic containers whichwill significantly reduce their shelf life. Notehow soft drinks begin to lose their carbonation

in unopened plastic bottles after a certain pe-riod of time. Other important considerationsare chilling characteristics, labeling, feel, aes-thetics, and ease of opening.

10.106 Using specific examples, discuss the design is-sues involved in various products made of plas-tics versus reinforced plastics.

By the student. Reinforced plastics are supe-rior to conventional plastics in terms of strengthand strength- and stiffness-to-weight ratios, butnot cost (see also Table 10.1 on p. 585. Conse-quently, their use is more common for criticalapplications. For example, the bucket support-ing power-line service personnel is made of rein-forced fiber, as are ladders and pressurized gasstorage tanks (for oxygen, nitrogen, etc.) onthe Space Shuttle.

10.107 Make a list of products, parts, or componentsthat are not currently made of plastics, and of-fer reasons why they are not.

By the student. Consider, as examples, the fol-lowing:

• Some products, such as machine guards orautomobile fenders, give an impression ofrobustness if made of a metal but not ifmade of a plastic.

165






• Plastics are generally not suitable for hightemperature applications, such as automo-bile pistons, cookware, or turbine blades.

• Plastics are too compliant (flexible) forapplications for machine elements suchas ball bearings, cams, or highly-loadedgears.

10.108 In order to use a steel or aluminum container tohold an acidic material, such as tomato juice orsauce, the inside of the container is coated witha polymeric barrier. Describe the methods ofproducing such a container. (See also Chapter7.)

By the student. The students are encouragedto section various cans and inspect their innersurfaces. The most common method is to (a)dissolve a thermosetting polymer in a chemicalliquid carrier, usually a ketone, (b) spraying itonto the interior of the can, and (c) boiling off,leaving an adherent polymer coating. A lesscommon approach is to laminate or coat the in-side surface of the sheet stock with a metallicmaterials.

10.109 Using the information given in this chapter, de-velop special designs and shapes for possiblenew applications of composite materials.

By the student. This is a challenging topicand suitable for a technical paper. Consider,for example, the following two possibilities: (a)A tough transparent polymer, such as poly-carbonate, that is reinforced with glass fibers.Strength will increase but transparency will bereduced. (b) A ceramic-matrix composite re-inforced with copper, thus help diminish ther-mal cracking of the ceramic. If continuouslydispersed throughout the matrix, the copperwould conduct heat evenly throughout the ma-trix and thus reduce the thermal gradients inthe composite. However, the composites op-erating temperature should be below the melt-ing point of copper, even though ceramics resisthigh temperatures.

10.110 Would a composite material with a strong andstiff matrix and soft and flexible reinforcementhave any practical uses? Explain.

By the student. This type of composite proba-bly will have a higher toughness than the ma-

trix alone, since the soft and flexible reinforce-ment material would blunt a propagating crack.However, such a composites usefulness will de-pend on whether or not it has a combination ofhigher strength and toughness than a compos-ite with a ductile matrix and a strong reinforce-ment.

10.111 Make a list of products for which the use of com-posite materials could be advantageous becauseof their anisotropic properties.

By the student. Consider the following ex-amples: cables, packaging tape, pressure ves-sels and tubing, tires (steel-belted radials), andsports equipment.

10.112 Name several product designs in which bothspecific strength and specific stiffness are im-portant.

Specific strength and specific stiffness are im-portant in applications where the materialshould be light and possess good strength andstiffness. A few specific applications are struc-tural airplane components, helicopter blades,and automobile body panels.

10.113 Describe designs and applications in whichstrength in the thickness direction of a com-posite is important.

By the student. The thickness direction is im-portant, for example, in thick-walled pressurevessels, with application for high-pressure ser-vice of hydraulic fluids as well as for residen-tial water service. Radial reinforcement can bedone with discontinuous fibers, provided theyare oriented in the optimum direction. The stu-dents are encouraged to search the literature toprovide various other examples.

10.114 Design and describe a test method to determinethe mechanical properties of reinforced plasticsin their thickness direction.

By the student. This is a challenging prob-lem, and a literature search will be useful asa guide to developing appropriate techniques.The mechanical properties in the thickness di-rection are difficult to measure because of thesmall thickness as compared with the surface

166






area of a specimen. Note, however, that ten-sion tests in the thickness direction can be car-ried out by adhesively bonding both surfaceswith metal plates, and then pulling the platesapart by some suitable means. A feasible andindirect approach may be to derive the prop-erties in the thickness direction by performingtests in the other two principal directions, andthen applying a failure criterion, as describedin applied mechanics texts.

10.115 We have seen that reinforced plastics can be ad-versely affected by environmental factors, suchas moisture, chemicals, and temperature varia-tions. Design and describe test methods to de-termine the mechanical properties of compositematerials under these conditions.

By the student. This is an important and chal-lenging topic. Note that simple experiments,such as tension tests, are suitable when con-ducted in a controlled environment. Chambersare commonly installed around test specimensfor such environmentally-controlled testing.

10.116 As with other materials, the mechanical prop-erties of composites are obtained by preparingappropriate specimens and testing them. Ex-plain what problems you might encounter inpreparing specimens for testing and in the ac-tual testing process itself.

By the student. Testing composite materialscan be challenging because of anisotropic be-havior, with significant warping possible, aswell as difficulties involved in preparing ap-propriate specimens and clamping them in thetest equipment. Other approaches would mea-sure deformation in more than one direction (asopposed to conventional tests where generallyonly the longitudinal strain is measured). Tra-ditional dogbone specimens can be used.

10.117 Add a column to Table 10.1, describing the ap-pearance of these plastics, including availablecolors and opaqueness.

By the student. Note that most thermoplasticscan be made opaque, but only a few (such asacrylics and polycarbonates) are transparent.Most plastics are available in a variety of colors,such as polyethylene and ABS. Thermosets are

opaque and come in very limited colors, such asblack or brown.

10.118 It is possible to weave fibers in three dimen-sions, and to impregnate the weave with a cur-able resin. Describe the property differencesthat such materials would have compared tolaminated composite materials.

By the student. This is a challenging topic,requiring literature search. An example of anorthogonal three-dimensional weave is shown inthe accompanying figure, to give a perspectiveto the items listed below.

In general, the following comments can be maderegarding three-dimensional weaves as com-pared to laminate composites:

• The through-thickness properties can betailored for a particular application andcan be superior for 3D-weaves.

• 3D woven composites have a higher delam-ination resistance and impact damage tol-erance than 2D laminated composites.

• Different materials can be blended into afiber prior to weaving. Indeed, most cloth-ing involves blends of polymers or of poly-mers and natural fibers such as cotton orlinen.

• The size of the weave can be varied moreeasily to allow for changes in the structureof such a material.

• 3D woven composites are more difficultand expensive to manufacture than 2Dcomposites produced from laminated ma-terials.

• 3D woven composites have lower mechani-cal properties than laminated composites.

167






10.119 Conduct a survey of various sports equipmentand identify the components that are made ofcomposite materials. Explain the reason forand the advantages of using composites in theseapplications.

By the student. Consider, for example, the fol-lowing:

• Tennis and racquetball racquets made offiber reinforced composites. The mainreasons are to reduce weight, improvethe stiffness-to-weight ratio, and increasedamping.

• Other examples with similar desired char-acteristics are softball bats, golf-clubshafts, skis and ski bindings, and hockeyand jai-alai sticks.

10.120 Instead of a having a constant cross section, itmay be possible to make fibers or whiskers witha varying cross section or a fiber with a wavysurface. What advantages would such fibershave?

By the student. Perhaps the most compellingreason for this approach is associated with therelatively poor adhesive bond that may developbetween fibers and the matrix in a compos-ite material. In discontinuous-fiber-reinforcedcomposites, especially, failure is associated withpull-out of the fiber (see Fig. 10.20). With awavy fiber, there is mechanical interlocking be-tween the matrix and fiber. Note, for example,steel bars for reinforced concrete with texturedsurfaces for better interfacial strength betweenthe bar and the concrete.

10.121 Polymers (either plain or reinforced) can be asuitable material for dies in sheet-metal form-ing operations described in Chapter 7. Describeyour thoughts, considering die geometry andany other factors that may be relevant.

By the student. See also p. 397. Recall thatthis is already a practice in operations such asrubber-pad forming and hydroforming (Section7.5.3). The polymers must have sufficient rigid-ity, strength, and wear resistance. Consideringthese desirable characteristics, the use of plasticdies is likely to be appropriate and economicalfor relatively short production runs, and light

forming forces. The main reason that poly-mer tooling has become of greater interest isthe availability of rapid prototyping technolo-gies that are capable of producing such toolsand die inserts with low cost and lead times.

10.122 For ease of sorting for recycling, all plastic prod-ucts are now identified with a triangular symbolwith a single-digit number at its center and twoor more letters under it. Explain what thesenumbers indicate and why they are used.

This information can be summarized as (seealso top of p. 607):

1 Polyethylene2 High-density polyethylene3 Vinyl4 Low-density polyethylene5 Polypropylene6 Polystyrene7 Other

10.123 Obtain different kinds of toothpaste tubes,carefully cut them across with a sharp razorblade, and comment on your observations re-garding the type of materials used and how thetube could be produced.

By the student. This is a topic suitable forsome research. It will be noted that some col-lapsible tubes are blow molded, others are in-jection molded at one end and the other end issealed by hot-tool welding (see Section 12.16.1).Another design is injection-molded rigid tubingwhere the toothpaste is pumped out during use.Note also that some collapsible tubes have wallsthat consist of multilayers of different materialsand sealed on the closed end.

10.124 Design a machine that uses rapid-prototypingtechnologies to produce ice sculptures. De-scribe its basic features, commenting on the ef-fect of size and shape complexity on your de-sign.

By the student. Consider the following sugges-tions:

• A machine based on the principles ofballistic particle manufacturing (such asthree-dimensional printing) to spray small

168






droplets of water onto a frozen base, pro-ducing a sculpture incrementally, layer bylayer.

• Sheets of ice can be produced and thencut with a heat source, such as a laser, toproduce different shapes. The individualpieces would then be bonded, such as witha thin layer of water which then freezes,thus producing a sculpture.

• Layers of shaved ice can be sprayed, us-ing a water jet, under controlled conditions(similar to three-dimensional printing).

Note that in all these processes, the outersurfaces of the sculpture will have to besmoothened for a better surface finish. Thiscan be done, for instance, using a heat source(just as it is done in rounding the sharp edgesof cut glass plates using a flame).

10.125 A manufacturing technique is being proposedthat uses a variation of fused-deposition model-ing, where there are two polymer filaments that

are melted and mixed before being extruded toproduce the workpiece. What advantages doesthis method have?

There are several advantages to this approach,including:

• If the polymers have different colors,blending the polymers can produce a partwith a built-in color scheme.

• If the polymers have different mechanicalproperties, then functionally-graded mate-rials can be produced, that is, materialswith a designed blend of mechanical prop-erties.

• Higher production rates and better work-piece properties may be achieved.

• If the second polymer can be leached,it can be developed into a technique forproducing porous polymers or ship-in-the-bottle type parts.

169






170






Chapter 11

Properties and Processing of MetalPowders, Ceramics, Glasses,Composites, and Superconductors

Questions

Powder metallurgy

11.1 Explain the advantages of blending differentmetal powders.

Metal powders are blended for the following ba-sic reasons:

(a) Powders can be mixed to obtain specialphysical, mechanical, and chemical char-acteristics.

(b) Lubricants and binders can be mixed withmetal powders.

(c) A uniform blend can impart better com-paction properties and shorter sinteringtimes.

11.2 Is green strength important in powder-metalprocessing? Explain.

Green strength is very important in powder-metal processing. When a P/M part has beenejected from the compaction die, it must havesufficient strength to prevent damage and frac-ture prior to sintering.

11.3 Give the reasons that injection molding of metalpowders has become an important process.

Powder-injection molding has become an im-portant process because of its versatility andeconomics. Complex shapes can be obtained athigh production rates using powder metals thatare blended with a polymer or wax. Also, theparts can be produced with high density to netor near-net shape.

11.4 Describe the events that occur during sintering.

In sintering, a green P/M part is heated to atemperature of 70-90% of the lowest meltingpoint in the blend. At these temperatures, twomechanisms of diffusion dominate: direct dif-fusion along an existing interface, and, moreimportantly, vapor-phase material transfer toconvergent geometries. The result is that theparticles that were loosely bonded become in-tegrated into a strong but porous media.

11.5 What is mechanical alloying, and what are itsadvantages over conventional alloying of met-als?

In mechanical alloying, a desired blend ofmetal powders is placed into a ball mill (seeFig. 11.26b). The powders weld together whentrapped between two or more impacting balls,and eventually are mechanically bonded andalloyed because of large plastic deformations

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they undergo. The main advantage of mechan-ical alloying is that the particles achieve a highhardness due to the large amount of cold work,and alloys which otherwise cannot be obtainedthrough solidification can be achieved.

11.6 It is possible to infiltrate P/M parts with vari-ous resins, as well as with metals. What possi-ble benefits would result from infiltration? Givesome examples.

The main benefits to infiltration of a metal P/Mpart with another metal or polymer resin are:

(a) There can be a significant increase instrength;

(b) the infiltration can protect the P/M partfrom corrosion in certain environments;

(c) the polymer resin can act as a solid lubri-cant;

(d) the infiltrated part will have a higher den-sity and mass in applications where this isdesired.

11.7 What concerns would you have when electro-plating P/M parts?

By the student. There are several concerns inelectroplating (pp. 159-160) P/M parts, includ-ing:

(a) electroplating solutions are toxic and dan-gerous;

(b) it may be difficult to remove the residueliquid from inside P/M parts;

(c) it will be very difficult to perform plat-ing in the interior of the part, as there islow current density. Thus, only the sur-face will be plated and it will be difficultto obtain a uniform surface finish.

11.8 Describe the effects of different shapes and sizesof metal powders in P/M processing, comment-ing on the magnitude of the shape factor of theparticles.

The shape, size, size distribution, porosity,chemical purity, and bulk and surface charac-teristics of metal particles are all important. Asexpected, they have significant effects on per-meability and flow characteristics during com-paction in molds, and in subsequent sintering

operations (Sections 11.2.20 and 11.3). It isbeneficial to have angular shapes with approxi-mately equally-sized particles to aid in bonding.

11.9 Comment on the shapes of the curves and theirrelative positions shown in Fig. 11.6.

At low compaction pressures, the density ofP/M parts is low and at high compactingpressures, it approaches the theoretical density(that of the bulk material). Note that the con-cavity of the curves in Fig. 11.6a is downward,because in order to increase the density, smallerand smaller voids must be closed. Clearly, it iseasier to shrink larger cavities in the materialthan smaller ones. Note that there is a mini-mum density at zero pressure. The results inFig. 11.6b are to be expected because as den-sity increases, there is less porosity and thusgreater actual area in a cross-section; this leadsto higher strength and electrical conductivity.The reason why elongation also increases withdensity is because of the lower number of poroussites that would reduce ductility (see Section3.8.1).

11.10 Should green compacts be brought up to thesintering temperature slowly or rapidly? Ex-plain.

Note that rapid heating can cause excessivethermal stresses in the part being sintered andcan lead to distortion or cracking; on the otherhand, it reduces cycle times and thus improveproductivity. Slow heating has the advantageof allowing heating and diffusion to occur moreuniformly.

11.11 Explain the effects of using fine vs. coarse pow-ders in making P/M parts.

Coarse powders will have larger voids for thesame compaction ratios, an analogy of which isthe voids between marbles or tennis balls in abox (see also Fig. 3.2). The larger voids resultin lower density, strength, stiffness, and ther-mal and electrical conductivity of P/M parts.The shape, size and distribution of particles,porosity, chemical purity, and bulk and surfacecharacteristics are also important because theyhave significant effects on permeability and flowcharacteristics during compaction and in subse-quent sintering operations.

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11.12 Are the requirements for punch and die mate-rials in powder metallurgy different than thosefor forging and extrusion, described in Chapter6? Explain.

In forging, extrusion, and P/M compaction,abrasive resistance is a major consideration indie and punch material selection. For that rea-son, the dies on these operations utilize simi-lar and sometimes identical materials (see Ta-ble 3.6 on p. 114). Processes such as isostaticpressing utilize flexible molds, which generallyis not used in forging and extrusion.

11.13 Describe the relative advantages and limita-tions of cold and hot isostatic pressing, respec-tively.

Cold isostatic pressing (CIP) and hot isostaticpressing (HIP) both have the advantages of pro-ducing compacts with effectively uniform den-sity (Section 11.3.3). Shapes can be made withuniform strength and toughness. The main ad-vantage of HIP is its ability to produce com-pacts with essentially 100% density, good met-allurgical bonding, and good mechanical prop-erties. However, the process is relatively expen-sive and is, therefore, used mainly for compo-nents in the aerospace industry or in makingspecial parts.

11.14 Why do mechanical and physical properties de-pend on the density of P/M parts? Explain.

The mechanical properties depend on densityfor a number of reasons. Not only is there lessmaterial in a given volume for less dense P/Mparts, hence lower strength, but voids are stressconcentrations. Thus, the less dense materialwill have more and larger voids. The modulusof elasticity decreases with increasing voids be-cause there is less material across a cross sectionand hence elongation is greater under the sameload, as compared to a fully dense part. Phys-ical properties such as electrical and thermalconductivity are also affected adversely becausethe less dense the P/M part is, the less materialis available to conduct electricity or heat.

11.15 What type of press is required to compact pow-ders by the set of punches shown in Fig. 11.7d?(See also Chapters 6 and 7.)

The operation shown in Fig. 11.7d would re-quire a double-action press, so that independentmovements of the two punches can be obtained.This is usually accomplished with a mechanicalpress.

11.16 Explain the difference between impregnationand infiltration. Give some applications foreach.

The main difference between impregnation andinfiltration is the media (see Section 11.5). Inimpregnation, the P/M part is immersed in aliquid, usually a lubricant, at elevated temper-atures. The liquid is drawn into the P/M partby surface tension and fills the voids in theporous structure of the part. The lubricantalso lowers the friction and prevents wear ofthe part in actual use. In infiltration, a lower-melting-point metal is drawn into the P/M partthrough capillary action. This is mainly done toprevent corrosion, although low-melting-pointmetals could be used for frictional considera-tions in demanding environments.

11.17 Explain the advantages of making tool steels byP/M techniques over traditional methods, suchas casting and subsequent metalworking tech-niques.

From a cost standpoint, there may not be amajor advantage because P/M itself requiresspecial tooling to produce the part. However,some tool steels are very difficult to machineto desired shapes. Thus, by producing a P/Mtooling, the machining difficulties are greatlyreduced. P/M also allows the blending of com-ponents appropriate for cutting tools.

11.18 Why do compacting pressure and sintering tem-perature depend on the type of powder metalused? Explain.

Different materials require different sinteringtemperatures basically because they have dif-ferent melting points. To develop good strengthbetween particles, the material must be raisedto a high enough temperature where diffusionand second-phase transport mechanisms canbecome active, which is typically around 90%of the material melting temperature on an ab-solute scale. As for the compacting pressure, itwill depend on the type of metal powder such

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as its strength and ductility, the shape of theparticles, and the interfacial frictional charac-teristics between the particles.

11.19 Name various methods of powder productionand describe the morphology of powders pro-duced.

By the student. Refer to Fig. 11.2. Briefly:

• Atomization: spherical (for gas atomized)or rounded (for water atomized).

• Reduction: spongy, porous, spherical orirregular

• Electrolytic deposition: dendritic

• Carbonyls: dense, spherical

• Comminution: irregular, flaky, angular

• Mechanical alloying: flaky, angular

11.20 Are there any hazards involved in P/M process-ing? If any, what are their causes?

There are several hazards in P/M processing;the major one is that powder metals can beexplosive (particularly aluminum, magnesium,titanium, zirconium, and thorium). Thus,dust, sparks, and heat from friction should beavoided. In pressing, there are general concernsassociated with closing dies, where a finger maybe caught.

11.21 What is screening of metal powders? Why is itdone?

In screening (Section 11.2.2), the metal pow-ders are placed in a container with a numberof screens; the top is coarsest, and the mesh isincreasingly fine towards the bottom of the con-tainer. As the container is vibrated, the parti-cles fall through the screens until their openingsize is smaller than the particle diameter. Thus,screening separates the particles into ranges orsizes. This is done in order to have good controlof particle size.

11.22 Why is there density variations in compactedmetal powders? How is it reduced?

The main reason for density variation in com-pacting of powders is associated with mechan-ical locking and friction among the particlesand the container walls. This leads to varia-tions in pressure depending on distance from

the punch and from the container walls (seeFig. 11.7). The variation can be reduced byusing double-acting presses, lowering the fric-tional resistance of the punch and die sur-faces, or by adding lubricants that reduce inter-particle friction among the powders.

11.23 It has been stated that P/M can be competitivewith processes such as casting and forging. Ex-plain why this is so, commenting on technicaland economic advantages.

By the student. Refer to Section 11.7. As an ex-ample, consider MIM which is commonly usedwith metals with high melting temperatures.This process requires fine metal powder that ismixed with a polymer and injection molded; thematerial costs are high. On the other hand, theapplications for magnesium and aluminum diecastings are in large volumes (camera frames,fittings, small toys) are economical and not aswell-suited for MIM.

11.24 Selective laser sintering was described in Sec-tion 10.12.4 as a rapid prototyping technique.What similarities does this process have withthe processes described in this chapter?

By the student. Recall that selective laser sin-tering uses the phenomena described in Section11.4 and illustrated in Fig. 11.14. However, thehigh temperatures required to drive the mate-rial transfer is obtained from a laser and not byheating in a furnace as in P/M. Selective lasersintering also has significant part shrinkage.

11.25 Prepare an illustration similar to Fig. 6.28,showing the variety of P/M manufacturing op-tions.

By the student.

Ceramics and other materials

11.26 Describe the major differences between ceram-ics, metals, thermoplastics, and thermosets.

By the student. This broad question will re-quire extensive answers that can be tabulatedby the student. Note, for example, that thechemistries are very different: ceramics arecombinations of metals and non-metals, andplastics and thermosets involve repeating mers,usually based on long chains. Mechanically,the stress-strain behavior is very different as

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well; metals are linearly elastic and generallyhave high ductility and lower strain-hardeningcoefficients than thermoplastics. Ceramics arelinearly elastic and brittle; thermoplastics flowabove a critical temperature, while thermosetsare elastic and brittle. Comparisons could alsobe made regarding various other mechanical,physical, and chemical properties, as well astheir numerous applications.

11.27 Explain why ceramics are weaker in tensionthan in compression.

Ceramics are very sensitive to cracks, impuri-ties, and porosity, and thus generally have lowtensile strength and toughness (see, for exam-ple, Table 8.6 on p. 454). In compression, how-ever, the flaws in the material do not causethe stress concentrations as they do in tension,hence compressive strength is high. (See alsoSection 3.8.)

11.28 Why do the mechanical and physical propertiesof ceramics decrease with increasing porosity?Explain.

Porosity can be considered microscopic airpockets in the ceramic. Thus, porosity will al-ways decrease the strength of the ceramic be-cause of the smaller cross-sectional area thathas to support the external load. The holes inthe material also act as stress concentrations tofurther lower the strength. The porosity alsoacts as crack initiation sites, thus decreasingtoughness. Physical properties are affected like-wise, in that pores in the ceramic are typicallyfilled with air, which has much lower thermaland no electrical conductivity as compared withceramics.

11.29 What engineering applications could benefitfrom the fact that, unlike metals, ceramics gen-erally maintain their modulus of elasticity atelevated temperatures?

By the student. Consider, for example, that byretaining their high stiffness at elevated tem-peratures (see, for example, Fig. 11.24), dimen-sional accuracy of the parts or of the mechan-ical system can be maintained. Some exam-ples are bearings, cutting tools, turbine blades,machine-tool components, and various high-temperature applications.

11.30 Explain why the mechanical-property datagiven in Table 11.7 have such a broad range.What is the significance of this wide range inengineering applications?

By the student. The mechanical propertiesgiven in Table 11.7 on p. 701 vary greatly be-cause the properties of ceramics depend on thequality of the raw material, porosity, and themanner of producing the parts. Engineeringapplications that require high and reliable me-chanical properties (e.g., aircraft and aerospacecomponents) must assure that the materialsand processing of the part are the best avail-able.

11.31 List the factors that you would consider whenreplacing a metal component with a ceramiccomponent. Give examples of such possiblesubstitutions.

By the student. Review Section 11.8. Consider,for example, the following factors:

• The main drawbacks of ceramics are lowtensile strength and toughness. Hence, theapplication of the metal component to bereplaced should not require high tensilestrength or impact resistance.

• If the ceramic part is subjected to wear,then the performance of the mating mate-rial is important. It could be that a three-body wear (see p. 147) would be intro-duced that could severely affect productlife.

• Ceramics are typically probabilistic ma-terials, that is, there is a wide range ofmechanical properties in ceramic parts,whereas metals are typically determinis-tic and have a smaller distribution ofstrength. Thus, a major concern iswhether or not a material is suitable forthe particular design.

• As with all engineering applications, costis a dominant consideration.

11.32 How are ceramics made tougher? Explain.

Ceramics may be made tougher by using high-purity materials, selecting appropriate pro-cessing techniques, embedding reinforcements,

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modifying surfaces and reducing surface de-fects, and by intentionally producing microc-racks (less than 1 mm in size) in the ceramic toreduce the energy of propagation of an advanc-ing crack tip. Another important technique isdoping (see pp. 159 and 605), resulting in two ormore phases, as in partially stabilized zirconia(PSZ) and transformation toughened zirconia(TTZ).

11.33 Describe situations and applications in whichstatic fatigue can be important.

Static fatigue (see top of p. 702) occurs undera constant load and in environments where wa-ter vapor is present. Applications such as load-bearing members and sewer piping are suscep-tible to static fatigue if a tensile stress is de-veloped in the pipe by bending or torsion. Thestudent is encouraged to describe other appli-cations.

11.34 Explain the difficulties involved in making largeceramic components. What recommendationswould you make to overcome these difficulties?

By the student. Large components are diffi-cult to make from ceramics, mainly becausethe ceramic must be fired to fuse the con-stituent particles. Firing leads to shrinkageof the part, resulting in significant warpage orresidual stresses. With large parts, these fac-tors become even greater, so that it is very dif-ficult to produce reliable large ceramic parts.Such parts may be made by reinforcing thestructure, or by producing the structure fromcomponents with a ceramic coating or from as-sembled ceramic components.

11.35 Explain why ceramics are effective cutting-toolmaterials. Would ceramics also be suitable asdie materials for metal forming? Explain.

There are many reasons, based on the top-ics covered Chapters 6 through 8. Ceramicsare very effective cutting materials, based espe-cially on their hot hardness (see Table 8.6 onp. 454 and Figs. 8.30 and 8.37), chemical inert-ness, and wear resistance. In ceramic dies forforming, the main difficulties are that (1) ce-ramics are brittle, so any tensile or shear loadwould lead to crack propagation and failure,

and (2) ceramics are generally difficult to ma-chine or form to the desired die shapes with therequired accuracy without additional finishingoperations.

11.36 Describe applications in which the use of a ce-ramic material with a zero coefficient of thermalexpansion would be desirable.

By the student. A ceramic material with anear-zero coefficient of thermal expansion (seeFig. 11.23 and Section 3.9.5) would have a muchlower probability of thermal cracking when ex-posed to extreme temperature gradients, suchas in starting an engine, contacting of two solidsurfaces at widely different temperatures, andtaking a frozen-food container and placing it ina hot oven. This property would thus be use-ful in applications where the ceramic is to besubjected to temperature ranges. Note also theproperties of glass ceramics (Section 11.10.4).

11.37 Give reasons for the development of ceramic-matrix components. Name some present andother possible applications for such large com-ponents.

By the student. Ceramic-matrix componentshave been developed for high-temperature andcorrosive applications where the strength-to-weight ratio of these materials is beneficial. Theapplications of interest include:

• aircraft engine components, such as com-bustors, turbines, compressors, and ex-haust nozzles;

• ground-based and automotive gas turbinecomponents, such as combustors, first andsecond stage turbine vanes and blades, andshrouds;

• engines for missiles and reusable space ve-hicles; and

• industrial applications, such as heat ex-changers, hot gas filters, and radiant burn-ers.

11.38 List the factors that are important in dryingceramic components, and explain why they areimportant.

Refer to Section 11.9.4. Since ceramic slur-ries may contain significant moisture content,

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resulting in 15-20% shrinkage, the removal ofmoisture is a critical concern. Recall that themoisture must be removed in order to fuse theceramic particles. Important factors are: therates at which moisture is removed (which canlead to cracking, if excessive), the initial mois-ture content (the higher it is, the greater thewarpage and residual stress), and the particu-lar material (as some materials will not warp asmuch as others and are more ductile and resis-tant to local defects).

11.39 It has been stated that the higher the coefficientof thermal expansion of glass and the lower itsthermal conductivity, the higher is the level ofresidual stresses developed during processing.Explain why.

Refer to Sections 3.9.4 and 3.9.5. The coeffi-cient of thermal expansion is important in thedevelopment of residual stresses because a giventemperature gradient will result in a higherresidual strain upon cooling. Thermal conduc-tivity is important because the higher the ther-mal conductivity, the more uniform the tem-perature in the glass, and the more uniformthe strains upon cooling. The more uniformthe strains, the less the magnitude of residualstresses developed.

11.40 What types of finishing operations are typicallyperformed on ceramics? Why are they done?

Ceramics are usually finished through abrasivemethods, and they may also be glazed (see Sec-tion 11.9.5). Abrasive machining, such as grind-ing, is done to assure good tolerances and to re-move surface flaws. Recall that tolerances maybe rather poor because of shrinkage. Glazingis done to obtain a nonporous surface, which isimportant for food and beverage applications;it may also be done for decorative purposes.

11.41 What should be the property requirements forthe metal balls used in a ball mill? Explain whythese properties are important.

The metal balls in a ball mill (see Fig. 11.26b)must have very high hardness, strength, wearresistance, and toughness so that they do notdeform or fracture during the milling operation.High stiffness and mass is desirable to maximizethe compaction force (see p. 553).

11.42 Which properties of glasses allow them to beexpanded and shaped into bottles by blowing?Explain.

The properties of glasses which allow them tobe shaped into bottles by blowing is their vis-coplasticity at elevated temperatures and theirhigh strain-rate sensitivity exponent, m. Thusvery large strains can be achieved as comparedto metals. The strains can exceed even the su-perplastic aluminum and titanium alloys (seep. 44).

11.43 What properties should plastic sheet have whenused in laminated glass? Explain.

A plastic sheet used in laminated glass (a) mustobviously be transparent, (b) have a strong, in-timate bond with the glass, and (c) have hightoughness and strain to failure (see Fig. 10.13).The reason for the need for high strain to failureis to prevent shards of glass from being ejected,and thus prevent serious or fatal injuries duringfrontal impact.

11.44 Consider some ceramic products that you arefamiliar with and outline a sequence of pro-cesses performed to manufacture each of them.

By the student. As an example of a sequenceof operations involved, consider the manufac-ture of a coffee cup:

• A ceramic slurry is mixed.

• The slurry is poured into the mold.

• The mold is allowed to rest, allowing thewater in the slurry to be absorbed by themold or to evaporate.

• The mold is opened and the green part iscarefully removed.

• The handle can be a separate piece that isformed and attached at this stage; in somedesigns, the handle is cast integrally withthe cup.

• The cup is then trimmed to remove theflash from the mold.

• It is then decorated and fired; it may beglazed and fired again.

11.45 Explain the difference between physical andchemical tempering of glass.

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By the student. Refer to Section 11.11.2. Notethat in both physical and chemical tempering,compressive stresses are developed on the sur-face of the glass. In physical tempering, this isachieved through rapid cooling of the surface,which is then stressed in compression as thebulk cools. In chemical tempering, the same ef-fect is achieved through displacement of smalleratoms at the glass surface with larger ones.

11.46 What do you think is the purpose of the oper-ation shown in Fig. 11.27d?

In this operation, a bur-like tool (see p. 493) re-moves excess material from the top of the bottleand gives the desired shape to the neck.

11.47 Injection molding is a process that is used forplastics and powder metals as well as for ceram-ics. Why is it suitable for all these materials?

Injection molding can be used for any material(brought to a fluid state by heating) that willmaintain its shape after forming and cooling.This is also the case with ceramic slurries andpowder metals (in a polymer carrier, as in MIM.

11.48 Are there any similarities between the strength-ening mechanisms for glass and those for othermetallic and nonmetallic materials describedthroughout this text? Explain.

There are similarities. For example, metal partsas well as glass parts can be stress relievedor annealed to relieve surface residual stresses,which is in effect a strengthening mechanism.The results may be the same for both typesof materials, even though the means of achiev-ing them may differ. Note, for example, thatcompressive residual stresses are induced onglass surfaces through tempering, while metalsare typically shot peened or surface rolled (seepp. 154-155).

11.49 Describe and explain the differences in the man-ner in which each of the following flat surfaceswould fracture when struck with a large piece ofrock: (a) ordinary window glass, (b) temperedglass, and (c) laminated glass.

By the student. Note that:

(a) When subjected to an impact load, ordi-nary window glass will shatter into numer-ous fragments or shards of various sizes.

(b) Tempered glass will shatter into smallfragments.

(c) Laminated glass will shatter, but will notfly apart because the polymer laminatewill hold the fragments in place and at-tached to the polymer.

11.50 Describe the similarities and the differences be-tween the processes described in this chapterand in Chapters 5 through 10.

By the student. This could be a challengingtask, as it requires a detailed knowledge of allthe processes involved. Note, for example, thatthere are certain similarities between (a) forg-ing and powder compaction, (b) slush castingand slip casting, (c) extrusion of metals andextruding polymers, and (d) drawing of metalwire and drawing of glass fibers. Students areencouraged to respond to this question with abroad perspective and giving several more ex-amples.

11.51 What is the doctor-blade process? Why was itdeveloped?

The doctor-blade process, shown in Fig. 11.28,produces thin sheets of ceramic. This processhas, for example, been very cost-effective forapplications such as making dielectrics in ca-pacitors.

11.52 Describe the methods by which glass sheet ismanufactured.

By the student. Glass sheet is produced bythe methods described in Section 11.11 and inFig. 11.32. Basically:

• In the drawing process (or the relatedrolling process), molten glass is pinchedand pulled through rolls and then drawndown to the desired thickness.

• In the float method, a glass sheet floats ona bath of molten tin, producing a supe-rior surface finish; the glass then cools ina lehr.

11.53 Describe the differences and similarities in pro-ducing metal and ceramic powders. Which ofthese processes would be suitable for producingglass powder?

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There are several methods of producing pow-ders, but only a few are applicable to both ce-ramics and metals. The similarities include:

• Both can be produced by chemical reduc-tion, mechanical milling, ball or hammermilling, and grinding.

• Both require screening to produce con-trolled distributions of particle sizes.

• Ball milling can be performed on eithermaterial to further reduce their particlesize.

The differences include:

• Atomization is common for metals but notpractical for ceramics, because of the highmelting temperature of ceramics.

• The shape of the powders is different; met-als are often atomized and hence sphericalin shape, whereas ceramics are angular.

• Ceramics cannot be produced throughelectrolytic deposition.

Glass powders are of limited industrial inter-est (other than as glass lubrication in hot ex-trusion; see bottom of p. 318), but could con-ceivably be produced through hammer milling,grinding, or mechanical comminution.

11.54 How are glass fibers made? What applicationdo these fibers have?

Glass fibers (see pp. 612-613) are bundle drawnusing platinum dies. They are used as reinforce-ments in polymer composite materials, and asthermal and electrical insulation, and as a lu-bricant in hot extrusion.

11.55 Would you consider diamond a ceramic? Ex-plain.

While diamond has many of the characteristicsof ceramics, such as high hardness, brittleness,and chemical inertness, diamond is not a ce-ramic. By definition, a ceramic is a combinationof a metal and a non-metal, whereas diamondis a form of carbon. (See Section 8.6.9.)

11.56 What are the similarities and differences be-tween injection molding, metal injection mold-ing, and ceramic injection molding?

By the student. The similarities betweenpolymer injection molding and metal injectionmolding (MIM) and ceramic injection molding(CIM) include:

• The tool and die materials used are simi-lar.

• Die design rules are similar.

• The pressures achieved and part sizes arethe same, as is the equipment used.

• Operator skill required is comparable.

The differences include:

• Tool and die life for MIM or CIM is lowerthan that in polymer injection molding,because of the abrasiveness of the mate-rials involved.

• Injection molding tooling requires heating(for reaction injection molding) or cool-ing (for injection molding) capabilities,whereas MIM and CIM do not require thiscapability.

• Cycle times for MIM and CIM are lowerat the molding machine because cooling orcuring cycles are not necessary.

• After molding, plastic parts have attainedtheir full strength, whereas MIM and CIMparts require a sintering or firing step.

11.57 Aluminum oxide and partially stabilized zirco-nia are normally white in appearance. Can theybe colored? If so, how would you accomplishthis?

Coloring can be accomplished in a number ofways. First, an impurity can be mixes with theceramic in order to change its color. Alterna-tively, a stain, paint, or dye can be applied afterfiring; some of the dyes may require a secondfiring step.

11.58 It was stated in the text that ceramics havea wider range of strengths in tension than dometals. List the reasons why this is so.

By the student. This question can be answeredin a variety of ways. The students are encour-aged to examine reasons for this characteristic,including the susceptibility of ceramics to flawsin tension and the range of porosity that ce-ramic parts commonly contain.

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Problems

11.59 Estimate the number of particles in a 500-gsample of iron powder, if the particle size is 50µm.

From Table 3.3 on p. 106 and Fig. 11.6a, thedensity of iron is found to be ρ = 7.86 g/cm3.The particle diameter, D, is 50 µm = 0.005 cm.The volume of each spherical powder is

V =4π3

(D

2

)3

=4π3

(0.005

2

)3

or V = 6.545× 10−8 cm3. Thus, its mass is

m = ρV = (7.86)(6.545×10−8) = 5.14×10−7 g

Therefore, the number of particles in the sam-ple is

N =500

5.14× 10−7= 9.73× 108 = 973 million

11.60 Assume that the surface of a copper particle iscovered with a 0.1-µm-thick oxide layer. Whatis the volume occupied by this layer if the cop-per particle itself is 75 µm in diameter? Whatwould be the role of this oxide layer in subse-quent processing of the powders?

The volume of the oxide layer can be estimatedas

V = 4πr2t = 4π(37.5 µm)2(0.1 µm)

or V = 1770 µm3. Oxide layers adversely affectthe bond strength between the particles duringcompaction and sintering, which, in turn, hasan adverse effect on the strength and ductilityof the P/M part. Its physical properties suchas electrical and thermal conductivity are alsoaffected.

11.61 Determine the shape factor for a flakelike par-ticle with a ratio of surface area to thicknessof 12 × 12 × 1, for a cylinder with dimensionalratios 1:1:1, and for an ellipsoid with an axialratio of 5× 2× 1.

The volume of the flakelike particle is, in arbi-trary units, V = (12)(12)(1) = 144. The equiv-alent diameter for a sphere is

V =π

6D3

or, solving for D,

D = 3

√6Vπ

= 3

√6(144)π

= 6.50

The total surface area, A, of the particle is

A = (2)(12)(12) + (4)(12)(1) = 336.

Therefore,A

V=

336144

= 2.33

Thus, the shape factor is SF=(6.50)(2.33) =15.17. For the ellipsoid particle, all cross sec-tions across the three major and minor axes areelliptical in shape. (This is in contrast to an el-lipsoid of revolution, where one cross section iscircular.) The volume of an ellipsoid is

V =43πabc

where a, b, and c are the three semi-axes of theellipsoid. Because of arbitrary units, we cancalculate the volume of an ellipsoid with axesratios of 5:2:1 as

V =43πabc =

43(5)(2)(1) = 41.89

The equivalent diameter for a sphere is

D = 3

√6Vπ

= 3

√6(41.89)

π= 4.3

It can be shown that the surface area of theellipsoid is given by the expression

A =(π2

2

)(a+ b) (c)

where c is the semi-axis of the longest dimen-sion of the ellipse. Thus, again using arbitraryunits,

A =(π2

2

)(2 + 1)(5) = 74.02

Therefore,

A

V=

74.0241.89

= 1.77

Hence, the shape factor is SF= (4.30)(1.77) =7.61.

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11.62 It was stated in Section 3.3 that the energy inbrittle fracture is dissipated as surface energy.We also noted that the comminution processfor powder preparation generally involves brit-tle fracture. What are the relative energies in-volved in making spherical powders of diame-ters 1, 10, and 100 µm, respectively?

First, note that the surface energy is propor-tional to the surface area generated during com-minution. The surface area of a spherical par-ticle is 4πr2 = πD2. Consequently, the relativeenergies will be proportional to the diametersquared, or 1, 100, and 10,000, respectively.

11.63 Referring to Fig. 11.6a, what should be the vol-ume of loose, fine iron powder in order to makea solid cylindrical compact 25 mm in diameterand 15 mm high?

The volume of the cylindrical compact is V =π[(25)2/4]15 = 7360 mm3. Loose, fine ironpowder has a density of about 1.40 g/cm3 (seeFig. 11.6a). Density of iron is 7.86 g/cm3 (seeTable 3.1). Therefore, the weight of iron neededis

W = ρV

= (7.86 g/cm3)(7360 mm3)(10−3 cm3

mm3)

or W = 57.8 g. Thus, the initial volume is

V =W

ρ=

57.81.40

= 41.3 cm3

11.64 In Fig. 11.7e, we note that the pressure is notuniform across the diameter of the compact.Explain the reasons for this variation.

Note in the figure that the pressure drops to-wards the center of the compact; this is becauseof the internal frictional resistance in the radialdirection. This situation is similar to forgingwith friction, as described in Section 6.2.2. Alsonote that the pressure drop is steepest at theupper (punch) surface, and that at the levelwhere the pressure is in the range of 200-300MPa, the pressure is rather uniform across thecross section of the compact.

11.65 Plot the family of pressure-ratio px/po curvesas a function of x for the following ranges of

process parameters: µ = 0 to 1, k = 0 to 1, andD = 5 mm to 50 mm.

The key equation is Eq. (11.2) on p. 680:

px = poe−4kx/D

The results are plotted in three graphs, forD = 5 µm, D = 25 µm, and D = 50 µm.

k=0.1

0.250.5

1

0 5 10 15 20x, m

px/p

0

1

0.8

0.6

0.4

0.2

0

D=5 m

x, m

px/p0

1

0.8

0.6

0.4

0.2

00 10 20 30 40 50

k=0.10.25

0.51

D=25 mpx/p

0

1

0.8

0.6

0.4

0.2

00 10 20 30 40 50

k=0.10.25

0.5

1

D=50 µm

x, m

11.66 Derive an expression, similar to Eq. (11.2), forcompaction in a square die with dimensions aby a.

Referring to Fig. 11.8 and taking an elementwith a square cross section, the following equa-tion represents equilibrium:

a2px − a2(px + dpx)− 4a(µσx) dx = 0

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which reduces to

adpx + 4µσx dx = 0

This equation is the same as that in Section11.3.1. Therefore, the expression for the pres-sure at any x is

px = poe−4µkx/a

11.67 For the ceramic described in Example 11.7, cal-culate (a) the porosity of the dried part if theporosity of the fired part is to be 9%, and (b)the initial length, Lo, of the part if the linearshrinkages during drying and firing are 8% and7%, respectively.

(a) For this case we have

Va = (1− 0.09)Vf = 0.91Vf

Because the linear shrinkage during firingis 7%, we can write

Vd = Vf/(1− 0.07)3 = 1.24Vf

Therefore,

Va

Vd=

0.911.24

= 0.73, or 73%

Consequently, the porosity of the driedpart is (1 - 0.73) = 0.27, or 27%.

(b) We can now write

(Ld − L)Ld

= 0.07

orL = (1− 0.07)Ld

Since L = 20 mm,

Ld = 20/0.93 = 21.51 mm

and thus

Lo = (1 + 0.08)Ld = (1.08)(21.51)

or Lo = 23.23 mm.

11.68 What would be the answers to Problem 11.67if the quantities given were halved?

(a) For this case, we have

Va = (1− 0.045)Vf = 0.955Vf

Because the linear shrinkage during firingis 3.5%, we write

Vd = Vf/(1− 0.035)3 = 1.112Vf

Therefore,

Va/Vd = 0.955/1.112 = 0.86, or86%

Consequently, the porosity of the driedpart is (1 - 0.86) = 0.14, or 14%.

(b) We can now write

(Ld − L)Ld

= 0.035

orL = (1− 0.035)Ld

Since L = 20 mm, we have

Ld =20

0.965= 20.73 mm

and thus

Lo = (1 + 0.04)Ld = (1.04)(20.73)

or Lo = 21.56 mm.

11.69 Plot the UTS, E, and k values for ceramics asa function of porosity, P , and describe and ex-plain the trends that you observe in their be-havior.

The plots are given below.

1

0.8

0.6

0.4

0.2

0

UTS

/UTS

0

10.80.60.40.20

Porosity

n=4n=5

n=7

1

0.8

0.6

0.4

0.2

0

E/E

0

10.80.60.40.20Porosity

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1

0.8

0.6

0.4

0.2

0

k/k 0

10.80.60.40.20Porosity

11.70 Plot the total surface area of a 1-g sample ofaluminum powder as a function of the naturallog of particle size.

The density of aluminum is 2.7 g/cm3. Themass of each particle is:

m = ρV =(2.7 g/cm3

)(43π

)(D

2

)3

hence the number of particles is given by

N =1 gm

=1

(2.7)(

π6D

3) =

(0.707 cm3

)(D−3)

The total surface area of these particles is

A = NπD2 = (0.707 cm3)(D−3)πD2

or A = (2.22 cm3)D−1, where D is in cm (µm× 10,000). The plot is shown below.

2.5

2.0

1.5

1.0

0.5

0

Sur

face

are

a, m

2

-10 -9 -8 -7 -6 -5 -4lnD (D in cm)

11.71 Conduct a literature search and determine thelargest size of metal powders that can be pro-duced in atomization chambers.

By the student. The answer will depend onthe material and desired particle morphology.

Spherical particles will be in the sub-mm sizes(tens to hundreds of microns), while othershapes can approach a few mm in average di-ameter.

11.72 A coarse copper powder is compacted in a me-chanical press at a pressure of 20 tons/in2. Dur-ing sintering, the green part shrinks an addi-tional 8%. What is the final density of the part?

From Fig. 11.6, the copper density after com-paction is around 7 g/cm3. Since the materialshrinks an additional 8% during sintering, thevolume is 1/(0.92)3 times the original volume.Thus, the density will be around 8.99 g/cm3.

11.73 A gear is to be manufactured from iron powder.It is desired that it have a final density that is90% of that of cast iron, and it is known thatthe shrinkage in sintering will be approximately5%. For a gear 2.5-in. in diameter and with a0.75-in. hub, what is the required press force?

From Table 3.3 on p. 106, the density of iron is7.86 g/cm3. For the final part to have a finaldensity of 90% of this value, the density aftersintering must be 7.07 g/cm3. Since the partcontracts 5% during sintering, the density be-fore sintering must be 6.06 g/cm3. Referring toFig. 11.6a, the required pressure for this den-sity is around 20 tons/in2. The projected areais A = π/4(2.52 − 0.752) = 4.47 in2. The re-quired force is then 89 tons, or approximately90 tons.

11.74 What volume of powder is needed to make thegear in Problem 11.73 if its thickness is 0.5 in?

Refer to the solution to Problem 11.73. The vol-ume of the gear is the product of the projectedarea and its thickness. The actual surface areaof the gear is not given, but we can estimate theamount of powder needed by assuming that thediameter given is the pitch diameter and thatthe part can be treated as a cylinder with a cir-cular cross section with a 2.5 in. diameter anda 0.75-in. hole. The projected area, as calcu-lated in Problem 11.73, is A = 4.47 in2. Thus,the volume is

V = Ah = (4.47)(0.5) = 2.235 in3 = 36.6 cm3

Therefore, the weight of the gear is (7.07g/cm3)(36.6 cm3)=260 g. From Fig. 11.6a,loose fine iron powder has a density of around

183






1.4 g/cm3. Therefore, the volume of powderrequired is

V =m

ρ=

260 g1.4 g/cm3 = 185 cm3

11.75 The axisymmetric part shown in the accompa-nying figure is to be produced from fine copperpowder and is to have a tensile strength of 200MPa. Determine the compacting pressure andthe initial volume of powder needed.

2520

1210

Dimensions in mm

25

From Fig. 11.6b, the density of the copperpart must be around 8.5 g/cm3 to achieve thestrength of 200 MPa. From Fig. 11.6a, the re-quired pressure is around 1000 MPa. The pressforce will be determined by the largest cross-sectional area, which has the 25 mm outer di-ameter. The cross-sectional area is

A =π

4(D2

o −D2i

)=

π

4(0.0252 − 0.0102

)or A = 4.123×10−4 m2. Therefore, the requiredforce is

F = pA = (1000 MPa)(4.123× 10−4 m2)

or F = 412 kN. From the given geometry, thefinal part volume is found to be 8.0 cm3, andhence the mass required is

m = ρV = (8.5 g/cm3)(8.0 cm3) = 68 g

From Fig. 11.6a, the apparent density of finecopper powder is 1.44 g/cm3, so that the re-quired powder volume is

V =m

ρ=

68 g1.44 g/cm3 = 45 cm3

11.76 What techniques, other than the powder-in-tube process, could be used to produce super-conducting monofilaments?

Other principal superconductor-shaping pro-cesses are:

(a) coating of silver wire with superconduct-ing material,

(b) deposition of superconductor films by laserablation,

(c) doctor-blade process (see Section 11.9.1),

(d) explosive cladding (see Section 12.11), and

(e) chemical spraying.

11.77 Describe other methods of manufacturing theparts shown in Fig. 11.1a. Comment on the ad-vantages and limitations of these methods overP/M.

By the student. Several alternative methodsfor manufacturing the parts can be discussed.For example, the parts could be machined, inwhich case the machined part may have betterdimensional accuracy and surface finish, andwould be less expensive for short productionruns. The P/M part would be less expensivefor large production runs and would likely beless dense. As another example, consider forg-ing of these parts; the forging would be denserand stronger, and have similar surface finish ifcold forged. However, a P/M part would likelyhave a lower density and would be producedwithout flash.

11.78 If a fully-dense ceramic has the properties ofUTSo = 180 MPa and Eo = 300 GPa, what arethese properties at 20% porosity for values ofn = 4, 5, 6, and 7, respectively?

Inserting the appropriate quantities intoEqs. (11.5) and (11.6) on p. 701, we obtain thefollowing:

n UTS (MPa) E (GPa)4 80.9 196.85 66.2 196.86 54.2 196.87 44.4 196.8

Note that the magnitude of n does not affectthe magnitude of E.

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11.79 Calculate the thermal conductivities for ceram-ics at porosities of 1%, 5%, 10%, 20%, and 30%for ko = 0.7 W/m-K.

Equation (11.7) is needed to solve this prob-lem, which gives the thermal conductivity as afunction of porosity as

k = ko(1− P )

Inserting the values into the equation, we ob-tain thermal conductivities of:

P = 1% k = 0.693 W/mK5 0.66510 0.63020 0.5630 0.49

11.80 A ceramic has ko = 0.65 W/m-K. If this ce-ramic is shaped into a cylinder with a porositydistribution of P = 0.1(x/L)(1 − x/L), wherex is the distance from one end of the cylinderand L is the total cylinder length, estimate theaverage thermal conductivity of the cylinder.

The plot of porosity is given below:

Por

osity

0.025

0.02

0.015

0.01

0.005

00 0.25 0.5 0.75 1

Position, x/L

For the remainder of the problem, use X =x/L. The average porosity is then given by

P =∫ 1

0

0.1X(1−X)dX

=∫ 1

0

(−0.1X2 + 0.1X

)dX

= 0.0167

Since the thermal conductivity is linearly re-lated to the porosity, the average porosity canbe used, so that the average thermal conductiv-ity is:

k = ko

(1− P

)= (0.65)(1− 0.0167)

or k = 0.639 W/mK.



Design

11.82 Make sketches of several P/M products in whichdensity variations would be desirable. Explainwhy, in terms of the function of these parts.

By the student. Any kind of minimum-weightdesign application would be appropriate, suchas aerospace and automotive, where lightlyloaded regions can be reduced in weight by

making them more porous. With bearing sur-faces, a greater density at the surface is desir-able, while a substrate need not be as dense.

11.83 Compare the design considerations for P/Mproducts with those for products made by (a)casting and (b) forging. Describe your observa-tions.

185






By the student. Note that design considera-tions for P/M parts (Section 11.6) are similarto those for casting (Section 5.12) and forging(Section 6.2.7). The similarities are due to thenecessity of removing the parts from the moldsor dies. Hence, tapers should be used wheneverpossible and internal cavities are difficult toproduce. Large flat surfaces should be avoided,and the section thickness should be uniform asmuch as possible. There are many similaritieswith casting and forging part design, mainlybecause P/M parts need to be ejected, just asin forging, and the pattern for casting need tobe removed. There are some differences. Forexample, engraved or embossed lettering is dif-ficult in P/M but can be done easily in casting.P/M parts should be easy to eject; casting de-signs are more flexible in this regard.

11.84 It is known that in the design of P/M gears,the distance between the outside diameter ofthe hub and the gear root should be as largeas possible. Explain the reasons for this designconsideration.

The reason for this is twofold. First, it is verydifficult to develop a sufficiently high pressurein the cross section containing the root if thedistance is small. Secondly, if the distance issmall, it acts as a high stress concentration,which could cause part failure prior to beingsintered, especially during ejection.

11.85 How are the design considerations for ceramicsdifferent, if any, than those for the other mate-rials described in this chapter?

By the student. Refer to Section 11.12. Con-sider, for example, the fact that ceramics arevery notch sensitive, hence brittle, and are notsuitable for impact or energy-dissipating typeloading, and also not usable where any defor-mation is foreseeable. On the other hand, ce-ramics have exceptional properties at high tem-peratures, are very strong in compression, andare resistant to wear because of their high hard-ness and inertness to most materials.

11.86 Are there any shapes or design features thatare not suitable for production by powder met-allurgy? By ceramics processing? Explain.

By the student. Neither ceramics nor P/M

parts are suitable for large parts (see, for ex-ample, Fig. 11.10). For example, bolts, ar-chitectural channels, and some biomedical im-plants are poor P/M applications. Also, fatigueapplications are generally not appropriate forP/M parts because cracks can propagate easierthrough the structure. The students are en-couraged to comment further.

11.87 What design modifications would you recom-mend for the part shown in Problem 11.75?

By the student. A number of design changeswould be advisable to increase manufacturabil-ity using P/M techniques. For example, it isadvisable that the steps have a taper to aidin ejection (see Fig. 11.17). The sharp radiishould be larger. The part is unbalanced, andthe flange, though probably acceptable, couldbe made smaller, if appropriate.

11.88 The axisymmetric parts shown in the accompa-nying figure are to be produced through P/M.Describe the design changes that you would rec-ommend.

(a) (b)

(c)

There are several design changes that couldbe advisable, and the students are encouragedto develop lists of their own recommendations.Some considerations are:

(a) Part (a):

• The part has very thin walls, and itwould be advisable to have less severeaspect ratios.

186






• The part cannot be pressed in its cur-rent shape, but could conceivably bemetal injection molded.

• Sharp corners are not advisable; radiishould be incorporated for metal in-jection molding, and chamfers forpressed parts.

(b) Part (b): Same as in (a)

(c) Part (c): Same as in (a), and the flangesshould comply with the recommendationsgiven in Fig. 11.19.

11.89 Assume that in a particular design, a metalbeam is to be replaced with a beam made ofceramics. Discuss the differences in the behav-ior of the two beams, such as with respect tostrength, stiffness, deflection, and resistance totemperature and to the environment.

By the student. This is an open-ended problemthat can be answered in a number of ways bythe students. They can, for example, considera cantilever, where the deflection at the end ofthe cantilever is,

y = − Pl3

3EI

and then compare materials that would givethe same deflection for different beam heights,widths, or volumes. One could also considerthe lightest weight cantilever that could sup-port the load or one that would have a smalldeflection under the load. Students are encour-aged to examine this problem in depth.

11.90 Describe your thoughts regarding designs of in-ternal combustion engines using ceramic pis-tons.

By the student. There are several difficultiesassociated with such designs. For example, lu-bricants (see Section 4.4.4) typically are formu-lated for use with aluminum and steel parts,and the boundary additives may not be effec-tive on a ceramic surface, so higher wear ratesmay occur. Ceramic wear particles will be muchharder than metal engine blocks or cylinder lin-ers, and will raise concerns of three-body wear(see p. 147). The entire engine needs to beredesigned to account for the reduced mass inthe pistons and other components. This can be

beneficial since higher speeds can be attained,but the engine may run rougher at low speeds.(See the discussion of coefficient of fluctuationin Hamrock, Jacobson, and Schmid, Fundamen-tals of Machine Elements, 2d ed., p. 464.)

11.91 Assume that you are employed in technicalsales. What applications currently using non-P/M parts would you attempt to develop?What would you advise your potential cus-tomers during your sales visits? What kind ofquestions do you think they would ask?

By the student. This is a challenging questionthat requires knowledge of parts that are andare not currently produced by P/M. It would beadvisable for the instructor to limit the discus-sion to a class of product, such as P/M gears.In this case, the questions that would be askedof customers include:

(a) Are you aware of the advantages of P/Mprocesses?

(b) Are you aware of the tribological advan-tages of P/M parts?

(c) Are you interested in unique alloys orblends that can only be achieved withP/M technologies?

(d) Is it beneficial to achieve a 5-10% weightsavings using porous P/M parts?

Typical anticipated questions from the cus-tomer could include:

(a) Is there a cost or performance advantage?Are there any disadvantages?

(b) We have had no failures, so why should wechange anything?

(c) Are the P/M materials compatible.

11.92 Pyrex cookware displays a unique phenomenon:it functions well for a large number of cyclesand then shatters into many pieces. Investigatethis phenomenon, list the probable causes, anddiscuss the manufacturing considerations thatmay alleviate or contribute to such failures.

By the student. This is a challenging question.The basic phenomenon appears to be that, witheach thermal stress cycle new flaws in the ma-terial may develop, and existing flaws begin togrow. When the flaws have reached a critical

187






size (see p. 102), the part fails under normaluse. Note that any manufacturing that resultin a larger initial flaw, or being subjected toless effective tempering, will contribute to theproblem.

11.93 It has been noted that the strength of brittlematerials, such as ceramics and glasses, are verysensitive to surface defects such as scratches(notch sensitivity). Obtain some pieces of thesematerials, make scratches on them, and testthem by carefully clamping in a vise and bend-ing them. Comment on your observations.

By the student. This experiment can be per-formed using a glass cutter to make a deep andsharp scratch on the glass. It can be demon-strated that glass with such a scratch can beeasily broken with bare hands. Note also the di-rection of the bending moment with respect tothe direction of the scratch. As a comparison,even a highly heat-treated aluminum plate (i.e.,brittle behavior) will not be nearly as weakenedwhen a similar scratch is made on its surface.Note that special care must be taken in per-forming these experiments, using work glovesand eye protection and the like.

11.94 Make a survey of the technical literature anddescribe the differences, if any, between thequality of glass fibers made for use in reinforcedplastics and those made for use in fiber-opticcommunications. Comment on your observa-tions.

By the student. The glass fibers in reinforcedplastics has a much smaller diameter and has tobe of high quality for high strength. The glassfibers for communications applications are for-mulated for optical properties and the strengthis not a major concern, although some strengthis needed for installation.

11.95 Describe your thoughts on the processes thatcan be used to make (a) small ceramic stat-

ues, (b) white ware for bathrooms, (c) commonbrick, and (d) floor tile.

By the student. The answers will vary becauseof the different manufacturing methods used forthese products. Some examples are:

(a) Small ceramic statues are usually made byslip casting, then fired to fuse the particlesand develop strength, followed by decorat-ing and glazing.

(b) White ware for bathrooms are either slipcast or pressed, then fired, and sometimesglazed and re-fired.

(c) Common brick is wet pressed or slip cast,then fired.

(d) Floor tile is hot pressed or dry pressed,fired, and sometimes glazed and re-fired.

11.96 As described in this chapter, one method ofproducing superconducting wire and strip is bycompacting powders of these materials, placingthem into a tube, and drawing them throughdies, or rolling them. Describe your thoughtsconcerning the possible difficulties involved ineach step of this production.

By the student. Concerns include fracture ofthe green part before or during drawing, and itsimplications; inhomogeneous deformation thatcan occur during drawing and rolling and itspossible effects as a fracture-causing process;the inability of the particles to develop suffi-cient strength during this operation; and pos-sible distortion of the part from its drawn orrolled shape during sintering.

11.97 Review Fig. 11.18 and prepare a similar figurefor constant-thickness parts, as opposed to theaxisymmetric parts shown.

By the student. The figure will be very similarto Fig. 11.18, as the design rules are not neces-sarily based on the axisymmetric nature of theparts.

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Chapter 12

Joining and Fastening Processes

Questions

12.1 Explain the reasons that so many differentwelding processes have been developed.

A wide variety of welding processes have beendeveloped for several reasons. There are manytypes of metals and alloys with a wide range ofmechanical, physical, and metallurgical prop-erties and characteristics. Also, there arenumerous applications involving a wide va-riety of component shapes and thicknesses.For example, small or thin parts that cannotbe arc welded can be resistance welded, andfor aerospace applications, where strength-to-weight ratio is a major consideration, laser-beam welding and diffusion bonding are attrac-tive processes. Furthermore, the workpiece maynot be suitable for in-plant welding, and thewelding process may have to be brought to thesite, such as pipelines and large structures. (Seealso Section 12.1.)

12.2 List the advantages and disadvantages of me-chanical fastening as compared with adhesivebonding.

By the student. Advantages of mechanical fas-tening over adhesive bonding:

(a) disassembly is easier (bolted connections);

(b) stronger in tension;

(c) preloading is possible; and

(d) no need for large areas of contact.

Limitations:

(a) often costlier;(b) requires assembly;(c) weaker in shear; and(d) more likely to loosen (bolted connections).

12.3 What are the similarities and differences be-tween consumable and nonconsumable elec-trodes?

By the student. Review Sections 12.3 and 12.4.Comment, for example, on factors such as therole of the electrodes, the circuitry involved, theelectrode materials, and the manner in whichthey are used.

12.4 What determines whether a certain weldingprocess can be used for workpieces in horizon-tal, vertical, or upside-down positions, or for alltypes of positions? Explain, giving appropriateexamples.

By the student. Note, for example, that somewelding operations (see Table 12.2 on p. 734)cannot take place under any conditions ex-cept horizontal, such as submerged arc weld-ing, where a granular flux must be placed onthe workpiece. If a process requires a shieldinggas, it can be used in vertical or upside-downpositions. Oxyacetylene welding would be dif-ficult upside-down because the flux may dripaway from the surface instead of penetratingthe joint.

12.5 Comment on your observations regardingFig. 12.5.

189






By the student. The students are encouragedto develop their answers considering the signif-icance of the layered weld beads and the qual-ity of their interfaces. For example, there maybe concerns regarding the weld strength sincethe interfaces between adjacent beads may havesome slag or surface contaminants that have notbeen removed. The heat-affected zone and fa-tigue implications of such welds are also a sig-nificant concern.

12.6 Discuss the need for and role of fixtures in hold-ing workpieces in the welding operations de-scribed in this chapter.

By the student. The reasons for using fixturesare basically to assure proper alignment of thecomponents to be joined, reduce warpage, andhelp develop good joint strength. The fixturescan also be a part of the electrical circuit in arcwelding, where a high clamping force reducesthe contact resistance. See also Section 14.11.1.

12.7 Describe the factors that influence the size ofthe two weld beads in Fig. 12.13.

The reason why electron-beam weld beads arenarrower than those obtained by arc welding isthat the energy source in the former is muchmore intense, confined, and controllable, allow-ing the heating and the weld bead to be morelocalized. Other factors that influence the sizeof the weld bead are workpiece thickness, mate-rial properties, such as melting point and ther-mal conductivity. See also pp. 749-751.

12.8 Why is the quality of welds produced by sub-merged arc welding very good?

Submerged arc welding (see Fig. 12.6) hasvery good quality because oxygen in the atmo-sphere cannot penetrate the weld zone wherethe shielding flux protects the weld metal. Also,there are no sparks, spatter, or fumes as inshielded metal arc and some other welding pro-cess.

12.9 Explain the factors involved in electrode selec-tion in arc welding processes.

By the student. Refer to Section 12.3.8. Elec-trode selection is guided by many factors, in-cluding the process used and the metals to bewelded.

12.10 Explain why the electroslag welding process issuitable for thick plates and heavy structuralsections.

Electroslag welding (see Fig. 12.8) can be per-formed with large plates because the tempera-tures attainable through electric arcs are veryhigh. A continuous and stable arc can beachieved and held long enough to melt thickplates.

12.11 What are the similarities and differences be-tween consumable and nonconsumable elec-trode arc welding processes?

By the student. Similarities: both require anelectric power source, arcing for heating, and anelectrically-conductive workpiece. Differences:the electrode is the source of the weld metal inconsumable-arc welding, whereas a weld metalmust be provided in nonconsumable-arc weld-ing processes.

12.12 In Table 12.2, there is a column on the distor-tion of welded components, ranging from lowestto highest. Explain why the degree of distortionvaries among different welding processes.

By the student. Refer to Table 12.2 on p. 734.The distortion of parts is mainly due to ther-mal warping because of temperature gradientsdeveloped within the component. Note thatthe lowest distortions are in electron beam andlaser beam processes, where the heat is highlyconcentrated in narrow regions and with deeperpenetration. This is unlike most other processeswhere the weld zones are large and distortioncan be extensive.

12.13 Explain why the grains in Fig. 12.16 grow inthe particular directions shown.

The grains grow in the directions shown inFig. 12.16 because of the same reasons grainsgrow away from the wall in casting process so-lidification, described in Section 5.2. Heat fluxis in the opposite direction as grain growth,meaning a temperature gradient exists in thatdirection, so only grains oriented in the direc-tion perpendicular from the solid-metal sub-strate will grow.

12.14 Prepare a table listing the processes describedin this chapter and providing, for each process,

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the range of welding speeds as a function ofworkpiece material and thickness.

By the student. This is a good assignment forstudents, although it can be rather demandingbecause such extensive data is rarely available,except with a wide range or only for a particu-lar group of materials. Consequently, it can bedifficult to compare the processes; they never-theless should be encouraged to develop such alist as best they can.

12.15 Explain what is meant by solid state welding.

As descried briefly on p. 733, in solid state weld-ing, the metals to be joined do not melt; there isno liquid state in the interface. Note that thereare six processes listed under this category.

12.16 Describe your observations concerningFigs. 12.19 through 12.21.

By the student. This is a challenging questionand students are encouraged to develop and listas many answers as they can. For example,they can consider the crack locations, developan ability to identify them through inspection,describe the causes of the defects, and the ef-fects of different workpiece materials and pro-cessing conditions.

12.17 What advantages does friction welding haveover the other joining methods described in thischapter?

By the student. As described in Section 12.9,the main advantages of friction welding are thatthe entire cross-sectional area can be welded, in-stead of a mere bead along the periphery, andis suitable for a wide variety of materials. Also,with proper process control, the weld zone canbe very small and thin, so that thermal distor-tions will be minimal.

12.18 Why is diffusion bonding, when combined withsuperplastic forming of sheet metals, an attrac-tive fabrication process? Does it have any lim-itations?

By the student. As shown in Fig. 12.41,diffusion bonding combined with superplasticforming can produce lightweight, rigid, andstrong aerospace structures with high stiffness-to-weight ratios. The main drawback is the long

production time and the high costs involved,which may be justified for many aerospace ap-plications. The students are encouraged to findother examples of applications for this impor-tant process.

12.19 Can roll bonding be applied to various part con-figurations? Explain.

Roll bonding (Fig. 12.28) is mainly used in flatrolling, although other applications may be pos-sible. The important consideration is that thepressure (normal stress) between the sheets tobe joined be sufficiently high and the interfacesare clean and free of oxide layers. To meet thiscondition for shapes other than flat is likely tobe a difficult task and involve complex tooling.Also, any significant variation in pressure dur-ing rolling can make the bonded structure be-come not uniform and unreliable. The studentis encouraged to search the literature and at-tempt to find examples of such applications.

12.20 Comment on your observations concerningFig. 12.40.

By the student. The explosion welding oper-ation results in wavy interfaces (as shown inthe figures) due to the very high interfacialvelocities and pressures involved. The ripplesobserved are actually due to stress waves inthe interface, and help improve joint strengthby mechanical interlocking of the mating sur-faces. Some students may wish to investigateand elaborate further as to how these wavesare developed and how they affect interfacialstrength.

12.21 If electrical components are to be attached toboth sides of a circuit board, what solderingprocess(es) would you use? Explain.

A challenging problem arises when a printedcircuit board (see Section 13.13) has bothsurface-mount and in-line circuits on the sameboard and it is desired to solder all the jointsvia a reliable automated process. An importantpoint is that all of the in-line circuits shouldbe restricted to insertion from one side of theboard. Indeed, there is no performance require-ment which would dictate otherwise, but thisrestriction greatly simplifies manufacturing.

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The basic steps in soldering the connections onsuch a board are as follows:

(a) Apply solder paste to one side.

(b) Place the surface-mount packages ontothe board; also, insert in-line packagesthrough the primary side of the board.

(c) Reflow the solder (see bottom of p. 777).

(d) Apply adhesive to the secondary side ofthe board.

(e) Attach the surface mount devices on thesecondary side, using the adhesive.

(f) Cure the adhesive.

(g) Perform a wave-soldering operation(p. 778) on the secondary side to pro-duce electrical attachment of the surfacemounts and the in-line circuits to theboard.

12.22 Discuss the factors that influence the strengthof (a) a diffusion bonded component and (b) acold welded component.

Diffusion bonded strength (Section 12.12) is in-fluenced by temperature (the higher the tem-perature, the more the diffusion), pressure,time, and the materials being joined. Thecleanliness of the surfaces is also important tomake sure no lubricants, oxides, or other con-taminants interfere with the diffusion process.For this reason, these joints are commonly pre-pared by solvent cleaning and/or pickling to re-move oxides. Cold welded components (Section12.7) involve similar considerations except thattemperature is not a relevant parameter.

12.23 Describe the difficulties you might encounter inapplying explosion welding in a factory environ-ment.

By the student. Explosives are very dangerous;after all, they are generally used for destructivepurposes. There are safety concerns such ashearing loss, damage resulting from explosions,and fires. The administrative burden is highbecause there are many federal, state, and mu-nicipal regulations regarding the handling anduse of explosives and the registration involvedin using explosives.

12.24 Inspect the edges of a U.S. quarter, and com-ment on your observations. Is the cross-section,i.e., the thickness of individual layers, symmet-rical? Explain.

By the student. This is an interesting assign-ment to demonstrate the significance of coldwelding. The side view of a U.S. quarter isshown below. The center of the coin is a copperalloy and the outer layers are a nickel-based al-loy. (Note that pennies and nickels are typicallymade of one material.) The following observa-tions may be made about the coins:

• The core is used to obtain the properweight and feel, as well as sound.

• The strength of roll-bonded joints is veryhigh, as confirmed by the fact that onenever encounters coins that have peeledapart (although during their developmentsuch separation did occur).

• The outer layers, which are made of themore expensive alloy, are thin for cost re-duction.

• The thicknesses of the two outer layers isnot the same. This is due to the smearingaction that occurs around the peripheryduring blanking of the coins, as can be re-called from Section 7.3.

12.25 What advantages do resistance welding pro-cesses have over others described in this chap-ter?

By the student. Recall that resistance weldingis a cleaner process for which electrodes, flux, orshielding environment are not needed; the met-als to be welded provide all of these inherently.The process is easily automated and productionrate is high.

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12.26 What does the strength of a weld nugget in re-sistance spot welding depend on?

By the student. The students are encour-aged to search the literature and collect pho-tographs and more details on weld nuggets (seeFig. 12.33b). This question can be answeredfrom different viewpoints. Thus, for exam-ple, one may consider this question as a stress-analysis problem, whereby the joint strengthdepends on the size of the nugget, its relation-ship to the surrounding bodies, and the typesof materials welded and their mechanical andphysical properties. Other factors to be con-sidered are the role of process parameters suchas current, pressure, time, and the nature ofthe faying surfaces. It would also be interestingand instructional to find weld nuggets that arepoorly made.

12.27 Explain the significance of the magnitude of thepressure applied through the electrodes duringresistance welding operations.

As can be seen in Fig. 12.33, the pressure is ap-plied after sufficient heat is generated. Pressureis maintained until the current is shut off. Thehigher the pressure, the higher the strength ofthe joint (although too high a pressure will ex-cessively indent the surfaces and cause damage;see Fig. 12.27). Lower pressures produce weakjoints. It should be remembered, however, thatthe higher the force the lower the resistance,hence the lower the resistance heating. Conse-quently, proper control of pressure is importantin resistance welding.

12.28 Which materials can be friction stir welded, andwhich cannot? Explain your answer.

Friction stir welding (p. 764) has been com-monly applied to aluminum and copper alloys,and some research is being conducted to extendthe process to others as well as thermoplas-tics and reinforced thermoplastics. The mainrequirements are that the workpiece be suffi-ciently soft and have a low melting point. Theformer requirement ensures that the rotatingtool (Fig. 12.32) will have appropriate strengthfor the operation being performed, and the lat-ter to ensure that the power requirements arereasonable.

12.29 List the joining methods that would be suitablefor a joint that will encounter high stresses andwill need to be disassembled several times dur-ing the product life, and rank the methods.

By the student. Refer also to Table 12.1 onp. 733. Disassembly can be a difficult featureto assess when selecting joining methods. If thepart has to be disassembled often, bolted con-nections are likely to be the best solution, orelse a quick-disconnect clamp or similar devicesshould be used. If the number of disassembliesover the lifetime of the part is limited (such asautomobile dashboards), integrated snap fas-teners (see Fig. 12.55) and even soldering orbrazing can be options. However, soldering andbrazing are only suitable if the filler metal canbe melted without damaging the joint, and ifthe joint can be resoldered.

12.30 Inspect Fig. 12.31, and explain why the par-ticular fusion-zone shapes are developed as afunction of pressure and speed. Comment onthe influence of the properties of the material.

By the student. Inspecting the fusion zonesin Fig. 12.31, it is obvious that higher forcesand speeds both result in more pronounced fu-sion zones. The relevant material properties arestrength at elevated temperatures and physi-cal properties such as thermal conductivity andspecific heat. Because all materials soften atelevated temperatures, the hotter the interface,the more pronounced the fusion zone. Note alsothat a uniform (optimum) zone can be obtainedwith proper control of the relevant parameters.

12.31 Which applications could be suitable for theroll spot welding process shown in Fig. 12.35c?Give specific examples.

By the student. The roll spot-welding opera-tion, shown in Fig. 12.35, is commonly used tofabricate all types of containers and sheet-metalproducts. They can be leak proof provided thatthe spacing of the weld nuggets are sufficientlyclose.

12.32 Give several examples concerning the bulleteditems listed at the beginning of Section 12.1.

By the student. A visit to various stores andobserving the products displayed, as well as

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equipment and appliances found in homes, of-fices, and factories will give ample opportunityfor students to respond comprehensively to thisquestion.

12.33 Could the projection welded parts shown inFig. 12.36 be made by any of the processes de-scribed in other parts of this text? Explain.

By the student. The projection-welded partsshown could possibly be made through resis-tance spot welding (although it would requireseveral strokes) and resistance projection weld-ing. Various other processes may be able to pro-duce the parts shown, but the joint strength de-veloped or the economics of the processes maynot be as favorable. The shape can also beachieved through arc or gas welding processes(followed by finishing such as grinding, if neces-sary), as well as brazing or soldering (see Sec-tion 12.13). With a modified interface, mechan-ical fastening and adhesive bonding also couldbe suitable processes.

12.34 Describe the factors that influence flattening ofthe interface after resistance projection weldingtakes place.

Review Fig. 12.36 and note that:

(a) The projections provide localized areas ofheating, so the material in the projectionsoften and undergo diffusion.

(b) The normal force between the parts flat-tens these softened projections by plasticdeformation.

(c) Important factors are the nature of themating surfaces, the materials involved,the shape of the projections, the tempera-tures developed, the magnitude of the nor-mal force, and length of time.

12.35 What factors influence the shape of the upsetjoint in flash welding, as shown in Fig. 12.37b?

The important factors are the amount of heatgenerated (if too little heat, the material willnot deform to the required extent), the natureof the contracting surfaces (oxide layers, con-taminants, etc.), the force applied (the higherthe force, the greater the upset volume), the ex-posed length between the pieces and the clamps

(if too long, the part may buckle instead of be-ing upset), thermal conductivity (the lower theconductivity, the smaller the upset length), andthe rate at which the force is applied (the higherthe rate, the greater the force required for up-setting, due to strain-rate sensitivity of the ma-terial at elevated temperatures).

12.36 Explain how you would fabricate the structuresshown in Fig. 12.41b with methods other thandiffusion bonding and superplastic forming.

By the student. These structures can be madethrough a combination of sheet-metal formingprocesses (Chapter 7) and resistance welding,brazing, mechanical joining, or adhesive bond-ing. Note, however, that such complex partsand interfaces may not allow easy implementa-tion of these various operations without exten-sive tooling.

12.37 Make a survey of metal containers used forhousehold products and foods and beverages.Identify those that have utilized any of the pro-cesses described in this chapter. Describe yourobservations.

By the student. This is an interesting projectfor students. It will be noted that some foodand beverage containers are three-piece cans,with a welded seam along the length of the can;others may be soldered or seamed (see, for ex-ample, Fig. 12.53). These containers are typ-ically used for shaving cream, laundry starchsprays, and various spray cans for paints andother products.

12.38 Which process uses a solder paste? What arethe advantages to this process?

Solder paste is used in reflow soldering, de-scribed in Section 12.13.3, which is also used forsoldering integrated circuits onto printed circuitboards (Section 13.13).

12.39 Explain why some joints may have to be pre-heated prior to welding.

Some joints may have to be preheated prior towelding in order to:

(a) control and reduce the cooling rate, espe-cially for metals with high thermal con-ductivity, such as aluminum and copper,

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(b) control and reduce residual stresses devel-oped in the joint, and

(c) for more effective wave soldering (p. 778).

12.40 What are the similarities and differences be-tween casting of metals (Chapter 5) and fusionwelding?

By the student. Casting and fusion weldingprocesses are similar in that they both involvemolten metals that are allowed to recrystallize,cool, and solidify. The mechanisms are similarin that solidification begins with the formationof columnar grains (Section 5,3). The cooledstructure is essentially identical to a cast struc-ture with coarse grains. However, the weld joint(Fig. 12.15) is different in that selection of fillersand heat treatment (after welding) influence thejoint’s properties.

12.41 Explain the role of the excessive restraint (stiff-ness) of various components to be welded onweld defects.

Refer to Section 12.6.1. The effect of stiffnesson weld defects is primarily through the stressesdeveloped during heating and cooling of theweld joint. Note, for example, that not allowingfor contraction (such as due to a very stiff sys-tem) will cause cracks in the joint due to highthermal stresses (see Fig. 12.22).

12.42 Discuss the weldability of several metals, andexplain why some metals are easier to weld thanothers.

By the student. This is a challenging assign-ment and will require considerable effort. Re-view Section 12.62 and note that, as expected,weldability depends on many factors. See alsoTable 3.8 and the Bibliography at the end ofthis chapter.

12.43 Must the filler metal be of the same composi-tion as that of the base metal to be welded?Explain.

It is not necessary for the filler metal, rod, orwire to be the same as the base metal to bewelded. Filler metals are generally chosen forthe favorable alloying properties that they im-part to the weld zone. The only function thefiller metal must fulfill is to fill in the gaps in

the joint. The filler metal is typically an al-loy of the same metal, due to the fact that theworkpiece and the filler should melt at reason-ably close temperatures. To visualize why thisis the case, consider a copper filler used with amaterial with a much higher melting tempera-ture, such as steel. When the copper melts, thesteel workpiece is still in a solid state, and theinterface will be one of adhesion, with no sig-nificant diffusion between the copper and thesteel. (See also bottom of p. 743 and p. 773.)

12.44 Describe the factors that contribute to the dif-ference in properties across a welded joint.

By the student. An appropriate response willrequire the students to carefully review Sec-tion12.6.

12.45 How does the weldability of steel change as thesteel’s carbon content increases? Why?

By the student. Review Section 12.6.2. As thecarbon content increases, weldability decreasesbecause of martensite formation, which is hardand brittle (see p. 238).

12.46 Are there common factors among the weldabil-ity, solderability, castability, formability, andmachinability of metals? Explain, with appro-priate examples.

By the student. This is an interesting, but verychallenging, assignment and appropriate for astudent paper. As to be expected, the rela-tionships are complex, as can also be seen byreviewing Table 3.8 on p. 117. Note that forsome aluminum alloys, for example, machin-ability and weldability are opposite (i.e., D-Cvs. A ratings). The students should analyze thecontents of the following: Weldability - Section12.6.2; solderability - p. 777; castability - Sec-tions 5.4.2 and 5.6; formability - Sections 6.2.6and 7.7; machinability - Section 8.5.

12.47 Assume that you are asked to inspect a weldfor a critical application. Describe the proce-dure you would follow. If you find a flaw duringyour inspection, how would you go about deter-mining whether or not this flaw is important forthe particular application?

By the student. This is a challenging task, re-quiring a careful review of Section 12.6.1. Note,

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for example, that visual examination can detectsome defects, such as undercuts and toe cracks;however, underbead cracks or incomplete fusioncannot be detected visually. There are nonde-structive techniques (Section 4.8) for evaluat-ing a weld, acoustic and X-ray techniques be-ing the most common for determining porosityand large inclusions. Proof stressing a weld isa destructive approach, but it certainly can besuitable since defective welds cannot be placedin service safely.

Some analysis on flaw behavior and crack prop-agation in metal structures can be attempted,probably with finite-element methods or by us-ing advanced concepts for crack propagation.An understanding of the loads and the result-ing stresses often determines whether or not aflaw is important. For example, if the defect ina weld in a beam is at the neutral axis in bend-ing, the flaw is not likely to be critical. Onthe other hand, a defect in a highly loaded areaor in a stress concentration would raise seriousconcerns.

12.48 Do you think it is acceptable to differentiatebrazing and soldering arbitrarily by tempera-ture of application? Comment.

By the student. The definition is somewhat ar-bitrary. The temperature classification differ-entiates between the filler metals that can beused in thhe two processes. Note also that, withsoldering, thermal distortions are not as criticalbecause of the lower temperatures involved.

12.49 LoctiteR© is an adhesive used to keep metal boltsfrom vibrating loose; it basically glues the boltto the nut once the bolt is inserted in the nut.Explain how this adhesive works.

LoctiteR© is an anaerobic adhesive (see Table12.6 on p. 782), meaning that it cures in theabsence of oxygen, hence it does not solidify inair. Such a situation exists in the interfaces be-tween threaded fasteners and their nuts, as wellas pins and sleeves, so that the adhesive can beapplied to the threaded fastener and it does notcure until assembled. The students are encour-aged to also review the company literature.

12.50 List the joining methods that would be suitablefor a joint that will encounter high stresses and

cyclic (fatigue) loading, and rank the methodsin order of preference.

By the student. This is a challenging topicwhere the answers will depend on the work-piece materials that are being considered (seealso Table 12.1 on p. 733). Students should notbe limited to the answers given here, but shouldbe encouraged to rely upon their experience andtraining. However, some of the suitable meth-ods for such loadings are:

(a) Riveting is well-suited for such applica-tions, since the rivet can expand uponheading and apply compression to thehole; this can help arrest fatigue cracks.

(b) Bolts can be used for such applications;the use of a preload on a nut can lead tostiff joints with good fatigue resistance.

(c) Welding can be suitable, so long as theweld and the members are properly sized;fatigue crack propogation through theheat-affected zone is a concern.

(d) Brazing can be suitable for such applica-tions, depending on the materials to bebrazed.

(e) Adhesive bonding can also be suitable, aslong as the joints are properly designed(see Fig. 12.60 on p. 793). The mechanicalproperties of the adhesive is an importantconsideration, as well as the strength ofbond with the workpiece.

(f) Combinations of these methods are alsosuitable, such as combining adhesion withriveting as shown in Fig. 12..60d on p. 793.

12.51 Why is surface preparation important in adhe-sive bonding?

By the student. See Section 12.4.2. Surfacepreparation is important because the adhesivestrength depends greatly on its ability to prop-erly bond to a surface (see also Section 4.5). If,for example, there are lubricant residues on asurface, this ability is greatly hindered. As anexample, try sticking masking tape on a dusty,moist or greasy surface, or to your finger coatedwith a very thin layer of oil or grease.

12.52 Why have mechanical joining and fasteningmethods been developed? Give several specificexamples of their applications.

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By the student. Mechanical joining methods,described in Section 12.15, date back to 3000-2000 B.C., as shown in Table 1.1 on p. 3. Thesemethods have been developed mainly becausethey impart design flexibility to products, theygreatly ease assembly (especially disassembly,thus simplifying repair and part replacement),and have economic advantages.

12.53 Explain why hole preparation may be impor-tant in mechanical joining.

By the student. See Section 12.15.1. Note,for example, that if a hole has large burrs(see Fig. 7.5) it can adversely affect joint qual-ity, and also possibly causing crevice corrosion(p. 109). If the hole is significantly larger thanthe rivet, no compressive stress will be devel-oped on its cylindrical surface when the rivet isupset.

12.54 What precautions should be taken in mechani-cal joining of dissimilar metals?

By the student. In joining dissimilar metals,one must be careful about their possible chemi-cal interaction. Often, two dissimilar metals re-act in a cathodic process, causing galvanic cor-rosion and corrosive wear (see Section 3.9.7).This is especially a concern in marine applica-tions, where sea salt can cause major degrada-tion, as well as in chemical industries.

12.55 What difficulties are involved in joining plas-tics? What about in joining ceramics? Why?

By the student. See Section 12.16. Plastics canbe difficult to join. The thermal conductivity isso low that, if melted, plastics will flow beforethey resolidify; thermosets will not melt, butwill degrade as temperature is increased. Ther-moplastics are generally soft and thus cannotbe compressed very much in threaded connec-tions, so the bonds with these processes will notbe very strong. Thermoplastics are usually as-sembled with snap fasteners when strength isnot a key concern, or with adhesives. Ceramicscan be joined by adhesive bonding, and also bymechanical means in which the brittleness andnotch sensitivity of these materials are impor-tant concerns.

12.56 Comment on your observations concerning the

numerous joints shown in the figures in Section12.17.

By the student. The students may respond tothis question in different ways. For example,they can compare and contrast adhesive bondedjoints with those of welded and mechanicallyassembled joints. Note also the projected areaof the joints, the type of materials used, theirgeometric features, and the locations and direc-tions of the forces applied.

12.57 How different is adhesive bonding from otherjoining methods? What limitations does ithave?

By the student. Review Section 12.14. Adhe-sive bonding is significantly different from otherjoining methods in that the workpiece materi-als are of various types, there is no penetrationof the workpiece surfaces, and bonding is doneat room temperature. Its main limitations arethe necessity for clean surfaces, tight clearances,and the longer times required.

12.58 Soldering is generally applied to thinner com-ponents. Why?

Solders have much lower strength than brazefillers or weld beads. Therefore, in joining mem-bers to be subjected to significant loads, whichis typical of members with large thickness, onewould normally consider brazing or welding,but not soldering. A benefit of soldering whenjoining thin components is that it takes place atmuch lower temperatures than brazing or weld-ing, so that one does not have to be concernedabout the workpiece melting due to localizedheating, or significant warping in the joint area.

12.59 Explain why adhesively bonded joints tend tobe weak in peeling.

Adhesives are weak in peeling because there isa concentrated, high tensile stress at the tip ofthe joint when being peeled (see Fig. 12.50);consequently, their low tensile strength reducesthe peeling forces. (Recall that this situationis somewhat analogous to crack initiation andpropagation in metals under tensile stresses; seeFig. 3.30.) Note, however, that tougher adhe-sives can require considerable force and energyto peel, as can be appreciated when trying topeel off some adhesive tapes.

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12.60 Inspect various household products, and de-scribe how they are joined and assembled. Ex-plain why those particular processes were used.

By the student. Metallic food containers aregenerally seamed from sheet. Knife blades areoften riveted/bonded to their handles. Somepots and pans have a number of cold-weldedlayers of sheet, which are then deep drawn andformed to desired shapes. The reason pots havea number of layers different materials is to com-bine their desirable qualities, such as high ther-mal conductivity of copper with the strengthand ease of cleaning of stainless steel. Handleson pots and pans are typically spot welded, riv-eted, or assembled with threaded fasteners. Allof these processes meet functional, technologi-cal, economic, or aesthetic requirements.

12.61 Name several products that have been assem-bled by (a) seaming, (b) stitching, and (c) sol-dering.

By the student. Note, for example:

(a) Products assembled by seaming are foodcontainers and tops of beverage cans.

(b) Products made through stitching arecardboard and wood boxes, insulationand other construction materials, andfootwear.

(c) Soldered parts include electrical compo-nents such as diodes attached to circuitboards, pipe fittings, and electrical termi-nals.

12.62 Suggest methods of attaching a round bar madeof thermosetting plastic perpendicularly to aflat metal plate.

By the student. Consider, for example, the fol-lowing methods:

(a) Threading the end of the rod, drilling andtapping a hole into the plate, and screwingthe rod in, using a sealer if necessary.

(b) Press fit.

(c) Riveting the rod in place.

(d) Fittings can be employed.

12.63 Describe the tooling and equipment that arenecessary to perform the double-lock seaming

operation shown in Fig. 12.53, starting with flatsheet. (See also Fig. 7.23.)

By the student. With some search of the techni-cal literature and the Bibliography given at theend of Chapter 7, the students should be ableto describe designs and equipment required forperforming this operation.

12.64 What joining methods would be suitable toassemble a thermoplastic cover over a metalframe? Assume that the cover has to be re-moved periodically.

By the student. Because the cover has to beremoved periodically, the most feasible joiningmethod is simply snapping the lid on, as is doneon numerous food products (such as polypropy-lene lids on shortening or coffee cans) which, af-ter opening, can easily be resealed. The sealingis due to the elastic recovery of the lid after it isstretched over the edge of the container. Note,however, that at low temperatures (even in therefrigerator) the lid may crack due to lack ofsufficient ductility and severe notch sensitivityof the plastic. The students are encouraged toelaborate further.

12.65 Repeat Question 12.64, but for a cover made of(a) a thermosetting plastic, (b) metal, and (c)ceramic. Describe the factors involved in yourselection of methods.

By the student. Consider the following sugges-tions:

(a) For part (a):

i. A method similar to Answer 12.64above, since thermosetting plasticsalso have some small elastic recovery.

ii. Some mechanical means.iii. Methods would include snap fits.iv. Threaded interfaces.

(b) For (b):

i. Similar to (a) above, especiallythreaded interfaces, such as screwcaps on bottles.

(c) For (c):

i. The generally low ductility of ceram-ics would be a significant concern asthe cover may crack under repeated

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tensile hoop stresses involved in theiruse. The students are encouraged torespond to the question as to why aceramic cover may even be necessaryif the container is made of metal.

12.66 Do you think the strength of an adhesivelybonded structure is as high as that obtainedby diffusion bonding? Explain.

By the student. Because they rely on bondstrength, the joint strength in adhesivelybonded joints is usually not as high as thatachieved through diffusion bonding. Diffusionbonding (Section 12.12) is exceptional in thatthe two components, typically metals, are dif-fused into each other, making the joint verystrong. Adhesives are generally not as strongas the material they bond (see Table 12.6 onp. 782), unless the materials are inherentlyweak, such as paper, cardboard, and some plas-tics. (See also Section 4.4.)

12.67 Comment on workpiece size limitations, if any,for each of the processes described in this chap-ter.

By the student. This is a good topic for aproject. Basically, large parts can be accom-modated in these processes by appropriate fix-turing. Small parts, on the other hand, maybe delicate and thin, hence will require carefulhandling. Leads for electronic components aregenerally soldered; the wires are typically muchsmaller than 1-mm in diameter. (See also Table12.1 on p. 733.)

12.68 Describe part shapes that cannot be joined bythe processes described in this chapter. Givesspecific examples.

By the student. A review of the various figuresand illustrations in this chapter will clearly in-dicate that part shape is not a significant diffi-culty in joining processes. The basic reason isthat there is such a very wide variety of pro-cesses and possibilities available. The studentis encouraged to think of specific illustrationsof parts that may negate this statement. Inrare cases, if a part shape, as designed, is notsuitable for joining with other components, itsshape could indeed be modified to enable its as-

sembly with other components (see also designfor assembly, Section 14.11).

12.69 Give several applications of electrically con-ducting adhesives.

By the student. See also Section 12.14.4 whereseveral examples are given.

12.70 Give several applications for fasteners in vari-ous household products, and explain why otherjoining methods have not been used instead.

By the student. The students are encour-aged to carefully inspect the variety of productsavailable and to review Sections 12.15, 12.17.4,and 14.10. Note that fasteners are commonlyused in many household products, such as cof-fee makers, electric irons, appliances, furniture,which greatly facilitate assembly, as well as dis-assembly.

12.71 Comment on workpiece shape limitations, ifany, for each of the processes described in thischapter.

By the student. See Question 12.67 and notethat it pertained to size limitations, whereasthis question concerns shapes. Refer also todesign variability in Table 12.1 on p. 733 andwelding position in Table 12.2. Although thereare some limitations, these are often associatedwith fixturing requirements. Consider the fol-lowing: Roll bonding is generally used withsheet metals, so parts that do not involve thinlayers are difficult to roll bond. Ultrasonic weld-ing is typically restricted to thin foils. Frictionwelding requires parts be mounted into chucksor similar fixtures in order to be able to rotateone of the comments to be joined. Spot weld-ing operations can handle complex shapes byappropriate design of electrode holders. Diffu-sion bonding can produce complex shapes, ascan brazing and mechanical fastening.

12.72 List and explain the rules that must be fol-lowed to avoid cracks in welded joints, such ashot tearing, hydrogen-induced cracking, lamel-lar tearing, etc.

By the student. See Section 12.6.1 where allrelevant parameters are discussed.

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12.73 If a built-up weld is to be constructed (seeFig. 12.5), should all of it be done at once, orshould it be done a little at a time, with suffi-cient time allowed for cooling between beads?

With proper welding techniques (see also slaginclusions in Section 12.6.1) and care, the weldjoint can be built continuously, as this proce-dure will prevent excessive oxidation betweenbead interfaces, as well as reducing weld timeand thus making the process economical.

12.74 Describe the reasons that fatigue failure gener-ally occurs in the heat-affected zone of weldsinstead of through the weld bead itself.

Fatigue failure and crack propagation (see Sec-tions 2.7 and 3.8) are complex phenomena. Re-call that the base metal of the workpiece is of-ten a wrought product with varying degrees ofcold work. Thus, the base metal usually hasgood fatigue resistance. The weld zone itselfis highly alloyed, with high strength and alsogood fatigue resistance. However, the heat af-fected zone adjacent to the weld does not havethe advantageous metallurgy of the weld northe microstructure of the worked base metal;it has a large grained, equiaxed strucure (seeFig. 12.15 on p. 749). In addition, there is astress concentration associated with the weld,and the heat affected zone is generally in a vol-ume that is highly stressed. Thus, it is not sur-prising that the heat affected zone is the usualsite of fatigue failure.

12.75 If the parts to be welded are preheated, is thelikelihood that porosity will form increased ordecreased? Explain.

Weld porosity arises from a number of sources,including micropores (similar to those foundin castings; see Section 5.12.1), entrained orevolved gases, and bridging and cracking. Ifthe part is preheated, bridging and cracking arereduced and the cooling rate is lower, there-fore large shrinkage pores are less likely. How-ever, since cooling is slower with preheat, sol-uble gases may be more likely to be entrainedunless effective shielding gases are used.

12.76 What is the advantage of electron-beam andlaser-beam welding, as compared to arc weld-ing?

The main advantages of these processes are as-sociated with the very small weld zone, and thelocalized energy input and small heat-affectedzone. Weld failures, especially by fatigue, oc-cur in the heat-affected zone; thus, minimiz-ing this volume reduces the likelihood of largeflaws and rapid crack growth. Also, the low en-ergy input means that thermal distortions andwarping associated with these processes is muchlower than with arc welding.

12.77 Describe the common types of discontinuities inwelds, explain the methods by which they canbe avoided.

By the student. Note that discontinuities inwelds are discussed in Section 12.6. Some ofthe common defects are porosity, inclusions, in-complete fusion/penetration, underfilling, un-dercutting, overlaps, and cracks. The methodsby which they can be avoided are discussed inSection 12.6.1.

12.78 What are the sources of weld spatter? How canspatter be controlled?

Weld spatter arises from a number of sources.If the filler metal is a powder, errant parti-cles can strike the surface and loosely adhereto the surface, similar to the thermal spray-ing process (pp. 156-157). Even a continuouselectrode will spatter, as a violently evolvingor pumped shielding gas can cause the moltenmetal to emit droplets, which then adhere tothe workpiece surface near the weld zone.

12.79 Describe the functions and characteristics ofelectrodes. What functions do coatings have?How are electrodes classified?

By the student. The functions of electrodes in-clude:

(a) Serve as part of the electrical circuit deliv-ering the power required for welding.

(b) Melt and provide a filler metal.

(c) Have a coating or core that provides ashielding gas and flux.

(d) Help stabilize the arc and make the pro-cess more robust.

There are many characteristics of electrodesand the student is encouraged to develop an

200






appropriate list, noting the two classes of elec-trodes in arc welding processes: Consumableand nonconsumable. Students will need to per-form a literature search to determine classifi-cation of electrodes. For example, the follow-ing is taken from Hamrock, Schmid, and Ja-cobson, Fundamentals of Machine Elements, 2ded., McGraw-Hill, 2004.

Ultimatetensile Yield Elonga-

Electrode strength, strength, tion, ek,number Su, ksi Sy, ksi percentE60XX 62 50 17-25E70XX 70 57 22E80XX 80 67 19E90XX 90 77 14-17E100XX 100 87 13-16E120XX 120 107 14

12.80 Describe the advantages and limitations of ex-plosion welding.

Explosion welding is discussed in Section 12.11.The main advantage is that very dissimilar ma-terials can be bonded, producing high jointstrength, as well as specialized applications.The basic limitation is that it is a basically verydangerous operation.

12.81 Explain the difference between resistance seamwelding and resistance spot welding.

By the student. The difference between resis-tance seam welding and resistance spot weldingis in the spacing of the weld nuggets (see Sec-tions 12.10.1 and 12.10.2). If the nuggets over-lap, it is a seam weld; if they do not overlap, itis a spot weld.

12.82 Could you use any of the processes described inthis chapter to make a large bolt by welding thehead to the shank? (See Fig. 6.17.) Explain theadvantages and limitations of this approach

By the student. Note that processes such asarc welding and gas welding, as well as frictionwelding, can be used to join the two compo-nents. However, the advantage of the latter isthat the weld is over the entire contact area be-tween the two joined components, instead of asmall bead along the periphery of the contactlocation. Brazing is another method of join-ing the two components. The advantages are

that unique designs can be incorporated andusing different materials; the process is econom-ical for relatively few parts. The limitations arethe higher production times required, includingsubsequent finishing operations, as compared toheading operations, which is a common processfor making bolt heads.

12.83 Describe wave soldering. What are the advan-tages and disadvantages to this process?

Wave soldering, described on p. 778, involvesmoving a circuit board with inserted compo-nents over a stationary wave of solder, as shownin Fig. 12.48. The basic advantage to thisprocess is that it can simultaneously producea number of high-quality joints inexpensively.The main drawback is that it places restrictionson the layout of integrated circuit packages ona circuit board.

12.84 What are the similarities and differences be-tween a bolt and a rivet?

By the student. Bolts and rivets are very simi-lar in that two or more components are joinedby a mechanical means. Both preload the com-ponents to function in highly stressed joints.The main difference is that a bolt uses a threadand can thus be disassembled; a rivet is up-set and disassembly requires destruction of therivet.

12.85 It is common practice to tin plate electrical ter-minals to facilitate soldering. Why is tin a suit-able material?

Note in Table 12.5 on p. 777 that solders thatare suitable for general purpose and for elec-tronics applications are lead-tin alloys. Thus,the surface tension of the molten solder withthe tin plate will be very low, thus allowinggood wetting by the solder and resulting in agood joint.

12.86 Review Table 12.3 and explain why some ma-terials require more heat than others to melt agiven volume.

Refer to Section 3.9.2. Recall that the meltingpoint of a metal depends on the energy requiredto separate its atoms, thus it is a characteristicof the individual metal. For an alloy, it dependson the melting points of the individual alloyingelements. Additional factors are:

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Problems

12.87 Two flat copper sheets (each 1.5 mm thick) arebeing spot welded by the use of a current of7000 A and a current flow time of 0.3 s. Theelectrodes are 5 mm in diameter. Estimate theheat generated in the weld zone. Assume thatthe resistance is 200 µΩ.

This problem is very similar to Example 12.5 onp. 765. Note in Eq. (12.6) that the quantitiesnow are I = 7000 A and t = 0.3 s. As in theexample, the resistance is 200 µΩ. Therefore,

H = (7000)2(0.0002)(0.3) = 2940 J

As in the example, we take the weld nuggetvolume to be the projected volume below theelectrode, or

V =π

4d2t =

π

4(5)2(3) = 58.9 mm3

From Table 12.3 on p. 737, the specific en-ergy needed to melt copper is u = 6.1 J/mm3.Therefore, the heat needed is

Hmelt = (58.9)(6.1) = 359 J

The remaining heat (that is, 2940-359 J = 2581J) is dissipated into the volume of metal sur-rounding the weld nugget.

12.88 Calculate the temperature rise in Problem12.87, assuming that the heat generated is con-fined to the volume of material directly betweenthe two electrodes and that the temperaturedistribution is uniform.

The volume of metal directly under the 5-mmelectrodes is

V =π

4d2t =

π

4(5)2(3) = 58.9 mm3,

and this volume has a mass of (58.9)(0.00897)= 0.53 g = 0.00053 kg. The specific heat forcopper is 385 J/kgK. Therefore, the theoreticaltemperature rise is

∆T =2940 J

(385 J/kgK)(0.00053 kg)= 14, 400 K

Note that the melting point of copper is 1082C(1355 K), thus much more energy has been

provided than is needed for this small volume.Clearly, in practice, very little of the heat is con-centrated in this small volume. A more elabo-rate model of temperature distributions is pos-sible, but beyond the scope of this book. Textssuch as Carslaw, H.S., and Jaeger, J.C., Con-duction of Heat in Solids, Oxford UniversityPress, 1959, address such problems in detail.

12.89 Calculate the range of allowable currents forProblem 12.87, if the temperature should bebetween 0.7 and 0.85 times the melting tem-perature of copper. Repeat this problem forcarbon steel.

This problem can be interpreted as between0.7 and 0.85 times the melting temperature onan absolute (Kelvin) or a Celsius temperaturescale. This solution will use a Celsius scale,so that the final target temperature is between765 and 925 C. Using the same approach as inProblem 12.87, the allowable energy for thesecases is 100 and 121 J, respectively. With a re-sistance of 200 µΩ, the currents are 1310 and1420 A, respectively. The solution for carbonsteel is left for the student to supply, but usesthe same approach.

12.90 In Fig. 12.24, assume that most of the top por-tion of the top piece is cut horizontally with asharp saw. Thus, the residual stresses will bedisturbed, and, as described in Section 2.10, thepart will undergo shape change. For this case,how will the part distort? Explain.

Inspecting Fig. 12.24 and recalling Answer 2.25regarding Fig. 2.30, we arrive at the followingobservations and conclusions: (1) The top por-tion of the top piece is subjected to longitudinalcompressive residual stresses. (2) If we cut thisportion with a sharp saw (so that we do not in-duce further residual stresses during cutting),stresses will rearrange themselves and the partwill bend downward, i.e., it will hold water, as-suming it will not warp in the plane of the page.For details, recall the spring analogy in Prob-lem 2.25.

202






12.91 The accompanying figure shows a metal sheavethat consists of two matching pieces of hot-rolled, low-carbon-steel sheets. These twopieces can be joined either by spot welding or byV-groove welding. Discuss the advantages andlimitations of each process for this application.

Spot weld

2 716 in.

0.135 in.

V-groove weld

(b) (c)

(a)

By the student. The original method of joiningthe two sheaves was by resistance spot weld-ing, with 16 welds equally spaced around theperiphery, as shown in (b). Although the weldquality was satisfactory, the welding time persheave was 1 minute. In order to increase pro-duction rate, an alternative process was chosen(gas-metal-arc welding, GMAW) with a contin-uous weld around the periphery of the sheave asshown in (c). With an automated welding pro-cess, the welding time per sheave was reducedto 40 s.

12.92 A welding operation takes place on analuminum-alloy plate. A pipe 50-mm in diam-eter with a 4-mm wall thickness and a 60-mmlength is butt-welded onto a section of 15 x 15x 5 mm angle iron. The angle iron is of an L-shape and has a length of 0.3 m. If the weldzone in a gas tungsten-arc welding process isapproximately 8 mm wide, what would be thetemperature increase of the entire structure dueto the heat input from welding only? What ifthe process were an electron-beam welding op-eration with a bead width of 6 mm? Assume

that the electrode requires 1500 J and the alu-minum alloy requires 1200 J to melt one gram.

For the first part of the problem, assume thatthe electrode is placed around the entire pipe,so that the weld length is πD = π(50 mm) =0.157 m. If the weld cross section is triangular,its volume is approximately

V =12bhL =

12(0.008 m)2(0.157 m)

or V = 5.02× 10−6 m3 = 5020 mm3. The elec-trode material should be matched to aluminum,so it will likely be an aluminum alloy in orderto approximately match melting temperaturesand compatibility. The density should thereforebe around 2700 kg/m3 (see Table 3.3 on p. 106,where it is also noted that C = 900 J/kg-K).The specific heat to melt aluminum alloys isgiven by Table 12.3 as 2.9 J/mm3. Therefore,the energy input is (2.9)(5020) = 14.5 kJ. Thetotal volume of the aluminum is

V =π

4(d2

o − d2i )L+ 2btl

=π

4(502 − 422)(60) + 2(15)(5)(300)

= 79, 683 mm3

or V = 7.968× 10−5 m3. The temperature riseis then calculated as:

E = ρV C∆T

Solving for ∆T ,

∆T =E

ρV C

=14, 500

(2700)(7.968× 10−5)(900)

Or ∆T = 75C. For the second part of the prob-lem, the change to be made is in the input en-ergy. Using the same approach as above, wehave

V =12bhL =

12(0.006 m)2(0.157 m)

or V = 2.826×10−6 m3=2826 mm3. The inputenergy is (2.9)(2826)=8.20 kJ. The temperaturerise is therefore

∆T =E

ρV C

=8200

(2700)(7.968× 10−5)(900)= 42C

203






12.93 A shielded metal arc welding operation is tak-ing place on carbon steel to produce a fillet weld(see Fig. 12.21b). The desired welding speed isaround 25 mm/sec. If the power supply is 10V, what current is needed if the weld width isto be 7 mm?

Since its width is 7 mm, the cross-sectional areaof the weld is A = 1

2 (7 in2) = 24.5 mm2 =2.45×10−5 m2. For shielded metal arc welding,we obtain from Section 12.3.1 that C = 75%.From Table 12.3, u is assigned a mean value of9.7 J/mm3. From Eq. (12.5), the weld speed istherefore calculated as

v = eV I

uA

Solving for the current, I,

I =uvA

eV=

(9.7)(25)(24.5)(0.75)(10)

or I = 792 A.

12.94 The energy applied in friction welding is givenby the formula E = IS2/C, where I is the mo-ment of inertia of the flywheel, S is the spindlespeed in rpm, and C is a constant of propor-tionality (5873, when the moment of inertia isgiven in lb-ft2). For a spindle speed of 600 rpmand an operation in which a steel tube (3.5 in.OD, 0.25 in. wall thickness) is welded to a flatframe, what is the required moment of inertiaof the flywheel if all of the energy is used toheat the weld zone (approximated as the mate-rial 0.25 in. deep and directly below the tube)?Assume that 1.4 ft-lbm is needed to melt theelectrode.

The flywheel moment of inertia can be calcu-lated as:

E =IS2

C

Solving for I,

I =EC

S2=

(1.4)(5873)(600)2

= 0.0228 lb-ft2

12.95 In oxyacetylene, arc, and laser-beam cutting,the processes basically involve melting of theworkpiece. If an 80 mm diameter hole is tobe cut from a 250 mm diameter, 12 mm thickplate, plot the mean temperature rise in the

plate as a function of kerf. Assume that one-half of the energy goes into the plate and one-half goes into the blank.

The volume melted is

V = (πD)th = π(80 mm)(12 mm)t = 30106t

where t is the kerf width in mm. The energyinput is then E = uV/2 = 1508ut, where u isthe specific energy required to melt the work-piece, as given in J/mm3 in Table 12.1. Notethat we have divided the energy by two becauseonly one-half of the energy goes into the blank.The volume of the blank is

V =π

4d2h

=π

4

[(250 mm)2 − (80 mm)2

](12 mm)

= 5.29× 10−4 m3

The temperature rise in the blank is ∆T =E/ρV Cp; substituting for the input energy,

∆T =1508ut

ρCp(5.29× 10−4)

= (2.85× 106)t(

u

ρCp

)It can be seen that the plot of temperature riseis a linear function of width, t. This is plottedbelow for selected materials.

500

400

300

200

100

0Avg

. Tem

p. I

ncre

ase

in b

lank

(°C

)

0 10 20 30Kerf width (mm)

AlCuSteelTi

12.96 Refer to the simple butt and lap joints shownin Fig. 12.1. (a) Assuming the area of the buttjoint is 3 mm × 20 mm and referring to the ad-hesive properties given in Table 12.6, estimatethe minimum and maximum tensile force thatthis joint can withstand. (b) Estimate theseforces for the lap joint assuming its area is 15mm × 15 mm.

204






Referring to Table 12.6 on p. 782, note thatthe lowest adhesive strength is for epoxy orpolyurethane at 15.4 MPa, and the highesttension-shear strength is for modified acrylic at25.9 MPa. These values are used in the solutionbelow.

(a) For a butt joint, assuming there is strongadhesion between the adhesive and work-piece, the full strength of the adhesive canbe developed. In this case, we can calcu-late the required load-bearing area as

A = (3)(20) = 60 mm2 = 6.0× 10−5 m2

Consequently, we have

Fmin = (15.4× 106)(6.0× 10−5) = 924 N

and

Fmax = (25.9×106)(6.0×10−5) = 1554 N

(b) For the lap joint, we similarly obtain A =(15)(15) = 225 mm2 = 2.25 × 10−4 m2.Note that in this case, the joint is loadedin shear, and the shear strength is one-half the tensile strength, as discussed incourses on mechanics of solids. Therefore,

Fmin =12(15.4× 106

) (2.25× 10−4

)or Fmin = 1730 N. Also,

Fmax =12(25.9× 106

) (2.25× 10−4

)or Fmax = 2910 N.

12.97 As shown in Fig. 12.61, a rivet can buckle if itis too long. Using information from solid me-chanics, determine the length-to-diameter ratioof a rivet that will not buckle during riveting.

The riveting process is very similar to head-ing (see Section 6.2.4). Basically, the designrequirement is that the length-to-diameter ra-tio should be 3 or less. If the heading tool has acontrolled geometry, a longer length can be ac-commodated if the head diameter is not morethan 1.5 times the shank diameter.

12.98 Repeat Example 12.2 if the workpiece is (a)magnesium, (b) copper or (c) nickel.

(a) For the magnesium workpiece, Table 12.3gives u = 2.9 J/mm3. Therefore, fromEq. (12.5),

v = eV I

uA= (0.75)

(20)(200)(2.9)(30)

= 34.5 mm/s.

(b) For the copper workpiece, we have u = 6.1J/mm3. Therefore, from Eq. (12.5),

v = eV I

uA= (0.75)

(20)(200)(6.1)(30)

= 16.4 mm/s.

(c) For the nickel workpiece, we have u = 9.8J/mm3. Therefore, from Eq. (12.5),

v = eV I

uA= (0.75)

(20)(200)(9.8)(30)

= 10.2 mm/s.

12.99 A submerged arc welding operation takes placeon 10 mm thick stainless steel, producing a buttweld as shown in Fig. 12.20c. The weld geom-etry can be approximated as a trapezoid with15 mm and 10 mm as the top and bottom di-mensions, respectively. If the voltage providedis 40 V at 400 A, estimate the welding speed ifa stainless steel filler wire is used.

A sketch of the weld cross section is shown be-low.

15

10

10

The area of the trapezoid is

A = (10)(10) + 2(

12

)(2.5)(10) = 125 mm2

For submerged arc welding, it is stated in Sec-tion 12.3.1 that e = 0.90. For a stainless steelworkpiece, the unit specific energy is obtainedfrom Table 12.3 as u = 9.4 J/mm3. Therefore,from Eq. (12.5),

v = eV I

uA

= 0.9(40)(400)(9.4)(125)

= 12.2 mm/s

205








Design

12.101 Design a machine that can perform frictionwelding of two cylindrical pieces, as well asremove the flash from the welded joint. (SeeFig. 12.30.)

By the student. Note that this machine can bevery similar to a lathe, where one-half of theworkpiece is held in a fixture attached to thetailstock, the other half is in the rotating chuck,and a cutting tool is used as in turning.

12.102 How would you modify your design in Problem12.101 if one of the pieces to be welded is non-circular?

By the student. The machine is more com-plicated, machining can become more difficult,with essentially a milling operation taking placeafter welding.

12.103 Describe product designs that cannot be joinedby friction welding processes.

By the student. Consider, for example, that ifone of the components is a very thin tube, it willnot be able to support the large axial loads in-volved in friction welding; likewise, if the othercomponent is very think and slender.

12.104 Make a comprehensive outline of joint designsrelating to the processes described in this chap-ter. Give specific examples of engineering ap-plications for each type of joint.

By the student. Refer to Section 12.17. Thisis a challenging problem, and would be suitablefor a project or a paper.

12.105 Review the two weld designs in Fig. 12.58a, and,based on the topics covered in courses on thestrength of materials, show that the design on

the right is capable of supporting a larger mo-ment, as shown.

The problem statement assumes that the failurein the part on the left will be in the weld itself;if the material strength determines the momentthat can be supported, then the weld designis irrelevant. Assuming that the weld zone isroughly square, it is better to place the welds asshown on the right because the strength arisesfrom the cube of the distance from the neutralaxis. In the design on the left, only the extremeends are fully loaded, and some material (at theneutral axis) is subjected to very little stress.

12.106 In the building of large ships, there is a needto weld large sections of steel together to forma hull. For this application, consider each ofthe welding operations described in this chap-ter, and list the benefits and drawbacks of thatoperation for this product. Which welding pro-cess would you select? Why?

By the student. This specialized topic is verysuitable for a student paper, requiring a searchof the technical literature in shipbuilding tech-nologies. For example, the following may besuggested:

206






Process Advantages DisadvantagesOxyfuel Inexpensive;

portable.Significant jointdistortion; weld-ing of thicksections is diffi-cult.

SMAW Inexpensive;portable.

Thick sectionsneed a built-up weld (seeFig. 12.5 onp. 738), possiblycompromisingjoint strength.

SAW Good weldstrength; can beautomated

Limitedworkspace;workpiece mustbe horizontal, adifficult restric-tion for boathulls

ESW Good weldstrength, well-suited for verticalwelds.

More compli-cated equipmentrequired.

12.107 Examine various household products, and de-scribe how they are joined and assembled. Ex-plain why those particular processes are usedfor these applications.

By the student. Consider the following: Metal-lic food containers that are seamed from sheet.Knife blades that are riveted/bonded to theirhandles. Pots with a number of cold-weldedlayers of sheet, which are then deep drawn andformed to desired shapes. All of these pro-cesses are used because other processes whichare technologically feasible may lack economic,functional, or aesthetic advantages.

12.108 A major cause of erratic behavior (hardwarebugs) and failure of computer equipment is fa-tigue failure of the soldered joints, especially insurface-mount devices and devices with bondwires. (See Fig. 12.48.) Design a test fixturefor cyclic loading of a surface-mount joint forfatigue testing.

By the student. This is a very demandingproject, and can be expanded into a groupdesign project. Students can consider if thetest should duplicate the geometry of a surfacemount or if an equivalent geometry can be ana-lyzed. They can determine loading cycle dura-tions and amplitudes, as well as various othertest parameters.

12.109 Using two strips of steel 1 in. wide and 8in. long, design and fabricate a joint that givesthe highest strength in a tension test in the lon-gitudinal direction.

By the student. This is a challenging problemand an experimental project, as well; it couldalso be made into a contest among students inclass. It must be noted, however, that the thick-ness of the strips is not given in the statement ofthe problem (although the word strip generallyindicates a thin material). The thickness is afactor that students should recognize and com-ment on, and supply their answers accordingly.It can also be seen that most of the processesdescribed in Chapter 12 can be used for such ajoint. Consequently, a wide variety of processesand designs should be considered, making theresponse to this question extensive.

In using a single bolt through the two strips,for example, it should be apparent that if thebolt diameter is too large, the stresses in therest of the cross section may be too high, caus-ing the strips to fail prematurely. If, on theother hand, the bolt diameter is too small, itwill easily shear off under the applied tensileforce. Thus, there has to be an optimum tobolt size. The students are encouraged to con-sider multiple-bolt designs, as well as a host ofother processes either singly or in combination.

12.110 Make an outline of the general guidelines forsafety in welding operations. For each of theoperations described in this chapter, prepare aposter which effectively and concisely gives spe-cific instructions for safe practices in welding(or cutting). Review the various publicationsof the National Safety Council and other simi-lar organizations.

By the student. This is a valuable study by thestudents, and the preparation of a poster or aflyer is a good opportunity for students. Safetyin Welding is a standard published by theAmerican National Standards Institute (ANSIZ49.1) and describes in detail the safety pre-cautions that must be taken. Most of the stan-dards are process-specific. As an example, somesafety guidelines for shielded metal-arc weldingare:

• The operator must wear eye and skin pro-

207






tection against radiation.

• Leather gloves and clothing should beworn to prevent burns from arc spatter.

• Welding should be done in properly venti-lated areas, where fresh air is available toworkers and the work area is not floodedby shielding gases.

• To prevent electric shock, the weldershould not weld while standing on a wetsurface.

• The workpiece should be positioned tominimize trauma to the back and arms.

12.111 A common practice for repairing expensive bro-ken or worn parts, such as may occur when, forexample, a fragment is broken from a forging,is to fill the area with layers of weld bead andthen to machine the part back to its originaldimensions. Make a list of the precautions thatyou would suggest to someone who uses thisapproach.

By the student. Considerations are that the in-terface between the forging and the filling maynot have sufficient strength. The weld beadwill have different properties than the substrate(the forging) and have an uneven surface, thusmachining may result in vibration and chat-ter. The weld material may cause the cuttingtools to wear more rapidly. The weld may frac-ture during machining and compromise part in-tegrity. The weld material may have insufficientductility and toughness for the application.

12.112 In the roll bonding process shown in Fig. 12.28,how would you go about ensuring that the in-terfaces are clean and free of contaminants, sothat a good bond is developed? Explain.

By the student. The students are encouragedto perform a literature search for particular ap-proaches. The basic procedure has been (a)wire brushing the surfaces, which removes ox-ide from the surfaces, and (b) solvent cleaning,which removes residues and organic films fromthe surface. (See also Section 4.5.2.)

12.113 Alclad stock is made from 5182 aluminum al-loy, and has both sides coated with a thin layerof pure aluminum. The 5182 provides high

strength, while the outside layers of pure alu-minum provide good corrosion resistance, be-cause of their stable oxide film. Alclad is com-monly used in aerospace structural applicationsfor these reasons. Investigate other commonroll bonded materials and their uses, and pre-pare a summary table.

By the student. This topic could be a chal-lenging project for students. Examples includecoinage (see also Question 12.24) and a thincoating of metals on workpieces where the coat-ing serves as a solid lubricant in metalworking(see p. 152).

12.114 Obtain a soldering iron and attempt to soldertwo wires together. First, try to apply the sol-der at the same time as you first put the solder-ing iron tip to the wires. Second, preheat thewires before applying the solder. Repeat thesame procedure for a cool surface and a heatedsurface. Record your results and explain yourfindings.

By the student. This is a valuable and inex-pensive laboratory experience, showing the im-portance of surface tension. With cold wires,molten solder has high surface tension againstthe wires, and thus the solder does not wet thesurface. At elevated temperatures, the solderhas low surface tension and the solder coats thewire surfaces very effectively. Students can beasked to examine this phenomenon further byplacing a small piece of solder of known volume(which can be measured with a precision scale)on a steel plate section. When heated, the sol-der spreads according to the surface tempera-ture of the steel. It will be noted that above athreshold value, the solder will flow freely andcoat the surface.

12.115 Perform a literature search to determine theproperties and types of adhesives used to affixartificial hips onto the human femur.

By the student. Sometimes an adhesive isused, but with some designs this is not nec-essary, as they rely upon osteointegration orbone-ingrowth to affix the implant. Usually thecement is polymethylmethacrylate, an acrylicpolymer often referred to as bone cement, orelse a hydroxyapetite polymer is used. Newmaterials are constantly being developed and

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a number of variations can be found in the lit-erature and through an Internet search. A com-mon trend is to develop cements from calciumphosphates, as these are closer matches to themineral content of bone.

12.116 Using the Internet, investigate the geometry ofthe heads of screws that are permanent fasten-ers (that is, ones that can be screwed in but notout).

By the student. These heads usually present astraight vertical surface for the screwdriver inone direction, but a curved surface in the op-posite direction, so that a screwdriver simplyslips when turned counterclockwise and is noteffective for unscrewing. The sketch on the leftwas obtained from www.k-mac-fasteners.com,while the photo on the right was obtained fromwww.storesonline.com.

12.117 Obtain an expression similar to Eq. (12.6), butfor electron beam and laser welding.

By the student. The heat input is generallygiven by

H = cIA

where c is a constant that indicates the portionof laser energy absorbed by the material, and Iis the intensity of light over the area A.

209






210






Chapter 13

Fabrication of Microelectronic,Micromechanical, andMicroelectromechanical Devices;Nanomanufacturing

Questions

13.1 Define the terms wafer, chip, device, integratedcircuit, and surface mount.

A wafer is a slice of a thin cylinder of silicon.A chip is a fragment of a wafer. A deviceis either a (1) micromechanical arrangementwithout any integrated electronic circuitry or a(2) simple electronic element such as a transis-tor. An integrated circuit is a semiconductor-based design, incorporating large amounts ofelectronic devices.

13.2 Why is silicon the most commonly used semi-conductor in IC technology? Explain.

The reason is its unique capabilities regardingthe growth of oxides and deposition of metalcoatings onto the oxides, so that metal on ox-ide semiconductors can be easily fabricated.

13.3 What do the terms VLSI, IC, CVD, CMP, andDIP stand for?

VLSI - Very large scale integration; IC - inte-grated circuit; CVD - chemical vapor deposi-tion; CMP - chemical mechanical planarizationor chemical mechanical polishing; DIP - dualin-line package.

13.4 How do n-type and p-type dopants differ? Ex-plain.

The difference is whether or not they donate ortake an electron from the (usually) silicon intowhich they are doped.

13.5 How is epitaxy different than other forms of filmdeposition?

Epitaxial layers are grown from the substrate,as described in Section 13.5. Other films areexternally applied without consuming the sub-strate.

13.6 Comment on the differences between wet anddry etching.

Wet etching involves liquid-based solutions intowhich the workpiece is immersed. The processis usually associated with high etch rates andisotropic etch patterns, and is relatively easy tomask. Dry etching usually involves placing theworkpiece into a chamber with gas or plasma,and the plasma drives the etching process. Theprocess is usually associated with low etch ratesand anisotropic etching, and is more difficult tomask.

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13.7 How is silicon nitride used in oxidation?

As described on in Section 13.6, silicon nitrideis used for selective oxidation, since silicon ni-tride inhibits the passage of oxygen and watervapor. Thus, the silicon nitride acts as a maskin oxidation.

13.8 What are the purposes of prebaking and post-baking in lithography?

As described on p. 817, prebaking of wafers isdone (prior to lithography) to remove solventfrom the photoresist, and to harden the pho-toresist. After lithography, the wafer is post-baked to improve the adhesion of the remainingphotoresist.

13.9 Define selectivity and isotropy and their impor-tance in relation to etching.

Selectivity describes the preferential etchingof some materials over others with a givenetchant. Isotropy describes the rate of etchingin different directions relative to the surface.These are important with respect to etchingbecause to obtain high-quality integrated cir-cuits, good definition and closely packed devicesare needed. This requires and understanding ofboth selectivity and isotropy in etching.

13.10 What do the terms linewidth and registrationrefer to?

Linewidth (p. 818) is the width of the small-est feature obtainable on the silicon surface.Current minimum linewidths are around 0.13µm. Registration refers to alignment of wafersin lithography. Both of these are interrelated,since highly resolved integrated circuits cannotbe obtained unless the linewidth is sufficientlysmall and the registration is performed prop-erly.

13.11 Compare diffusion and ion implantation.

Diffusion and ion implantation are similar. Dif-fusion refers to the process of atom migration,and is closely related to temperature. Ion im-plantation involves accelerating ions and direct-ing them to a surface where they are incorpo-rated. Thus, both diffusion and ion implanta-tion can be used to drive dopants into semicon-ductor materials.

13.12 What is the difference between evaporation andsputtering?

In evaporation, the coating is heated until it isa vapor, which then deposits from vapor phaseonto the cooler workpiece surface. In sputter-ing, ions impact the coating material and causeatoms to be ejected or sputtered. These atomsthen condense on the workpiece. A further de-scription is in Section 4.5.1.

13.13 What is the definition of yield? How importantis yield? Comment on its economic significance.

Yield is the ratio of functional chips to the num-ber of chips produced. Obviously, yield is ex-tremely important because of its major influ-ence on the economics of chip manufacture.

13.14 What is accelerated life testing? Why is it prac-ticed?

In accelerated life testing, the test subject is ex-posed to a harsher environment than its work-ing environment. For example, the test subjectmay be exposed to higher temperatures, higherstresses, or higher temperature variations. Thetime until failure is then measured, and infer-ences are made as to its expected life in theworking environment. Accelerated life testingis essential because many products last a verylong time, and thus it would not be practical totest under normal conditions.

13.15 What do BJT and MOSFET stand for?

BJT: Bipolar junction transistor, and MOS-FET: Metal on oxide field effect transistor.

13.16 Explain the basic processes of (a) surface mi-cromachining and (b) bulk micromachining.

In surface micromachining, the selectivity ofwet etching is exploited to produce small me-chanical features on silicon or on other surfaces.As shown in Fig. 13.34, surface micromachininginvolves production of a desired feature throughfilm deposition and etching; a spacer layer isthen removed through wet etching, where thespacer layer is easily etched while the structuralmaterial is not etched.

13.17 What is LIGA? What are its advantages overother processes?

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LIGA is an acronym from the German termsX ray Lithographie, Galvanoformung und Ab-formung, or x-ray lithography, electroformingand molding, as shown in Fig. 13.44 on p. 852.LIGA has the capability of producing MEMSand micromechanical devices with very largeaspect ratios. The operation also allows theproduction of polymer MEMS devices and themass production of MEMS devices, since theLIGA-produced structure is a mold for furtherprocessing.

13.18 What is the difference between isotropic andanisotropic etching?

In isotropic etching, material is chemically ma-chined in all directions at the same rate, asshown in Fig. 13.17a on p. 824. Anisotropicetching involves chemical machining where onedirection etches faster than another, with theextreme being vertical etching (Fig. 13.23f onp. 831) where material is only removed in onedirection.

13.19 What is a mask? What is its composition?

A mask is a protective layer that contains all ofthe geometric information desired for an etch-ing or ion implantation step; it can be consid-ered to be a protective coating. Masking pre-vents machining where the mask is present. Amask can be produced from a variety of mate-rials, although typically they are polymers.

13.20 What is the difference between chemically as-sisted ion etching and dry plasma etching?

As described in Section 13.8.2, chemically as-sisted ion etching is one type of dry etching.Dry etching involves etching in a plasma, andchemically assisted ion etching uses chemical re-active species in the plasma to remove material,and the ion bombardment is used to help re-move the chemical species attached to the sur-face.

13.21 Which process(es) in this chapter allow(s) fab-rication of products from polymers? (See alsoChapter 10.)

By the student. It will be noted that poly-mers are most easily produced from LIGA andsolid freeform fabrication processes. They canbe produced through surface micromachining,

but in practice, it is very difficult because ofthe presence of surface residual stresses and thelack of high selectivity in etchants.

13.22 What is a PCB?

PCB is a printed circuit board (see Section13.13).

13.23 With an appropriate sketch, describe the ther-mosonic stitching process.

As shown in Fig 13.28 on p. 836, in thermosonicstitching, a gold wire is welded to a bond padon a printed circuit board. The wire is thenfed from a spool through a nozzle, so that awire thread’ is stretched to a lead on the inte-grated circuit package. The gold wire is thenthermosonically welded to the package, similarto ultrasonic welding described in Section 12.7.

13.24 Explain the difference between a die, a chip,and a wafer.

A die is a completed integrated circuit. A chipis the portion of the wafer used to construct in-tegrated circuits. A wafer is a slice of a singlecrystal silicon cylinder. It should be noted thatthere are many dice on a chip, and a chip ispart of a wafer.

13.25 Why are flats or notches machined onto siliconwafers? Explain.

The flats are machined to assist in registration,and to also indicate the crystallographic orien-tation of the silicon in the wafer. Anisotropicetching processes require that designers accountfor the crystallographic orientation.

13.26 What is a via? What is its function?

A via is an electrical connections between layersof a printed circuit board, as depicted in Fig.13.31 on p. 840. With a large number of cir-cuits on a board, it is clear that the requiredelectrical connections are very difficult to makeif all of the connections must lie within a singleplane. A via allows the designer to make elec-trical connections on a number of planes, thusgreatly simplifying layout on a board.

13.27 What is a flip chip? Describe its advantagesover a surface-mount device.

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A flip chip, shown in Fig. 13.30 on p. 838, hasa large number of solid metal balls that mateto bond pads on the printed circuit board. Itgets its name from the process used to removethe chip from its supply tape and assembling itto the circuit board. The main advantage of aflip chip over a surface mount device is that alarger number of connections can be made on asmaller area, allowing a greater density of inte-grated circuits on a printed circuit board.

13.28 Explain how IC packages are attached to aprinted circuit board if both sides will containICs.

If both sides are to contain ICs, the designermust first make sure that all through-holemount packages are placed on one side of theboard. This same side will then have all remain-ing components attached through reflow (paste)soldering. The opposite side of the board willhave components glued in place and then a wavesoldering operation takes place.

13.29 In a horizontal epitaxial reactor (see the accom-panying figure), the wafers are placed on a stage(susceptor) that is tilted by a small amount,usually 1-3. Why is this procedure done?

The stage in the horizontal epitaxial reactor isusually tilted by a small amount to provideequal amounts of reactant gases in both thefront and back of the chamber. If the stageis not tilted, the reactant gases would be par-tially used up (on the wafers in the front of thechamber) before they reach the wafers at theback end of the chamber, causing nonuniformi-ties in the film deposited.

13.30 The accompanying table describes threechanges in the manufacture of a wafer: in-crease of the wafer diameter, reduction of thechip size, and increase of the process complex-ity. Complete the table by filling in the wordsincrease, decrease, or no change to indicate theeffect that each change would have on waferyield and on the overall number of functionalchips.

Effects of manufacturing changesNumber of

Wafer functionalChange yield chipsIncrease wafer

diameterReduce chip

sizeIncrease

processcomplexity

The completed table is shown below:

Effects of manufacturing changesNumber of

Wafer functionalChange yield chipsIncrease wafer No change Increase

diameterReduce chip Increase Increase

sizeIncrease Decrease Decrease

processcomplexity

13.31 The speed of a transistor is directly propor-tional to the width of its polysilicon gate, witha narrower gate resulting in a faster transistorand a wider gate resulting in a slower transistor.Knowing that the manufacturing process has acertain variation for the gate width, say ±0.1µm, how might a designer alter the gate size ofa critical circuit in order to minimize its speedvariation? Are there any penalties for makingthis change? Explain.

In order to minimize the speed variation of crit-ical circuits, gate widths are typically designedat larger than the minimum allowable size. Asan example, if a gate width is 0.5 µm and theprocess variation is ±0.1 µm, a ±20% variationin speed would be expected. However, if thegate width is increased to 0.8 µm, the speedvariation reduces to ±12.5%. The penalty forthis technique is a larger transistor size (and, inturn, a larger die area) and also a slower tran-sistor.

13.32 A common problem in ion implantation is chan-neling, in which the high-velocity ions travel

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deep into the material through channels alongthe crystallographic planes before finally beingstopped. What is one simple way to stop thiseffect?

A simple and common method of stopping ionchanneling during implantation is to tilt thecrystal material by a few degrees (4-7) so thatthe incident ion beam is not coincident with thecrystallographic planes of the material.

13.33 The MEMS devices described in this chapteruse macroscale machine elements, such as spurgears, hinges, and beams. Which of the follow-ing machine elements can or cannot be appliedto MEMS, and why?

(a) ball bearing;

(b) helical springs;

(c) bevel gears;

(d) rivets;

(e) worm gears;

(f) bolts;

(g) cams.

All of the devices can be manufactured, but ballbearings, helical springs, worm gears, and boltsare extremely difficult to manufacture, as wellas use, in micromechanical systems. The mainreason these components cannot be easily man-ufactured is that they are three dimensional,whereas the MEMS manufacturing processes,as currently developed, are best suited for 2D,or at most, 2 1

2D devices.

13.34 Figure 13.7b shows the Miller indices on a waferof (100) silicon. Referring to Fig. 13.5, identifythe important planes for the other wafer typesillustrated in Fig. 13.7a.

The crystal orientation doesn’t depend onwhether the silicon is n- or p-type, so that whatis shown in part (b) fits equally well for either100 wafer. The proper structure for a 111type is as follows:

Primaryflat

100 plane111 plane

54.74°

13.35 Referring to Fig. 13.23, sketch the holes gener-ated from a circular mask.

The challenge to this problem is that conicalsections are difficult to sketch. Note, however,that some etching processes will expose crys-tallographic planes, resulting in an undercut ofthe circular mask in places. The sketches aregiven below:

(a) (b) (c)

(hemisphericalshape) Note: undercuts!

(d) (e) (f)

Scalloping

13.36 Explain how you would produce a spur gear ifits thickness were one tenth its diameter andits diameter were (a) 10 µm, (b) 100 µm, (c) 1mm, (d) 10 mm, and (e) 100 mm.

215






The answer depends on the material, but letsassume the material is silicon.

(a) 10-µm spur gear could be producedthrough surface micromachining.

(b) 100 µm spur gear could be producedthrough micromachining; if silicon is notthe desired material, LIGA is also an op-tion.

(c) 1-mm gear can be produced throughLIGA, chemical blanking, or chemicaletching from foil.

(d) 10 mm gear can be blanked or chemicallyblanked.

(e) 100 mm gear could best be machined (seeSection 8.10.7).

13.37 Which clean room is cleaner, a Class-10 or aClass-1?

Recall that the class of a clean room is definedas the number of 0.5 µm or larger particleswithin a cubic foot of air (see Section 13.2).Thus, a Class-1 room is cleaner than a Class-10room.

13.38 Describe the difference between a microelec-tronic device, a micromechanical device andMEMS.

By the student. A microelectronic device is anyintegrated circuit. Literally, the device is micro-scale, so that a microscope is needed to see thisdevice, but practically, it refers to a device pro-duced through the processes described in thischapter. A micromechanical device is a con-struct that is mechanical, and not electronic,in nature; it uses gears, mirrors, actuators, andother mechanical systems in their operation. Itcould be argued that a micromechanical deviceis a microelectronic device that has at least onemoving part. MEMS is a special class, contain-ing a micromechanical device and integrated mi-croelectronic control circuitry, thus it is an in-tegrated microelectronic and micromechanicaldevice.

13.39 Why is silicon often used with MEMS andMEMS devices?

See also the answer to Problem 13.2. ForMEMS and MEMS devices, silicon is used

widely because the manufacturing strategies forsilicon have been extensively investigated anddeveloped, and silicon has a unique capabilityto grow epitaxial layers.

13.40 Explain the purpose of a spacer layer in surfacemicromachining.

Recall that a spacer layer is a layer of an easy-to-wet etch material. It can separate mechan-ical devices as they are being built in a layer-by-layer approach, and then removed in a wetetching step that leaves the structural materialunchanged. Borophosphosilicate glasses are themost common spacer layer materials.

13.41 What do the terms SIMPLE and SCREAMstand for?

As described in Section 13.14.2 on p. 847,SIMPLE stands for silicon micromachiningby single-step plasma etching and SCREAMstands for single-crystal silicon reactive etch-ing and metallization. These are shown inFigs. 13.39 and 13.40 on p. 848.

13.42 Which process(es) in this chapter allow the fab-rication of products from ceramics? (See alsoChapter 11.)

By the student. Note that ceramic products aredifficult to manufacture on a microscale. Theonly processes that would work are slip cast-ing from a LIGA-produced mold or equivalentmold from microstereolithography.

13.43 What is HEXSIL?

HEXSIL, shown in Fig. 13.49 on p. 856, com-bines hexagonal honeycomb structures, siliconmicromachining, and thin-film deposition. Thisprocess produces high aspect-ratio structuressuch as the microtweezers shown in Fig. 13.50on p. 857.

13.44 Describe the differences between stereolithogra-phy and microstereolithography.

Microstereolithography uses the same mecha-nisms as stereolithography, but the laser is fo-cused on a much smaller area. As describedin Section 13.16, microstereolithography usesa laser focused onto a diameter as small as1 µm, whereas conventional stereolithographytypically uses laser diameters of 250 µm.

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13.45 Lithography produces projected shapes; conse-quently, true three-dimensional shapes are moredifficult to produce. Which of the processes de-scribed in this chapter are best able to producethree-dimensional shapes, such as lenses?

Making three dimensional shapes is very dif-ficult. A shape with a smooth surface is es-pecially challenging, since a stepped surfaceresults from multilayer lithography. Three-dimensional objects can be produced byisotropic etching, but the surface will not nec-essarily have the desired contour. The bestlithography-based process for producing three-dimensional surfaces is stereolithography or mi-crostereolithography, which can be combinedwith electroforming or other processes, such asLIGA.

13.46 List and explain the advantages and limitationsof surface micromachining as compared to bulkmicromachining.

By the student. Review Section 13.14.2 andconsider the following partial list:

Advantages of surface micromachining:

• Not restricted to single-crystal materials.• Multilayer objects can be produced.• Very good dimensional tolerances.• Complex shapes in multiple layers.• A mature technology which is fairly ro-

bust.

Disadvantages of surface micromachining:

• Additional manufacturing steps are re-quired to deposit and remove spacer lay-ers.

• The process is effectively limited to siliconas the substrate material.

• Wet etchants can result in structures thatfail to separate from surfaces, as shown inFig. 13.36 on p. 845.

13.47 What are the main limitations to the LIGA pro-cess?

LIGA has the capability of producing MEMSand micromechanical devices with very largeaspect ratios. It also allows the production ofpolymer MEMS devices and the mass produc-tion of these devices (since the LIGA-producedstructure is a mold for further processing). Themain limitations of LIGA are economic, as col-limated x-rays are obtained only with specialequipment, currently available only at selectedU.S. National Laboratories. Thus, the cost ofparts produced is very high.

13.48 Describe the process(es) that can be used tomake the microtweezers shown in Fig. 13.49other than HEXSIL.

The HEXSIL tweezers shown in Fig. 13.49 onp. 856 are difficult, although not impossible,to produce through other processes. The im-portant features to be noted in these tweezersare the high aspect ratios and the presence oflightening holes in the structure, resulting in acompliant and lightweight structure. Althoughprocesses such as SCREAM (pp. 855-857) canbe used, the required aspect ratio will be diffi-cult to achieve. LIGA also can be used, but itis expensive. For each of these processes, thetweezers shown would require redesign of themicrotweezers. For example, in LIGA, it wouldbe desirable to have a draft in the vertical mem-bers to aid in molding. However, a structurethat serves the same function can be produced,even though vertical sidewalls cannot be pro-duced.

Problems

13.49 A certain wafer manufacturer produces twoequal-sized wafers, one containing 500 chips andthe other containing 300 chips. After testing, itis observed that 50 chips on each type of wafer

are defective. What are the yields of the twowafers? Can any relationship be established be-tween chip size and yield?

217






The yield for the 500 chip wafer is (500-50)/500= 90.0%, and for the 300 chip wafer, it is (300-50)/300 = 83.3%. Thus, given the same num-ber of defects per wafer, the wafer with smallerchips (i.e., more chips per wafer) will have ahigher yield. This is because the same amountof defects are spread out over a larger numberof chips, thus making the number of defectivechips a smaller percentage.

13.50 A chlorine-based polysilicon etch process dis-plays a polysilicon:resist selectivity of 4:1 and apolysilicon:oxide selectivity of 50:1. How muchresist and exposed oxide will be consumed inetching 350 nm of polysilicon? What shouldthe polysilicon:oxide selectivity be in order toremove only 4 nm of exposed oxide?

The etch rate of the resist is 1/4 that of polysil-icon. Therefore, etching 350 nm of polysiliconwill result in (350)(1/4) = 87.5 nm of resist be-ing etched. Similarly, the amount of exposedoxide etched away will be (350)(1/50) = 7 nm.To remove only 4 nm of exposed oxide, thepolysilicon:oxide selectivity would be 350/4 =88:1.

13.51 During a processing sequence, four silicon-dioxide layers are grown by oxidation: 400 nm,150 nm, 40 nm, and 15 nm. How much of thesilicon substrate is consumed?

The total oxide thickness is 400 nm + 150 nm+ 40 nm + 15 nm = 605 nm. From Section13.6, the ratio of oxide to the amount of siliconconsumed is 1:0.44. Hence, to grow 605 nm ofoxide, approximately (0.44)(605 nm) = 266 nmof silicon will be consumed.

13.52 A certain design rule calls for metal lines to beno less than 2 µm wide. If a 1 µm-thick metallayer is to be wet etched, what is the minimumphotoresist width allowed? (Assume that thewet etch is perfectly isotropic.) What would bethe minimum photoresist width if a perfectlyanisotropic dry-etch process were used?

A perfectly isotropic wet-etch process will etchequally in the vertical and horizontal directions.Therefore, the wet-etch process requires a min-imum photoresist width of 2 µm, plus 1 µm perside, to allow for the undercutting, hence a to-tal width of 4 µm. The perfectly anisotropic

dry-etch process displays no undercutting and,therefore, requires a photoresist width of only2 µ.

13.53 Using Fig. 13.18, obtain mathematical expres-sions for the etch rate as a function of temper-ature.

The following data points are obtained fromFig. 13.18:

Direc- 1/T Etch rate ln(Etchtion (×10−3K−1) (µ/hr) rate)〈110〉 2.55 70 4.248

3.3 4 1.386〈100〉 2.55 70 4.248

3.3 2 0.6931〈111〉 2.55 2 0.6931

3.3 0.015 -4.200

Figure 13.18 suggests that a plot of ln(Etchrate) vs. 1/T will be linear. Therefore, we ex-pect a relationship of the form

ln(y) = a

(1T

)+ b

ory = ea(1/T )+b = ebeα/T

where y is the etch rate, a is the slope of theln(y) vs. 1/T curve, and b is the y-intercept.From the data in the table above, we can ob-tain the following

Direction a b〈110〉 -3.816 13.98〈100〉 -4.74 16.33〈111〉 -6.524 17.33

Therefore, the equation in the 〈110〉 directionis:

y =(1.179× 106

)e−3.816/T

in the 〈100〉 direction:

y =(1.236× 107

)e−4.74/T

in the 〈111〉 direction:

y =(3.3× 107

)e−6.524/T

13.54 If a square mask of side length 100 µm isplaced on a 100 plane and oriented with aside in the 〈110〉 direction, how long will ittake to etch a hole 4 µm deep at 80C usingethylene-diamine/pyrocatechol? Sketch the re-sulting profile.

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Etching will take place in the 〈100〉 direction,but it should be noted that there will be angledetch fronts at the sides of the square. However,since the width is much larger than the depth,we can ignore these fronts. From Fig. 13.18on p. 825, the etch rate at 80C is around 22µm/hr, so the time required to etch 4 µm is(4/22)(60)=10.91 min.

The resulting profile will look like Fig. 13.23con p. 831, but the angles of the side walls willbe different on the opposite sides of the square.

13.55 Obtain an expression for the width of the trenchbottom as a function of time for the mask shownin Fig. 13.17b.

If L is the original trench width and l is thewidth at the bottom of the trench, then the re-lationship between l and L is

L = l + 2x tan 54.7

where x is the depth of the trench. The depthof the trench is related to time by the etch rate,or

x = ht

where h is the etch rate, as given in Tables13.2 and 13.3 on p. 824 for various etchants andworkpiece materials. Therefore,

L = l + 2h tan 54.7

or

l = L− 2ht tan 54.7 = L− 2.824ht

13.56 Estimate the time of contact and average forcewhen a fluorine atom strikes a silicon surfacewith a velocity of 1 mm/s. Hint: See Eqs. (9.11)and (9.13).

It should be noted that, at this scale, continuumapproaches are no longer valid, and the appli-cation of Eqs. (9.11) and (9.13) on p. 553 areuseful only for illustrative purposes. However,if we use the properties for silicon (ρ = 2330kg/m3, E = 190 GPa) and we note that flu-orine has an atomic radius of 0.119 nm, andan atomic weight of 18.998, so that one atomweighs

W =18.998g/mole

6.023× 1023atoms/mole

or W = 3.154 × 10−23 g/atom. Therefore, thewave speed in silicon is

co =

√E

ρ=

√190 GPa

2330 kg/m3 = 9030 m/s

The time of contact is, from Eq. (9.11),

to =5rco

(cov

)1/5

=5(0.119× 10−9

)9030

(90300.001

)1/5

= 1.62× 10−12 s

The contact force is given by Eq. (9.13) as

F =2mvto

=2(3.54× 10−26 kg

)(0.001 m/s)

1.62× 10−12 s= 4.37× 10−17 N

13.57 Calculate the undercut in etching a 10-µm-deeptrench if the anisotropy ratio is (a) 200, (b) 2,and (c) 0.5. Calculate the sidewall slope forthese three cases.

For a trench depth is 10 µm, then fromEq. (13.4) on p. 807, we obtain the undercutx as

AR =E1

E2=

10 µm/tx/t

=10 µmx

The sidewall slope, θ, is given by

tan θ =x

10 µm

The following table can now be constructed:

Anisotropy Undercut, Side wall slope,ratio x, (µm) θ ()200 0.05 0.282 5 26.60.5 20 63.4

13.58 Calculate the undercut in etching a 10-µm-deeptrench for the wet etchants listed in Table 13.3.What would the undercut be if the mask weremade of silicon oxide?

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The same approach as in Problem 13.57 is used.Note that the dry etchants do not have an un-dercut, but they also etch through silicon diox-ide very quickly. While, in theory, a thick maskcould be used, in practice it will be difficult tomask a dry etchant with silicon dioxide becausethe silicon dioxide mask will be removed. Theundercut for a perfect mask is:

UndercutEtchant Selectivity x (µm)

HF:HNO3: — 10CH3COOH

KOH 100:1 0.1EDP 35:1 0.28N(CH3)4OH 50:1 0.02SF6 — 0

If the mask material is silicon dioxide, it will beetched at a rate as given in Table 13.3 on p. 824.While the undercut can be defined based on theoriginal mask dimensions, and therefore yieldthe same answers as above, we can also cal-culate the undercut that would be observed inetching a 10m-deep hole. The results are asfollows:

Tim

eto

etch

Etc

h10

µm

SiO

2O

bse

rved

rate

tren

chre

mov

edR

ef.

under

cut

Etc

hant

(µm

/m

in)

(min

)(n

m)

(nm

)(n

m)

HF:H

NO

3:

20

0.5

15

10000

9985

CH

3C

OO

HK

OH

25

50

100

50

ED

P0.7

513.3

32.6

6280

277

N(C

H3) 4

OH

1.5

6.6

70.6

67

20

19.3

SF

60.5

20

00

0

13.59 Estimate the time required to etch a spur-gearblank from a 75-mm-thick slug of silicon.

Note that the answer will depend on theetchant used and the ability to replenish theetchant. Using the highest etch rate for sili-con in Table 13.2 on p. 823, of 310 nm/min for126HNO3:60H2O:5NH4F, the time required is

t =75 mm

310 nm/s= 2.42× 105 s

or almost three days. This problem demon-strates that etching processes are useful onlyfor thin parts or for shallow (micron-sized) fea-tures.

13.60 A resist is applied in a resist spinner spun oper-ating at 2000 rpm, using a polymer resist withviscosity of 0.05 N-s/m. The measured resistthickness is 1.5 µm. What is the expected resistthickness at 6000 rpm? Let α=1.0 in Eq. (13.3).

Equation (13.3) on p. 816 gives the resist thick-ness as

t =kCβηγ

ωα

We can now compare the two conditions andwrite

t1t2

=k1C

β1 η

γ1ω

α2

k2Cβ2 η

γ2ω

α1

Note that the only variable which changes is ω.Therefore,

t1t2

=k1C

β1 η

γ1ω

α2

k2Cβ2 η

γ2ω

α1

=ωα

2

ωα1

=(6000)1.0

(2000)1.0= 3

Hence, the final thickness will be 1/3 the refer-ence film thickness of 1.5 µm, or 0.5 µm.

13.61 Examine the hole profiles in the accompanyingfigure and explain how they might be produced.

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It will be helpful to refer to Example 13.2 andFig. 13.23 on p. 831 to understand this solutionbelow. We can state the following:

• For the top left profile, there is undercutbeneath the mask. It is difficult to tellfrom the dimensions given, but there is ei-ther isotropic etching or preferential etch-ing in the horizontal direction compared tothe vertical direction. This situation couldoccur if:

(a) An isotropic wet etchant is used (seeFig. 13.17a on p. 824).

(b) The crystal workpiece is aligned sothat there is preferential wet etchingin the horizontal direction

(c) An etch-stop has been used.

• The top right figure shows a profile thatmatches Fig. 13.23d, which is caused byorientation-dependant etching.

• The bottom left figure shows a materialthat has undercut the mask. Comparedto the top right figure, this figure suggestseither isotropic etching or etching that ispreferential in the thickness direction.

• The bottom right cross section has a ver-tical wall and a pointed trench. This pro-file could be produced by a deep reactiveion etching operation or a chemically reac-tive ion etching operation, followed by anorientation-dependent etching operation.

13.62 A polyimide photoresist requires 100 mJ/cm2

per µm of thickness in order to develop prop-erly. How long does a 150 µm film need todevelop when exposed by a 1000 W/m2 lightsource?

It is useful to convert units to avoid confusionin making the calculations. Note that the poly-imide photoresist requires the following powerdensity:

P = 100mJ

cm3(µm)th =

1000 Nmt m2

h

(µm)

where t is the exposure time and h is the filmthickness. Since the power available is 1000W/m2, we can calculate the time from

1000Wm2

=1000 Nmt m2

h

(µm)

or, solving for t,

t =(

1000 Nm1000 W

)h = 150 s

13.63 How many levels are required to produce themicromotor shown in Fig. 13.22d?

At a minimum, the following layers are needed:

• Base for rotor.

• Rotor.

• Pin or bearing (it must protrude past therotor).

• Lip on bearing to retain rotor.

This list assumes that the electrical connectionscan be made on the same layers as the MEMSfeatures, as otherwise an additional layer is re-quired.

13.64 It is desired to produce a 500µm by 500 µmdiaphragm, 25 µm thick, in a silicon wafer 250µm thick. Given that you will use a wet etchingtechnique with KOH in water and with an etchrate of 1 µm/min, calculate the etching timeand the dimensions of the mask opening thatyou would use on a (100) silicon wafer.

Diaphragms can be produced in a number ofways. This problem and solution merely ad-dress the wet etching portion of the process,and assumes a diaphragm can be placed overa proper opening with a diffusion bonding stepsuch as shown in part 3 of Fig. 13.41a on p. 849.Using a wet etchant, a cavity as shown in part2 of Fig. 13.41a will be produced, with an in-clined sidewall. From Table 13.3 on p. 824,note that KOH has a 111 / 100 selectivityof 100:1. Thus, there will be a slight under-cut of the mask, and the sidewalls will have aslope of tan−1 0.01 = 0.57 from the vertical asshown.

A 25 µm hole needs 25 min at the prescribedetch rate of 1 µ/min. The undercut of the side-walls will be 0.25 µm at a selectivity of 100:1.Thus, if the top dimensions of the diaphragmare critical, a mask that is 450 × 450 µm willproduce a 500×500 µm dimension. If the meandimension of the diaphragm is critical, then a475× 475 µm mask is needed.

221






13.65 If the Reynolds number for water flow througha pipe is 2000, calculate the water velocity ifthe pipe diameter is (a) 10 mm; (b) 100 µm.Do you expect the flow in MEMS devices to beturbulent or laminar? Explain.

As given by Eq. (5.10) on p. 202, the Reynoldsnumber is

Re =vDρ

η

where v is the velocity, D is the channel diame-ter, ρ is the density of water (1000 kg/m3), andη is the viscosity of water (8.90×10−4 Ns/m2).Solving for the velocity,

v =(Re)ηDρ

If the channel diameter is 10 mm, then

v =(2000)(8.9× 10−4)

(0.01)(1000)= 0.178 m/s

If the diameter is 100 µm, then

v =(2000)(8.9× 10−4)

(0.0001)(1000)= 17.8 m/s

This is a very high velocity, and probably wouldnever be achieved in a MEMS device. As dis-cussed in Section 5.4.1, laminar flow takes placefor Reynolds numbers below 2000. Clearly,MEMS devices will most likely be laminar.

Design

13.66 The accompanying figure shows the cross sec-tion of a simple npn bipolar transistor. Developa process flow chart to fabricate this device.

n+

n

n

n

p

p AlSiO2

The steps in the production of a simple bipolartransistor are as follows:

n

1. Oxidation 2. Lithography

n

n n

3 Oxide etch 4. Resist removal

np

np

np

np

5. p region implant 6. Oxidation

7. Lithography 8. Oxide etch

np

npn+

npn+

npn+

9. Resist removal 10. n+ region implant

11. Oxidation 12. Lithography

npn+

npn+

npn+

npn+

13. Oxide etch 14. Resist removal

15. Al deposition 16. Lithography

npn+

npn+

17. Aluminum etch 18. Resist removal

222






13.67 Referring to the MOS transistor cross section inthe accompanying figure and the given table ofdesign rules, what is the smallest transistor sizeW obtainable? Which design rules, if any, haveno impact on the magnitude of W? Explain.

R1

R3 R2

R4 R6 R5

W

Rule ValueNo. Rule name (µm)R1 Minimum polysilicon

width0.50

R2 Minimum poly-to-contact spacing

0.15

R3 Minimum enclosure ofcontact by diffusion

0.10

R4 Minimum contact width 0.60R5 Minimum enclosure of

contact by metal0.10

R6 Minimum metal-to-metal spacing

0.80

The smallest transistor size, W , that can beobtained using the given design rule is:

W = R3 +R4 +R5 +R6 +R5 +R4 +R3

or W = 0.10+0.60+0.10+0.80+0.10+0.60+0.10 = 2.40 µm. Design rules R1 and R2 haveno impact on the smallest obtainable W .

13.68 The accompanying figure shows a mirror thatis suspended on a torsional beam; it can be in-clined through electrostatic attraction by ap-plying a voltage on either side of the mirror atthe bottom of the trench. Make a flow chart ofthe manufacturing operations required to pro-duce this device.

The device shown in the problem was producedat the University of California at Berkeley Sen-sor and Actuator Center. As can be noted,the layer below the mirror is very deep andhas near-vertical sidewalls; hence, this devicewas clearly produced through a dry (plasma)etching approach. Also note that it was ma-chined from the top since the sidewall slope isslightly inclined. However, a high-quality mir-ror could not be produced in this manner. Theonly means of producing this micromirror is(a) to perform deep reactive ion etching on thelower portion, (a) traditional surface microma-chining on the top layer, and (c) then joiningthe two layers through silicon fusion bonding.(See Fig. 13.48 on p. 856 for further examplesof this approach.)

13.69 Referring to Fig. 13.36, design an experiment tofind the critical dimensions of an overhangingcantilever that will not stick to the substrate.

By the student. There are several possible so-lutions and approaches to this problem. An ex-perimental investigation by K. Komvopolous,Department of Mechanical Engineering at theUniversity of California at Berkeley, involvesproducing a series of cantilevers of differentaspect ratios on a wafer. After productionthrough surface micromachining, followed byrinsing, some of the cantilevers attach them-selves to the substrate while others remain sus-pended. The figure below shows the transition.Based on beam theory from the mechanics ofsolids, a prediction of the adhesive forces canbe determined.

223






13.70 Explain how you would manufacture the deviceshown in Fig. 13.32.

By the student. This device would involve avery elaborate series of surface micromachin-ing operations. The solution for Problem 13.66should be studied and understood before at-tempting this more complicated problem.

13.71 Inspect various electronic and computer equip-ment, take them apart as much as you can, andidentify components that may have been man-ufactured by the techniques described in thischapter.

This is a good assignment that can be inexpen-sively performed, as most schools and individ-uals have obsolete electronic devices that canbe harvested for their components. Some in-teresting projects also can arise from this ex-periment. One project, for example, wouldbe to microscopically examine the chips toobserve the manufacturers logos, as graphi-cal icons are often imprinted on chip sur-faces. See http://www.microscopy.fsu.edu/mi-cro/gallery.html.

13.72 Do any aspects of this chapter’s contents andthe processes described bear any similarity tothe processes described throughout previouschapters in this book? Explain and describewhat they are.

By the student. There are, as to be expected,some similarities. For example, the principles ofetching processes are the same as in chemicalmachining (see Section 9.10). Also, there arepolishing and grinding applications (as in fin-ishing the wafers and grinding the sides) that

are similar to the polishing and grinding pro-cesses, described in Chapter 9. Producing sil-icon wafers involves the Czochralski (CZ) pro-cess (see Fig. 5.30 on p. 235). Printed circuitboards are stamped and the holes are drilled,as described in previous chapters. Packaginginvolves potting and encapsulation of polymers(p. 636). The students are encouraged to com-ment further.

13.73 Describe your understanding of the importantfeatures of clean rooms, and how they are main-tained.

Clean rooms are described in Section 13.2. Stu-dents are encouraged to search for additional in-formation, such as the design features of HEPAfilters, the so-called bunny suits, and humid-ity controls. It should also be noted that anydiscussion of clean rooms has to recognize thesources of contaminants (mostly people andtheir clothing) and the strategies used to con-trol them.

13.74 Describe products that would not exist withoutthe knowledge and techniques described in thischapter. Explain.

By the student. This topic would be a goodproject. Clearly, a wide variety of modern prod-ucts could not exist without using the processesdescribed in this chapter. Certainly, the pres-ence of the integrated circuit has had a pro-found impact on our lives, and any productthat contains an integrated circuit would ei-ther not exist or it would be more expensiveand less reliable. Personal computers, televi-sion sets, and cellular phones are other majorexamples of products that could not exist, orexist in a vastly different form, without inte-grated circuits are televisions, automobiles, andmusic players. The students are encouraged tocomment further, with numerous examples oftheir own.

13.75 Review the technical literature and give moredetails regarding the type and shape of theabrasive wheel used in the wafer-cutting oper-ation shown in Step 2 in Fig. 13.6 on p. 810.

By the student. The main source for such in-formation would be manufacturers and distrib-utors of abrasive wheels. It should be noted

224






that the wheel is contoured, hence the waferdoes not have a vertical wall. This means thatthe wafer will have a barrel shape, which is ben-eficial for avoiding chipping.

13.76 It is well known that microelectronic devicesmay be subjected to hostile environments (suchas high temperature, humidity, and vibration)as well as physical abuse (such as being droppedon a hard surface). Describe your thoughts onhow you would go about testing these devicesfor their endurance under these conditions.

By the student. This is a good topic for stu-dents to investigate and develop testing meth-ods for electronic devices. It will be helpful tohave students refer to various ASTM standardsand other sources to find standardized test pro-cedures, and evaluate if they are sufficient forthe difficulties encountered.

13.77 Conduct a literature search and determine thesmallest diameter hole that can be produced by(a) drilling; (b) punching; (c) water-jet cutting;(d) laser machining; (e) chemical etching and(f) EDM.

By the student. This is an interesting topic fora web-based research project. Specific dimen-sions depend on the desired depth of the hole.As examples of solutions for thin foils, there are10 µm diameter drills available commercially.Laser machining is limited to the focus diam-eter of the laser, and is usually as small as 1

µm.

13.78 Design an accelerometer similar to the oneshown in Fig. 13.32 using the (a) SCREAM pro-cess and (b) HEXSIL process, respectively.

By the student. The students should draw uponthe manufacturing sequence shown in Fig. 13.54on p. 862, and consider the capability of theSCREAM and HEXSIL processes to producelarge, overhanging structures.

13.79 Conduct a literature search and write a one-page summary of applications in biomems.

By the student. This is an interesting topic for aliterature search, and it can be easily expandedinto an assignment for a paper. There are manymore proposed applications for biomems thanrealized in commercial products, but sensorsused in medicine are widespread, including labon a chip devices for rapid and simultaneousscreening for many conditions. In-vivo applica-tions of MEMS are relatively few in number asof today.

13.80 Describe the crystal structure of silicon. Howdoes it differ from the structure of FCC? Whatis the atomic packing factor?

This topic is described in Section 13.3. Cal-culating the atomic packing factor of silicon iscomplex. It can be shown that the structure issurprisingly very open, with a packing densityof 34% as compared to 74% for fcc crystals.

225






manufacturing processes for engineering materials kalpakjian solution

Education

true stress

true strains

necking strain true

truestrain curve

true fracture strain

tensile truestress

phenomenon true

specimen height