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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A caliper, for example, that can only measure to the nearest 0.001 cm is not precise enough to measure a 0.0005 cm press-fi t clearance between two mating gears.

controlled to obtain as good a performance as is possible. The student is encouraged to give several examples.

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 =4a

lx

so that, from Eq. (4.1),

Ra =4l

∫ l/4

0

4ax

ldx

Evaluating the integral,

Ra =16a

l2

(12x2

)l/4

0

=16a

l212

(l2

16− 0

)=

a

2

From Eq. (4.2),

R2q =

4l

∫ l/4

0

y2 dx =4l

∫ l/4

0

16a2

l2x2 dx

Evaluating the integral,

R2q =

64a2

l3

(13x3

)l/4

0

=64a2

l313

l3

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 =0.105

d2o − d2

i

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

V =

πh4

(do2 – di

2 ) = π (0.64)

4 (192 – 9.52)

or V = 1.36 cm3. Note that the volume must remain constant, so that

hπ4

(do2 – di

2 ) = 1.36 cm3

or

h =

5.44

do2 – di

2

di do h ID (cm) (cm) (cm)

1 0.9525 1.905 0.635 2 1.2116 2.46 0.373 3 0.716 0.264 0.264 4 0.446 0.264 0.254

Refer to Fig. 4.7b and make measurements of the external and internal diameters (in the horizontal direction in the photograph) of the four specimens shown. Remembering that in plastic deformation the volume of the rings remains constant, calculate (a) the reduction in height and (b) the coeffi cient of friction for each of the three compressed specimens.

Note that the friction coeffi cient increases as the test progresses. This is commonly observed in lubricated specimens, where the lubricant fi lm is initially thick, but breaks down as the contact area between the ring and die increases.

Specimen 1 has not been deformed, so its dimensions are taken from Fig. 4.8. The re-

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; R̄ = 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

UCLx̄ = ¯̄x+A2R̄ = 50 + (0.419)(7) = 52.933

LCLx̄ = ¯̄x−A2R̄ = 50− (0.419)(7) = 47.067

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

UCLR = D4R̄ = (1.924)(7) = 13.468

LCRR = D3R̄ = (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 = D4R̄

solving for R̄,

R̄ =UCLR

D4=

4.851.924

= 2.521

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

UCLx̄ = ¯̄x+A2R̄ = (40.5)+(0.419)(2.521) = 41.556

LCLx̄ = ¯̄x−A2R̄ = (40.5)−(0.419)(2.521) = 39.444

Equation (4.14) gives:

LCLR = D3R̄ = (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

UCLx̄ = ¯̄x+A2R̄ = (75)+(0.308)(10) = 78.080

LCLx̄ = ¯̄x−A2R̄ = (75)− (0.308)(10) = 71.920

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

UCLR = D4R̄ = (1.777)(10) = 17.77

LCRR = D3R̄ = (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 x̄ 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 R̄ = 0.07.Eqs. (4.11) and (4.12) give the upper and lowercontrol limits for the averages as

UCLx̄ = ¯̄x+A2R̄ = (0.6733) + (0.729)(0.07)

54

or UCLx̄=0.7243.

LCLx̄ = ¯̄x−A2R̄ = (0.6733)− (0.729)(0.07)

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

UCLR = D4R̄ = (2.282)(0.07) = 0.1600

LCRR = D3R̄ = (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

x̄ =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

UCLx̄ = ¯̄x+A2R̄ = (125)+(0.419)(17.82) = 132.46

LCLx̄ = ¯̄x−A2R̄ = (125)−(0.419)(17.82) = 117.53

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

UCLR = D4R̄ = (1.924)(17.82) = 34.28

LCRR = D3R̄ = (0.078)(17.82) = 1.390

For the sample shown, the average is x̄ = 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.

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

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

55

56