Ceramics, Graphite, and Diamond:Structure, General Properties, andApplications
QUALITATIVE PROBLEMS
8.19 Explain why ceramics are weaker in tension than in compression.
Ceramics are very sensitive to cracks, impurities, and porosity, and thus generally havelow toughness. In compression, however, the flaws in the material do not cause stressconcentrations or crack propagation, as they do in tension.
8.20 What are the advantages of cermets? Suggest applications in addition tothose given in this chapter.
High-temperature oxidation resistance and toughness are the advantageous propertiesof cermets. They also may be useful for components in furnaces (such as moving belts)and automotive-engine parts (such as pistons and cylinders).
8.21 Explain why the electrical and thermal conductivity of ceramics decreaseswith increasing porosity.
Pores in the ceramic are usually filled with air, and air has much lower thermal andelectrical conductivity than ceramics. Increasing porosity will thus reduce the ceramic’sthermal and electrical conductivities.
8.22 Explain why the mechanical property data given in Table 8.2 have such abroad range. What is the significance of this in engineering practice?
The properties given in Table 8.2 on p. 201 vary greatly because the mechanical prop-erties of ceramics depend greatly on the quality of the raw material, porosity in theproduct, and the manner in which the product is made. Engineering applications thatrequire high mechanical properties must ensure that the material quality and processingof the part are the best available.
8.23 Describe the reasons that have encouraged the development of syntheticdiamond.
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Ceramics, Graphite, and Diamond: Structure, General Properties, and Applications 93
By the student. Synthetic diamonds were first used exclusively in industrial applications;they have superior properties to natural diamonds because of the absence of impurities.The hardness of diamond makes it useful in machining and grinding hard metallic andnonmetallic materials. For widespread and economical use, diamonds are mass producedsynthetically.
8.24 Explain why the mechanical properties of ceramics generally differ from thoseof metals.
Metals and ceramics have different types of bonding, metallic and ionic, respectively.Ionic bonds are stronger than metallic bonds, hence more force must be applied tothe material to break the bonds. The crystal structure for ceramics is generally morecomplex than metals, and requires higher forces to cause slip. Also, the high-temperatureproperties of ceramics are very attractive in many applications.
8.25 Explain how ceramics can be made tougher.
Ceramics may be made tougher by using high-purity raw materials, controlled processingtechniques, and adding various reinforcements. The presence of tougher second phasesand microcracks (less than one µm in size) in the ceramic can also reduce the energyof propagation of an advancing crack tip. (See also partially stabilized zirconia, Section8.2.2 on p. 196.)
8.26 List and describe situations in which static fatigue can be important.
By the student. There are numerous possible answers. Static fatigue occurs in environ-ments where water vapor is present and the part is under a constant load. In applicationssuch sewer piping, if a tensile stress is developed in the pipe by bending or torsion, staticfatigue can be a significant problem. Any situation where a tension member is exposedto water vapor (such as mounting brackets or structural members outdoors) are alsosubject to static fatique. Some MEMS devices see failure of borosilicate glass layers dueto static fatigue as well.
8.27 What properties are important in making heat-resistant ceramics for use onoven tops? Why?
Heat-resistant ceramics for oven tops should possess hot corrosion resistance to maintaintheir appearance, hardness for wear and scratch resistance, and toughness for impactloading. Because the heating of oven top is uneven, thermal gradients will be set upacross the surface. To minimize the possibility for thermal cracking, the oven top shouldhave low coefficient of thermal expansion and high thermal conductivity.
8.28 A large variety of glasses is now available. Why is this so?
By the student. The cost of each type of glass varies, and a high-quality glass (high cost)shouldn’t be used in a low-quality application. However, high quality glasses are requiredfor fiber optics, which need to transmit light with high efficiency. Other applicationsmay require high strength, low cost, chemical resistance, impact resistance, abrasionresistance or resistance to thermal shock, which would be best exhibited by 96% silicaor fused-silica type glass. Thus, in summary, there are a large variety of glasses availablebecause of the large variety of applications where glass is used.
8.29 What is the difference between the structure of graphite and that of dia-mond? Is it important? Explain.
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Ceramics, Graphite, and Diamond: Structure, General Properties, and Applications 94
Graphite has a crystalline form of a layered structure of basal planes of close-packedcarbon atoms (see Fig. 1.5 on p. 41). Diamond, on the other hand, has a covalentlybonded structure. The structure of graphite permits easy slip on basal planes, while thecovalent bonding of diamond greatly restricts slip, thus making it very hard.
8.30 List and explain materials that are suitable for use as a coffee cup.
By the student. A coffee cup must obviously be made of a material that is safe forfood contact, so it should not affect or be affected by coffee. In addition, it should beinexpensive. Furthermore, it must be thermally insulating, have a melting point higherthan the boiling point of water, and be easy to clean. Ceramics, polymers, porouspolymers (foams), and multi-layered metals (with air or a vacuum between layers) arethus suitable.
8.31 Aluminum oxide and PSZ are described as white in appearance. Can theybe colored? If so, how would you accomplish this?
Both aluminum oxide and partially stabilized zirconia can be produced in a variety ofcolors. This can be done by using impurities in small concentrations in the ceramic,which develop a color. There is, of course, also the option of enameling, painting, orcoloring the surfaces.
8.32 Why does the strength of a ceramic part depend on its size?
Ceramic strength is mainly compromised by the presence of flaws where cracks caninitiate. In a small volume, there is less likelihood that a large flaw or a number of flawscan exist, while the reverse is true for larger parts, i.e., in a large volume, it is morelikely that a large flaw will exist (see also the discussion regarding Fig. 2.11 on p. 67).
8.33 In old castles and churches in Europe, the glass windows display pronouncedripples and are thicker at the bottom than at the top. Explain.
The silica in the glass is not really solid; it is merely a supercooled liquid. Over centuries,the glass has creeps or flows due to the force of gravity. This situation depends greatlyon the type of glass, as some will creep faster than others.
8.34 Is a carbide an example of a composite material? Explain.
This depends on the extent to which the definition is applied. Cobalt is often thematrix, with tungsten-carbide particles serving as a discontinuous phase. Two-phasedmetal allows, where the second phase is insoluble in the first, is a similar structure, butalloys are not generally considered to be composite materials.
8.35 Ceramics are hard and strong in both compression and shear. Why, then,are they not used as nails or other fasteners? Explain.
Ceramics aren’t used as nails or other fasteners because they have poor impact resistanceand thus would not be able to be driven into a workpiece with impact forces as appliedby a hammer. Other fasteners such as bolts depend on a tensile proof stress (the proofstrength of a bolt is a common design specification) for a well-designed joint, and thiswould be a poor design application for a ceramic which is weak and has a wide range ofstrengths in tension.
8.36 Perform an Internet search and determine the chemistry of glass used for (a)fiber-optic communication lines, (b) crystal glassware, and (c) high-strengthglass fibers.
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Ceramics, Graphite, and Diamond: Structure, General Properties, and Applications 95
By the student. There are a large number of answers possible. For example, fiber-optic communication lines will be graded according to their intended distance for signaltransmission, and crystal glassware will have a variety of formulations.
8.37 Investigate and list the ceramics used for high-temperature superconductorapplications.
By the student. This is an area of significant ongoing development. In recent years, sig-nificant progress has been made in understanding high-temperature superconducting ma-terials and their potential use as conductors. Two bismuth based oxides (Bi2Sr2CaCu2Ox
, also known as Bi-2212, and Bi2Sr2CaCu3Ox or Bi-2223) are superconducting, ceramicmaterials of choice for various military and commercial applications, such as electricalpropulsion for ships and submarines, shallow-water and ground minesweeping systems,transmission cable generators, and superconducting magnetic-energy storage (SMES).However, wide variations in the electrical performance of test magnet coils using monofil-ament superconducting strips indicate that multifilament conductors are essential in re-liably achieving the required critical current densities, so that property variations areaveraged over a length. Tape stacking is a new approach to manufacturing multifila-ment tape. In this method, 7 to 10 monofilament tapes or strips are stacked, heated,and pressed together; the resulting laminate is then rolled to final thickness. The qualityand performance of multifilament tapes are directly related to that of the constituentsi.e. the monofilament tape. A variety of processes have been explored to produce wiresand tapes, using Bi-2212 and Bi-2223.
8.38 Explain why synthetic diamond gemstones are not appreciably less expensivethan natural diamond gemstones.
This is surprising at first. There is significant labor involved in mining natural diamond,and then an addition cost to ship them to markets. However, there is also significant costinvolved in growing artificial diamond. However, gemstones require significant polishingbefore they are useful, and facets in the stone must be ground and polished with greatprecision. The vast majority of the cost associated with gemstones is associated withthe polishing processes involved.
QUANTITATIVE PROBLEMS
8.39 In a fully dense ceramic, UTSo = 200 MPa and Eo = 330 GPa. What arethese properties at 15% porosity for values of n = 4, 5, 6, and 7, respec-tively?
Inserting the appropriate quantities into the Eqs. (8.1) and (8.2) on p. 202, we obtainthe following:
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Ceramics, Graphite, and Diamond: Structure, General Properties, and Applications 96
Note that the magnitude of n does not affect the magnitude of E.
8.40 Plot the UTS, E, and k values for ceramics as a function of porosity P .Describe and explain the trends that you observe in their behavior.
Equations (8.1) through (8.3) on pp. 201-202 are needed to solve this problem. Thecurves can be obtained using an assumed value at zero porosity, or else they can benon-dimensionalized, as shown below. The plots are as follows:
Porosity (%)0 10 20 30 40 50
UT
S/U
TS o
1
0
0.2
0.4
0.6
0.8
n=7
n=4n=5
n=6
Porosity (%)0 10 20 30 40 50
E/E o
0.2
0.4
0.6
0.8
1.0
Porosity (%)0 10 20 30 40 50
k/k o
0.5
0.6
0.7
0.8
0.91.0
There are several observations that can be made. Most striking is the near-linear be-havior of stiffness and thermal conductivity with respect to porosity, and the highlynonlinear behavior of strength. Thus, to produce high-strength ceramics, reduction ofporosity is especially important.
8.41 What would be the tensile strength and the modulus of elasticity of theceramic in Problem 8.39 for porosities of 25% and 50%, for the four n valuesgiven?
Equations (8.1) and (8.2) on p. 202 are needed to solve this problem. Inserting theappropriate quantities into these equations, we obtain the following:
n UTS (MPa)P = 0.25 P = 0.50
4 73.6 27.05 57.3 16.46 44.6 9.957 34.7 6.09
The modulus of elasticity values are as follows: for P = 0.25, E = 191 GPa, and forP = 0.50, E = 90.7 GPa.
Note that the porosities that are examined in this problem are far outside those nor-mally encountered with ceramics. The equations for strength and stiffness are not veryapplicable to this extreme range. However, students can be encouraged to discuss thevalidity of the approach and whether or not the values obtained are realistic.
8.42 Calculate the thermal conductivities for ceramics at porosities of 10%, 20%,and 40% for ko = 0.7 W/mK.
Equation (8.3) on p. 202 is needed to solve this problem. Inserting the values into theequation, we obtain thermal conductivities of:
P = 10% k = 0.63 W/mKP = 20% k = 0.56 W/mKP = 40% k = 0.42 W/mK
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Ceramics, Graphite, and Diamond: Structure, General Properties, and Applications 97
8.43 A ceramic has ko = 0.80 W/mK. If this ceramic is shaped into a cylinderwith a porosity distribution given by P = 0.1(x/L)(1 − x/L), where x isthe distance from one end of the cylinder and L is the total cylinder length,plot the porosity as a function of distance, evaluate the average porosity, andcalculate the average thermal conductivity.
The plot of porosity is as follows:
Poro
sity
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 given by
P̄ =
∫ 1
00.1X(1−X)dX =
∫ 1
0
(−0.1X2 + 0.1X
)dX = 0.0167
Since the thermal conductivity is linearly related to the porosity, the average porositycan be used, so that the average thermal conductivity is:
k̄ = ko(1− P̄
)= (0.80)(1− 0.0167) = 0.787 W/mK
8.44 It can be shown that the minimum weight of a column which will support agiven load depends on the ratio of the material’s stiffness to the square rootof its density. Plot this property for a ceramic as a function of porosity.
The stiffness of a ceramic is given by Eq. (8.2) on p. 201 as E = Eo(1− 1.9P + 0.9P 2).The density is given by ρ = ρo(1− P ). Therefore, the desired quantity is:
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Ceramics, Graphite, and Diamond: Structure, General Properties, and Applications 98
E/Eo
ρ/ρo
1
0.9
0.8
0.7
0.6
0.5
0.4
0.30 10 20 30 40 50
Porosity (%)
SYNTHESIS, DESIGN AND PROJECTS
8.45 Make a list of the ceramic parts that you can find around your house or inyour car. Give reasons why those parts are made of ceramics.
By the student. Some examples are: Bathroom fixtures: they will not discolor or corrodein normal use, are hard enough to resist the abrasive action of cleaning powders, and arerelatively inexpensive. Coffee mugs: their smooth finish makes them easy to clean, arecorrosion resistant, and are inexpensive. Light-fixture components: electric insulatorsand resistant to heat. Spark plugs: thermal and electrical insulation and corrosionresistance.
8.46 Assume that you are working in technical sales and are fully familiar with allthe advantages and limitations of ceramics. Which of the markets tradition-ally using nonceramic materials do you think ceramics can penetrate? Whatwould you like to talk about to your potential customers during your salesvisits? What questions do you think they may ask you about ceramics?
By the student. There are a number of acceptable answers to this question, and studentsshould not be restricted to the answer given here. Applications that require high hotstrength and wear and corrosion resistance in components such as car engines. Beneficialfeatures of properties should be pointed out, along with recent trends in the improvementof properties. All the advantages that ceramics have over the current material shouldalso be pointed out. Questions concerning the brittleness and low toughness would alsobe raised, as well as questions concerning the consistency of the quality of ceramic parts,particularly as their size and shape complexity increases.
8.47 Describe applications in which a ceramic material with a near-zero coefficientof thermal expansion would be desirable.
By the student. A ceramic material with a near-zero coefficient of thermal expansionwill have a lower tendency of thermal cracking when exposed to temperature gradi-ents. This property would be useful in applications where the ceramic would be cycledthrough temperature ranges, as with space shuttle tiles and cutting tools in machining,particularly in interrupted cutting operations such as milling.
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Ceramics, Graphite, and Diamond: Structure, General Properties, and Applications 99
8.48 The modulus of elasticity of ceramics is typically maintained at elevatedtemperatures. What engineering applications could benefit from this char-acteristic?
By the student. Note that by retaining their high stiffness at elevated temperatures,dimensional accuracy can be maintained. Some examples are bearings, cutting tools,turbine blades, machine-tool components, and electronics. The student is encouraged toexpand on this answer.
8.49 List and discuss the factors that you would take into account when replacinga metal component with a ceramic component in a specific product.
By the student. Note, for example, that the main limitations of ceramics are low tensilestrength and low toughness. The application of the metal component that would bereplaced should, therefore, not require high tensile strength or high impact resistance.
8.50 Obtain some data from the technical literature in the Bibliography of thischapter, and show quantitatively the effects of temperature on the strengthand the modulus of elasticity of several ceramics. Comment on how theshape of these curves differs from those for metals.
By the student. The general trend that will be seen is that such properties as strengthand stiffness do not change as drastically as in metals (see, for example, Fig. 22.1 onp. 592).
8.51 Conduct a literature search and write a brief paper summarizing the prop-erties and potential applications of graphene.
By the student. Graphene is a single sheet of graphite, and it has been suggested thatit has many properties, especially in MEMS and microelectronics.
8.52 It was noted in Section 8.4.1 that there are several basic types of glassesavailable. Make a survey of the technical literature and prepare a table forthese glasses, indicating various mechanical, physical, and optical properties.
By the student. A large variety of answers are possible. For example, the following tableis given in Schey, J., Introduction to Manufacturing Processes, 3d ed., p. 500, based ondata from D.C. Boyd and D.A. Thompson:
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Ceramics, Graphite, and Diamond: Structure, General Properties, and Applications 100
Property Corning glass works code number and type7940 E glass 7740 1720 0800 8871 8830fused boro- alumino- soda-lime- potash- boro-silica silicate silicate silica lead silicate
8.53 Ceramic pistons are being considered for high-speed combustion engines.List the benefits and concerns that you would have regarding this application.
By the student. Ceramic pistons would be advantageous in that they would have ahigh strength and potentially low wear. In addition, the inertial forces associated witha ceramic piston would be much lower than for a metal one, and the need for coolingthe piston becomes less imperative. The main drawbacks are that the ceramic couldexcessively wear the cylinder liner; with three-body wear, any ceramic wear particlescould cause severely damage in the engine. Also, the low fracture toughness of theceramic may cause catastrophic failure of the engine.
8.54 It has been noted that the strength of brittle materials (such as ceramics andglasses) is very sensitive to surface defects, such as scratches (known as notchsensitivity). Obtain several pieces of these materials, scratch them, and testthem by carefully clamping them in a vise and bending them. Comment onyour observations.
By the student. Note that special care must be taken in performing these experiments,and eye protection and the like are necessary. This experiment can be performed using aglass cutter to make a deep and sharp scratch on the glass. It can be demonstrated thatglass, with such a scratch, can be easily broken with bare hands (using work gloves). Notealso the direction of the bending moment with respect o the direction of the scratch. As
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Ceramics, Graphite, and Diamond: Structure, General Properties, and Applications 101
a comparison, even a highly heat-treated aluminum plate will not be nearly as weakenedwhen a similar scratch is made on its surface.
8.55 Electric space heaters for home use commonly utilize a ceramic filament asthe heating element. List the required properties for this filament, explainwhy a ceramic is a suitable material, and perform an Internet search todetermine the specific ceramic material actually utilized in this application.
This is an open-ended problem, and students should be encouraged to obtain their ownsolutions with as specific of answers as possible. Some of the mechanical propertiesrequired for the filament are:
• High melting point
• Sufficient rigidity so that it does not deflect in its retainer.
• Sufficient strength to withstand thermal strains.
• Low coefficient of thermal expansion so that excessive stresses or deflections do notoccur during operation.
There are many types of ceramics used in such heaters, such as aluminisilicates of varioustypes.
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