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Classification of Solid Materials

Dalam dokumen Fundamentals of Machine Elements (Halaman 72-75)

Introduction to Materials and Manufacturing

3.3 Classification of Solid Materials

Engineering materials fall into four major classes: metals, ce- ramics (including glasses), polymers (including elastomers), and composites. The members of each class generally have the following common features:

1. Similar properties, chemical makeup, and atomic struc- ture

2. Similar processing routes 3. Similar applications

3.3.1 Metals

Metals are combinations of metallic elements. They have large numbers of nonlocalized electrons (i.e., electrons not bound to particular atoms). Metals are extremely good con- ductors of electricity and heat and are not transparent to visi- ble light; a polished metal surface has a lustrous appearance.

Furthermore, metals are strong and usually deformable, mak- ing them extremely important materials in machine design.

Metals are usually ductile and can be made stronger by alloying and by mechanical and heat treatment. High- strength alloys can have a percent elongation as low as 2%, but even this is enough to ensure that the material yields be- fore it fractures. Some cast metals have very low ductility, however. Metals are often used in circumstances where cyclic loading is encountered (see Chapter 7), and they are gener- ally resistant to corrosion. Ductile materials, such as steel, ac- commodate stress concentrations by deforming in a way that redistributes the load more evenly.

An isotropicmaterial has properties that are the same in all directions; a material with directional properties is anisotropic. On a microscopic scale, metals form well- defined crystals with ordered packing of atoms. The crystals in a metal are vary small and are randomly oriented. Thus,

while a crystal may be anisotropic, a metal should be consid- ered polycrystalline, and the averaged results of many crys- tals leads to a reasonable assumption of isotropy. However, the crystals, or grains, in a metal may be elongated or ori- ented, leading to anisotropic behavior, especially with sheet metals. Thus, manufacturing history will determine whether a metal is isotropic or anisotropic.

Most metals are initially cast and then can be further pro- cessed to achieve the desired shape. Further processes in- clude secondary casting (such as sand, shell, investment or die casting), bulk forming (forging, extrusion, rolling, draw- ing), sheet forming (deep drawing, stretch forming, stamp- ing) or machining (milling, turning, grinding, polishing). An additional option for metals is to produce metal powders and form desired shapes through powder metallurgy techniques.

Manufacturing processes are addressed in Section 3.8.

3.3.2 Ceramics and Glasses

Ceramicsare compounds of metallic and nonmetallic ele- ments, most frequently oxides, nitrides, and carbides. For ex- ample, the ceramic material aluminum oxide (also known as alumina, carborundum, or in single crystal form, sapphire), is Al2O3. Glassesare made up of metallic and nonmetal- lic elements just as are ceramics, but glasses typically have no clear crystal structure. A typical soda-lime glass consists of approximately 70 wt% silicon dioxide (SiO2), the balance being mainly soda (Na2O) and lime (CaO). Both ceramics and glasses typically insulate against the passage of electric- ity and heat and are more resistant to high temperatures and harsh chemical environments than are metals and polymers.

Ceramics and glasses, like metals, have high density.

However, instead of being ductile like metals, ceramics and glasses are brittle at room temperature. They are typically 15 times stronger in compression than in tension. They are stiff, hard, and abrasion resistant (hence their use for bearings and cutting tools). Thus, they must be considered an important class of engineering material for use in machine elements.

Ceramics are usually obtained by forming a ceramic slurry (a suspension of ceramic powders in water) with binders, and then firing the ceramics to develop a strong bond between particles. Some machining operations are pos- sible with ceramics, but they are often too brittle to be suc- cessfully machined. However, grinding and polishing are commonly performed successfully with ceramics.

Example 3.2: Thermal Expansion

Given: A piece of stabilized zirconia (ZrO2) has a thermal expansion coefficient of12×10−6/C, and is to be implanted into a steel ring. The fit between the steel and the zirconia is a medium press fit at room temperature. When the tempera- ture fluctuates from room temperature to 500C, the zirconia should not loosen.

Find: The correct class of steel to be used from those given in Table 3.1.

Solution:Since the zirconia should not loosen from the steel when the temperature is increased, a slightly smaller coef- ficient of thermal expansion, but very close to the zirconia value, is desired for the stainless steel. From Table 3.1, a low- or medium alloy has a coefficient of thermal expansion of 11×10−6/C and would therefore be the preferred classes of steel.

3.3.3 Polymers and Elastomers

Polymersare organic compounds composed of carbon, hy- drogen, and other nonmetallic elements. Polymers have large and complex molecular structures.

Polymers, also called plastics, are of two basic types: thermoplastics and thermosets. Thermoplastics are long-chain molecules, sometimes with branches, where the strength arises from interference between chains and branches. Thermosetshave a higher degree of cross-linking so that molecular chains are linked together and cannot slide over each other. In general, thermoplastics are more ductile than thermosets, and at elevated temperatures they soften significantly and melt. Thermosets are more brittle, do not soften as much as thermoplastics, and usually degrade chem- ically before melting.

Elastomershave a networked cross-linked structure, but not as extensive as that for more rigid thermosets, so they produce large elastic deformations at relatively light loads. A common example of an elastomer is the material in a rubber band, which displays the typical characteristics of large elas- tic deformation followed by brittle fracture without any plas- tic deformation. Further, rubber bands are highly nonlinearly elastic, which is typical of elastomers.

Polymers and elastomers can be extremely flexible with large elastic deformations. Polymers are roughly five times less dense than metals but have nearly equivalent strength-to-weight ratios. Because polymers creep (the time- dependent permanent deformation that occurs under static stress) even at room temperature, a polymer machine element under load may, with time, acquire a permanent set. The properties of polymers and elastomers change greatly with variations in temperature. For example, a polymer that is tough and flexible at 20C may be brittle at the 4C environ- ment of a household refrigerator and yet creep rapidly at the 100C of boiling water.

The mechanical properties of polymers are characterized using the same parameters as for metals (i.e., modulus of elas- ticity and tensile, impact, and fatigue strengths). However, polymers vary much more in strength, stiffness, etc, than do metals. This variation can be explained by the difference in chain lengths and the amount of polymer that is in a crys- talline or amorphous state. Further, a thermoplastic that is deformed plastically will have its molecules aligned in the direction of strain, leading to higher strength and anisotropy.

Thus, two polymers with the identical chemical constituents can have very different microstructures and associated vari- ation in mechanical and physical properties. In addition, the mechanical characteristics of polymers, for the most part, are highly sensitive to the rate of deformation, temperature, and chemical nature of the environment (the presence of water, oxygen, organic solvents, etc.). Therefore, the particular val- ues given for polymer mechanical properties should be used with caution.

Polymers are easy to shape: complicated parts perform- ing several functions can be molded from a polymer in a sin- gle operation (see Section 3.8.2). However, injection molding operations have high tooling costs and can only be justified for large production runs. Large elastic deflections allow the design of polymer components that snap together, making assembly fast and inexpensive. Polymers are corrosion resis- tant and have low coefficients of friction.

Thermoplastics and thermosets have very different man- ufacturing options and strategies. Thermoplastics are gen- erally heated to a temperature above their melting point, formed into a desired shape and then cooled. Examples of common manufacturing processes for thermoplastics in- clude extrusion, injection molding, blow molding, and ther-

Classification of Solid Materials 57

0 200 400 600 x 104

Lead Epoxy Pure aluminum Wood

Steel Nylon Graphite

Ratio of strength to density, N-m/kg Figure 3.3: Strength/density ratio for various materials.

moforming. Thermosets are blended from their constituents, formed to a desired shape and then cured at elevated tem- perature and/or pressure to develop cross links. Common manufacturing methods used for thermosets include reaction injection molding, compression molding, and potting (simi- lar to casting).

3.3.4 Composites

Figure 3.3 compares a number of materials from a minimum- weight design standpoint (i.e., a larger strength-to-density ra- tio leads to a lighter design). Fibers can have much better strength-to-weight ratios than conventional extruded bars, molded plastics, and sintered ceramics. However, fibers are often susceptible to corrosion, even in air. For example, graphite fibers will oxidize readily in air and cannot provide their exceptional strength for long.

Many modern technologies require machine elements with demanding combinations of properties that cannot be met by conventional metal alloys, ceramics, and polymeric materials. Present-day technologies require solid materials that have low density, high strength, stiffness and abrasion resistance, and that are not easily corroded. This combina- tion of characteristics is rather formidable, considering that strong materials are usually relatively dense and that increas- ing stiffness generally decreases impact strength.

Composite materialscombine the attractive properties of two or more material classes while avoiding some of their drawbacks. A composite is designed to display a combina- tion of the best characteristics of each component material.

For example, graphite-reinforced epoxy acquires strength from the graphite fibers while the epoxy protects the graphite from oxidation. The epoxy also helps support shear stresses and provides toughness.

The three main types of composite material are:

1. Particle reinforced, which contain particles with approxi- mately the same dimensions in all directions distributed in a matrix, such as concrete.

2. Discontinuous fiber reinforced, which use fibers of limited length-to-diameter ratio in a matrix, such as fiberglass.

3. Continuous fiber reinforced, where continuous fibers are incorporated, such as seen in graphite tennis rackets.

Figure 3.4 shows a cross section of a continuous fiber- reinforced composite material. Most such composites contain glass, polymer or carbon fibers, and a polymer matrix. These composites cannot be used at elevated temperatures because the polymer matrix softens or degrades, but at room temper- ature their performance can be outstanding. Some disadvan-

σ2

1

2

Fiber Matrix

σ1 τ12

Figure 3.4: Cross section of fiber-reinforced composite mate- rial.

tages of composites are that they are expensive and relatively difficult to form and join.

Composite materials have many characteristics that are different from those of the other three classes of material con- sidered. Whereas metals, polymers, and ceramics are ho- mogeneous(properties are not a function of position in the solid),isotropic(properties are the same in all directions at a point in the solid), oranisotropic(properties are different in all directions at a point in the solid), composites arenon- homogeneousandorthotropic. An orthotropic material has properties that are different in three mutually perpendicular directions at a point in the solid but has three mutually per- pendicular planes of material symmetry. Consideration in this text is limited to simple, unidirectional, fiber-reinforced orthotropic composite materials, such as shown in Fig. 3.4.

An important parameter in discontinuous fiber- reinforced composites is the fiber length. Some critical fiber length is necessary for effective strengthening and stiffening of the composite material. The critical length,lcr, of the fiber depends on the fiber diameter, d, its ultimate strength, Su, and the fiber-matrix bond strength,τf, according to

lcr= Sud

f . (3.2)

The two in the denominator of Eq. (3.2) accounts for the fact that the fiber is embedded in the matrix and splits into two parts at failure. For a number of glass- and carbon-fiber- reinforced composites this critical length is about 1 mm, or 20 to 150 times the fiber diameter.

Discontinuous and particle-reinforced composites share many of the manufacturing methods with thermosetting polymers and metals, depending on the matrix materials.

Polymer matrix materials are commonly molded or placed onto a form (lay-up) and then cured in an oven. Metal matrix composites use variants of casting or powder metal- lurgy techniques. Continuous fiber reinforced polymers have unique manufacturing approaches, including tape layup, pultrusion, pulforming, and filament winding.

P

0 0’ Y

U

R

0.002 Sy

Su

Stress, σ =

P A

Slope, E =σε

Strain, ε = δ l

Figure 3.5: Typical stress-strain curve for a ductile material.

Example 3.3: Strength of a Composite Material

Given:A fiber-reinforced plastic contains carbon fibers hav- ing an ultimate strength of 1 GPa and a modulus of elasticity of 150 GPa. The fibers are 3 mm long with a diameter of 30 µm.

Find: Determine the required fiber-matrix bond strength in order to fully develop the strength of the fiber reinforcement.

Solution:From Eq. (3.2), the fiber-matrix bond strength can be expressed as

τf =Sud 2lcr

=

109 30

×10−6

2(0.003) = 5MPa.

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