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METAL MATRIX COMPOSITES (MMC) Introduction Metal matrix composites (MMCs), as the name implies, have a metal matrix. Examples of matrices in such composites include aluminum, magnesium, and titanium. Typical fibers include carbon and silicon carbide. Metals are mainly reinforced to increase or decrease their properties to suit the needs of design. For example, the elastic stiffness and strength of metal scan be increased and large coefficients of thermal expansion and thermal and electric conductivities of metals can be reduced, by the addition of fibers such as silicon carbide. Advantages Metal matrix composites are mainly used to provide advantages over monolithic metals such as steel and aluminum. These advantages include higher specific strength and modulus by reinforcing low‐density metals, such as aluminum and titanium; lower coefficients of thermal expansion by reinforcing with fibers with low coefficients of thermal expansion, such as graphite; and maintaining properties such as strength at high temperatures. MMCs have several advantages over polymer matrix composites. These include higher elastic properties; higher service temperature; insensitivity to moisture; higher electric and thermal conductivities; and better wear, fatigue, and flaw resistances. The drawbacks of MMCs over PMCs include higher processing temperatures and higher densities. Do any properties degrade when metals are reinforced with fibers? Yes, reinforcing metals with fibers may reduce ductility and fracture toughness. Ductility of aluminum is 48% and it can decrease to below 10% with simple reinforcements of silicon carbide whiskers. The fracture toughness ofaluminum alloys is 18.2 to 36.4 (20 to 40 ) and it reduces by 50% or more when reinforced with silicon fibers. MMC‐ Manufacturing Fabrication methods for MMCs are varied. One method of manufacturing them is diffusion bonding, which is used in manufacturing boron/aluminum composite parts (Figure). A fiber mat of boron is placed between two thin aluminum foils about 0.002 in. (0.05 mm) thick. A polymer binder or an acrylic adhesive holds the fibers together in the mat. Layers of these metal foils are stacked at angles as required by the design. The laminate is first heated in a
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vacuum bag to remove the binder. The laminate is then hot pressed with a temperature of about 932°F (500°C) and pressure of about 5 ksi (35 MPa) in a die to form the required machine element.
Application (MMC) Space: The space shuttle uses boron/aluminum tubes to support its fuselage frame. In addition to decreasing the mass of the space shuttle by more than 320 lb (145 kg), boron/aluminum also reduced the thermal insulation requirements because of its low thermal conductivity. The mast of the Hubble Telescope uses carbon‐reinforced aluminum. Military: Precision components of missile guidance systems demand dimensional stability — that is, the geometries of the components cannot change during use. Metal matrix composites such as SiC/aluminum composites satisfy this requirement because they have high micro yield strength.* In addition, the volume fraction of SiC can be varied to have a coefficient of thermal expansion compatible with other parts of the system assembly. Transportation: Metal matrix composites are finding use now in automotive engines that are lighter than their metal counterparts. Also, because of their high strength and low weight, metal matrix composites are the material of choice for gas turbine engines.
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MICROMECHANICS Basic definitions Structural materials possess a great number of physical, chemical and other types of properties, but at least two principal characteristics are of primary importance. These characteristics are the stiffness and strength that provide the structure with the ability to maintain its shape and dimensions under loading or any other external action. High stiffness means that material exhibits low deformation under loading. However, by saying that stiffness is an important property we do not mean that it should be necessarily high. The ability of a structure to have controlled deformation (compliance) can also be important for some applications (e.g., springs; shock absorbers; pressure, force, and displacement gauges). Lack of material strength causes an uncontrolled compliance, i.e., in failure after which a structure does not exist anymore. Usually, we need to have as high strength as possible, but there are some exceptions (e.g., controlled failure of explosive bolts is used to separate rocket stages). Thus, without controlled stiffness and strength the structure cannot exist. Naturally, both properties depend greatly on the structure’s design but are determined by the stiffness and strength of the structural material because a good design is only a proper utilization of material property. It is natural to characterize material strength by the ultimate stress Stress = σ = F A Where F is force causing the failure
Stress is measured as force divided by area, i.e., according to international (SI) units, in pascals (Pa) so that 1 Pa =1N/m2. Because the loading of real structures induces relatively high stresses, we also use kilopascals (1 kPa = 103
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Pa), megapascals (1MPa = 106 Pa), and gigapascals (1GPa = 109 Pa). Conversion of old metric (kilogram per square centimeter)and English (pound per square inch) units to pascals can be done using the following relations: 1 kg/cm2 = 98 kPa and 1 psi = 6.89 kPa. Specific strength of a material is defined as
Kσ =
σ ρ
If we use old metric units, i.e., measure force and mass in kilograms and dimensions in meters, substitution of Eq. (1.1) into Eq. (1.2) yields kσ in meters. Thisresult has a simple physical sense, namely kσ is the length of the vertically hanging fiberunder which the fiber will be broken by its own weight. The stiffness of the bar shown in Figure can be characterized by an elongation ∆ corresponding to the applied force F or acting stress σ = F/A. However, ∆ is proportional to the bar’s length L0. To evaluate material stiffness, we introduce strain
ε=
∆ L0
Since ε is very small for structural materials, the ratio is normally multiplied by 100, and ε is expressed as a percentage. Naturally, for any material, there should be some interrelation between stress and strain, i.e., ε = f(σ) or σ = f(ε) These equations specify the so‐called constitutive law and are referred to as constitutive equations. The work performed by force F in Figure is accumulated in the bar as potential energy, which is also referred to as strain energy or elastic energy. Consider some infinitesimal elongation d ∆ and calculate the elementary work performed by the force F in Fig. 1.1 as dW = Fd ∆ . Then, work corresponding to point 1 of the curve in Fig. 1.2 is ∆1
W = ∫ Fd ∆ 0
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where ∆1 is the elongation of the bar corresponding to point 1 of the curve. The work W is equal to elastic energy of the bar which is proportional to the bar’s volume and can bepresented as ε1
E = L0 A∫ σ d ε 0
Where σ = F A , ε = ∆ L , and ε1 = ∆1 L 0 0 ε1
Now the integral U = ∫0 σ d ε is a specific elastic energy (energy accumulated in a unit volume of the bar) that is referred to as an elastic potential. It is important that U does not depend on the history of loading. This means that irrespective of the way we reach point 1 of the curve in Fig. 1.2 (e.g., by means of continuous loading, increasing force F step by step, or using any other loadingprogram), the final value of U will be the same and will depend only on the value of final strain ε1 for the given material. A very important particular case of the elastic model is the linear elastic model described by the well‐known Hooke’s law (see Fig. 1.3) σ = Eε Here, E is the modulus of elasticity. It follows that E = σif ε = 1, i.e., if L = L0. Thus, the modulus can be interpreted as the stress causing elongation of the bar in Fig. 1.1 to be the same as the initial length. Since the majority ofstructural materials fail before such a high elongation can occur, the modulus is usuallymuch higher than the ultimate stress σ.
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Similar to specific strength kσ in Eq. (1.2), we can introduce the corresponding specific modulus
KE = E ρ which describes a material’s stiffness with respect to its material density. Hooke’s law describes rather well the initial part of stress–strain diagram for the majority of structural materials. However, under a relatively high level of stress or strain, materials exhibit nonlinear behavior. One of the existing models is the nonlinear elastic material model introduced above (see Fig. 1.2). This model allows us to describe the behavior of highly deformable rubber‐type materials. Another model developed to describe metals is the so‐called elastic–plastic material model. The corresponding stress–strain diagram is shown in Fig. 1.5. In contrast to an elastic material (see Fig. 1.2), the processes of active loading and unloading are described with different laws in this case. In addition to elastic strain, εe, which disappears after the load is taken off, the residual strain (for the bar shown in Fig. 1.1, it is plastic strain, εp) remains in the material. As for an elastic material, the stress–strain curve in Fig. 1.5 does not depend on the rate of loading (or time of loading). However, in contrast to an elastic material, the final strain of an elastic–plastic material can depend on the history of loading, i.e., on the law according to which the final value of stress was reached. Thus, for elastic or elastic–plastic materials, constitutive equations do not include time. However, under relatively high temperature practically all the materials demonstrate time‐dependent behavior (some of them do it even
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under room temperature). If we apply some force F to the bar shown in Fig. 1.1 and keep it constant, we can see that for a time‐sensitive material the strain increases under a constant force. This phenomenon is called the creep of the material. So, the most general material model that is used in this book can be described with a constitutive equation of the following type: ε = f(σ,t,T)
where t indicates the time moment, whereas σ and T are stress and temperature, corresponding to this moment. In the general case, constitutive equation, ε = f(σ,t,T) , specifies strain that can be decomposed into three constituents corresponding to elastic, plastic and creep deformation, i.e.,
ε = εe + εp + εc However, in application to particular problems, this model can be usually substantially simplified. To show this, consider the bar in Fig. 1.1 and assume that a force F is applied at the moment t = 0 and is taken off at moment t = t1 as shown in Fig. 1.6a.
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At the moment t = 0, elastic and plastic strains that do not depend on time appear, and while time is running, the creep strain is developed. At the moment t = t1, the elastic strain disappears, while the reversible part of the creep strain, ε c , disappears with time. Residual strain consists of the plastic t
strain, εp, and residual part of the creep strain, ε c . r
The mechanics of fiber‐reinforced composite materials are studied at two levels: 1. The micromechanics level, in which the interaction of the constituent materials is examined on a microscopic scale. Equations describing the elastic and thermal characteristics of a lamina are, in general, based on micromechanics formulations. An understanding of the interaction between various constituents is also useful in delineating the failure modes in a fiber‐ reinforced composite material. 2. The macromechanics level, in which the response of a fiber‐reinforced composite material to mechanical and thermal loads is examined on a macroscopic scale. The material is assumed to be homogeneous. Equations of orthotropic elasticity are used to calculate stresses, strains, and deflections. Volume Fractions Consider a composite consisting of fiber and matrix. Take the following symbol notations:
vc , v f , vm = volume of composite, fiber, and matrix, respectively
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ρc , ρ f , ρ m = density of composite, fiber, and matrix, respectively. Now define the fiber volume fraction Vf and the matrix volume fraction Vm as
‐‐‐‐ (1)
Note that the sum of volume fractions is
‐‐‐‐ (2)
And we have from (1) and (2)
‐‐‐‐ (3)
Mass Fractions Consider a composite consisting of fiber and matrix and take the following symbol notation: mc , m f , mm = mass of composite, fiber, and matrix, respectively. The mass fraction (weight fraction) of the fibers (Wf) and the matrix (Wm) are defined as
Mf =
mf mc
and M m =
mm
mc
‐‐‐‐ (4)
Note that the sum of mass fractions is
M f + Mm =1
‐‐‐‐ (5)
And from (4) and (5), we have
m f + mm = mc
‐‐‐‐ (6)
From the definition of density, we have
mc = ρ c vc , m f = ρ f v f and mm = ρ m vm Substituting (7) in (4), we get
‐‐‐‐ (7)
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Mf =
ρf ρ V f and M m = m Vm ρc ρc
‐‐‐‐ (8)
in terms of the fiber and matrix volume fractions.
Density The sum of mass fractions is given from equation (6) as
m f + mm = mc
‐‐‐‐ (9)
Using (7) in the above equation, we have
‐‐‐‐ (10)
‐‐‐‐ (11)
and
Now, consider that the volume of a composite vc is the sum of the volumes of the fiber vf and matrix (vm): Using (10), The density of the composite in terms of mass fractions can be found as
‐‐‐‐ (12)
Problem A glass/epoxy lamina consists of a 70% fiber volume fraction. The density of the fiber is ρf = 2500 kg/m3, density of the matrix is ρm=1200 kg/m3. Determine the following 1. Density of lamina 2. Mass fractions of the glass and epoxy
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3. (a) Volume of composite lamina if the mass of the lamina is 4 kg and (b)Volume and (c) mass of glass and epoxy. 1) 2) The fiber and matrix mass fractions are
Mf =
ρf 2500 Vf = ( 0.7 ) = 0.8294 2100 ρc
Mm =
ρm 1200 Vm = ( 0.3) = 0.1706 ρc 2110
Note that the sum of the mass fractions,
M m + M m = 0.8294 + 0.1706 = 1.0 3)The volume of composite is
vc =
mc
ρc
=
4 = 1.896 ×10−3 m3 2110
The volume of the fiber is The volume of the matrix is
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FIBER–MATRIX INTERACTIONS IN A UNIDIRECTIONAL LAMINA We consider the mechanics of materials approach [1] in describing fiber– matrix interactions in a unidirectional lamina owing to tensile and compressive loadings. The basic assumptions in this vastly simplified approach are as follows: 1. Fibers are uniformly distributed throughout the matrix. 2. Perfect bonding exists between the fibers and the matrix. 3. The matrix is free of voids. 4. The applied force is either parallel to or normal to the fiber direction. 5. The lamina is initially in a stress‐free state (i.e., no residual stresses are present in the fibers and the matrix). 6. Both fibers and matrix behave as linearly elastic materials. Theoretical formulas for finding the stiffness, strength, and hygrothermal properties of a unidirectional lamina are a function of fiber volume fraction. Measurements of the constituents are generally based on their mass, so fiber mass fractions must also be defined. Moreover, defining the density of a composite also becomes necessary because its value is used in the experimental determination of fiber volume and void fractions of a composite. LONGITUDINAL TENSILE LOADING In this case, the load on the composite lamina is a tensile force applied parallel to the longitudinal direction of the fibers. Unidirectional Continuous Fibers: Assuming a perfect bonding between fibers and matrix, we can write
‐‐‐‐ (1)
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Comparing Equation 3.2 with Equation 3.3 and noting that
Ef>Em,
we
conclude that the fiber stress σ f is always greater than the matrix stress σ m . The tensile force Pc applied on the composite lamina is shared by the fibers and the matrix so that
Pc = Pf + Pm
‐‐‐(4)
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Equ ation (7) is called the rule of mixtures. This equation shows that the longitudinal modulus of a unidirectional continuous fiber composite is intermediate between the fiber modulus and the matrix modulus; it increases linearly with increasing fiber volume fraction; and since Ef>Em, it is influenced more by the fiber modulus than the matrix modulus. The fraction of load carried by fibers in longitudinal tensile loading is
‐‐‐‐‐(8)
Equation (8) is plotted in Figure 3.2 as a function ofEf/Em ratio and fiber volume fraction. In polymer matrix composites, the fiber modulus is much greater than the matrix modulus. In most polymer matrix composites,Ef/Em> 10. Thus, evenfor vf =0.2, fibers carry >70% of the composite load. Increasing the fiber volume fraction increases the fiber load fraction as well as the composite load. Although cylindrical fibers can be theoretically packed to almost 90%volume fraction, the practical limit is close to ~80%. Over this limit, the matrix will not be able to wet the fibers.
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POLYMER MATRIX COMPOSITES Polymer matrix composites (PMCs) consist of a polymer (e.g., epoxy, polyester, urethane) reinforced by thin diameter fibers (e.g., graphite, aramids, boron). For example, graphite/epoxy composites are approximately five times stronger than steel on a weight‐for‐weight basis. The reasons why they are the most common composites include their low cost, high strength, and simple manufacturing principles. Drawbacks of polymer matrix composites: The main drawbacks of PMCs include low operating temperatures, high coefficients of thermal and moisture expansion and low elastic properties in certain directions. Fiber materials (reinforcement phase) used in PMC 1. Glass Fiber Glass is the most common fiber used in polymer matrix composites. Its advantages include its high strength, low cost, high chemical resistance, and good insulating properties. The drawbacks include low elastic modulus, poor adhesion to polymers, high specific gravity, sensitivity to abrasion (reduces tensile strength), and low fatigue strength. Manufacturing: Glass fibers are made generally by drawing from a melt as shown in Figure 1.9. The melt is formed in a refractory furnace at about 2550°F (1400°C) from a mixture that includes sand, limestone, and alumina. The melt is stirred and maintained at a constant temperature. It passes through as many as 250 heated platinum alloy nozzles of about 10μm diameter, where it is drawn into filaments of needed size at high speeds of about 25 m/s.
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These fibers are sprayed with an organic sizing solution before they are drawn. The sizing solution is a mixture of binders, lubricants, and coupling and antistatic agents; binders allow filaments tobe packed in strands, lubricants prevent abrasion of filaments, and coupling agents give better adhesion between the inorganic glass fiber and the organic matrix. Fibers are then drawn into strands and wound on a forming tube. Strands are groups of more than 204 filaments. The wound array of strands is then removed and dried in an oven to remove any water or sizing solutions. The glass strand can then be converted into several forms as shown in Figure.
Schematic of manufacturing glass fibers and available glass forms. (From Bishop, W., in Advanced Composites, Partridge, I.K., Ed., Kluwer Academic Publishers, London, 1990, Figure 4, p.177.
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2. Graphite fiber Graphite fibers are very common in high‐modulus and high‐strength applications such as aircraft components, etc. The advantages of graphite fibers include high specific strength and modulus, low coefficient of thermal expansion, and high fatigue strength. The drawbacks include high cost, low impact resistance, and high electrical conductivity. Manufacturing: Graphite fibers are generally manufactured from three precursor materials: rayon, polyacrylonitrile (PAN), and pitch. PAN is the most popular precursor. The precursor is drawn into long strands or fibers and then heated to a very high temperature with‐out allowing it to come in contact with oxygen. Without oxygen, the fiber cannot burn.
In the first process, called stabilization, the fiber is passed through a furnace between 392 and 572°F (200 and 300°C) to stabilize its dimensions during the subsequent high‐temperature processes. In the second process, called carbonization, it is pyrolized in an inert atmosphere of nitrogen or argon between 1832 and 2732°F (1000 and 1500°C). The high temperature causes the atoms in the fiber to vibrate violently until most of the non‐carbon atoms are expelled. This process is called carbonization. In the last process, called graphitization, it is heat treated above 4532°F (2500°C). The graphitization yields a microstructure that is more graphitic than that produced by carbonization. The fibers may also be subjected to tension in the last two heating processes to develop fibers with a higher degree of orientation.
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At the end of this three‐step heat treatment process, the fibers are surface treated to develop fiber adhesion and increase laminar shear strength when they are used in composite structures. They are then collected on a spool. 3. Aramid fiber An aramid fiber is an aromatic organic compound made of carbon, hydrogen, oxygen, and nitrogen. Its advantages are low density, high tensile strength, low cost, and high impact resistance. Its drawbacks include low compressive properties and degradation in sunlight. Types: The two main types of aramid fibers are Kevlar 29® and Kevlar49®. Both types of Kevlar fibers have similar specific strengths, but Kevlar49 has a higher specific stiffness. Kevlar 29 is mainly used in bulletproof vests, ropes, and cables. High performance applications in the aircraft industry use Kevlar 49. Manufacturing: The fiber is produced by making a solution of proprietary polymers and strong acids such as sulfuric acid. The solution is then extruded into hot cylinders at 392°F (200°C), washed, and dried on spools. The fiber is then stretched and drawn to increase its strength and stiffness.
MATRIX phase in PMC Polyesters: The advantages are low cost and the ability to be made translucent; drawbacks include service temperatures below 170°F (77°C), brittleness, and high shrinkage of as much as 8% during curing. Phenolics: The advantages are low cost and high mechanical strength; drawbacks include high void content. Epoxies: The advantages are high mechanical strength and good adherence to metals and glasses; drawbacks are high cost and difficulty in processing. Epoxy Epoxy resins are the most commonly used resins. They are low molecular weight organic liquids containing epoxide groups. Epoxide has three members in its ring: one oxygen and two carbon atoms. The reaction of epichlorohydrin with phenols or aromatic amines makes most epoxies. Hardeners, plasticizers, and fillers are also added to produce epoxies with a wide range of properties of viscosity, impact, degradation, etc. The main reasons why epoxy is the most used polymer matrix material are
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¾ High strength ¾ Low viscosity and low flow rates, which allow good wetting of fibers and prevent misalignment of fibers during processing ¾ Low volatility during cure ¾ Low shrink rates, which reduce the tendency of gaining large shear stresses of the bond between epoxy and its reinforcement ¾ Available in more than 20 grades to meet specific property and processing requirements
Manufacturing of PMC Polymer‐matrix composites are much easier to fabricate than metal‐matrix, carbon‐matrix, and ceramic‐matrix composites, whether the polymer is a thermoset or a thermoplastic. This is because of the relatively low processing temperatures required to fabricate polymer‐matrix composites. For thermosets, such as epoxy, phenolic, and furfuryl resin, the processing temperature typically ranges from room temperature to about 200°C; for thermoplastic polymers, such as polyimide (PI), polyethersulfone (PES), polyetheretherketone (PEEK), polyetherimide (PEI), and polyphenyl sulfide (PPS), the processing temperature typically ranges from 300 to 400°C. Autoclave forming: . This method of manufacturing a high‐performance polymer‐matrix composite laminate containing continuous fibers commonly involves prepreg sheets. Aprepreg is a sheet of continuous oriented fibers that have been impregnated with a polymer or a polymer precursor. First, a peel ply made out of nylon or cellophane coated with Teflon is placed on the mold. Teflon is used for easy removal of the part and the peel ply achieves a desired finish that is smooth and wrinkle free. Replacing Teflon by mold releasing powders and liquids can also accomplish removal of the part. Prepregs of the required number are laid up one ply at a time by automated means or by hand. Each ply is pressed to remove any entrapped air and wrinkles. The lay‐up is sealed at the edges to form a vacuum seal. Now one establishes the bleeder system to get rid of the volatiles and excess resin during the heating and vacuum process that follows later. The bleeder system consists of several bleeder sheets made of glass cloth. These are placed on the edges and the top of the lay‐up. Then, vacuum connections are placed over the bleeders and the lay‐up is bagged. A partial vacuum is developed to smooth the bag surface. The whole assembly is put in an auto clave (Figure 1.18), where heat and pressure are applied with an inert
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gas such as nitrogen. The vacuum system is kept functioning to remove volatiles during the cure cycle and to keep the part conformed to the mold. The cure cycle may last more than 5 h. One method of forming unidirectional fiber composite parts with a constant cross‐section (e.g., round, rectangular, pipe, plate, I‐shaped) is pultrusion, in which fibers are drawn from spools, passed through a polymer resin bath for impregnation, and gathered together to produce a particular shape before entering a heated die. One method of forming continuous fiber composites in the shape of cylinders or related objects is filament winding, which involves wrapping continuous fibers from a spool around a (commonly cylindrical) mandrel. The fibers are wound in various predetermined directions (e.g., 90°) relative to the axis of the mandrel. The winding pattern is a part of the composite design. Since the composite is very strong in the fiber direction, filament winding results in a cylindrical article that resists radial expansion, as needed for pressure vessels. The fibers can be impregnated with a resin before or after winding. Filament winding is used to make pressure tanks and pipes. The temperature of the mandrel, the impregnation temperature of the resin, the impregnation time, the tension of the fibers, and the pressure of the fiber winding are processing parameters that need to be controlled. Difference in processing thermoset and thermoplastic Polymers The processing of polymer‐matrix composites typically requires heating. In the case of a thermosetting resin, the heating is to cause the completion of polymerization (cross‐linking) of the resin. In the case of a thermoplastic matrix, the heating is done to soften or melt the thermoplastic matrix (The melting temperature is higher than the softening temperature but it allows more extensive flow). As the polymerization process is a reaction, it takes time. In contrast, the softening or melting is a phase transition that occurs once the appropriate temperature is reached. As a result, the processing time tends to be considerably longer for a thermoset‐matrix composite than a thermoplastic‐ matrix composite.
