Basics of Composite Material (QnA)
- 1. Mukuldev Khunte-MME 6th Semester 01UG15050023 Name-Mukuldev Khunte Dept. of Metallurgical & Materials Engineering (6th Semester) Roll No-01UG15050023 Subject- Engineering Polymers & Composite Assignment Q. Define composite material. Classify this material as per their parametric level. What are the fundamental properties of a composite material. Answer –Definition-A material which is composed of two or more materials at a microscopic scale and have chemically distinct phases. And Heterogeneous at a microscopic scale but statically homogeneous at macroscopic scale. Constituent materials have significantly different properties. Composite materials are commonly classified at following two distinct levels: • The first level of classification is usually made with respect to the matrix constituent. The major composite classes include Organic Matrix Composites (OMCs), Metal Matrix Composites (MMCs) and Ceramic Matrix Composites (CMCs). The term organic matrix composite is generally assumed to include two classes of composites, namely Polymer Matrix Composites (PMCs) and carbon matrix composites commonly referred to as carbon-carbon composites. • The second level of classification refers to the reinforcement form - fibre reinforced composites, laminar composites and particulate composites. Fibre Reinforced composites (FRP) can be further divided into those containing discontinuous or continuous fibres. • Fibre Reinforced Composites are composed of fibres embedded in matrix material. Such a composite is a discontinuous fibre or short fibre composite if its properties vary with fibre length. On the other hand, when the length of the fibre is such that any further increase in length does not further increase, the elastic modulus of the composite, the composite is continuous fibre reinforced • Laminar Composites are composed of layers of materials held together by matrix. Sandwich structures fall under this category. • Particulate Composites are composed of particles distributed or embedded in a matrix body. The particles may be flakes or in powder form. Concrete and wood particle boards are examples of this category. fundamental properties of a composite material 1. Specific Strength-This is simply the rigidity or hardness of a material regarding its weight. For example, several composites such as fiberglass share comparable impact resistance (bangability) with steel and titanium at a fraction of the weight employed. 2. Expense-Many composites can be manufactured with less cost than their traditional metal counterparts. 3. Application-Because composites are composed of 2 or more "phases", they can be formulated to meet the needs of a specific application with considerable ease. 4. Processability-As most of you know, metal processing requires high amounts of thermal energy (heat). Plastics and plastic based composites require less heat to mould or process the products. There is a constant desire to produce composites which can be processed at low temperatures but when cured or set-up (paint drying or a mould cooling), they are very impact resistant and very heat resistant or fire retardant. Q-Advantage and Limitation of Composite Material.
- 2. Mukuldev Khunte-MME 6th Semester 01UG15050023 Answer- Advantages of Composite Materials 1. Composites reduced the overall structural member weight by 20-50%. 2. Composites are very corrosion and fatigue resistance. 3. Composites have tolerable mechanical properties. 4. Composites have lower assembly costs because it requires very few fasteners, bolts etc. Disadvantages of Composite Materials 1. Composites have high recurring costs. 2. Composites are higher non-recurring costs. 3. Composites have higher material costs. 4. Composites have very expensive repairs and maintenance. 5. Composites needed isolation to prevent adjacent aluminium part galvanic corrosion Q-State the law of mixture and stablish this. What is the significance of this law. Answer- Composite stiffness can be predicted using a micro-mechanics approach termed the rule of mixtures. Assumptions 1. Fibres are uniformly distributed throughout the matrix. 2. Perfect bonding between fibres and matrix. 3. Matrix is free of voids. 4. Applied loads are either parallel or normal to the fibre direction. 5. Lamina is initially in a stress-free state (no residual stresses). 6. Fibre and matrix behave as linearly elastic materials Rule of mixture is the simple’s relation between the properties of a composite and those of its constituents. The simplest is the linear mixture rule, e.g. Ec = v1E1 +v2E2 for the modulus. Ec is the composite modulus v1 is the volume fraction of component 1 with modulus E1 (say the matrix) and v2, E2 are the corresponding quantities for component 2 (say, the reinforcement). This is true physically is you have parallel phases and the force is parallel with the phases. The other extreme is the reciprocal mixture rule: 1/Ec = v1/E1 + v2/E2 in this case the elements are in series. The real situation is somewhere in-between. Formally the two extremes can be combined Q-Determine the young’s Modulus in Transvers and Longitudinal direction in an isostrain condition. Discuss Critical Length of fiber. Answer- Elastic Behaviour—Longitudinal Loading Let us now consider the elastic behaviour of a continuous and oriented fibrous composite that is loaded in the direction of fiber alignment. First, it is assumed that the fiber–matrix interfacial bond is very good, such that deformation of both matrix and fibers is the same (an isostrain situation). Under these conditions, the total load sustained by the composite Fc is equal to the sum of the loads carried by the matrix phase Fm and the fiber phase Ff, or Fc = Fm + Ff (16.4) From the definition of stress, Equation 6.1, F sA; and thus expressions for Fc, Fm, and Ff in terms of their respective stresses (sc, σm, and σf) and cross-sectional areas (Ac, Am, and Af) are possible. Substitution of these into Equation 16.4 yields σσcAc = σmAm + σfAf (16.5) and then, dividing through by the total cross-sectional area of the composite, Ac, we have where AmAc and AfAc are the area fractions of the matrix and fiber phases, respectively. If the composite, matrix, and fiber phase lengths are all equal, AmAc is equivalent to the volume fraction of the matrix, Vm, and likewise for the fibers, Vf AfAc. Equation 16.6 now becomes
- 3. Mukuldev Khunte-MME 6th Semester 01UG15050023 σc = σmVm + σfVf The previous assumption of an isostrain state means that €c =€m+ €f (16.7) (16.8) 16.9 yields an expression for the modulus of elasticity of a continuous and aligned fibrous composite in the direction of alignment (or longitudinal direction), Ecl, as since the composite consists of only matrix and fiber phases; that is, Vm Vf 1. Thus, Ecl is equal to the volume-fraction weighted average of the moduli of elasticity of the fiber and matrix phases. Other properties, including density, also have this dependence on volume fractions. Equation 16.10a is the fiber analogue of Equation 16.1, the upper bound for particle-reinforced composites. It can also be shown, for longitudinal loading, that the ratio of the load carried by the fibers to that carried by the matrix is Elastic Behavior—Transverse Loading A continuous and oriented fiber composite may be loaded in the transverse direction; that is, the load is applied at a 90 angle to the direction of fiber alignment as shown in Figure 16.8a. For this situation the stress s to which the composite as well as both phases are exposed is the same, or σc = σm= σf= σ (16.12) This is termed an isostress state. Also, the strain or deformation of the entire composite c is €c= €mVm + €fVf (16.13) but, since sE, where Ect is the modulus of elasticity in the transverse direction. Now, dividing through by s yields (16.15) which reduces to (16.16) Equation 16.16 is analogous to the lower-bound expression for particulate composites, Equation 16.2. Ecl= EmVm + EfVf Ecl = Em(1 – Vf )+ EfVf Ff Fm EfVf = EmVm 1/Ect =Vm/Em+vfEf
- 4. Mukuldev Khunte-MME 6th Semester 01UG15050023 Figure 16.6 The deformation pattern in the matrix surrounding a fiber that is subjected to an applied tensile load. Critical Length of fiber- The mechanical characteristics of a fiber-reinforced composite depend not only on the properties of the fiber, but also on the degree to which an applied load is transmitted to the fibers by the matrix phase. Important to the extent of this load transmittance is the magnitude of the interfacial bond between the fiber and matrix phases. Under an applied stress, this fiber–matrix bond ceases at the fiber ends, yielding a matrix deformation pattern as shown schematically in Figure 16.6;in other words, there is no load transmittance from the matrix at each fiber extremity. Some critical fiber length is necessary for effective strengthening and stiffening of the composite material. This critical length lc is dependent on the fiber diameter d and its ultimate (or tensile) strength s*f , and on the fiber–matrix bond strength (or the shear yield strength of the matrix, whichever is smaller) tc according to ss When a stress equal to s*f is applied to a fiber having just this critical length, the stress–position profile shown in Figure 16.7a results; that is, the maximum fiber load is achieved only at the axial center of the fiber.As fiber length l increases, the fiber reinforcement becomes more effective; this is demonstrated in Figure 16.7b, a stress–axial position profile for l 7 lc when the applied stress is equal to the fiber strength. Figure 16.7c shows the stress–position profile for l 6 lc. Fibers for which l W lc (normally l 7 15lc) are termed continuous; discontinuous or short fibers have lengths shorter than this.For discontinuous fibers of lengths s*f dlc = 2tc
- 5. Mukuldev Khunte-MME 6th Semester 01UG15050023 Figure 16.7 Stress–position profiles when fiber length l (a) is equal to the critical length lc, (b) is greater than the critical length, and (c) is less than the critical length for a fiber-reinforced composite that is subjected to a tensile stress equal to the fiber tensile strength s*f . 0 l 0 significantly less than lc, the matrix deforms around the fiber such that there is virtually no stress transference and little reinforcement by the fiber.These are essentially the particulate composites as described above. To affect a significant improvement in strength of the composite, the fibers must be continuous. Q- Discuss stress Strain diagram of a composite material. Answer- Tensile Stress–Strain Behavior—Longitudinal Loading Mechanical responses of this type of composite depend on several factors to include the stress–strain behaviors of fiber and matrix phases, the phase volume fractions, and, in addition, the direction in which the stress or load is applied. Furthermore, the properties of a composite having its fibers aligned are highly anisotropic, that is, dependent on the direction in which they are measured. us first consider the stress–strain behavior for the situation wherein the stress is * * c ccc * * * * * cc c
- 6. Mukuldev Khunte-MME 6th Semester 01UG15050023 Figure (a) Schematic stress–strain curves for brittle fiber and ductile matrix materials. Fracture stresses and strains for both materials are noted. (b) Schematic stress–strain curve for an aligned fiber-reinforced composite that is exposed to a uniaxial stress applied in the direction of alignment; curves for the fiber and matrix materials shown in part (a) are also superimposed. in this figure are fracture strengths in tension for fiber and matrix, s*f and sm*, respectively, and their corresponding fracture strains, *f and *;m furthermore, it is assumed that m* 7 *f , which is normally the case.A fiber-reinforced composite consisting of these fiber and matrix materials will exhibit the uniaxial stress–strain response illustrated in Figure 16.9b; the fiber and matrix behaviors from Figure 16.9a are included to provide perspective. In the initial Stage I region, both fibers and matrix deform elastically; normally this portion of the curve is linear. Typically, for a composite of this type, the matrix yields and deforms plastically (b) while the fibers continue to stretch elastically, inasmuch as the tensile strength of the fibers is significantly higher than the yield strength of the matrix.This process constitutes Stage II as noted in the figure; this stage is ordinarily very nearly linear, but of diminished slope relative to Stage I. Furthermore, in passing from Stage I to Stage II, the proportion of the applied load that is borne by the fibers increases. The onset of composite failure begins as the fibers start to fracture, which corresponds to a strain of approximately *f as noted in Figure b. Composite failure is not catastrophic for a couple of reasons. First, not all fibers fracture at the same time,since there will always be considerable variations in the fracture strength of brittle fiber materials . In addition, even after fiber failure, the matrix is still intact inasmuch as *f 6 m* a. Thus, these fractured fibers, which are shorter than the original ones, are still embedded within the intact matrix, and consequently are capable of sustaining a diminished load as the matrix continues to plastically deform. Q-What are fibers and Whiskers ? Answer- A fiber has: – •High length‐to‐diameter diameter ratio. – Its diameter approximates its crystal size. • Modern composites exploit the fact that small scale samples of most of the materials are much stronger than bulk materials. Thus, thin fibers of glass are 200‐500 times stronger than bulk glass. • Several types of fibers are available commercially. Some of the more commonly used fibers are made from materials such as carbon, glass, Kevlar, steel, and other metals. • Glass is the most popular fiber used in composites since it is relatively inexpensive. It comes in two principal varieties; E‐glass, and S‐glass. The latter is stronger than the former. applied along the direction of alignment, the longitudinal direction, which is indicated in Figure a To begin, assume the stress versus strain behaviors for fiber and matrix phases that are represented schematically in Figure 16.9a; in this treatment we consider the fiber to be totally brittle and the matrix phase to be reasonably ductile.Also indicated I II
- 7. Mukuldev Khunte-MME 6th Semester 01UG15050023 • Fibers are significantly stronger than bulk materials because: – They have a far more “perfect” structure, i.e. their crystals are aligned along the fiber axis. – There are fewer internal defects, especially in direction normal to fiber orientation, and hence there are lesser number of dislocations. • At larger scales, the degree of structural perfection within a material sample is far less that what is present at small (micro and Nano) scales. For this reason fibers of several engineering materials are far stronger than their equivalent bulk material samples. Whiskers • Whiskers are similar in diameter to fibers, but in general, they are short and have low length‐to‐ diameter ratios, barely exceeding a few hundreds. • Thus, the difference in mechanical properties of a whiskers vis‐à‐vis bulk material is even more pronounced. This is because the degree of perfection in whiskers whiskers is even higher vis‐à‐vis that in fibers. – Whiskers are produced by crystallizing materials on a very small scale. – Internal alignment within each whisker is extremely high. • Modern composites composites derive much of their desired desired properties properties by using fibers and whiskers as one of the constituent materials. • Fibers made from carbon, E‐glass, S‐glass, and Kevlar are commonly used in modern composite structures Q-Classify Fibers as per individual parameters. Answer- Textile fibers have been used to make cloth for several thousand years. First manufactured fiber was produced commercially on 1885 and was produced from fibers of plants and animals. Wool, flax, cotton and silk were commonly used textile fibers. Textile fibers are characterized by the flexibility, fineness
- 8. Mukuldev Khunte-MME 6th Semester 01UG15050023 and large length in relation to the maximum transverse dimension. On the basis of origin fibers can be classified broadly into three types: I. Natural fibers: Fibers which grow or develop and come from natural resources like plant and animals. II. Manufactured (or man-made) fibers: Fibers produced by industrial processes, whether from natural polymers transformed upon the action of chemical reagents or through polymers obtained by chemical synthesis (synthetic fibers). III. Mineral Fibers: Asbestos is the only naturally occurring mineral fiber that was used extensively for making industrial products but is now restricted due to its suspected carcinogenic effect. Q- What are matrix? classify as per dimensional cases. Answer- Fibers or particles embedded in matrix of another material are the best example of modern-day composite materials, which are mostly structural In matrix-based structural composites, the matrix serves two paramount purposes viz., binding the reinforcement phases in place and deforming to distribute the stresses among the constituent reinforcement materials under an applied force. Q-What are the significance of fiber and matrix ? Answer-A fiber is characterized geometrically not only by its very high length-to-diameter ratio but by its near- crystal-sized diameter. Strength sand stiffnesses of a few selected fiber materials are arranged in increasing average S/p and E/p. The common structural materials, aluminum, titanium, and steel, are listed for the purpose of comparison. However, a direct comparison between fibers and structural metals is not valid because fibers must have a surrounding matrix to perform in a structural member, whereas structural metals are Yeady-to-use'. A whisker has essentially the same near-crystal-sized diameter as a fiber, but generally is very short and stubby, although the length-to- diameter ratio can be in the hundreds. Thus, a whisker is an even more obvious example of the crystal-bulk-material-property-difference paradox. That is, a whisker is even more perfect than a fiber and therefore exhibits even higher properties. Whiskers are obtained by crystallization on a very small scale resulting in a nearly perfect alignment of crystals. Materials such as iron have crystalline structures with a theoretical strength of 2,900,000
- 9. Mukuldev Khunte-MME 6th Semester 01UG15050023 psi (20 GPa), yet commercially available structural steels, which are mainly iron, have strengths ranging from 75,000 psi to about 100,000 psi (570 to 690 MPa). Q-Discontinous and randomly oriented fiber composite. Answer-Fiber Geometry Some common geometries for fiber-reinforced composites: • Aligned The properties of aligned fiber-reinforced composite materials are highly anisotropic. The longitudinal tensile strength will be high whereas the transverse tensile strength can be much less than even the matrix tensile strength. It will depend on the properties of the fibers and the matrix, the interfacial bond between them, and the presence of voids. There are 2 different geometries for aligned fibers: 1. Continuous & Aligned The fibers are longer than a critical length which is the minimum length necessary such that the entire load is transmitted from the matrix to the fibers. If they are shorter than this critical length, only some of the load is transmitted. Fiber lengths greater that 15 times the critical length are considered optimal. Aligned and continuous fibers give the most effective strengthening for fiber composites. 2. Discontinuous & Aligned The fibers are shorter than the critical length. Hence discontinuous fibers are less effective in strengthening the material, however, their composite modulus and tensile strengths can approach 50-90% of their continuous and aligned counterparts. And they are cheaper, faster and easier to fabricate into complicated shapes. • Random This is also called discrete, (or chopped) fibers. The strength will not be as high as with aligned fibers, however, the advantage is that the material will be istropic and cheaper • Woven The fibers are woven into a fabric which is layered with the matrix material to make a laminated structure. Q-How does a fiber orientation related with stress distribution? Answer-When considering the effect of fibre orientation on the strength of a composite material made up of a continuous aligned fibres embedded in a matrix, it should be recognised that there are 3 possible modes of failure... 1. Tensile fracture parallel to the fibres (whether the fibres fail or the matrix fails will depend on the particular combination of fibre and matrix materials as well as the volume fraction of fibres), 2. Shear failure of the matrix as a result of a large shear stress acting parallel to the fibres , 3. Tensile failure of the matrix or fibre/matrix interface when stressed perpendicular to the fibres.
- 10. Mukuldev Khunte-MME 6th Semester 01UG15050023 We have already determined suitable expressions for the strength of a composite when tested parallel to the fibres, We'll call this strength X. We also know the tensile strength of the matrix material which we'll call Y. The shear strength of the matrix can be determined using the Tresca criteria and is simply Y/2. In order to examine the effect of orientation on strength we need to make use of Mohr's Circle to establish the state of stress aligned parallel and perpendicular to the fibres and then to equate these stresses with the appropriate failure stress of the composite in each those directions. For failure to occur, the applied stress must be increase until either These equations are plotted out below and since failure is a "weakest link" phenomenon, fracture will occure at whichever criterion is reached first and so the mechanism of failure changes from tensile failure of the fibres to shear of teh matrix to tensile failure of the matrix as the fibre angle is increased from 0 to 90°.
- 11. Mukuldev Khunte-MME 6th Semester 01UG15050023 Q-What is Transformational Toughening in case of ceramic matrix composite (CMC). Why toughening is required for CMC and what are the characteristics of it? Answer-Ceramic Matrix Composites Significant research effort has been directed toward the development of tough ceramics, since such materials have the potential to open up a large range of specialised engineering applications. Toughness can be significantly improved with the production of ceramics with a very small flaw size (< 50µm) in accordance with the Griffith theorem. However, producing such ceramics, and non-destructive testing for flaw size, is very difficult. An alternative solution is to engineer the microstructure of ceramics in such a way as to inhibit crack propagation. To this end, a family of toughened ceramics known as ceramic matrix composites (CMCs) is now undergoing development. Toughness values as high as 10 to 20 MPa.m½ have been achieved, which is comparable with some metals. Fibre reinforcement combines crack bridging, fibre pull-out, and crack deflection mechanisms. As an overall toughening technique, it appears to give the greatest improvement. Further, the use of metal fibres adds the toughening mechanism that comes from the plasticity of the metallic phase. Several toughening mechanisms have been engineered into ceramics. These include: 1. Crack deflection 2. Crack pinning 3. Crack bowing 4. Plasticity in metallic phase 5. Transformation toughening 6. Compressive matrix residual stresses 7. Matrix microcracking 8. Frictional interlocking 9. Crack bridging 10. Fibre pull-out