Nigeria's free digital learning hub. Videos, lecture notes, practice questions and leaderboards aligned with CCMAS, NBTE, NCCE, WAEC, NECO, IJMB and JAMB curricula.
Loading
Nigeria's free digital learning hub. Videos, lecture notes, practice questions and leaderboards aligned with CCMAS, NBTE, NCCE, WAEC, NECO, IJMB and JAMB curricula.
This is the complete lecture note for MME507 Composite Materials. You can read it online below, open it in the full viewer, or download the PDF for offline study. It pairs with the course's video lectures, so you can read and watch together.
Course code: MME 507 Course title: Composite Materials Course credit load: 3 Units Series 1: Fundamental Aspects and Types of Composite Materials: Fibre Reinforced, Particulate, Dispersion Strengthened and Laminar Composites • Part 1.1: Fundamental Aspects of Composite Materials o Sub Part 1.1.1: Definition, Constituents and Interface o Sub Part 1.1.2: Volume and Weight Fractions o Sub Part 1.1.3: Why Composites Are Used: Specific Properties • Part 1.2: Fibre Reinforced Composites o Sub Part 1.2.1: Reinforcing Fibres o Sub Part 1.2.2: Continuous Aligned Fibre Composites o Sub Part 1.2.3: Discontinuous and Random Fibres • Part 1.3: Particulate Composites o Sub Part 1.3.1: Structure and Types of Particulate Composites o Sub Part 1.3.2: Modulus Bounds for Particulate Composites o Sub Part 1.3.3: Properties and Applications of Particulate Composites • Part 1.4: Dispersion Strengthened Composites o Sub Part 1.4.1: Principle of Dispersion Strengthening o Sub Part 1.4.2: Particle Size, Spacing and Volume Fraction o Sub Part 1.4.3: Stability at High Temperature • Part 1.5: Laminar Composites o Sub Part 1.5.1: Laminae and Laminates
o Sub Part 1.5.2: Sandwich Structures o Sub Part 1.5.3: Bonding and Applications of Laminar Composites Introduction A tennis racket, a wind turbine blade, a jet's wing and the bone in your own arm share one idea: two or more materials, each doing the job it does best, working as one. Nature has used the trick for millions of years, and engineers have copied it since bricks were first reinforced with straw. This Series introduces what a composite is and the four families of composites on which the rest of the course builds. The aim of this Series is to equip you to define a composite material, calculate its basic properties, and describe the structure, behaviour and uses of fibre reinforced, particulate, dispersion strengthened and laminar composites. We shall commence with the fundamental aspects of composite materials, including constituents, fractions and the reasons for using composites. Next, we will explore fibre reinforced composites, from reinforcing fibres to aligned and short fibre behaviour. Afterwards, we will turn to particulate composites, followed by dispersion strengthened composites. We will conclude with laminar composites, covering laminates, sandwich structures and their bonding and uses. Series Learning Outcomes Upon completing this Series, you should be able to: • SLO 1.1: define a composite material and describe its constituents, interface and advantages, • SLO 1.2: calculate volume fractions, weight fractions and densities of composites, • SLO 1.3: compare the properties of the common reinforcing fibres, • SLO 1.4: calculate the moduli of continuous and discontinuous fibre composites, • SLO 1.5: describe the structure and applications of particulate composites, • SLO 1.6: apply the rule-of-mixtures bounds to estimate the modulus of a particulate composite, • SLO 1.7: explain the strengthening mechanism of dispersion strengthened composites,
• SLO 1.8: explain the effect of particle size, spacing and temperature on dispersion strengthening, • SLO 1.9: describe the structure of laminates and sandwich structures, and • SLO 1.10: describe the bonding and applications of laminar composites. 1.1 Fundamental Aspects of Composite Materials 1.1.1 Definition, Constituents and Interface A composite material is a material made from two or more chemically distinct constituents that are combined so that the result has properties better than those of any constituent alone. The constituents stay separate and are divided by an interface. The matrix is the continuous phase that binds the material together and passes load to the other phase, and the reinforcement is the phase, in the form of fibres, particles or sheets, that gives strength or stiffness. The interface between matrix and reinforcement decides how well the two phases work together. If bonding is good, load passes from the matrix into the reinforcement, and the composite is stronger than the matrix. If bonding is poor, the reinforcement pulls out or the interface cracks, and the benefit is lost. Composite design therefore considers the choice of matrix and reinforcement, the geometry and amount of the reinforcement, and the quality of the interface. Worked example: A composite of 60 vol% E-glass fibre (2.54 g/cm³) in epoxy (1.20 g/cm³) has density ρc = 0.60 × 2.54 + 0.40 × 1.20 = 1.524 + 0.480 = 2.00 g/cm³, which is about one quarter that of steel (7.85 g/cm³). Application: Naming the matrix, the reinforcement and the interface is the first step in describing any composite, from a brick with straw to an aircraft panel. Fill in the blank: The continuous phase of a composite that binds the reinforcement together and passes load to it is called the __________. 1.1.2 Volume and Weight Fractions The amounts of matrix and reinforcement are described by fractions. The volume fraction of the reinforcement, Vf, is its volume divided by the total volume of the composite, and the weight fraction, Wf, is its weight divided by the total weight. The matrix fraction follows from Vm = 1 − Vf if the composite has no voids. Volume fractions govern most mechanical properties, but weight fractions are easier to measure in the laboratory, so both are needed. The two are linked through the densities. If ρf and ρm are the densities of fibre and matrix, the composite density is ρc = Vf ρf + Vm ρm, and the weight fraction is Wf = Vf ρf/ρc. In the opposite direction, Vf = (Wf/ρf)/[Wf/ρf + (1 − Wf)/ρm]. Voids, which are pores trapped during
manufacture, lower the density and weaken the composite, so their volume is found by comparing measured and calculated densities. Worked example: For the glass-epoxy composite with Vf = 0.60 and ρc = 2.00 g/cm³, Wf = 0.60 × 2.54/2.00 = 0.76. Conversely, for Wf = 0.70, Vf = (0.70/2.54)/(0.70/2.54 + 0.30/1.20) = 0.2756/0.5256 = 0.52. Application: Converting between weight and volume fractions is a routine step in specifying, ordering and testing composite materials. Fill in the blank: The ratio of the volume of the reinforcement to the total volume of the composite is the volume __________. 1.1.3 Why Composites Are Used: Specific Properties Composites are used because they give a combination of properties that no single material can match. The most important is high strength and stiffness at low density. The specific modulus, the modulus divided by density, and the specific strength, the strength divided by density, measure how well a material resists load for its weight. Composites also offer properties that can be tailored, since the fibre direction, amount and type can be chosen to suit the loads on a part. Other advantages include good fatigue resistance, corrosion resistance, the ability to mould complex shapes in one piece and low thermal expansion in some designs. There are also disadvantages: composites are often costly, their properties differ with direction, they can be damaged by impact and their failure is complex. Recycling is harder than for metals. A designer therefore uses composites where the gain in performance justifies these costs. Worked example: Steel has E = 210 GPa and ρ = 7.85 g/cm³, so E/ρ = 26.8 GPa per g/cm³, and a strength of 500 MPa gives 63.7 MPa per g/cm³. A carbon-epoxy laminate with E = 140 GPa, ρ = 1.6 g/cm³ and strength 1500 MPa gives 87.5 and 937 respectively, which are 3.3 and 14.7 times higher. Application: Specific properties explain why composites are chosen for aircraft, racing cars and sports equipment, where every kilogram saved is valuable. Fill in the blank: The modulus of a material divided by its density is called the specific __________.
Figure 1.1: Classification of composite materials by reinforcement and matrix. Pilot questions 1.1 1. Define a composite material. (Linked to SLO 1.1) 2. What are the roles of the matrix and the reinforcement? (Linked to SLO 1.1) 3. State two advantages of composites over metals. (Linked to SLO 1.1) 4. Write the expression for the density of a composite. (Linked to SLO 1.2) 5. How is the weight fraction related to the volume fraction? (Linked to SLO 1.2) 1.2 Fibre Reinforced Composites 1.2.1 Reinforcing Fibres A fibre reinforced composite contains fibres embedded in a matrix. Fibres are so strong because they are thin: a thin fibre has few flaws, and its structure is aligned along its length. The common reinforcing fibres are glass, carbon and aramid. E-glass is cheap and gives good strength, carbon fibres are very stiff and light, and aramid fibres such as Kevlar are tough and have high specific strength.
Fibres are supplied as continuous strands called rovings, woven fabrics, mats of short fibres and chopped strands. Continuous fibres carry load best, but they are harder to place. Fibres are coated with a thin layer called a size, which protects them and improves bonding to the matrix. The choice of fibre depends on the stiffness, strength, density, cost and temperature resistance needed. Worked example: Specific modulus is E/ρ. For carbon, 230/1.76 = 131, for aramid, 131/1.44 = 91, and for E-glass, 72/2.54 = 28, all in GPa per g/cm³. Specific strength for the same fibres is 3.5/1.76 = 2.0, 3.6/1.44 = 2.5 and 3.4/2.54 = 1.3 GPa per g/cm³. Application: Comparing fibres by
This is the opening of the note. Read the full note or download the PDF to continue.
