Reinforcement
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The document discusses different types of reinforcements that can be used to improve the mechanical properties of composite materials, including fibers like carbon, aramid, and stainless steel. It notes advantages like increased strength, stiffness, heat resistance, and lower weight, as well as limitations such as higher costs. The arrangement and volume fraction of fibers significantly impacts the mechanical, chemical, electrical and thermal properties of the composite. Unidirectional fiber reinforcement can allow for up to 80% fiber loading by weight and maximize strength in the longitudinal direction.
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Reinforcement
- 6. HIGH PERFORMANCE REINFORCEMENT OTHER REINFORCEMENTS • Glass Beads – Does not increase viscosity • Asbestos – Health hazard • Confined to Brake lining & Clutch facing • Carbon Fibers – Higher Strength Reduced Coefficient of Friction Higher Thermal & Electrical Conductivity Inert Surface • Aramid – High Temperature Resistance, Tough Bonding difficult • UHMWPE – Low Melting Point Unreactive Surface • Short Stainless Steel Fibers –Less build up in Viscosity Conductive Applications (RFI & EMI Shield)
- 10. REINFORCEMENT
- 11. REINFORCEMENT
- 12. REINFORCEMENT
- 13. HIGH PERFORMANCE REINFORCEMENT ADVANTAGES • High Strength & Stiffness • Light Weight • Design Flexibility • Dimensional Stability • Higher Heat Deflection Temperature • High Dielectric Strength • Corrosion Resistance • Less Finishing • Moderate Tooling Cost
- 14. HIGH PERFORMANCE REINFORCEMENT LIMITATIONS • Higher Processing Temperatures & Pressures • Machine Wear • Lower Impact Strength • Anisotropic Properties • Loss of Transparency • Finish • Cost • Higher Specific Gravity
- 16. REINFORCEMENT •High Performance Reinforcement Continuous Fibers Volume Loading •Low Performance Reinforcement Short Fibers Particulate Reinforcing Fillers Volume Loading
- 17. HIGH PERFORMANCE REINFORCEMENT MECHANICAL STRENGTH Depends on amount, type & arrangement of fibers CHEMICAL,ELECTRICAL & THERMAL PROPERTY Depends on choice of matrix, formulation & other additives
- 18. HIGH PERFORMANCE REINFORCEMENT ARRANGEMENT OF FIBERS • Unidirectional - 80% loading by weight possible Continuous Pultrusion • Bidirectional - 75% loading by weight possible Hand Lay up • Multidirectional – 10 to 50% loading by weight possible Compression & Injection Molding Spray, Preform, Pressure Bag
- 21. HIGH PERFORMANCE REINFORCEMENT Unidirectional Reinforcement Longitudinal Response Assumptions: • Isostrain conditions (ε of composite = ε of fiber = ε of matrix) • Matrix & Fibers are elastic • Poisson ratio of matrix = Poisson Ratio of Fiber F Composite =ⁿ∑F Fiber + F Matrix Force = Stress (σ) x Area (A) Ασ(Composite) =n Ασ(fiber) + Ασ(Matrix) Volume (Vc) Composite = n Volume (Vf) Fiber + Volume (Vm) Matrix Volume Fraction of Matrix Φm =Vm ∕ Vc Volume fraction of Fiber Φf = nVf ∕ Vc Φm = 1 - Φf
- 22. HIGH PERFORMANCE REINFORCEMENT Volume = Length x Area Length of composite = fiber = matrix σc = Φfσf + (1 - Φf )σm Under isostrain conditions ε of composite = ε of fiber = ε of matrix σc∕ εc = Φf σf ∕ εf+ (1 - Φf )σm ∕ εm Ec = Φf Ef + (1 - Φf )Em Limitation of Assumptions Micro cracks, Voids Perfect adhesion between fiber & matrix Practical: k Fiber utilization efficiency K Critical Volumetric Fraction σc =K{Φf}σf + (1 - Φf )σm
- 23. HIGH PERFORMANCE REINFORCEMENT Transverse Direction Assumption: • Isostress Conditions • Fiber & Matrix are elastic • Poisson Ratio of Fiber & Matrix are equal Strain terms are additive Vcεc = nVf εf + Vmεm εc = Φfεf + (1 - Φf )εm Under isostress conditions σc = σf =σm Jc = ΦfJf + (1 - Φf )Jm 1/Ec = Φf/Ef + (1 - Φf )/Em Ec = Ef Em/Φf/Em + (1 - Φf )/Ef
- 24. HIGH PERFORMANCE REINFORCEMENT Angular Orientation of the Fibers Three possible modes of failure 1.Fracture at right angles to the fiber axis [σc]θ= [σc]n Cos2 θ 2.Failure in the Shear plane ║ to Fiber due to debonding or matrix failure [σc]θ = τm ∕sin θ cos θ 3.Tension failure of matrix ║ to fiber [σc]θ = [σc]90 ∕ sin2 θ
- 25. HIGH PERFORMANCE REINFORCEMENT Randomly Oriented Fibers [Eθ]c = ∫[Eθ]c dθ ∕ ∫ dθ Integrate 0 to π ∕ 2 By simple law of mixture Eθ = FΦf Ef + (1 - Φf )Em F = Fiber efficiency factor – function of fiber volumetric fraction & ratio of modulus of fibers to that of matrix F is in the range of 0.15 – 0.60