Introduction to Composite Materials 11 400 Plastic/general yielding Kelcy=2.5 in.1/2 到 300 Elastic-plastic/mixed mode Steel Kc/oy =0.6 in.1/2 200 Titanium Elastic/plane strain 100 Aluminum Composites Ceramics 100 200 300 400 500 Polymers Yield strength,10psi FIGURE 1.4 Fracture toughness as a function of yield strength for monolithic metals,ceramics,and metal-ceramic composites.(Source:Eager,T.W.,Whither advanced materials?Adv.Mater.Pro- cesses,ASM International,June 1991,25-29.) Strength 图 Ceramic Metal Composite Affordability Toughness Corrosion resistance Formability Joinability FIGURE 1.5 Primary material selection parameters for a hypothetical situation for metals,ceramics,and metal-ceramic composites.(Source:Eager,T.W.,Whither advanced materials?Adv.Mater.Pro- cesses,ASM International,June 1991,25-29.) 2006 by Taylor Francis Group,LLC
Introduction to Composite Materials 11 FIGURE 1.4 Fracture toughness as a function of yield strength for monolithic metals, ceramics, and metal–ceramic composites. (Source: Eager, T.W., Whither advanced materials? Adv. Mater. Processes, ASM International, June 1991, 25–29.) FIGURE 1.5 Primary material selection parameters for a hypothetical situation for metals, ceramics, and metal–ceramic composites. (Source: Eager, T.W., Whither advanced materials? Adv. Mater. Processes, ASM International, June 1991, 25–29.) Plastic/general yielding Kc/σy = 2.5 in.1/2 Kc/σy = 0.6 in.1/2 Elastic/plane strain Ceramics Composites Elastic-plastic/mixed mode Aluminum Yield strength, ×103 psi Fracture toughness, ksi.in 1/2 Polymers 400 300 200 100 100 200 300 400 500 Titanium Steel Strength Ceramic Metal Composite Affordability Corrosion resistance Joinability Formability Toughness 1343_book.fm Page 11 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC
12 Mechanics of Composite Materials,Second Edition joinability,corrosion resistance,and affordability-are plotted.3 If the values at the circumference are considered as the normalized required property level for a particular application,the shaded areas show values provided by ceramics,metals,and metal-ceramic com- posites.Clearly,composites show better strength than metals,but lower values for other material selection parameters. Why are fiber reinforcements of a thin diameter? The main reasons for using fibers of thin diameter are the following: Actual strength of materials is several magnitudes lower than the theoretical strength.This difference is due to the inherent flaws in the material.Removing these flaws can increase the strength of the material.As the fibers become smaller in diameter,the chances of an inherent flaw in the material are reduced.A steel plate may have strength of 100 ksi(689 MPa),while a wire made from this steel plate can have strength of 600 ksi(4100 MPa).Figure 1.6 shows how the strength of a carbon fiber increases with the decrease in its diameter. 2.5 1.5 7.5 10 12.5 15 Fiber diameter(um) FIGURE 1.6 Fiber strength as a function of fiber diameter for carbon fibers.(Reprinted from Lamotte,E.De, and Perry,A.J.,Fibre Sci.Technol.,3,159,1970.With permission from Elsevier.) 2006 by Taylor Francis Group,LLC
12 Mechanics of Composite Materials, Second Edition joinability, corrosion resistance, and affordability — are plotted.3 If the values at the circumference are considered as the normalized required property level for a particular application, the shaded areas show values provided by ceramics, metals, and metal–ceramic composites. Clearly, composites show better strength than metals, but lower values for other material selection parameters. Why are fiber reinforcements of a thin diameter? The main reasons for using fibers of thin diameter are the following: • Actual strength of materials is several magnitudes lower than the theoretical strength. This difference is due to the inherent flaws in the material. Removing these flaws can increase the strength of the material. As the fibers become smaller in diameter, the chances of an inherent flaw in the material are reduced. A steel plate may have strength of 100 ksi (689 MPa), while a wire made from this steel plate can have strength of 600 ksi (4100 MPa). Figure 1.6 shows how the strength of a carbon fiber increases with the decrease in its diameter.6 FIGURE 1.6 Fiber strength as a function of fiber diameter for carbon fibers. (Reprinted from Lamotte, E. De, and Perry, A.J., Fibre Sci. Technol., 3, 159, 1970. With permission from Elsevier.) 3 2.5 2 1.5 1 5 7.5 10 Fiber diameter (μm) Fiber strength (GPa) 12.5 15 1343_book.fm Page 12 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC
Introduction to Composite Materials 13 For higher ductility*and toughness,and better transfer of loads from the matrix to fiber,composites require larger surface area of the fiber-matrix interface.For the same volume fraction of fibers in a composite,the area of the fiber-matrix interface is inversely propor- tional to the diameter of the fiber and is proved as follows. Assume a lamina consisting of N fibers of diameter D.The fiber- matrix interface area in this lamina is A=NπDL. (1.8) If one replaces the fibers of diameter,D,by fibers of diameter,d, then the number of fibers,n,to keep the fiber volume the same would be 2 (1.9) Then,the fiber-matrix interface area in the resulting lamina would be An=nπdL. NπD2L d 4 (Volume of fibers) (1.10) d This implies that,for a fixed fiber volume in a given volume of composite,the area of the fiber-matrix interface is inversely pro- portional to the diameter of the fiber. Fibers able to bend without breaking are required in manufacturing of composite materials,especially for woven fabric composites.Abil- ity to bend increases with a decrease in the fiber diameter and is measured as flexibility.Flexibility is defined as the inverse of bend- ing stiffness and is proportional to the inverse of the product of the elastic modulus of the fiber and the fourth power of its diameter;it can be proved as follows. Bending stiffness is the resistance to bending moments.According to the Strength of Materials course,if a beam is subjected to a pure bending moment,M, Ductility is the ability of a material to deform without fracturing.It is measured by extending a rod until fracture and measuring the initial(A and final (A cross-sectional area.Then ductil- ity is defined as R=1-(AA). 2006 by Taylor Francis Group,LLC
Introduction to Composite Materials 13 • For higher ductility* and toughness, and better transfer of loads from the matrix to fiber, composites require larger surface area of the fiber–matrix interface. For the same volume fraction of fibers in a composite, the area of the fiber–matrix interface is inversely proportional to the diameter of the fiber and is proved as follows. Assume a lamina consisting of N fibers of diameter D. The fiber– matrix interface area in this lamina is AI = N π D L. (1.8) If one replaces the fibers of diameter, D, by fibers of diameter, d, then the number of fibers, n, to keep the fiber volume the same would be . (1.9) Then, the fiber–matrix interface area in the resulting lamina would be AII = n π d L. = = . (1.10) This implies that, for a fixed fiber volume in a given volume of composite, the area of the fiber–matrix interface is inversely proportional to the diameter of the fiber. • Fibers able to bend without breaking are required in manufacturing of composite materials, especially for woven fabric composites. Ability to bend increases with a decrease in the fiber diameter and is measured as flexibility. Flexibility is defined as the inverse of bending stiffness and is proportional to the inverse of the product of the elastic modulus of the fiber and the fourth power of its diameter; it can be proved as follows. Bending stiffness is the resistance to bending moments. According to the Strength of Materials course, if a beam is subjected to a pure bending moment, M, * Ductility is the ability of a material to deform without fracturing. It is measured by extending a rod until fracture and measuring the initial (Ai ) and final (Af ) cross-sectional area. Then ductility is defined as R = 1 – (Af/Ai ). n=N D d ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 N DL d π 2 4 (Volume of fibers) d 1343_book.fm Page 13 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC
14 Mechanics of Composite Materials,Second Edition dvM dx?=EI (1.11) where v=deflection of the centroidal line (in.or m) E=Young's modulus of the beam(psi or Pa) I second moment of area (in.4 or m) x=coordinate along the length of beam (in.or m) The bending stiffness,then,is EI and the flexibility is simply the inverse of EI.Because the second moment of area of a cylindrical beam of diameter d is I=ndi 64, (1.12) then Flexibility1 Ed4 (1.13) For a particular material,unlike strength,the Young's modulus does not change appreciably as a function of its diameter.Therefore, the flexibility for a particular material is inversely proportional to the fourth power of the diameter. What fiber factors contribute to the mechanical performance of a composite? Four fiber factors contribute to the mechanical performance of a composite?: Length:The fibers can be long or short.Long,continuous fibers are easy to orient and process,but short fibers cannot be controlled fully for proper orientation.Long fibers provide many benefits over short fibers.These include impact resistance,low shrinkage,improved surface finish,and dimensional stability.However,short fibers pro- vide low cost,are easy to work with,and have fast cycle time fab- rication procedures.Short fibers have fewer flaws and therefore have higher strength. Orientation:Fibers oriented in one direction give very high stiffness and strength in that direction.If the fibers are oriented in more than one direction,such as in a mat,there will be high stiffness and strength in the directions of the fiber orientations.However,for the same volume of fibers per unit volume of the composite,it cannot match the stiffness and strength of unidirectional composites. 2006 by Taylor Francis Group,LLC
14 Mechanics of Composite Materials, Second Edition , (1.11) where v = deflection of the centroidal line (in. or m) E = Young’s modulus of the beam (psi or Pa) I = second moment of area (in.4 or m4) x = coordinate along the length of beam (in. or m) The bending stiffness, then, is EI and the flexibility is simply the inverse of EI. Because the second moment of area of a cylindrical beam of diameter d is , (1.12) then . (1.13) For a particular material, unlike strength, the Young’s modulus does not change appreciably as a function of its diameter. Therefore, the flexibility for a particular material is inversely proportional to the fourth power of the diameter. What fiber factors contribute to the mechanical performance of a composite? Four fiber factors contribute to the mechanical performance of a composite7: • Length: The fibers can be long or short. Long, continuous fibers are easy to orient and process, but short fibers cannot be controlled fully for proper orientation. Long fibers provide many benefits over short fibers. These include impact resistance, low shrinkage, improved surface finish, and dimensional stability. However, short fibers provide low cost, are easy to work with, and have fast cycle time fabrication procedures. Short fibers have fewer flaws and therefore have higher strength. • Orientation: Fibers oriented in one direction give very high stiffness and strength in that direction. If the fibers are oriented in more than one direction, such as in a mat, there will be high stiffness and strength in the directions of the fiber orientations. However, for the same volume of fibers per unit volume of the composite, it cannot match the stiffness and strength of unidirectional composites. d v dx M EI 2 2 = I d = π 4 64 Flexibility Ed ∝ 1 4 1343_book.fm Page 14 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC
Introduction to Composite Materials 15 Shape:The most common shape of fibers is circular because han- dling and manufacturing them is easy.Hexagon and square- shaped fibers are possible,but their advantages of strength and high packing factors do not outweigh the difficulty in handling and processing. Material:The material of the fiber directly influences the mechanical performance of a composite.Fibers are generally expected to have high elastic moduli and strengths.This expectation and cost have been key factors in the graphite,aramids,and glass dominating the fiber market for composites. What are the matrix factors that contribute to the mechanical performance of composites? Use of fibers by themselves is limited,with the exceptions of ropes and cables.Therefore,fibers are used as reinforcement to matrices.The matrix functions include binding the fibers together,protecting fibers from the environment,shielding from damage due to handling,and distributing the load to fibers.Although matrices by themselves generally have low mechan- ical properties compared to those of fibers,the matrix influences many mechanical properties of the composite.These properties include transverse modulus and strength,shear modulus and strength,compressive strength, interlaminar shear strength,thermal expansion coefficient,thermal resis- tance,and fatigue strength. Other than the fiber and the matrix,what other factors influence the mechanical performance of a composite? Other factors include the fiber-matrix interface.It determines how well the matrix transfers the load to the fibers.Chemical,mechanical,and reaction bonding may form the interface.In most cases,more than one type of bonding occurs. Chemical bonding is formed between the fiber surface and the matrix.Some fibers bond naturally to the matrix and others do not. Coupling agents*are often added to form a chemical bond. The natural roughness or etching of the fiber surface causing inter- locking may form a mechanical bond between the fiber and matrix. If the thermal expansion coefficient of the matrix is higher than that of the fiber,and the manufacturing temperatures are higher than the operating temperatures,the matrix will radially shrink more than the fiber.This causes the matrix to compress around the fiber. *Coupling agents are compounds applied to fiber surfaces to improve the bond between the fiber and matrix.For example,silane finish is applied to glass fibers to increase adhesion with epoxy matrix. 2006 by Taylor Francis Group,LLC
Introduction to Composite Materials 15 • Shape: The most common shape of fibers is circular because handling and manufacturing them is easy. Hexagon and squareshaped fibers are possible, but their advantages of strength and high packing factors do not outweigh the difficulty in handling and processing. • Material: The material of the fiber directly influences the mechanical performance of a composite. Fibers are generally expected to have high elastic moduli and strengths. This expectation and cost have been key factors in the graphite, aramids, and glass dominating the fiber market for composites. What are the matrix factors that contribute to the mechanical performance of composites? Use of fibers by themselves is limited, with the exceptions of ropes and cables. Therefore, fibers are used as reinforcement to matrices. The matrix functions include binding the fibers together, protecting fibers from the environment, shielding from damage due to handling, and distributing the load to fibers. Although matrices by themselves generally have low mechanical properties compared to those of fibers, the matrix influences many mechanical properties of the composite. These properties include transverse modulus and strength, shear modulus and strength, compressive strength, interlaminar shear strength, thermal expansion coefficient, thermal resistance, and fatigue strength. Other than the fiber and the matrix, what other factors influence the mechanical performance of a composite? Other factors include the fiber–matrix interface. It determines how well the matrix transfers the load to the fibers. Chemical, mechanical, and reaction bonding may form the interface. In most cases, more than one type of bonding occurs. • Chemical bonding is formed between the fiber surface and the matrix. Some fibers bond naturally to the matrix and others do not. Coupling agents* are often added to form a chemical bond. • The natural roughness or etching of the fiber surface causing interlocking may form a mechanical bond between the fiber and matrix. • If the thermal expansion coefficient of the matrix is higher than that of the fiber, and the manufacturing temperatures are higher than the operating temperatures, the matrix will radially shrink more than the fiber. This causes the matrix to compress around the fiber. * Coupling agents are compounds applied to fiber surfaces to improve the bond between the fiber and matrix. For example, silane finish is applied to glass fibers to increase adhesion with epoxy matrix. 1343_book.fm Page 15 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC