3 Mechanics The mechanics of materials deal with stresses,strains,and deformations in engineering structures subjected to mechanical and thermal loads.A common assumption in the mechanics of conventional materials,such as steel and aluminum,is that they are homogeneous and isotropic continua.For a homo- geneous material,properties do not depend on the location,and for an iso- tropic material,properties do not depend on the orientation.Unless severely cold-worked,grains in metallic materials are randomly oriented so that,on a statistical basis,the assumption of isotropy can be justified.Fiber-reinforced composites,on the other hand,are microscopically inhomogeneous and non- isotropic (orthotropic).As a result,the mechanics of fiber-reinforced composites are far more complex than that of conventional materials. 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.Equa- tions of orthotropic elasticity are used to calculate stresses,strains,and deflections. In this chapter,we look into a few basic concepts as well as a number of simple working equations used in the micro-and macromechanics of fiber-reinforced composite materials.Detailed derivations of these equations are given in the references cited in the text. 2007 by Taylor&Franeis Group.LLC
3 Mechanics The mechanics of materials deal with stresses, strains, and deformations in engineering structures subjected to mechanical and thermal loads. A common assumption in the mechanics of conventional materials, such as steel and aluminum, is that they are homogeneous and isotropic continua. For a homogeneous material, properties do not depend on the location, and for an isotropic material, properties do not depend on the orientation. Unless severely cold-worked, grains in metallic materials are randomly oriented so that, on a statistical basis, the assumption of isotropy can be justified. Fiber-reinforced composites, on the other hand, are microscopically inhomogeneous and nonisotropic (orthotropic). As a result, the mechanics of fiber-reinforced composites are far more complex than that of conventional materials. 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. In this chapter, we look into a few basic concepts as well as a number of simple working equations used in the micro- and macromechanics of fiber-reinforced composite materials. Detailed derivations of these equations are given in the references cited in the text. 2007 by Taylor & Francis Group, LLC
3.1 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 load- ings.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. A review of other approaches to the micromechanical behavior of a composite lamina is given in Ref.[2]. 3.1.1 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. 3.1.1.1 Unidirectional Continuous Fibers Assuming a perfect bonding between fibers and matrix,we can write Ef =Em Ec, (3.1) where sf,sm,and sc are the longitudinal strains in fibers,matrix,and compos- ite,respectively (Figure 3.1). Since both fibers and matrix are elastic,the respective longitudinal stresses can be calculated as Of=Et8f=Er8c, (3.2) Om EmEm EmEc. (3.3) Comparing Equation 3.2 with Equation 3.3 and noting that Er Em,we conclude that the fiber stress of is always greater than the matrix stress om. The tensile force Pe applied on the composite lamina is shared by the fibers and the matrix so that Pe=Pt+Pm (3.4) 2007 by Taylor Francis Group,LLC
3.1 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. A review of other approaches to the micromechanical behavior of a composite lamina is given in Ref. [2]. 3.1.1 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. 3.1.1.1 Unidirectional Continuous Fibers Assuming a perfect bonding between fibers and matrix, we can write «f ¼ «m ¼ «c, (3:1) where «f, «m, and «c are the longitudinal strains in fibers, matrix, and composite, respectively (Figure 3.1). Since both fibers and matrix are elastic, the respective longitudinal stresses can be calculated as sf ¼ Ef«f ¼ Ef«c, (3:2) sm ¼ Em«m ¼ Em«c: (3:3) Comparing Equation 3.2 with Equation 3.3 and noting that Ef > Em, we conclude that the fiber stress sf is always greater than the matrix stress sm. The tensile force Pc applied on the composite lamina is shared by the fibers and the matrix so that Pc ¼ Pf þ Pm: (3:4) 2007 by Taylor & Francis Group, LLC
Pe O和 E Fiber Longitudinal direction u Composite C Omu -Matrix Pe Efu Emu &=Em=Ec Strain (e) FIGURE 3.1 Longitudinal tensile loading of a unidirectional continuous fiber lamina. Since force=stress X area,Equation 3.4 can be rewritten as OcAc =OfAf+OmAm or e=f Ac Am (3.5) where oe =average tensile stress in the composite Af =net cross-sectional area for the fibers Am=net cross-sectional area for the matrix Ac Af+Am Since vA and v(v)Equation 3.5 gives A Oc =ofVf +omVm ofVf +om(1 -Vf). (3.6 Dividing both sides of Equation 3.6 by sc,and using Equations 3.2 and 3.3,we can write the longitudinal modulus for the composite as EL EfVf EmVm ErVf Em(1-Vf)=Em+Vi(Ef-Em). (3.7) 2007 by Taylor&Francis Group.LLC
Since force ¼ stress 3 area, Equation 3.4 can be rewritten as scAc ¼ sfAf þ smAm or sc ¼ sf Af Ac þ sm Am Ac , (3:5) where sc ¼ average tensile stress in the composite Af ¼ net cross-sectional area for the fibers Am ¼ net cross-sectional area for the matrix Ac ¼ Af þ Am Since vf ¼ Af Ac and vm ¼ (1 vf) ¼ Am Ac , Equation 3.5 gives sc ¼ sfvf þ smvm ¼ sfvf þ sm(1 vf): (3:6) Dividing both sides of Equation 3.6 by «c, and using Equations 3.2 and 3.3, we can write the longitudinal modulus for the composite as EL ¼ Efvf þ Emvm ¼ Efvf þ Em(1 vf) ¼ Em þ vf(Ef Em): (3:7) Strain (e) Pc Pc Longitudinal direction sf sm sc Fiber Matrix Composite sfu Stress ( s) efu emu sLtu smu ef=em =ec FIGURE 3.1 Longitudinal tensile loading of a unidirectional continuous fiber lamina. 2007 by Taylor & Francis Group, LLC.
Equation 3.7 is called the rule of mixtures.This equation shows that the longitudinal modulus of a unidirectional continuous fiber composite is inter- mediate between the fiber modulus and the matrix modulus;it increases linearly with increasing fiber volume fraction;and since Er>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 P OfVE Erve Pe oeV:om(1 Vr)ErV;Eu(1 Vr) (3.8) Equation 3.8 is plotted in Figure 3.2 as a function ofratio and fiber volume fraction.In polymer matrix composites,the fiber modulus is much greater than the matrix modulus.In most polymer matrix composites,>10.Thus,even for ve=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. In general,the fiber failure strain is lower than the matrix failure strain,that is,Sru FIGURE 3.2 Fraction of load shared by fibers in longitudinal tensile loading of a unidirectional continuous fiber lamina. 2007 by Taylor Francis Group,LLC
Equation 3.7 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 Pf Pc ¼ sfvf sfvf þ sm(1 vf) ¼ Efvf Efvf þ Em(1 vf) : (3:8) Equation 3.8 is plotted in Figure 3.2 as a function of Ef 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, even for vf ¼ 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. In general, the fiber failure strain is lower than the matrix failure strain, that is, «fu < «mu. Assuming all fibers have the same tensile strength and the tensile rupture of fibers immediately precipitates a tensile rupture of the composite, the 100 0.9 0.7 0.6 0.5 0.3 0.1 = vf 50 10 5 1 0.1 0.5 1 5 Pf /Pc (%) 10 50 100 Ef /Em FIGURE 3.2 Fraction of load shared by fibers in longitudinal tensile loading of a unidirectional continuous fiber lamina. 2007 by Taylor & Francis Group, LLC.
longitudinal tensile strength oLtu of a unidirectional continuous fiber composite can be estimated as OLtu gfuV+om(1-Vf), (3.9) where ofu =fiber tensile strength (assuming a single tensile strength value for all fibers,which is not actually the case) m-matrix stress at the fiber failure strain,that is,at sm=(Figure 3.1) For effective reinforcement of the matrix,that is,for oLtu >omu,the fiber volume fraction in the composite must be greater than a critical value.This critical fiber volume fraction is calculated by setting oLtu=omu.Thus,from Equation 3.9, Critical vrmu (3.10a) 0u-0n Equation 3.9 assumes that the matrix is unable to carry the load transferred to it after the fibers have failed,and therefore,the matrix fails immediately after the fiber failure.However,at low fiber volume fractions,it is possible that the matrix will be able to carry additional load even after the fibers have failed.For this to occur, Omu(1-Vf)>ofuv+m(1-Vf), from which the minimum fiber volume fraction can be calculated as Minimum Vf= Omu-m (3.10b) 0mm十0u-0m If the fiber volume fraction is less than the minimum value given by Equation 3.10b,the matrix will continue to carry the load even after the fibers have failed at r=fu As the load on the composite is increased,the strain in the matrix will also increase,but some of the load will be transferred to the fibers.The fibers will continue to break into smaller and smaller lengths,and with decreas- ing fiber length,the average stress in the fibers will continue to decrease. Eventually,the matrix will fail when the stress in the matrix reaches omu, causing the composite to fail also.The longitudinal tensile strength of the composite in this case will be omu(1-v). Figure 3.3 shows the longitudinal strength variation with fiber volume fraction for a unidirectional continuous fiber composite containing an elastic, brittle matrix.Table 3.1 shows critical fiber volume fraction and minimum fiber 2007 by Taylor&Francis Group.LLC
longitudinal tensile strength sLtu of a unidirectional continuous fiber composite can be estimated as sLtu ¼ sfuvf þ s0 m(1 vf), (3:9) where sfu ¼ fiber tensile strength (assuming a single tensile strength value for all fibers, which is not actually the case) s0 m ¼ matrix stress at the fiber failure strain, that is, at «m ¼ «fu (Figure 3.1) For effective reinforcement of the matrix, that is, for sLtu > smu, the fiber volume fraction in the composite must be greater than a critical value. This critical fiber volume fraction is calculated by setting sLtu ¼ smu. Thus, from Equation 3.9, Critical vf ¼ smu s0 m sfu s0 m : (3:10a) Equation 3.9 assumes that the matrix is unable to carry the load transferred to it after the fibers have failed, and therefore, the matrix fails immediately after the fiber failure. However, at low fiber volume fractions, it is possible that the matrix will be able to carry additional load even after the fibers have failed. For this to occur, smu(1 vf) > sfuvf þ s0 m(1 vf), from which the minimum fiber volume fraction can be calculated as Minimum vf ¼ smu s0 m smu þ sfu s0 m : (3:10b) If the fiber volume fraction is less than the minimum value given by Equation 3.10b, the matrix will continue to carry the load even after the fibers have failed at sf ¼ sfu. As the load on the composite is increased, the strain in the matrix will also increase, but some of the load will be transferred to the fibers. The fibers will continue to break into smaller and smaller lengths, and with decreasing fiber length, the average stress in the fibers will continue to decrease. Eventually, the matrix will fail when the stress in the matrix reaches smu, causing the composite to fail also. The longitudinal tensile strength of the composite in this case will be smu(1vf). Figure 3.3 shows the longitudinal strength variation with fiber volume fraction for a unidirectional continuous fiber composite containing an elastic, brittle matrix. Table 3.1 shows critical fiber volume fraction and minimum fiber 2007 by Taylor & Francis Group, LLC.
Multiple u fracture Single fracture of fibers of fibers u Omu 60 0 (0%) Vf minimum Vf critical (100%) FIGURE 3.3 Longitudinal tensile strength variation with fiber volume fraction in a unidirectional continuous fiber composite in which the matrix failure strain is greater than the fiber failure strain. volume fraction for unidirectional continuous fiber-reinforced epoxy.For all practical applications,fiber volume fractions are much greater than these values. There are other stresses in the fibers as well as the matrix besides the longitudinal stresses.For example,transverse stresses,both tangential and radial,may arise due to the difference in Poisson's ratios,vr and vm,between the fibers and matrix.If vr vm,the matrix tends to contract more in the transverse directions than the fibers as the composite is loaded in tension in the longitudinal direction.This creates a radial pressure at the interface and,as a result,the matrix near the interface experiences a tensile stress in the tangential TABLE 3.1 Critical and Minimum Fiber Volume Fractions in E-glass,Carbon, and Boron Fiber-Reinforced Epoxy Matrix2 Composite Property E-Glass Fiber Carbon Fiber Boron Fiber E 10×10psi 30×10°psi 55×10psi 0和0细 250.000psi 400.,000psi 450.000psi 8和二F 0.025 0.0133 0.0082 Om=Em Efu 2,500psi 1,330psi 820 psi Critical vr 3.03% 2.17% 2.04% Minimum Vr 2.9% 2.12% 2% a Matrix properties:mu=10.000 psi.Em=0.1 X 10 psi,and mu=0.1. 2007 by Taylor Francis Group,LLC
volume fraction for unidirectional continuous fiber-reinforced epoxy. For all practical applications, fiber volume fractions are much greater than these values. There are other stresses in the fibers as well as the matrix besides the longitudinal stresses. For example, transverse stresses, both tangential and radial, may arise due to the difference in Poisson’s ratios, nf and nm, between the fibers and matrix. If nf < nm, the matrix tends to contract more in the transverse directions than the fibers as the composite is loaded in tension in the longitudinal direction. This creates a radial pressure at the interface and, as a result, the matrix near the interface experiences a tensile stress in the tangential vf sLtu sfu smu s 9 m vf minimum vf critical 0 (0%) 1 (100%) Single fracture of fibers Multiple fracture of fibers FIGURE 3.3 Longitudinal tensile strength variation with fiber volume fraction in a unidirectional continuous fiber composite in which the matrix failure strain is greater than the fiber failure strain. TABLE 3.1 Critical and Minimum Fiber Volume Fractions in E-glass, Carbon, and Boron Fiber-Reinforced Epoxy Matrixa Composite Property E-Glass Fiber Carbon Fiber Boron Fiber Ef 10 3 106 psi 30 3 106 psi 55 3 106 psi sfu 250,000 psi 400,000 psi 450,000 psi «fu ¼ sfu Ef 0.025 0.0133 0.0082 sm 0 ¼ Em «fu 2,500 psi 1,330 psi 820 psi Critical vf 3.03% 2.17% 2.04% Minimum vf 2.9% 2.12% 2% a Matrix properties: smu ¼ 10,000 psi, Em ¼ 0.1 3 106 psi, and «mu ¼ 0.1. 2007 by Taylor & Francis Group, LLC
direction and a compressive stress in the radial direction.Tangential and radial stresses in the fibers are both compressive.However,all these stresses are relatively small compared with the longitudinal stresses. Another source of internal stresses in the lamina is due to the difference in thermal contraction between the fibers and matrix as the lamina is cooled down from the fabrication temperature to room temperature.In general,the matrix has a higher coefficient of thermal expansion (or contraction),and,therefore, tends to contract more than the fibers,creating a"squeezing"effect on the fibers.A three-dimensional state of residual stresses is created in the fibers as well as in the matrix.These stresses can be calculated using the equations given in Appendix A.2. 3.1.1.2 Unidirectional Discontinuous Fibers Tensile load applied to a discontinuous fiber lamina is transferred to the fibers by a shearing mechanism between fibers and matrix.Since the matrix has a lower modulus,the longitudinal strain in the matrix is higher than that in adjacent fibers.If a perfect bond is assumed between the two constituents, the difference in longitudinal strains creates a shear stress distribution across the fiber-matrix interface.Ignoring the stress transfer at the fiber end cross sections and the interaction between the neighboring fibers,we can calculate the normal stress distribution in a discontinuous fiber by a simple force equi- librium analysis(Figure 3.4). Consider an infinitesimal length dx at a distance x from one of the fiber ends(Figure 3.4).The force equilibrium equation for this length is (得d)o+do)-(匠)-(mddx=0, P :dor leuipnjbuo uonoellp Pe FIGURE 3.4 Longitudinal tensile loading of a unidirectional discontinuous fiber lamina. 2007 by Taylor Francis Group.LLC
direction and a compressive stress in the radial direction. Tangential and radial stresses in the fibers are both compressive. However, all these stresses are relatively small compared with the longitudinal stresses. Another source of internal stresses in the lamina is due to the difference in thermal contraction between the fibers and matrix as the lamina is cooled down from the fabrication temperature to room temperature. In general, the matrix has a higher coefficient of thermal expansion (or contraction), and, therefore, tends to contract more than the fibers, creating a ‘‘squeezing’’ effect on the fibers. A three-dimensional state of residual stresses is created in the fibers as well as in the matrix. These stresses can be calculated using the equations given in Appendix A.2. 3.1.1.2 Unidirectional Discontinuous Fibers Tensile load applied to a discontinuous fiber lamina is transferred to the fibers by a shearing mechanism between fibers and matrix. Since the matrix has a lower modulus, the longitudinal strain in the matrix is higher than that in adjacent fibers. If a perfect bond is assumed between the two constituents, the difference in longitudinal strains creates a shear stress distribution across the fiber–matrix interface. Ignoring the stress transfer at the fiber end cross sections and the interaction between the neighboring fibers, we can calculate the normal stress distribution in a discontinuous fiber by a simple force equilibrium analysis (Figure 3.4). Consider an infinitesimal length dx at a distance x from one of the fiber ends (Figure 3.4). The force equilibrium equation for this length is p 4 d2 f (sf þ dsf) p 4 d2 f sf ð Þ pdf dx t ¼ 0, dx sf + dsf sf t lf df x Pc Pc Longitudinal direction FIGURE 3.4 Longitudinal tensile loading of a unidirectional discontinuous fiber lamina. 2007 by Taylor & Francis Group, LLC.
which on simplification gives dor 4r dxd' (3.11) where or=longitudinal stress in the fiber at a distance x from one of its ends T=shear stress at the fiber-matrix interface de=fiber diameter Assuming no stress transfer at the fiber ends,that is,or=0 at x=0,and integrating Equation 3.11,we determine the longitudinal stress distribution in the fiber as T dx. (3.12) For simple analysis,let us assume that the interfacial shear stress is constant and is equal to Ti.With this assumption,integration of Equation 3.12 gives 4Ti 0f= -X. (3.13) d From Equation 3.13,it can be observed that for a composite lamina containing discontinuous fibers,the fiber stress is not uniform.According to Equation 3.13,it is zero at each end of the fiber(i.e.,x=0)and it increases linearly with x.The maximum fiber stress occurs at the central portion of the fiber (Figure 3.5).The maximum fiber stress that can be achieved at a given load is 0ain=2n音 (3.14) where x=/2=load transfer length from each fiber end.Thus,the load transfer length,/is the minimum fiber length in which the maximum fiber stress is achieved. For a given fiber diameter and fiber-matrix interfacial condition,a critical fiber length le is calculated from Equation 3.14 as 6=器4 (3.15) where ofu=ultimate tensile strength of the fiber le=minimum fiber length required for the maximum fiber stress to be equal to the ultimate tensile strength of the fiber at its midlength(Figure 3.6b) Ti=shear strength of the fiber-matrix interface or the shear strength of the matrix adjacent to the interface,whichever is less 2007 by Taylor Francis Group,LLC
which on simplification gives dsf dx ¼ 4t df , (3:11) where sf ¼ longitudinal stress in the fiber at a distance x from one of its ends t ¼ shear stress at the fiber–matrix interface df ¼ fiber diameter Assuming no stress transfer at the fiber ends, that is, sf ¼ 0 at x ¼ 0, and integrating Equation 3.11, we determine the longitudinal stress distribution in the fiber as sf ¼ 4 df ðx 0 t dx: (3:12) For simple analysis, let us assume that the interfacial shear stress is constant and is equal to ti. With this assumption, integration of Equation 3.12 gives sf ¼ 4ti df x: (3:13) From Equation 3.13, it can be observed that for a composite lamina containing discontinuous fibers, the fiber stress is not uniform. According to Equation 3.13, itis zero at each end of the fiber(i.e., x ¼ 0) and itincreaseslinearly with x. The maximum fiber stress occurs at the central portion of the fiber (Figure 3.5). The maximum fiber stress that can be achieved at a given load is (sf)max ¼ 2ti lt df , (3:14) where x ¼ lt=2 ¼ load transfer length from each fiber end. Thus, the load transfer length, lt, is the minimum fiber length in which the maximum fiber stress is achieved. For a given fiber diameter and fiber–matrix interfacial condition, a critical fiber length lc is calculated from Equation 3.14 as lc ¼ sfu 2ti df, (3:15) where sfu ¼ ultimate tensile strength of the fiber lc ¼ minimum fiber length required for the maximum fiber stress to be equal to the ultimate tensile strength of the fiber at its midlength (Figure 3.6b) ti ¼ shear strength of the fiber–matrix interface or the shear strength of the matrix adjacent to the interface, whichever is less 2007 by Taylor & Francis Group, LLC
Increasing oc (G)max 2 2 (a) Increasing oc (b) FIGURE 3.5 Idealized(a)longitudinal stress and (b)shear stress distributions along a discontinuous fiber owing to longitudinal tensile loading. Fiber tensile strength(ofu) k (a) (b) (c) FIGURE 3.6 Significance of critical fiber length on the longitudinal stresses of a discon- tinuous fiber. 2007 by Taylor Francis Group.LLC
Increasing sc Increasing sc (sf )max lf lt 2 (a) (b) lt 2 ti ti lt 2 lt 2 FIGURE 3.5 Idealized (a) longitudinal stress and (b) shear stress distributions along a discontinuous fiber owing to longitudinal tensile loading. lf lc Fiber tensile strength (sfu) 2 lc 2 lc 2 lc 2 FIGURE 3.6 Significance of critical fiber length on the longitudinal stresses of a discontinuous fiber. 2007 by Taylor & Francis Group, LLC
From Equations 3.14 and 3.15,we make the following observations: 1.For ll,the maximum fiber stress may reach the ultimate fiber strength over much of its length(Figure 3.6c).However,over a distance equal to l/2 from each end,the fiber remains less effective. 3.For effective fiber reinforcement,that is,for using the fiber to its potential strength,one must select. 4.For a given fiber diameter and strength,le can be controlled by increas- ing or decreasing Ti.For example,a matrix-compatible coupling agent may increase Ti,which in turn decreases le.If l can be reduced relative to /r through proper fiber surface treatments,effective reinforcement can be achieved without changing the fiber length. Although normal stresses near the two fiber ends,that is,at xl,the longitudinal tensile strength of a unidirectional discontinu- ous fiber composite is calculated by substituting(max=ru and=l(Figure 3.6c).Thus, OLu =Gfuv +om(1-Ve) =1-2立) vi+om (1-vp). (3.17) In Equation 3.17,it is assumed that all fibers fail at the same strength level of ofu.Comparison of Equations 3.9 and 3.17 shows that discontinuous fibers always strengthen a matrix to a lesser degree than continuous fibers.However, 2007 by Taylor Francis Group,LLC
From Equations 3.14 and 3.15, we make the following observations: 1. For lf lc, the maximum fiber stress may reach the ultimate fiber strength over much of its length (Figure 3.6c). However, over a distance equal to lc=2 from each end, the fiber remains less effective. 3. For effective fiber reinforcement, that is, for using the fiber to its potential strength, one must select lf lc. 4. For a given fiber diameter and strength, lc can be controlled by increasing or decreasing ti. For example, a matrix-compatible coupling agent may increase ti, which in turn decreases lc. If lc can be reduced relative to lf through proper fiber surface treatments, effective reinforcement can be achieved without changing the fiber length. Although normal stresses near the two fiber ends, that is, at x lc is lc 2 . For lf > lc, the longitudinal tensile strength of a unidirectional discontinuous fiber composite is calculated by substituting (sf)max ¼ sfu and lt ¼ lc (Figure 3.6c). Thus, sLtu ¼ sfuvf þ s0 m(1 vf) ¼ sfu 1 lc 2lf vf þ s0 m(1 vf): (3:17) In Equation 3.17, it is assumed that all fibers fail at the same strength level of sfu. Comparison of Equations 3.9 and 3.17 shows that discontinuous fibers always strengthen a matrix to a lesser degree than continuous fibers. However, 2007 by Taylor & Francis Group, LLC.
