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<l 4=e 4> (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 (a) (b) (c) lc lf = lc 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 l<le,the maximum fiber stress may never reach the ultimate fiber strength (Figure 3.6a).In this case,either the fiber-matrix interfacial bond or the matrix may fail before fibers achieve their potential strength. 2.For lf>l,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 x<h/2,are lower than the maximum fiber stress,their contributions to the total load-carrying capacity of the fiber cannot be completely ignored.Including these end stress distributions,an average fiber stress is calculated as which gives =om(-) (3.16) le Note that the load transfer length foriswhereas that foris For l>l,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 never reach the ultimate fiber strength (Figure 3.6a). In this case, either the fiber–matrix interfacial bond or the matrix may fail before fibers achieve their potential strength. 2. 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 < lt=2, are lower than the maximum fiber stress, their contributions to the total load-carrying capacity of the fiber cannot be completely ignored. Including these end stress distributions, an average fiber stress is calculated as sf ¼ 1 lf ðlf 0 sf dx, which gives sf ¼ (sf)max 1 lt 2lf : (3:16) Note that the load transfer length for lf < lc is lf 2 , whereas that for lf > 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.