Micromechanical Analysis of a Lamina 223 Substituting Equation (3.38)and Equation(3.39)in Equation (3.37)and using Equation (3.36)gives 1-1+1 互E+E。 (3.40) Because the thickness fractions are the same as the volume fractions as the other two dimensions are equal for the fiber and the matrix(see Equa- tion3.28): 1_y+ E2 Ef Em (3.41) Equation(3.41)is based on the weighted mean of the compliance of the fiber and the matrix. Example 3.4 Find the transverse Young's modulus of a glass/epoxy lamina with a fiber volume fraction of 70%.Use the properties of glass and epoxy from Table 3.1 and Table 3.2,respectively. Solution From Table 3.1,the Young's modulus of the fiber is E=85 GPa. From Table 3.2,the Young's modulus of the matrix is E'=3.4 GPa. Using Equation(3.41),the transverse Young's modulus,E2,is 10.7.0.3 E2853.4' E2=10.37GPa. Figure 3.8 plots the transverse Young's modulus as a function of fiber volume fraction for constant fiber-to-matrix elastic moduli ratio,E/E For metal and ceramic matrix composites,the fiber and matrix elastic moduli are of the same order.(For example,for a SiC/aluminum metal matrix composite,E/E=4 and for a SiC/CAS ceramic matrix composite,Er/Em= 2).The transverse Young's modulus of the composite in such cases changes more smoothly as a function of the fiber volume fraction. 2006 by Taylor Francis Group,LLC
Micromechanical Analysis of a Lamina 223 Substituting Equation (3.38) and Equation (3.39) in Equation (3.37) and using Equation (3.36) gives (3.40) Because the thickness fractions are the same as the volume fractions as the other two dimensions are equal for the fiber and the matrix (see Equation 3.28): (3.41) Equation (3.41) is based on the weighted mean of the compliance of the fiber and the matrix. Example 3.4 Find the transverse Young’s modulus of a glass/epoxy lamina with a fiber volume fraction of 70%. Use the properties of glass and epoxy from Table 3.1 and Table 3.2, respectively. Solution From Table 3.1, the Young’s modulus of the fiber is Ef = 85 GPa. From Table 3.2, the Young’s modulus of the matrix is Em = 3.4 GPa. Using Equation (3.41), the transverse Young’s modulus, E2, is Figure 3.8 plots the transverse Young’s modulus as a function of fiber volume fraction for constant fiber-to-matrix elastic moduli ratio, Ef /Em. For metal and ceramic matrix composites, the fiber and matrix elastic moduli are of the same order. (For example, for a SiC/aluminum metal matrix composite, Ef /Em = 4 and for a SiC/CAS ceramic matrix composite, Ef /Em = 2). The transverse Young’s modulus of the composite in such cases changes more smoothly as a function of the fiber volume fraction. 11 1 E E 2 t t E t t f f c m m c = + . 1 E2 V E V E f f m m = + . 1 07 85 0 3 3 4 10 37 2 2 E E GPa = + = . . . , . . 1343_book.fm Page 223 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC
224 Mechanics of Composite Materials,Second Edition % E/Em=1 30 E/E.=5 E/Em=25 20 E/En=125 0 0.2 0.4 0.6 0.8 Fiber volume fraction,V FIGURE 3.8 Transverse Young's modulus as a function of fiber volume fraction for constant fiber to matrix moduli ratio. For polymeric composites,the fiber-to-matrix moduli ratio is very high. (For example,for a glass/epoxy polymer matrix composite,E/E=25).The transverse Young's modulus of the composite in such cases changes appre- ciably only for large fiber volume fractions.Figure 3.8 shows that,for high E/E ratios,the contribution of the fiber modulus only increases substantially for a fiber volume fraction greater than 80%.These fiber volume fractions are not practical and in many cases are physically impossible due to the geometry of fiber packing.Figure 3.9 shows various possibilities of fiber packing.Note that the ratio of the diameter,d,to fiber spacing,s,d/s varies with geometrical packing.For circular fibers with square array packing(Figure 3.9a), (3.42a) This gives a maximum fiber volume fraction of 78.54%as s >d.For circular fibers with hexagonal array packing(Figure 3.9b), 3V (3.42b) 2006 by Taylor Francis Group,LLC
224 Mechanics of Composite Materials, Second Edition For polymeric composites, the fiber-to-matrix moduli ratio is very high. (For example, for a glass/epoxy polymer matrix composite, Ef /Em = 25). The transverse Young’s modulus of the composite in such cases changes appreciably only for large fiber volume fractions. Figure 3.8 shows that, for high Ef /Em ratios, the contribution of the fiber modulus only increases substantially for a fiber volume fraction greater than 80%. These fiber volume fractions are not practical and in many cases are physically impossible due to the geometry of fiber packing. Figure 3.9 shows various possibilities of fiber packing. Note that the ratio of the diameter, d, to fiber spacing, s, d/s varies with geometrical packing. For circular fibers with square array packing (Figure 3.9a), (3.42a) This gives a maximum fiber volume fraction of 78.54% as s ≥ d. For circular fibers with hexagonal array packing (Figure 3.9b), (3.42b) FIGURE 3.8 Transverse Young’s modulus as a function of fiber volume fraction for constant fiber to matrix moduli ratio. 40 Ef / Em = 1 Ef / Em = 5 Ef / Em = 25 Ef / Em = 125 30 20 10 0 0 0.2 0.4 0.6 Fiber volume fraction, Vf Transverse Young’s modulus ratio, E2/E m 0.8 1 d s Vf = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 4 1 2 π / . d s Vf = ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ 2 3 1 2 π / . 1343_book.fm Page 224 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC
Micromechanical Analysis of a Lamina 225 d (a) (6) FIGURE 3.9 Fiber to fiber spacing in (a)square packing geometry and(b)hexagonal packing geometry. This gives a maximum fiber volume fraction of 90.69%because s2d.These maximum fiber volume fractions are not practical to use because the fibers touch each other and thus have surfaces where the matrix cannot wet out the fibers. In Figure 3.10,the transverse Young's modulus is plotted as a function of fiber volume fraction using Equation(3.41)for a typical boron/epoxy lamina. Also given are the experimental data points.In Figure 3.10,the experimental and analytical results are not as close to each other as they are for the longitudinal Young's modulus in Figure 3.6. 2006 by Taylor Francis Group,LLC
Micromechanical Analysis of a Lamina 225 This gives a maximum fiber volume fraction of 90.69% because s ≥ d. These maximum fiber volume fractions are not practical to use because the fibers touch each other and thus have surfaces where the matrix cannot wet out the fibers. In Figure 3.10, the transverse Young’s modulus is plotted as a function of fiber volume fraction using Equation (3.41) for a typical boron/epoxy lamina. Also given are the experimental data points.4 In Figure 3.10, the experimental and analytical results are not as close to each other as they are for the longitudinal Young’s modulus in Figure 3.6. FIGURE 3.9 Fiber to fiber spacing in (a) square packing geometry and (b) hexagonal packing geometry. s s d d (b) (a) 1343_book.fm Page 225 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC
226 Mechanics of Composite Materials,Second Edition 40 (d9) 30 Experimental data points 'snmnpow s,Sunox 00 20 Mechanics of materials approach asJaAsueJL 10 0 0 0.2 0.4 0.6 0.8 Fiber volume fraction,V 40 O Experimental data points 30 'snmnpow sSunox asiansueL 20 10 Mechanics of materials approach 0.45 0.55 0.65 0.75 Fiber volume fraction,V FIGURE 3.10 Theoretical values of transverse Young's modulus as a function of fiber volume fraction for a Boron/Epoxy unidirectional lamina (E,414 GPa,V=0.2,E=4.14 GPa,V 0.35)and comparison with experimental values.Figure (b)zooms figure (a)for fiber volume fraction between 0.45 and 0.75.(Experimental data from Hashin,Z.,NASA tech.rep.contract no.NAS1- 8818,November 1970.) 2006 by Taylor Francis Group,LLC
226 Mechanics of Composite Materials, Second Edition FIGURE 3.10 Theoretical values of transverse Young’s modulus as a function of fiber volume fraction for a Boron/Epoxy unidirectional lamina (Ef = 414 GPa, νf = 0.2, Em = 4.14 GPa, νm = 0.35) and comparison with experimental values. Figure (b) zooms figure (a) for fiber volume fraction between 0.45 and 0.75. (Experimental data from Hashin, Z., NASA tech. rep. contract no. NAS1- 8818, November 1970.) 40 30 20 10 0.2 0.4 0.6 Fiber volume fraction, Vf Transverse Young’s modulus, E2 (GPa) Experimental data points Mechanics of materials approach 0.8 1 0 0 40 30 20 10 0 0.45 0.55 Experimental data points Mechanics of materials approach Fiber volume fraction, Vf Transverse Young’s modulus, E2 (GPa) 0.65 0.75 1343_book.fm Page 226 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC
Micromechanical Analysis of a Lamina 227 t/2 te (a) (b) FIGURE 3.11 A longitudinal stress applied to a representative volume element to calculate Poisson's ratio of unidirectional lamina. 3.3.1.3 Major Poisson's Ratio The major Poisson's ratio is defined as the negative of the ratio of the normal strain in the transverse direction to the normal strain in the longitudinal direction,when a normal load is applied in the longitudinal direction Assume a composite is loaded in the direction parallel to the fibers,as shown in Figure 3.11.The fibers and matrix are again represented by rectangular blocks.The deformations in the transverse direction of the composite(?)is the sum of the transverse deformations of the fiber(8)and the matrix()as 8=8+8 (3.43) Using the definition of normal strains, = 8 (3.44a) tr Em= (3.44b) 2006 by Taylor Francis Group,LLC
Micromechanical Analysis of a Lamina 227 3.3.1.3 Major Poisson’s Ratio The major Poisson’s ratio is defined as the negative of the ratio of the normal strain in the transverse direction to the normal strain in the longitudinal direction, when a normal load is applied in the longitudinal direction. Assume a composite is loaded in the direction parallel to the fibers, as shown in Figure 3.11. The fibers and matrix are again represented by rectangular blocks. The deformations in the transverse direction of the composite is the sum of the transverse deformations of the fiber and the matrix as (3.43) Using the definition of normal strains, (3.44a) (3.44b) FIGURE 3.11 A longitudinal stress applied to a representative volume element to calculate Poisson’s ratio of unidirectional lamina. σ1 σ1 h (a) (b) t r t c t m/2 t m/2 t c+ δc T t f + δf T t f Lc t c t m/2 t m/2 ( ) δc T ( ) δ f T ( ) δm T δδδ c T f T m T = + . ε δ f T f T f t = , ε δ m T m T tm = , 1343_book.fm Page 227 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC