11.1 RULE OF MIXTURES 441 L+△L1 Figure 11.5:Deformation of Element 1 subjected to an axial force in the x direction. L-△L L-△L 11.1.5 Longitudinal Poisson Ratio v12 Element 1,shown in Figure 11.2,is subjected to a force Fi in the longitudinal x direction.Because of this force the element deforms,as illustrated in Figure 11.5. The sides of the deformed element are(L+△Li),(L-△L2),and(L-△Ls). The change in the cross-sectional area A of the face ijkl is △A=A-(L-△L2)(L-△L3) (11.30) By definition,the normal strains in the x2 and x3 directions are L2 2=-L 3=- △L (11.31) Owing to symmetry,AL2 =AL3 and,consequently,we have E2=e3. (11.32) By neglecting higher-order terms,we find that Egs.(11.30)-(11.32)give △A=A(2+∈3)=2Ae2. (11.33) By similar arguments,the changes in the matrix and fiber areas are △Ar=2Are2 △Am=2Amem2: (11.34) The change in the total area is the sum of the changes of the fiber and matrix areas: △A=△A+△Am· (11.35) Under the action of the longitudinal normal stress o1 F/A,the transverse normal strains are related to the longitudinal normal strain by e2=-112e1 e3=-13e1. (11.36) By virtue of Egs.(11.32)and (11.36),we have 12=13 (11.37) From Eqs.(11.36)and (11.33)we obtain z=-2=-A41 2AE1 (11.38) E1
11.1 RULE OF MIXTURES 441 F1 L L + ∆L1 L − ∆L2 L L i j k l L ∆– L3 Figure 11.5: Deformation of Element 1 subjected to an axial force in the x1 direction. 11.1.5 Longitudinal Poisson Ratio ν12 Element 1, shown in Figure 11.2, is subjected to a force F1 in the longitudinal x1 direction. Because of this force the element deforms, as illustrated in Figure 11.5. The sides of the deformed element are (L+ L1), (L− L2), and (L− L3). The change in the cross-sectional area Aof the face ijkl is A = A− (L− L2) (L− L3). (11.30) By definition, the normal strains in the x2 and x3 directions are 2 = − L2 L 3 = − L3 L . (11.31) Owing to symmetry, L2 = L3 and, consequently, we have 2 = 3. (11.32) By neglecting higher-order terms, we find that Eqs. (11.30)–(11.32) give A = A(2 + 3) = 2A2. (11.33) By similar arguments, the changes in the matrix and fiber areas are Af = 2Aff2 Am = 2Amm2. (11.34) The change in the total area is the sum of the changes of the fiber and matrix areas: A = Af + Am. (11.35) Under the action of the longitudinal normal stress σ1 = F1/A, the transverse normal strains are related to the longitudinal normal strain by 2 = −ν121 3 = −ν131. (11.36) By virtue of Eqs. (11.32) and (11.36), we have ν12 = ν13. (11.37) From Eqs. (11.36) and (11.33) we obtain ν12 = −2 1 = − A 2A 1 1 . (11.38)
442 MICROMECHANICS By combining Egs.(11.35)and (11.38)and introducing Eq.(11.34)into the resulting expression,we obtain 12=- 2Aen+2Amm1=-4-4如 (11.39) 2A E1 AE1 A E1 The changes in the lengths of the fiber,the matrix,and the composite are equal. Consequently,in the x direction the normal strains in the composite,in the fiber, and in the matrix are equal as follows: E1 Ef1 Eml. (11.40) With the definitions u=A Um Am A h2÷、电 hn-、2 (11.41) Efl Eml Egs.(11.39)-(11.41)give the following expression for the longitudinal Poisson ratio: V12 Ufvn2 +UmVm. (11.42) 11.1.6 Transverse Poisson Ratio v23 For a transversely isotropic material (which is under consideration here),the transverse shear and the Young moduli are related by the following expression (Eq.2.34) E G2s=20+3) (11.43) By rearranging this expression,we obtain the transverse Poisson ratio 3= 一1. (11.44) 2G3 More accurate expressions for E2 and G23 are obtained by the modified rule of mixtures(Section 11.2).These expressions are included in Table 11.1,below. Table 11.1.Expressions for the engineering constants Longitudinal Young modulus E =vrEn +vm Em Transverse Young modulus 6=(瓷+) where E2=√Ea+(1-√)Em Longitudinal shear modulus where Gb12=√Gn2+(1-√/)Gm Transverse shear modulus Gm=(+) where Gb23=√GG3+(1-√)Gm Longitudinal Poisson ratio V12 VfVn2 UmVm Transverse Poisson ratio =忌-1
442 MICROMECHANICS By combining Eqs. (11.35) and (11.38) and introducing Eq. (11.34) into the resulting expression, we obtain ν12 = −2Aff2 + 2Amm2 2A 1 1 = − Af A f2 1 − Am A m2 1 . (11.39) The changes in the lengths of the fiber, the matrix, and the composite are equal. Consequently, in the x1 direction the normal strains in the composite, in the fiber, and in the matrix are equal as follows: 1 = f1 = m1. (11.40) With the definitions vf = Af A vm = Am A νf12 = −f2 f1 νm = −m2 m1 , (11.41) Eqs. (11.39)–(11.41) give the following expression for the longitudinal Poisson ratio: ν12 = vfνf12 + vmνm. (11.42) 11.1.6 Transverse Poisson Ratio ν23 For a transversely isotropic material (which is under consideration here), the transverse shear and the Young moduli are related by the following expression (Eq. 2.34): G23 = E2 2(1 + ν23) . (11.43) By rearranging this expression, we obtain the transverse Poisson ratio ν23 = E2 2G23 − 1. (11.44) More accurate expressions for E2 and G23 are obtained by the modified rule of mixtures (Section 11.2). These expressions are included in Table 11.1, below. Table 11.1. Expressions for the engineering constants Longitudinal Young modulus E1 = vfEf1 + vm Em Transverse Young modulus E2 = √vf Eb2 + 1−√vf Em −1 where Eb2 = √vfEf2 + (1 − √vf)Em Longitudinal shear modulus G12 = √vf Gb12 + 1−√vf Gm −1 where Gb12 = √vfGf12 + (1 − √vf)Gm Transverse shear modulus G23 = √vf Gb23 + 1−√vf Gm −1 where Gb23 = √vfGf23 + (1 − √vf)Gm Longitudinal Poisson ratio ν12 = vfνf12 + vmνm Transverse Poisson ratio ν23 = E2 2G23 − 1