Copyrighted Materials rU Press frw CHAPTER ELEVEN Micromechanics Micromechanics is used to estimate the mechanical and hygrothermal properties of composite materials from the known values of the properties of the fiber and the matrix.There are three major categories of micromechanical approaches: (i)mechanics of material models based on simplifying assumptions that make it unnecessary to specify in detail the stress-strain distributions,(ii)elasticity models requiring that the stresses and strains be determined at the micromechanical level, and(iii)empirical expressions resulting from curve-fitting elasticity solutions or data. It is not our intent to discuss the numerous available models.Instead,we focus on two mechanics of materials models,namely on the "rule of mixtures"and the "modified rule of mixtures."The rule of mixtures is the simplest and most intuitive approach and is useful for introducing concepts.However,it fails to represent some of the properties with reasonable accuracy.The modified rule of mixtures is an improvement over the rule of mixtures,and predicts the properties with better accuracies. Methods to predict strength-though available-are not presented because they are less accurate than the models predicting elastic and hygrothermal properties. 11.1 Rule of Mixtures We consider the volume V of an element shown in Figure 11.1.The volume of this element is V=V+Vn+V, (11.1) where the subscripts f,m and v refer to the fiber,the matrix,and the void.It is convenient to introduce the volume fractions as follows: Vm (11.2) 436
CHAPTER ELEVEN Micromechanics Micromechanics is used to estimate the mechanical and hygrothermal properties of composite materials from the known values of the properties of the fiber and the matrix. There are three major categories of micromechanical approaches: (i) mechanics of material models based on simplifying assumptions that make it unnecessary to specify in detail the stress–strain distributions, (ii) elasticity models requiring that the stresses and strains be determined at the micromechanical level, and (iii) empirical expressions resulting from curve-fitting elasticity solutions or data. It is not our intent to discuss the numerous available models. Instead, we focus on two mechanics of materials models, namely on the “rule of mixtures” and the “modified rule of mixtures.” The rule of mixtures is the simplest and most intuitive approach and is useful for introducing concepts. However, it fails to represent some of the properties with reasonable accuracy. The modified rule of mixtures is an improvement over the rule of mixtures, and predicts the properties with better accuracies. Methods to predict strength – though available – are not presented because they are less accurate than the models predicting elastic and hygrothermal properties. 11.1 Rule of Mixtures We consider the volume V of an element shown in Figure 11.1. The volume of this element is V = Vf + Vm + Vv, (11.1) where the subscripts f, m and v refer to the fiber, the matrix, and the void. It is convenient to introduce the volume fractions as follows: vf ≡ Vf V vm ≡ Vm V vv ≡ Vv V . (11.2) 436
11.1 RULE OF MIXTURES 437 fiber void matrix Figure 11.1:Illustration of the matrix,fiber,and void volumes. Equations (11.1)and (11.2)give Ur vm +vv =1. (11.3) When the void fraction is negligible (v=0),we have Um =1-Uf. (11.4) The mass of the element in Figure 11.1 is M=M+Mm My. (11.5) By neglecting the mass of the void,we can write this equation as M=PrVi+Pm Vm, (11.6) where Pr and pm are the fiber and the matrix densities,respectively.The density of the composite is M Pcomp==UrPr+UmPm- (11.7) In the following,we treat unidirectional,fiber-reinforced composites without voids.We derive the properties using two types of elements(Fig.11.2).Element 1 contains a single fiber bundle of circular cross section.Element 2 consists of a fiber layer sandwiched between two layers of matrix material.In Element 2,the òδd fiber fiber Lm f m matrix L=vL 斗长一 r= L2 Ur=rL)L Figure 11.2:Representative elements.Element 1(left)and Element 2(right)
11.1 RULE OF MIXTURES 437 fiber void matrix Figure 11.1: Illustration of the matrix, fiber, and void volumes. Equations (11.1) and (11.2) give vf + vm + vv = 1. (11.3) When the void fraction is negligible (vv = 0), we have vm = 1 − vf. (11.4) The mass of the element in Figure 11.1 is M = Mf + Mm + Mv. (11.5) By neglecting the mass of the void, we can write this equation as M = ρfVf + ρmVm, (11.6) where ρf and ρm are the fiber and the matrix densities, respectively. The density of the composite is ρcomp = M V = vfρf + vmρm. (11.7) In the following, we treat unidirectional, fiber-reinforced composites without voids. We derive the properties using two types of elements (Fig. 11.2). Element 1 contains a single fiber bundle of circular cross section. Element 2 consists of a fiber layer sandwiched between two layers of matrix material. In Element 2, the L L L matrix fiber Af L L L m f m fiber Lf = v Lf 2 f f )( L LLv v = 2 f f L A v = Figure 11.2: Representative elements. Element 1 (left) and Element 2 (right)
438 MICROMECHANICS F 4L米 Figure 11.3:Element 1 subjected to a force in the fiber direction(left),and Element 2 subjected to a force in the transverse direction (right). fiber and matrix volumes are the same as in Element 1.Hence,the thickness of the fiber layer in Element 2 is vrL. 11.1.1 Longitudinal Young Modulus E1 Element 1 is subjected to a force F in the fiber direction.The force,distributed over the surface,is(Fig.11.3) F1=01A, (11.8) where o1 is the average normal stress across the entire cross-sectional area A (A=L2).Part of the force is carried by the fibers and part by the matrix.Thus, we write 01A=Arof1+Amom1, (11.9) Ar and Am are the cross-sectional areas of the fiber bundle and the matrix,respec- tively.When the Poisson effect is neglected,the normal stresses in the composite o1,in the fiber bundle ofi,and in the matrix omi are 01 =E1E1 of ef En Oml =Eml Em, (11.10) where E and En are the composite and the fiber longitudinal Young moduli and Em is the matrix Young modulus,respectively.Equations(11.9)and (11.10)give a点=头aa+六mEe (11.11) The elongations AL and,hence,the longitudinal normal strains of the com- posite e1.the matrix em1.and the fiber en are equal: E1 Ef1=Eml. (11.12) The volume fractions are = m mA (11.13) Equations(11.11)-(11.13)result in the following expression for the longitud- inal Young modulus of the composite in terms of the fiber and matrix moduli: E1=vrEn vm Em=vrEa +(1-vr)Em. (11.14)
438 MICROMECHANICS F1 F1 L ∆L L F2 F2 m m f ∆L L L L Lf Figure 11.3: Element 1 subjected to a force in the fiber direction (left), and Element 2 subjected to a force in the transverse direction (right). fiber and matrix volumes are the same as in Element 1. Hence, the thickness of the fiber layer in Element 2 is vfL. 11.1.1 Longitudinal Young Modulus E 1 Element 1 is subjected to a force F1 in the fiber direction. The force, distributed over the surface, is (Fig. 11.3) F1 = σ1A, (11.8) where σ1 is the average normal stress across the entire cross-sectional area A (A = L2). Part of the force is carried by the fibers and part by the matrix. Thus, we write σ1A = Afσ f 1 + Amσm1, (11.9) Af and Am are the cross-sectional areas of the fiber bundle and the matrix, respectively. When the Poisson effect is neglected, the normal stresses in the composite σ1, in the fiber bundle σ f 1, and in the matrix σm1 are σ1 = 1E1 σf1 = f1Ef1 σm1 = m1Em, (11.10) where E1 and Ef1 are the composite and the fiber longitudinal Young moduli and Em is the matrix Young modulus, respectively. Equations (11.9) and (11.10) give 1E1 = Af A f1Ef1 + Am A m1Em. (11.11) The elongations L and, hence, the longitudinal normal strains of the composite 1, the matrix m1, and the fiber f1 are equal: 1 = f1 = m1. (11.12) The volume fractions are vf = Af A vm = Am A . (11.13) Equations (11.11)–(11.13) result in the following expression for the longitudinal Young modulus of the composite in terms of the fiber and matrix moduli: E1 = vfEf1 + vm Em = vfEf1 + (1 − vf) Em. (11.14)
11.1 RULE OF MIXTURES 439 11.1.2 Transverse Young Modulus E2 We consider a rectangular element with sides Lmade up of three layers(Element 2, Fig.11.3).The inner layer is a sheet of fiber with the same volume as the fiber bundle in Element 1.The length and the transverse (x2 direction)Young modulus of this inner layer are Lr and Ep2,respectively,and the length and the Young mod- ulus of the outer layers are Lm =(L-Lr)/2 and Em.The element is subjected to a force F2.The force is distributed uniformly across the surface A(A=L2).The normal stress in the transverse direction is o2=F2/A,which can be expressed as 02=e2E2, (11.15) where E2 is the transverse Young modulus of the element and e2 is the average transverse normal strain △L e2= L (11.16) The change in the length of the element is △L=2Lmem2+Lfe2, (11.17) where em2 and er2 are the transverse normal strains in the matrix and fiber layers. By neglecting the Poisson effect,we have .0m2 0f2 Em2= Em 0二E2 (11.18) By introducing Eqs.(11.17)and (11.18)into Eq.(11.16),we obtain -光管+光器 2=1 L=L Em+ (11.19) The transverse normal stresses in the composite o2,the matrix om2,and the fiber or layers are equal as follows: 02=0m2=0f2. (11.20) The matrix and fiber volume fractions are w= L 2Lm Um=L (11.21) By substituting Eqs.(11.20),(11.21),and e2 =02/E2 into Eq.(11.19),we obtain the transverse Young modulus E2= 品+)=（+） (11.22) 11.1.3 Longitudinal Shear Modulus G12 We consider Element 2.The length and the longitudinal shear modulus of the inner layer are Lr and Gn2 respectively,and the length and the shear modulus of the outer layers are Lm =(L-Lr)/2 and Gm.The element is subjected to a
11.1 RULE OF MIXTURES 439 11.1.2 Transverse Young Modulus E 2 We consider a rectangular element with sides Lmade up of three layers (Element 2, Fig. 11.3). The inner layer is a sheet of fiber with the same volume as the fiber bundle in Element 1. The length and the transverse (x2 direction) Young modulus of this inner layer are Lf and Ef2, respectively, and the length and the Young modulus of the outer layers are Lm = (L− Lf) /2 and Em. The element is subjected to a force F2. The force is distributed uniformly across the surface A(A= L2). The normal stress in the transverse direction is σ2 = F2/A, which can be expressed as σ2 = 2E2, (11.15) where E2 is the transverse Young modulus of the element and 2 is the average transverse normal strain 2 = L L . (11.16) The change in the length of the element is L = 2Lmm2 + Lff2, (11.17) where m2 and f2 are the transverse normal strains in the matrix and fiber layers. By neglecting the Poisson effect, we have m2 = σm2 Em f2 = σf2 Ef2 . (11.18) By introducing Eqs. (11.17) and (11.18) into Eq. (11.16), we obtain 2 = L L = 2Lm L σm2 Em + Lf L σf2 Ef2 . (11.19) The transverse normal stresses in the composite σ2, the matrix σm2, and the fiber σf2 layers are equal as follows: σ2 = σm2 = σf2. (11.20) The matrix and fiber volume fractions are vf = Lf L vm = 2Lm L . (11.21) By substituting Eqs. (11.20), (11.21), and 2 = σ2/E2 into Eq. (11.19), we obtain the transverse Young modulus E2 = vf Ef2 + vm Em −1 = vf Ef2 + 1 − vf Em −1 . (11.22) 11.1.3 Longitudinal Shear Modulus G12 We consider Element 2. The length and the longitudinal shear modulus of the inner layer are Lf and Gf12 respectively, and the length and the shear modulus of the outer layers are Lm = (L− Lf)/2 and Gm. The element is subjected to a
440 MICROMECHANICS /m 12 /m Figure 11.4:Element 2 subjected to a shear force (left)and deformation of the "top"(ijkl) surface(right). shear force F2(Fig.11.4)distributed uniformly across the surface A(A=L2). The shear stress t12 Fi2/Ais T12=12G12, (11.23) where G2 is the longitudinal shear modulus of the element and yi2 is the average shear strain, △L h2兰tanh2= (11.24) where AL is due to the shear deformations of the matrix and fiber layers, △L=2Lmym12+Lf12, (11.25) where ymi2 and yn2 are the shear strains in the matrix and fiber layers as follows: Ym12 Tm12 T12 Gm 12= (11.26) Gn2 The shear stresses in the composite,the matrix,and the fiber layers are equal: T12=Tm12=Tf12, (11.27) By combining Eqs.(11.21)and (11.23)-(11.27),we obtain the longitudinal shear modulus: G12= U (11.28) G12 Gm G12 Gm 11.1.4 Transverse Shear Modulus G23 We again consider Element 2 made up of three layers(Fig.11.4).The transverse shear modulus G23 of this element is derived in the same way as the longitudinal shear modulus G12.The result is G3= 1- (11.29) G3
440 MICROMECHANICS L L F12 ∆L m m f j k l i F12 Lf F12 L j k i l F12 F12 F12 f m m L Figure 11.4: Element 2 subjected to a shear force (left) and deformation of the “top” (ijkl) surface (right). shear force F12 (Fig. 11.4) distributed uniformly across the surface A (A = L2). The shear stress τ12 = F12/Ais τ12 = γ12G12, (11.23) where G12 is the longitudinal shear modulus of the element and γ12 is the average shear strain, γ12 ∼= tan γ12 = L L , (11.24) where L is due to the shear deformations of the matrix and fiber layers, L = 2Lmγm12 + Lfγf12, (11.25) where γm12 and γf12 are the shear strains in the matrix and fiber layers as follows: γm12 = τm12 Gm γf12 = τf12 Gf12 . (11.26) The shear stresses in the composite, the matrix, and the fiber layers are equal: τ12 = τm12 = τf12. (11.27) By combining Eqs. (11.21) and (11.23)–(11.27), we obtain the longitudinal shear modulus: G12 = vf Gf12 + vm Gm −1 = vf Gf12 + 1 − vf Gm −1 . (11.28) 11.1.4 Transverse Shear Modulus G 23 We again consider Element 2 made up of three layers (Fig. 11.4). The transverse shear modulus G23 of this element is derived in the same way as the longitudinal shear modulus G12. The result is G23 = vf Gf23 + 1 − vf Gm −1 . (11.29)
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
11.1 RULE OF MIXTURES 443 △L3 △L2 AL Figure 11.6:Deformation of Element 1 subjected to a uniform temperature change AT. 11.1.7 Thermal Expansion Coefficients When the unrestrained Element 1 is subjected to a AT temperature change,its volume changes(Fig.11.6).The new length,width,and height are (L+AL), (L+AL2),and (L+AL3),and the corresponding normal strains are LM e=L AL e2=L △L (11.45) The longitudinal and transverse @2,@3 thermal expansion coefficients are defined as a=1 T 应= △T a=3 (11.46) T Owing to symmetry,the transverse thermal expansion coefficients are equal (2=@3),and the transverse normal strains are equal (e2 =e3). Equations(11.45)and (11.46)refer to the element. For the fiber and the matrix,the strain-stress relationships are(Eqs.2.162 and 2.163 and Table 2.7,page 15) -ai△T=o (11.47) Ea em1-am△T=oml Em R-△T=E置ma-AT= 0m1 (11.48) Ea Em These equations are written with the assumptions that the longitudinal normal stress o dominates and is the only stress needed to be considered. Longitudinal thermal expansion coefficient.When Element 1 is subjected to a temperature change A T,in the x direction the fiber and the matrix elongate the same amount as the composite.Hence,the normal strains in the x direction are equal e1=ef1=eml· (11.49) From Eq.(11.47)the longitudinal normal stress (in the x direction)in the fiber and the matrix may be written as (Eqs.11.47 and 11.49) o0=(e1-a△T)Eom1=(e1-am△T)Em (11.50)
11.1 RULE OF MIXTURES 443 L ∆L1 L i j k L l ∆L2 ∆L3 x1 x3 x2 Figure 11.6: Deformation of Element 1 subjected to a uniform temperature change T. 11.1.7 Thermal Expansion Coefficients When the unrestrained Element 1 is subjected to a T temperature change, its volume changes (Fig. 11.6). The new length, width, and height are (L+ L1), (L+ L2), and (L+ L3), and the corresponding normal strains are 1 = L1 L 2 = L2 L 3 = L3 L . (11.45) The longitudinal α1 and transverse α2,α3 thermal expansion coefficients are defined as α1 = 1 T α2 = 2 T α3 = 3 T . (11.46) Owing to symmetry, the transverse thermal expansion coefficients are equal (α2 = α3), and the transverse normal strains are equal (2 = 3). Equations (11.45) and (11.46) refer to the element. For the fiber and the matrix, the strain–stress relationships are (Eqs. 2.162 and 2.163 and Table 2.7, page 15) f1 − αf1 T = σf1 Ef1 m1 − αm T = σm1 Em (11.47) f2 − αf2 T = −νf12 Ef1 σf1 m2 − αm T = −νm Em σm1. (11.48) These equations are written with the assumptions that the longitudinal normal stress σ1 dominates and is the only stress needed to be considered. Longitudinal thermal expansion coefficient. When Element 1 is subjected to a temperature change T, in the x direction the fiber and the matrix elongate the same amount as the composite. Hence, the normal strains in the x1 direction are equal 1 = f1 = m1. (11.49) From Eq. (11.47) the longitudinal normal stress (in the x1 direction) in the fiber and the matrix may be written as (Eqs. 11.47 and 11.49) σf1 = (1 − αf1 T)Ef1 σm1 = (1 − αm T)Em. (11.50)
444 MICROMECHANICS The total force acting on the ijkl face(Fig.11.6)is the sum of the forces acting on the fiber and the matrix: F=F+Fm. (11.51) Since there is no net force on the ijkl face,we write F+Fm of At om1 Am =0. (11.52) By introducing Eq.(11.50)into (11.52)and by noting that vr=Ar/A and vm Am/A,we obtain E1(Ervr+Emvm)-(an Envr+am Emvm)AT=0. (11.53) By combining Eqs.(11.46,left)and (11.53),and by noting that Envr EmUm E1(Eq.11.14),we obtain d1=En an VmEm dm (11.54) E E Transverse thermal expansion coefficient.When the temperature of Element 1 is changed by AT,the cross-sectional area of the face ijkl as well as the fiber and matrix areas change (Fig.11.5).The changes in these areas are(Eqs.11.33-11.34) △A=2Ae2△A=2Are2 △Am=2Amem2. (11.55) The change in the total area is the sum of the changes of the fiber and matrix areas: △A=△Ar+△Am (11.56) By combining Eqs.(11.55)and (11.56)and by noting that vr=Ar/Aand vm Am/A,we obtain △A E2-=Urep+Umem2. (11.57) From Eq.(11.48),the transverse normal strains in the fiber and the matrix may be written as (x2 direction) +a△T =-12E oml+am△T. ∈m2=-mEm (11.58) By introducing Egs.(11.47)into these expressions and observing that e1= en =Emi =AT,we obtain e2=-12(a1-an)△T+af2△Tem2=-m(a1-am)△T+am△T. (11.59) Equations (11.46,middle),(11.57),and (11.59)yield a2 vrar2 Umam vrvr2 (an-a1)+vmVm (am-a1), (11.60) where @i is given by Eq.(11.54)
444 MICROMECHANICS The total force acting on the ijkl face (Fig. 11.6) is the sum of the forces acting on the fiber and the matrix: F1 = Ff + Fm. (11.51) Since there is no net force on the ijkl face, we write Ff + Fm = σf1Af + σm1Am = 0. (11.52) By introducing Eq. (11.50) into (11.52) and by noting that vf = Af/A and vm = Am/A, we obtain 1 (Ef1vf + Emvm) − (αf1Ef1vf + αm Emvm) T = 0. (11.53) By combining Eqs. (11.46, left) and (11.53), and by noting that Ef1vf + Emvm = E1 (Eq. 11.14), we obtain α1 = vfEf1 E1 αf1 + vm Em E1 αm. (11.54) Transverse thermal expansion coefficient. When the temperature of Element 1 is changed by T, the cross-sectional area of the face ijkl as well as the fiber and matrix areas change (Fig. 11.5). The changes in these areas are (Eqs. 11.33–11.34) A= 2A2 Af = 2Aff2 Am = 2Amm2. (11.55) The change in the total area is the sum of the changes of the fiber and matrix areas: A = Af + Am. (11.56) By combining Eqs. (11.55) and (11.56) and by noting that vf = Af/Aand vm = Am/A, we obtain 2 = A 2A = vff2 + vmm2. (11.57) From Eq. (11.48), the transverse normal strains in the fiber and the matrix may be written as (x2 direction) f2 = −νf12 σf1 Ef1 + αf2 T m2 = −νm σm1 Em + αm T. (11.58) By introducing Eqs. (11.47) into these expressions and observing that 1 = f1 = m1 = α1 T, we obtain f2 = −νf12(α1 − αf1) T + αf2 T m2 = −νm(α1 − αm) T + αm T. (11.59) Equations (11.46, middle), (11.57), and (11.59) yield α2 = vfαf2 + vmαm + vfνf12 (αf1 − α1) + vmνm (αm − α1), (11.60) where α1 is given by Eq. (11.54).
11.1 RULE OF MIXTURES 445 11.1.8 Moisture Expansion Coefficients We assume that the moisture content of the interface is negligible.Further,we consider composites in which the fibers do not absorb moisture and the moisture is uniformly distributed throughout the matrix.The moisture content of Element 1 is the sum of the moisture contents of the fiber and the matrix cV crV+cmVm, (11.61) where Vis the volume,c the moisture concentration,and,as before,f and m refer to the fiber and the matrix.Since here we treat composites in which the fiber moisture content is zero(cr =0),we have c=CmV V .=CmUm. (11.62) The moisture-induced strains of Element 1 are 61=B1c2=B2ce3=B3c, (11.63) where B is the longitudinal and B2 and B3 are the transverse moisture expression coefficients.Owing to symmetry,the transverse moisture expansion coefficients are equal(B2=序). The moisture expansion coefficients of the fiber and the matrix are(i =1,2,3) Cr Cr c 序m-=m (11.64) Cm Cm c We now compare the definitions of the thermal and moisture expansion coef- ficients as follows: @i= △T 月，=与 du-AT (11.65) c @m= △7 2=0 From these equations we note the analogy between the thermal and moisture expansion coefficients.The temperature change AT corresponds to the moisture concentration c,and the thermal expansion coefficients correspond to the moisture expansion coefficients as follows: a→B i→Ba (11.66 am→
11.1 RULE OF MIXTURES 445 11.1.8 Moisture Expansion Coefficients We assume that the moisture content of the interface is negligible. Further, we consider composites in which the fibers do not absorb moisture and the moisture is uniformly distributed throughout the matrix. The moisture content of Element 1 is the sum of the moisture contents of the fiber and the matrix cV = cfVf + cmVm, (11.61) where V is the volume, c the moisture concentration, and, as before, f and m refer to the fiber and the matrix. Since here we treat composites in which the fiber moisture content is zero (cf = 0), we have c = cmVm V = cmvm. (11.62) The moisture-induced strains of Element 1 are 1 = β1c 2 = β2c 3 = β3c, (11.63) where β1 is the longitudinal and β2 and β3 are the transverse moisture expression coefficients. Owing to symmetry, the transverse moisture expansion coefficients are equal (β2 = β3). The moisture expansion coefficients of the fiber and the matrix are (i = 1, 2, 3) βfi = fi cf = c cf fi c βmi = mi cm = c cm mi c . (11.64) We now compare the definitions of the thermal and moisture expansion coef- ficients as follows: αi = i T βi = i c αfi = fi T cf c βfi = fi c αm = mi T cm c βm = mi c . (11.65) From these equations we note the analogy between the thermal and moisture expansion coefficients. The temperature change T corresponds to the moisture concentration c, and the thermal expansion coefficients correspond to the moisture expansion coefficients as follows: αi =⇒ βi αfi =⇒ cf c βfi αm =⇒ cm c βm. (11.66)