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)