228 Mechanics of Composite Materials,Second Edition and (3.44c） where m=transverse strains in composite,fiber,and matrix,respectively. Substituting Equation (3.44)in Equation(3.43), teee =trer+tnem. (3.45) The Poisson's ratios for the fiber,matrix,and composite,respectively,are Vi=-- (3.46a) (3.46b) and V12=- (3.46c) Substituting in Equation (3.45), -tevizec =-tivjer-tmVmem (3.47 where V.=Poisson's ratio of composite,fiber,and matrix,respectively e.m=longitudinal strains of composite,fiber and matrix,respec- tively However,the strains in the composite,fiber,and matrix are assumed to be the equal in the longitudinal direction(e==),which,from Equation (3.47),gives tevn2=tivf+tmVmr Vn=Vs te t (3.48) 2006 by Taylor Francis Group,LLC
228 Mechanics of Composite Materials, Second Edition and (3.44c) where εc,f,m = transverse strains in composite, fiber, and matrix, respectively. Substituting Equation (3.44) in Equation (3.43), (3.45) The Poisson’s ratios for the fiber, matrix, and composite, respectively, are (3.46a) (3.46b) and . (3.46c) Substituting in Equation (3.45), (3.47) where v12,f,m = Poisson’s ratio of composite, fiber, and matrix, respectively = longitudinal strains of composite, fiber and matrix, respectively However, the strains in the composite, fiber, and matrix are assumed to be the equal in the longitudinal direction , which, from Equation (3.47), gives (3.48) ε δ c T c T c t = , ttt c c T f f T m m T εεε = + . ν ε ε f f T f L = − , ν ε ε m m T m L = − , ν ε ε 12 = − c T c L − =− − t tt c c L fff L m mm L ν ε νε νε 12 , ν12, , f m εcfm L , , ( ) εεε c L f L m L = = t tt c f f mm ν νν 12 = + , νν ν 12 = + f f c m m c t t t t . 1343_book.fm Page 228 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC
Micromechanical Analysis of a Lamina 229 Because the thickness fractions are the same as the volume fractions,per Equation (3.28), V12=ViVt+VmVm. (3.49) Example 3.5 Find the major and minor Poisson's ratio of a glass/epoxy lamina with a 70%fiber volume fraction.Use the properties of glass and epoxy from Table 3.1 and Table 3.2,respectively. Solution From Table 3.1,the Poisson's ratio of the fiber is V=0.2. From Table 3.2,the Poisson's ratio of the matrix is Vm=0.3. Using Equation(3.49),the major Poisson's ratio is v12=(0.2)0.7)+(0.3)0.3) =0.230. From Example 3.3,the longitudinal Young's modulus is E1=60.52GPa and,from Example 3.4,the transverse Young's modulus is E2=10.37GPa. Then,the minor Poisson's ratio from Equation(2.83)is E2 Va=V2 Er 10.37 =0.230 60.52 =0.03941. 3.3.1.4 In-Plane Shear Modulus Apply a pure shear stress te to a lamina as shown in Figure 3.12.The fibers and matrix are represented by rectangular blocks as shown.The resulting 2006 by Taylor Francis Group,LLC
Micromechanical Analysis of a Lamina 229 Because the thickness fractions are the same as the volume fractions, per Equation (3.28), (3.49) Example 3.5 Find the major and minor Poisson’s ratio of a glass/epoxy lamina with a 70% fiber volume fraction. Use the properties of glass and epoxy from Table 3.1 and Table 3.2, respectively. Solution From Table 3.1, the Poisson’s ratio of the fiber is νf = 0.2. From Table 3.2, the Poisson’s ratio of the matrix is νm = 0.3. Using Equation (3.49), the major Poisson’s ratio is From Example 3.3, the longitudinal Young’s modulus is E1 = 60.52 GPa and, from Example 3.4, the transverse Young’s modulus is E2 = 10.37 GPa. Then, the minor Poisson’s ratio from Equation (2.83) is 3.3.1.4 In-Plane Shear Modulus Apply a pure shear stress τc to a lamina as shown in Figure 3.12. The fibers and matrix are represented by rectangular blocks as shown. The resulting νν ν 12 = + f f mm V V . ν12 02 07 03 03 0 230 = + = ( . )( . ) ( . )( . ) . . ν ν 21 12 2 1 0 230 10 37 60 52 0 03941 = = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = E E . . . . . 1343_book.fm Page 229 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC
230 Mechanics of Composite Materials,Second Edition t/2 t/2 FIGURE 3.12 An in-plane shear stress applied to a representative volume element for finding in-plane shear modulus of a unidirectional lamina. shear deformations of the composite 8 the fiber 8,and the matrix 8m are related by δe=δf+δm (3.50) From the definition of shear strains, δe=Yte, (3.51a) δr=Yy (3.51b) and δm=Ymtm (3.51c) where Y=shearing strains in the composite,fiber,and matrix,respec- tively =thickness of the composite,fiber,and matrix,respectively. From Hooke's law for the fiber,the matrix,and the composite, Yc= (3.52a) G12 Y-G (3.52b) and 2006 by Taylor Francis Group,LLC
230 Mechanics of Composite Materials, Second Edition shear deformations of the composite δc the fiber δf, and the matrix δm are related by . (3.50) From the definition of shear strains, , (3.51a) , (3.51b) and , (3.51c) where γc,f,m = shearing strains in the composite, fiber, and matrix, respectively tc,f,m = thickness of the composite, fiber, and matrix, respectively. From Hooke’s law for the fiber, the matrix, and the composite, (3.52a) (3.52b) and FIGURE 3.12 An in-plane shear stress applied to a representative volume element for finding in-plane shear modulus of a unidirectional lamina. t m/2 t m/2 t r t c h τc τc δδ δ cfm = + δ γ c cc = t δ γ f ff = t δ γ m mm = t γ τ c c G= 12 , γ τ f f Gf = , 1343_book.fm Page 230 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC
Micromechanical Analysis of a Lamina 231 Ym= (3.52c) where Gi2m=shear moduli of composite,fiber,and matrix,respectively. From Equation(3.50)through Equation(3.52), Gm (3.53) The shear stresses in the fiber,matrix,and composite are assumed to be equal (te=t=tm),giving 1=1上+1血 (3.54) G12 Gy te Gm te Because the thickness fractions are equal to the volume fractions,per Equation (3.28), 1-y+y. (3.55) G12 G1 Gm Example 3.6 Find the in-plane shear modulus of a glass/epoxy lamina with a 70%fiber volume fraction.Use properties of glass and epoxy from Table 3.1 and Table 3.2,respectively. Solution The glass fibers and the epoxy matrix have isotropic properties.From Table 3.1,the Young's modulus of the fiber is E=85 GPa and the Poisson's ratio of the fiber is y=0.2 The shear modulus of the fiber G Er 2(1+vr) 85 =21+0.2) =35.42GPa. 2006 by Taylor Francis Group,LLC
Micromechanical Analysis of a Lamina 231 (3.52c) where G12,f,m = shear moduli of composite, fiber, and matrix, respectively. From Equation (3.50) through Equation (3.52), (3.53) The shear stresses in the fiber, matrix, and composite are assumed to be equal (τc = τf = τm), giving (3.54) Because the thickness fractions are equal to the volume fractions, per Equation (3.28), (3.55) Example 3.6 Find the in-plane shear modulus of a glass/epoxy lamina with a 70% fiber volume fraction. Use properties of glass and epoxy from Table 3.1 and Table 3.2, respectively. Solution The glass fibers and the epoxy matrix have isotropic properties. From Table 3.1, the Young’s modulus of the fiber is Ef = 85 GPa and the Poisson’s ratio of the fiber is νf = 0.2. The shear modulus of the fiber γ τ m m Gm = , τ τ τ c c f f f m m m G t G t G t 12 = + . 11 1 G G 12 t t G t t f f c m m c = + . 1 G12 V G V G f f m m = + . G E GPa f f f = + = + = 2 1 85 21 02 35 42 ( ) ( .) . . ν 1343_book.fm Page 231 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC
232 Mechanics of Composite Materials,Second Edition From Table 3.2,the Young's modulus of the matrix is E=3.4 GPa and the Poisson's ratio of the fiber is Vm=0.3. The shear modulus of the matrix is Em Gm=21+vm） 3.40 2(1+0.3) =1.308GPa. From Equation (3.55),the in-plane shear modulus of the unidirectional lamina is 10.70,0.30 GL2 35.421.308 G2=4.014GPa. Figure 3.13a and Figure 3.13b show the analytical values from Equation (3.55)of the in-plane shear modulus as a function of fiber volume fraction for a typical glass/epoxy lamina.Experimental valuest are also plotted in the same figure. 3.3.2 Semi-Empirical Models The values obtained for transverse Young's modulus and in-plane shear modulus through Equation (3.41)and Equation(3.55),respectively,do not agree well with the experimental results shown in Figure 3.10 and Figure 3.13.This establishes a need for better modeling techniques.These tech- niques include numerical methods,such as finite element and finite differ- ence,and boundary element methods,elasticity solution,and variational principal models.5 Unfortunately,these models are available only as compli- cated equations or in graphical form.Due to these difficulties,semi-empirical models have been developed for design purposes.The most useful of these models include those of Halphin and Tsai because they can be used over a wide range of elastic properties and fiber volume fractions. Halphin and Tsai developed their models as simple equations by curve fitting to results that are based on elasticity.The equations are semi-empirical in nature because involved parameters in the curve fitting carry physical meaning 2006 by Taylor Francis Group,LLC
232 Mechanics of Composite Materials, Second Edition From Table 3.2, the Young’s modulus of the matrix is Em = 3.4 GPa and the Poisson’s ratio of the fiber is νm = 0.3. The shear modulus of the matrix is From Equation (3.55), the in-plane shear modulus of the unidirectional lamina is Figure 3.13a and Figure 3.13b show the analytical values from Equation (3.55) of the in-plane shear modulus as a function of fiber volume fraction for a typical glass/epoxy lamina. Experimental values4 are also plotted in the same figure. 3.3.2 Semi-Empirical Models The values obtained for transverse Young’s modulus and in-plane shear modulus through Equation (3.41) and Equation (3.55), respectively, do not agree well with the experimental results shown in Figure 3.10 and Figure 3.13. This establishes a need for better modeling techniques. These techniques include numerical methods, such as finite element and finite difference, and boundary element methods, elasticity solution, and variational principal models.5 Unfortunately, these models are available only as complicated equations or in graphical form. Due to these difficulties, semi-empirical models have been developed for design purposes. The most useful of these models include those of Halphin and Tsai6 because they can be used over a wide range of elastic properties and fiber volume fractions. Halphin and Tsai6 developed their models as simple equations by curve fitting to results that are based on elasticity. The equations are semi-empirical in nature because involved parameters in the curve fitting carry physical meaning. G E GPa m m m = + = + = 2 1 3 40 21 03 1 308 ( ) . ( .) . . ν 1 0 70 35 42 0 30 1 308 4 014 12 12 G G GPa = + = . . . . . . 1343_book.fm Page 232 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC