3 Elastic Constants Based on Micromechanics 3.1 Basic Equations The purpose of this chapter is to predict the material constants(also called elastic constants)of a composite material by studying the micromechanics of the problem,i.e.by studying how the matrix and fibers interact.These are the same material constants used in Chap.2 to calculate the compliance and stiffness matrices.Computing the stresses within the matrix,within the fiber, and at the interface of the matrix and fiber is very important for understand- ing some of the underlying failure mechanisms.In considering the fibers and surrounding matrix,we have the following assumptions[1]: 1.Both the matrix and fibers are linearly elastic. 2.The fibers are infinitely long. 3.The fibers are spaced periodically in square-packed or hexagonal packed arrays. There are three different approaches that are used to determine the elastic constants for the composite material based on micromechanics.These three approaches are [1]: 1.Using numerical models such as the finite element method. 2.Using models based on the theory of elasticity. 3.Using rule-of-mixtures models based on a strength-of-materials approach. Consider a unit cell in either a square-packed array (Fig.3.1)or a hexagonal-packed array (Fig.3.2)-see [1].The ratio of the cross-sectional area of the fiber to the total cross-sectional area of the unit cell is called the fiber volume fraction and is denoted by Vf.The fiber volume fraction satisfies the relation 0<Vf<1 and is usually 0.5 or greater.Similarly,the matrir volume fraction Vm is the ratio of the cross-sectional area of the matrix to the total cross-sectional area of the unit cell.Note that Vm also satisfies
3 Elastic Constants Based on Micromechanics 3.1 Basic Equations The purpose of this chapter is to predict the material constants (also called elastic constants) of a composite material by studying the micromechanics of the problem, i.e. by studying how the matrix and fibers interact. These are the same material constants used in Chap. 2 to calculate the compliance and stiffness matrices. Computing the stresses within the matrix, within the fiber, and at the interface of the matrix and fiber is very important for understanding some of the underlying failure mechanisms. In considering the fibers and surrounding matrix, we have the following assumptions [1]: 1. Both the matrix and fibers are linearly elastic. 2. The fibers are infinitely long. 3. The fibers are spaced periodically in square-packed or hexagonal packed arrays. There are three different approaches that are used to determine the elastic constants for the composite material based on micromechanics. These three approaches are [1]: 1. Using numerical models such as the finite element method. 2. Using models based on the theory of elasticity. 3. Using rule-of-mixtures models based on a strength-of-materials approach. Consider a unit cell in either a square-packed array (Fig. 3.1) or a hexagonal-packed array (Fig. 3.2) – see [1]. The ratio of the cross-sectional area of the fiber to the total cross-sectional area of the unit cell is called the fiber volume fraction and is denoted by V f . The fiber volume fraction satisfies the relation 0 < V f < 1 and is usually 0.5 or greater. Similarly, the matrix volume fraction V m is the ratio of the cross-sectional area of the matrix to the total cross-sectional area of the unit cell. Note that V m also satisfies
26 3 Elastic Constants Based on Micromechanics 3 Unit Cell Fig.3.1.A unit cell in a square-packed array of fiber-reinforced composite material 0<Vm<1.The following relation can be shown to exist between Vf and vm: vf+Vm =1 (3.1) In the above,we use the notation that a superscript m indicates a matrix quantity while a superscript f indicates a fiber quantity.In addition,the matrix material is assumed to be isotropic so that Em=E2=Em and v2=".However,the fiber material is assumed to be only transversely isotropic such that E=E2,vis =v12,and vis=vs2 =vf. Using the strength-of-materials approach and the simple rule of mixtures, we have the following relations for the elastic constants of the composite ma- terial (see [1).For Young's modulus in the 1-direction (also called the longi- tudinal stiffness),we have the following relation: E=EV+Emvm (3.2) where Ef is Young's modulus of the fiber in the 1-direction while Em is Young's modulus of the matrix.For Poisson's ratio v12,we have the following relation: h2=v2V时+vmVm (3.3)
26 3 Elastic Constants Based on Micromechanics Fig. 3.1. A unit cell in a square-packed array of fiber-reinforced composite material 0 < V m < 1. The following relation can be shown to exist between V f and V m: V f + V m = 1 (3.1) In the above, we use the notation that a superscript m indicates a matrix quantity while a superscript f indicates a fiber quantity. In addition, the matrix material is assumed to be isotropic so that Em 1 = Em 2 = Em and νm 12 = νm. However, the fiber material is assumed to be only transversely isotropic such that Ef 3 = Ef 2 , νf 13 = νf 12, and νf 23 = νf 32 = νf . Using the strength-of-materials approach and the simple rule of mixtures, we have the following relations for the elastic constants of the composite material (see [1]). For Young’s modulus in the 1-direction (also called the longitudinal stiffness), we have the following relation: E1 = Ef 1 V f + EmV m (3.2) where Ef 1 is Young’s modulus of the fiber in the 1-direction while Em is Young’s modulus of the matrix. For Poisson’s ratio ν12, we have the following relation: ν12 = νf 12V f + νmV m (3.3)
3.1 Basic Equations 27 Unit Cell Fig.3.2.A unit cell in a hexagonal-packed array of fiber-reinforced composite material where and are Poisson's ratios for the fiber and matrix,respectively. For Young's modulus in the 2-direction (also called the transverse stiffness), we have the following relation: 1 vf vm 瓦可+D (3.4) where E is Young's modulus of the fiber in the 2-direction while Em is Young's modulus of the matrix.For the shear modulus G12,we have the following relation: 1 vf vm (3.5) where G and Gare the shear moduli of the fiber and matrix,respectively. For the coefficients of thermal expansion a and a2 (see Problem 2.9),we have the following relations: afEiv/+amEmym (3.6) EfVf+EmVm g=g-()4a-av网 +e+()a-v (3.7) where of and ag are the coefficients of thermal expansion for the fiber in the 1- and 2-directions,respectively,and am is the coefficient of thermal expansion
3.1 Basic Equations 27 Fig. 3.2. A unit cell in a hexagonal-packed array of fiber-reinforced composite material where νf 12 and νm are Poisson’s ratios for the fiber and matrix, respectively. For Young’s modulus in the 2-direction (also called the transverse stiffness), we have the following relation: 1 E2 = V f Ef 2 + V m Em (3.4) where Ef 2 is Young’s modulus of the fiber in the 2-direction while Em is Young’s modulus of the matrix. For the shear modulus G12, we have the following relation: 1 G12 = V f Gf 12 + V m Gm (3.5) where Gf 12 and Gm are the shear moduli of the fiber and matrix, respectively. For the coefficients of thermal expansion α1 and α2 (see Problem 2.9), we have the following relations: α1 = αf 1Ef 1 V f + αmEmV m Ef 1 V f + EmV m (3.6) α2 = αf 2 − Em E1 νf 1 (αm − αf 1 )V m V f + αm + Ef 1 E1 νm(αm − αf 1 )V f V m (3.7) where αf 1 and αf 2 are the coefficients of thermal expansion for the fiber in the 1- and 2-directions, respectively, and αm is the coefficient of thermal expansion
28 3 Elastic Constants Based on Micromechanics for the matrix.However,we can use a simple rule-of-mixtures relation for a2 as follows: a2 agvf+amym (3.8) A similar simple rule-of-mixtures relation for o1 cannot be used simply because the matrix and fiber must expand or contract the same amount in the 1-direction when the temperature is changed. While the simple rule-of-mixtures models used above give accurate results for E and v12,the results obtained for E2 and Gi2 do not agree well with finite element analysis and elasticity theory results.Therefore,we need to modify the simple rule-of-mixtures models shown above.For E2,we have the following modified rule-of-mixtures formula: + Vf 1 Vf+nvm (3.9) where n is the stress-partitioning factor (related to the stress o2).This factor satisfies the relation 0<n<1 and is usually taken between 0.4 and 0.6. Another alternative rule-of-mixtures formula for E2 is given by: 1 nvf nmym EE以 +Em (3.10) where the factors n and nm are given by: Ev时+[1-以)Em+m以E]vm (3.11) ElV/+Emvm [（-vm）E-（1-wm)EmVJ+Emvm (3.12) ElV+Emvm The above alternative model for E2 gives accurate results and is used whenever the modified rule-of-mixtures model of(3.9)cannot be applied,i.e. when the factor n is not known. The modified rule-of-mixtures model for G12 is given by the following formula: vf 1 + G12 VI+nVm (3.13) where n'is the shear stress-partitioning factor.Note that n'satisfies the re- lation 0<n<1 but using n=0.6 gives results that correlate with the elasticity solution. Finally,the elasticity solution gives the following formula for Gi2: G12=Gm (Gm+G2)-V时(Gm-G2) (3.14) (Gm+Gi2)+V5(Gm-Gf2)
28 3 Elastic Constants Based on Micromechanics for the matrix. However, we can use a simple rule-of-mixtures relation for α2 as follows: α2 = αf 2V f + αmV m (3.8) A similar simple rule-of-mixtures relation for α1 cannot be used simply because the matrix and fiber must expand or contract the same amount in the 1-direction when the temperature is changed. While the simple rule-of-mixtures models used above give accurate results for E1 and ν12, the results obtained for E2 and G12 do not agree well with finite element analysis and elasticity theory results. Therefore, we need to modify the simple rule-of-mixtures models shown above. For E2, we have the following modified rule-of-mixtures formula: 1 E2 = V f Ef 2 + ηV m Em V f + ηV m (3.9) where η is the stress-partitioning factor (related to the stress σ2). This factor satisfies the relation 0 <η< 1 and is usually taken between 0.4 and 0.6. Another alternative rule-of-mixtures formula for E2 is given by: 1 E2 = ηfV f Ef 2 + ηmV m Em (3.10) where the factors ηf and ηm are given by: ηf = Ef 1 V f + 1 − νf 12νf 21 Em + νmνf 21Ef 1 V m Ef 1 V f + EmV m (3.11) ηm = 1 − νm2 Ef 1 − 1 − νmνf 12 Em V f + EmV m Ef 1 V f + EmV m (3.12) The above alternative model for E2 gives accurate results and is used whenever the modified rule-of-mixtures model of (3.9) cannot be applied, i.e. when the factor η is not known. The modified rule-of-mixtures model for G12 is given by the following formula: 1 G12 = V f Gf 12 + η V m Gm V f + η V m (3.13) where η is the shear stress-partitioning factor. Note that η satisfies the relation 0 < η < 1 but using η = 0.6 gives results that correlate with the elasticity solution. Finally, the elasticity solution gives the following formula for G12: G12 = Gm (Gm + Gf 12) − V f (Gm − Gf 12) (Gm + Gf 12) + V f (Gm − Gf 12) (3.14)
3.2 MATLAB Functions Used 29 3.2 MATLAB Functions Used The six MATLAB functions used in this chapter to calculate the elastic ma- terial constants are: E1(Vf,Elf,Em)-This function calculates the longitudinal Young's modulus E for the lamina.Its input consists of three arguments as illustrated in the listing below. NU12(Vf,NU12f,NUm)-This function calculates Poisson's ratio v12 for the lamina.Its input consists of three arguments as illustrated in the listing below. E2(Vf,E2f,Em,Eta,NU12f,NU21f,NUm,Elf,p)-This function calcu- lates the transverse Young's modulus E2 for the lamina.Its input consists of nine arguments as illustrated in the listing below.Use the value zero for any argument not needed in the calculations. G12(Vf,G12f,Gm,EtaPrime,p)-This function calculates the shear mod- ulus Gi2 for the lamina.Its input consists of five arguments as illustrated in the listing below.Use the value zero for any argument not needed in the calculations. Alphal(Vf,Elf,Em,Alphalf,Alpham)-This function calculates the co- efficient of thermal expansion a for the lamina.Its input consists of five arguments as illustrated in the listing below. Alpha2(Vf,Alpha2f,Alpham,E1,Elf,Em,NUIf,NUm,Alphalf,p)-This function calculates the coefficient of thermal expansion o2 for the lamina.Its input consists of ten arguments as illustrated in the listing below.Use the value zero for any argument not needed in the calculations. The following is a listing of the MATLAB source code for each function: function y E1(Vf,Eif,Em) %E1 This function returns Young's modulus in the % longitudinal direction.Its input are three values: % Vf -fiber volume fraction % Eif -longitudinal Young's modulus of the fiber 名 Em -Young's modulus of the matrix This function uses the simple rule-of-mixtures formula 名 of equation (3.2) Vm =1-Vf; y Vf*E1f Vm+Em; function y NU12(Vf,NU12f,NUm) %NU12 This function returns Poisson's ratio NU12 % Its input are three values: Vf -fiber volume fraction 名 NU12f -Poisson's ratio NU12 of the fiber % NUm-Poisson's ratio of the matrix
3.2 MATLAB Functions Used 29 3.2 MATLAB Functions Used The six MATLAB functions used in this chapter to calculate the elastic material constants are: E1 (Vf, E1f, Em) – This function calculates the longitudinal Young’s modulus E1 for the lamina. Its input consists of three arguments as illustrated in the listing below. NU12 (Vf, NU12f, NUm) – This function calculates Poisson’s ratio ν12 for the lamina. Its input consists of three arguments as illustrated in the listing below. E2 (Vf, E2f, Em, Eta, NU12f, NU21f, NUm, E1f, p) – This function calculates the transverse Young’s modulus E2 for the lamina. Its input consists of nine arguments as illustrated in the listing below. Use the value zero for any argument not needed in the calculations. G12 (Vf, G12f, Gm, EtaPrime, p) – This function calculates the shear modulus G12 for the lamina. Its input consists of five arguments as illustrated in the listing below. Use the value zero for any argument not needed in the calculations. Alpha1 (Vf, E1f, Em, Alpha1f, Alpham) – This function calculates the coefficient of thermal expansion α1 for the lamina. Its input consists of five arguments as illustrated in the listing below. Alpha2 (Vf, Alpha2f, Alpham, E1, E1f, Em, NU1f, NUm, Alpha1f, p) – This function calculates the coefficient of thermal expansion α2 for the lamina. Its input consists of ten arguments as illustrated in the listing below. Use the value zero for any argument not needed in the calculations. The following is a listing of the MATLAB source code for each function: function y = E1(Vf,E1f,Em) %E1 This function returns Young’s modulus in the % longitudinal direction. Its input are three values: % Vf - fiber volume fraction % E1f - longitudinal Young’s modulus of the fiber % Em - Young’s modulus of the matrix % This function uses the simple rule-of-mixtures formula % of equation (3.2) Vm = 1 - Vf; y = Vf*E1f + Vm*Em; function y = NU12(Vf,NU12f,NUm) %NU12 This function returns Poisson’s ratio NU12 % Its input are three values: % Vf - fiber volume fraction % NU12f - Poisson’s ratio NU12 of the fiber % NUm - Poisson’s ratio of the matrix
30 3 Elastic Constants Based on Micromechanics This function uses the simple rule-of-mixtures 名 formula of equation (3.3) Vm =1-Vf; y Vf*NU12f Vm*NUm; function y E2(Vf,E2f,Em,Eta,NU12f,NU21f,NUm,Eif,p) %E2 This function returns Young's modulus in the % transverse direction.Its input are nine values: % Vf fiber volume fraction 名 E2f- transverse Young's modulus of the fiber ” Em - Young's modulus of the matrix % Eta stress-partitioning factor % NU12f Poisson's ratio NU12 of the fiber % NU21f Poisson's ratio NU21 of the fiber % NUm -Poisson's ratio of the matrix % Eif -longitudinal Young's modulus of the fiber % ? parameter used to determine which equation to use: % p =1 -use equation (3.4) % p =2-use equation (3.9) % p =3 -use equation (3.10) % Use the value zero for any argument not needed % in the calculations. Vm 1-Vf; if p ==1 y 1/(Vf/E2f Vm/Em) elseif p ==2 y 1/((Vf/E2f Eta*Vm/Em)/(Vf Eta*Vm)); elseif p =3 deno Eif*Vf Em*Vm; etaf (E1f*Vf ((1-NU12f*NU21f)*Em NUm*NU21f*E1f)*Vm)/deno; etam =(((1-NUm*NUm)*E1f -(1-NUm*NU12f)*Em)*Vf Em*Vm)/deno; y =1/(etaf*Vf/E2f etam*Vm/Em); end function y =G12(Vf,G12f,Gm,EtaPrime,p) %G12 This function returns the shear modulus G12 Its input are five values: 名 Vf fiber volume fraction % G12f -shear modulus G12 of the fiber % Gm shear modulus of the matrix EtaPrime-shear stress-partitioning factor % P parameter used to determine which equation to use: % p=1-use equation (3.5) % p=2-use equation (3.13) % p=3-use equation (3.14) % Use the value zero for any argument not needed % in the calculations. Vm 1 Vf;
30 3 Elastic Constants Based on Micromechanics % This function uses the simple rule-of-mixtures % formula of equation (3.3) Vm = 1 - Vf; y = Vf*NU12f + Vm*NUm; function y = E2(Vf,E2f,Em,Eta,NU12f,NU21f,NUm,E1f,p) %E2 This function returns Young’s modulus in the % transverse direction. Its input are nine values: % Vf - fiber volume fraction % E2f - transverse Young’s modulus of the fiber % Em - Young’s modulus of the matrix % Eta - stress-partitioning factor % NU12f - Poisson’s ratio NU12 of the fiber % NU21f - Poisson’s ratio NU21 of the fiber % NUm - Poisson’s ratio of the matrix % E1f - longitudinal Young’s modulus of the fiber % p - parameter used to determine which equation to use: % p = 1 - use equation (3.4) % p = 2 - use equation (3.9) % p = 3 - use equation (3.10) % Use the value zero for any argument not needed % in the calculations. Vm = 1 - Vf; if p == 1 y = 1/(Vf/E2f + Vm/Em); elseif p == 2 y = 1/((Vf/E2f + Eta*Vm/Em)/(Vf + Eta*Vm)); elseif p == 3 deno = E1f*Vf + Em*Vm; etaf = (E1f*Vf + ((1-NU12f*NU21f)*Em + NUm*NU21f*E1f)*Vm) /deno; etam = (((1-NUm*NUm)*E1f - (1-NUm*NU12f)*Em)*Vf + Em*Vm) /deno; y = 1/(etaf*Vf/E2f + etam*Vm/Em); end function y = G12(Vf,G12f,Gm,EtaPrime,p) %G12 This function returns the shear modulus G12 % Its input are five values: % Vf - fiber volume fraction % G12f - shear modulus G12 of the fiber % Gm - shear modulus of the matrix % EtaPrime - shear stress-partitioning factor % p - parameter used to determine which equation to use: % p = 1 - use equation (3.5) % p = 2 - use equation (3.13) % p = 3 - use equation (3.14) % Use the value zero for any argument not needed % in the calculations. Vm = 1 - Vf;
3.2 MATLAB Functions Used 吧 if p==1 y =1/(Vf/G12f Vm/Gm); elseif p ==2 y =1/((Vf/G12f EtaPrime*Vm/Gm)/(Vf EtaPrime*Vm)); elseif p ==3 y Gm*((Gm +G12f)-Vf*(Gm -G12f))/((Gm G12f)+ Vf*(Gm G12f)) end function y Alpha1(Vf,Eif,Em,Alphaif,Alpham) lpha1 This function returns the coefficient of thermal % expansion in the longitudinal direction. Its input are five values: Vf -fiber volume fraction 名 E1f longitudinal Young's modulus of the fiber Em - Young's modulus of the matrix % Alphalf -coefficient of thermal expansion in the % 1-direction for the fiber % Alpham coefficient of thermal expansion for the matrix Vm 1-Vf; y (Vf+E1f+Alpha1f Vm*Em*Alpham)/(E1f*Vf Em*Vm); function y Alpha2(Vf,Alpha2f,Alpham,E1,E1f,Em,NU1f,NUm, Alphaif,p) %Alpha2 This function returns the coefficient of thermal % expansion in the transverse direction. % Its input are ten values: Vf fiber volume fraction % Alpha2f coefficient of thermal expansion in the % 2-direction for the fiber % Alpham -coefficient of thermal expansion for the matrix % E1 -longitudinal Young's modulus of the lamina E1f -longitudianl Young's modulus of the fiber 名 Em -Young's modulus of the matrix % NU1f -Poisson's ratio of the fiber NUm -Poisson's ratio of the matrix % Alphaif -coefficient of thermal expansion in the 名 1-direction p parameter used to determine which equation to use % p =1 -use equation (3.8) p =2 -use equation (3.7) Use the value zero for any argument not needed in % the calculation Vm 1 Vf; 1fp=1 y Vf*Alpha2f Vm*Alpham; elseif p ==2 y =(Alpha2f -(Em/E1)*NU1f*(Alpham -Alphalf)+Vm)*Vf+ (Alpham (Eif/E1)*NUm*(Alpham Alphaif)*Vf)*Vm; end
3.2 MATLAB Functions Used 31 if p == 1 y = 1/(Vf/G12f + Vm/Gm); elseif p == 2 y = 1/((Vf/G12f + EtaPrime*Vm/Gm)/(Vf + EtaPrime*Vm)); elseif p == 3 y = Gm*((Gm + G12f) - Vf*(Gm - G12f))/((Gm + G12f) + Vf*(Gm - G12f)); end function y = Alpha1(Vf,E1f,Em,Alpha1f,Alpham) %Alpha1 This function returns the coefficient of thermal % expansion in the longitudinal direction. % Its input are five values: % Vf - fiber volume fraction % E1f - longitudinal Young’s modulus of the fiber % Em - Young’s modulus of the matrix % Alpha1f - coefficient of thermal expansion in the % 1-direction for the fiber % Alpham - coefficient of thermal expansion for the matrix Vm = 1 - Vf; y = (Vf*E1f*Alpha1f + Vm*Em*Alpham)/(E1f*Vf + Em*Vm); function y = Alpha2(Vf,Alpha2f,Alpham,E1,E1f,Em,NU1f,NUm, Alpha1f,p) %Alpha2 This function returns the coefficient of thermal % expansion in the transverse direction. % Its input are ten values: % Vf - fiber volume fraction % Alpha2f - coefficient of thermal expansion in the % 2-direction for the fiber % Alpham - coefficient of thermal expansion for the matrix % E1 - longitudinal Young’s modulus of the lamina % E1f - longitudianl Young’s modulus of the fiber % Em - Young’s modulus of the matrix % NU1f - Poisson’s ratio of the fiber % NUm - Poisson’s ratio of the matrix % Alpha1f - coefficient of thermal expansion in the % 1-direction % p - parameter used to determine which equation to use % p = 1 - use equation (3.8) % p = 2 - use equation (3.7) % Use the value zero for any argument not needed in % the calculation Vm = 1 - Vf; if p == 1 y = Vf*Alpha2f + Vm*Alpham; elseif p == 2 y = (Alpha2f - (Em/E1)*NU1f*(Alpham - Alpha1f)*Vm)*Vf + (Alpham + (E1f/E1)*NUm*(Alpham - Alpha1f)*Vf)*Vm; end
32 3 Elastic Constants Based on Micromechanics Example 3.1 Derive the simple rule-of-mixtures formula for the calculation of the longitu- dinal modulus E given in (3.2). Solution Consider a longitudinal cross-section of length L of the fiber and matrix in a lamina as shown in Fig.3.3.Let Af and Am be the cross-sectional areas of the fiber and matrix,respectively.Let also F and Fm be the longitudinal forces in the fiber and matrix,respectively.Then we have the following relations: fiber Am matrix L Fig.3.3.A longitudinal cross-section of fiber-reinforced composite material for Example 3.1 Ff=afA! (3.15a) Fi=om Am (3.15b) where of and o are the longitudinal normal stresses in the fiber and matrix, respectively.These stresses are given in terms of the longitudinal strainsf and em as follows: of=Efef (3.16a) am =Emem (3.16b) where Ef is the longitudinal modulus of the fiber and Em is the modulus of the matrix
32 3 Elastic Constants Based on Micromechanics Example 3.1 Derive the simple rule-of-mixtures formula for the calculation of the longitudinal modulus E1 given in (3.2). Solution Consider a longitudinal cross-section of length L of the fiber and matrix in a lamina as shown in Fig. 3.3. Let Af and Am be the cross-sectional areas of the fiber and matrix, respectively. Let also Ff 1 and F m 1 be the longitudinal forces in the fiber and matrix, respectively. Then we have the following relations: Fig. 3.3. A longitudinal cross-section of fiber-reinforced composite material for Example 3.1 Ff 1 = σf 1 Af (3.15a) F m 1 = σm 1 Am (3.15b) where σf 1 and σm 1 are the longitudinal normal stresses in the fiber and matrix, respectively. These stresses are given in terms of the longitudinal strains ε f 1 and εm 1 as follows: σf 1 = Ef 1 ε f 1 (3.16a) σm 1 = Emεm 1 (3.16b) where Ef 1 is the longitudinal modulus of the fiber and Em is the modulus of the matrix
3.2 MATLAB Functions Used 33 Let F be the total longitudinal force in the lamina where F is given by: F1=01A (3.17） where o1 is the total longitudinal normal stress in the lamina and A is the total cross-sectional area of the lamina.The total longitudinal normal stress o is given by: 01=E1E1 (3.18) However,using force equilibrium,it is clear that we have the following relation between the total longitudinal force and the longitudinal forces in the fiber and matrix: F=F+Fm (3.19) Substituting (3.15a,b)and (3.17)into (3.19),then substituting (3.16a,b)and (3.18)into the resulting equation,we obtain the following relation: E1E1A=EfEAf+Emem Am (3.20) Next,we use the compatibility condition e=em=since the matrix,fiber, and lamina all have the same strains.Equation(3.20)is simplified as follows: EA=E:A+Em Am (3.21) Finally,we divide (3.21)by A and note that Af/A=Vf and A"/A Vm to obtain the required formula for E as follows: E1=EV+Emvm (3.22) MATLAB Example 3.2 Consider a graphite-reinforced polymer composite lamina with the following material properties for the matrix and fibers [1]: Em=4.62GPa,ym=0.360 E=233GPa,2=0.200 E=23.1GPa,=0.400 G2=8.96GPaG3=8.27GPa Use MATLAB and the simple rule-of-mixtures formulas to calculate the values of the four elastic constants E1,v12,E2,and G12 for the lamina.Use Vf=0.6
3.2 MATLAB Functions Used 33 Let F1 be the total longitudinal force in the lamina where F1 is given by: F1 = σ1A (3.17) where σ1 is the total longitudinal normal stress in the lamina and A is the total cross-sectional area of the lamina. The total longitudinal normal stress σ1 is given by: σ1 = E1ε1 (3.18) However, using force equilibrium, it is clear that we have the following relation between the total longitudinal force and the longitudinal forces in the fiber and matrix: F1 = Ff 1 + F m 1 (3.19) Substituting (3.15a,b) and (3.17) into (3.19), then substituting (3.16a,b) and (3.18) into the resulting equation, we obtain the following relation: E1ε1A = Ef 1 ε f 1Af + Emεm 1 Am (3.20) Next, we use the compatibility condition ε f 1 = εm 1 = ε1 since the matrix, fiber, and lamina all have the same strains. Equation (3.20) is simplified as follows: E1A = Ef 1 Af + EmAm (3.21) Finally, we divide (3.21) by A and note that Af /A = V f and Am/A = V m to obtain the required formula for E1 as follows: E1 = Ef 1 V f + EmV m (3.22) MATLAB Example 3.2 Consider a graphite-reinforced polymer composite lamina with the following material properties for the matrix and fibers [1]: Em = 4.62 GPa, νm = 0.360 Ef 1 = 233 GPa, νf 12 = 0.200 Ef 2 = 23.1 GPa, νf 23 = 0.400 Gf 12 = 8.96 GPa Gf 23 = 8.27 GPa Use MATLAB and the simple rule-of-mixtures formulas to calculate the values of the four elastic constants E1, ν12, E2, and G12 for the lamina. Use V f = 0.6
34 3 Elastic Constants Based on Micromechanics Solution This example is solved using MATLAB.First,the MATLAB function E1 is used to calculate the longitudinal modulus E in GPa as follows: >E1(0.6,233,4.62) ans 141.6480 Poisson's ratio v2 is then calculated using the MATLAB function NU12 as follows: >>NU12(0.6,0.200,0.360) ans 0.2640 The transverse modulus E2 is then calculated in GPa using the MATLAB function E2 as follows (note that we use the value zero for each parameter not needed in the calculations): >E2(0.6,23.1,4.62,0,0,0,0,0,1) ans 8.8846 The shear modulus for the matrix Gm is calculated in GPa using (2.8)as follows: >Gm=4.62/(2*(1+0.360)) Gm 1.6985 Finally,the shear modulus Gi2 of the lamina is calculated in GPa using the MATLAB function G12 as follows: >G12(0.6,8.96,Gm,0,1) ans 3.3062 Note that and G are not used in this example
34 3 Elastic Constants Based on Micromechanics Solution This example is solved using MATLAB. First, the MATLAB function E1 is used to calculate the longitudinal modulus E1 in GPa as follows: >> E1(0.6, 233, 4.62) ans = 141.6480 Poisson’s ratio ν12 is then calculated using the MATLAB function NU12 as follows: >> NU12(0.6, 0.200, 0.360) ans = 0.2640 The transverse modulus E2 is then calculated in GPa using the MATLAB function E2 as follows (note that we use the value zero for each parameter not needed in the calculations): >> E2(0.6, 23.1, 4.62, 0, 0, 0, 0, 0, 1) ans = 8.8846 The shear modulus for the matrix Gm is calculated in GPa using (2.8) as follows: >> Gm = 4.62/(2*(1 + 0.360)) Gm = 1.6985 Finally, the shear modulus G12 of the lamina is calculated in GPa using the MATLAB function G12 as follows: >> G12(0.6, 8.96, Gm, 0, 1) ans = 3.3062 Note that νf 23 and Gf 23 are not used in this example