9 Effective Elastic Constants of a Laminate 9.1 Basic Equations In this chapter,we introduce the concept of effective elastic constants for the lam- inate.These constants are the effective extensional modulus in the x direction Er, the effective extensional modulus in the y direction Ey,the effective Poisson's ratios Dy and Du,and the effective shear modulus in the r-y plane Gry The effective elastic constants are usually defined when considering the inplane loading of symmetric balanced laminates.In the following equations,we consider only symmetric balanced or symmetric cross-ply laminates.We therefore define the following three average laminate stresses [1]: 1 CH/2 二百J-HP (9.1) 1 H/2 y=百J-H2 Oudz (9.2 1 H/2 y=百J-H2 (9.3) where H is the thickness of the laminate.Comparing (9.1),(9.2),and (9.3)with (7.13),we obtain the following relations between the average stresses and the force resultants: -言N (9.4) ,=疗N (9.5) fy=五 (9.6) Solving (9.4),(9.5),and (9.6)for N:,Ny,and Nry,and substituting the results into (8.11)and(8.12)for symmetric balanced laminates,we obtain:
9 Effective Elastic Constants of a Laminate 9.1 Basic Equations In this chapter, we introduce the concept of effective elastic constants for the laminate. These constants are the effective extensional modulus in the x direction E¯x, the effective extensional modulus in the y direction E¯y, the effective Poisson’s ratios ν¯xy and ¯νyx, and the effective shear modulus in the x-y plane G¯xy. The effective elastic constants are usually defined when considering the inplane loading of symmetric balanced laminates. In the following equations, we consider only symmetric balanced or symmetric cross-ply laminates. We therefore define the following three average laminate stresses [1]: σ¯x = 1 H H/2 −H/2 σxdz (9.1) σ¯y = 1 H H/2 −H/2 σydz (9.2) τ¯xy = 1 H H/2 −H/2 τxydz (9.3) where H is the thickness of the laminate. Comparing (9.1), (9.2), and (9.3) with (7.13), we obtain the following relations between the average stresses and the force resultants: σ¯x = 1 H Nx (9.4) σ¯y = 1 H Ny (9.5) τ¯xy = 1 H Nxy (9.6) Solving (9.4), (9.5), and (9.6) for Nx, Ny, and Nxy, and substituting the results into (8.11) and (8.12) for symmetric balanced laminates, we obtain:
170 9 Effective Elastic Constants of a Laminate anH a12H 0 a12H a22H 0 (9.7) 0 0 a66 H The above 3 x 3 matrix is defined as the laminate compliance matrir for sym- metric balanced laminates.Therefore,by analogy with (4.5),we obtain the following effective elastic constants for the laminate: E:= 1 anH (9.8a) Ev= 1 a22H (9.8b) Gav= 1 (9.8c) a66H y=-012 (9.8d) a11 iyx=一 412 (9.8e) a22 It is clear from the above equations that Dry and Dyr are not independent and are related by the following reciprocity relation: 型=华 E Ev (9.9) Finally,we note that the expressions of the effective elastic constants of(9.8) can be re-written in terms of the components Aij of the matrix [A]as shown in Example 9.1. 9.2 MATLAB Functions Used The five MATLAB function used in this chapter to calculate the average laminate elastic constants are: Ebarr(A,H)-This function calculates the average laminate modulus in the r- direction Ez.There are two input arguments to this function-they are the thickness of the laminate H and the 3 x 3 stiffness matrix [A]for balanced symmetric lami- nates.The function returns a scalar quantity which the desired modulus. Ebary(A,H)-This function calculates the average laminate modulus in the y- direction Ey.There are two input arguments to this function-they are the thickness of the laminate H and the 3 x 3 stiffness matrix [A]for balanced symmetric lami- nates.The function returns a scalar quantity which the desired modulus. NUbarry(A,H)-This function calculates the average laminate Poisson's ratio Dy. There are two input arguments to this function-they are the thickness of the laminate H and the 3 x 3 stiffness matrix [A]for balanced symmetric laminates. The function returns a scalar quantity which the desired Poisson's ratio. NUbaryr(A,H)-This function calculates the average laminate Poisson's ratio uz. There are two input arguments to this function-they are the thickness of the
170 9 Effective Elastic Constants of a Laminate ⎧ ⎪⎨ ⎪⎩ ε0 x ε0 y γ0 xy ⎫ ⎪⎬ ⎪⎭ = ⎡ ⎢ ⎣ a11H a12H 0 a12H a22H 0 0 0 a66H ⎤ ⎥ ⎦ ⎧ ⎪⎨ ⎪⎩ σ¯x σ¯y τ¯xy ⎫ ⎪⎬ ⎪⎭ (9.7) The above 3 × 3 matrix is defined as the laminate compliance matrix for symmetric balanced laminates. Therefore, by analogy with (4.5), we obtain the following effective elastic constants for the laminate: E¯x = 1 a11H (9.8a) E¯y = 1 a22H (9.8b) G¯xy = 1 a66H (9.8c) ν¯xy = −a12 a11 (9.8d) ν¯yx = −a12 a22 (9.8e) It is clear from the above equations that ¯νxy and ¯νyx are not independent and are related by the following reciprocity relation: ν¯xy E¯x = ν¯yx E¯y (9.9) Finally, we note that the expressions of the effective elastic constants of (9.8) can be re-written in terms of the components Aij of the matrix [A] as shown in Example 9.1. 9.2 MATLAB Functions Used The five MATLAB function used in this chapter to calculate the average laminate elastic constants are: Ebarx(A, H) – This function calculates the average laminate modulus in the xdirection E¯x. There are two input arguments to this function – they are the thickness of the laminate H and the 3 × 3 stiffness matrix [A] for balanced symmetric laminates. The function returns a scalar quantity which the desired modulus. Ebary(A, H) – This function calculates the average laminate modulus in the ydirection E¯y. There are two input arguments to this function – they are the thickness of the laminate H and the 3 × 3 stiffness matrix [A] for balanced symmetric laminates. The function returns a scalar quantity which the desired modulus. NUbarxy(A, H) – This function calculates the average laminate Poisson’s ratio ¯νxy. There are two input arguments to this function – they are the thickness of the laminate H and the 3 × 3 stiffness matrix [A] for balanced symmetric laminates. The function returns a scalar quantity which the desired Poisson’s ratio. NUbaryx(A, H) – This function calculates the average laminate Poisson’s ratio ¯νyx. There are two input arguments to this function – they are the thickness of the
9.2 MATLAB Functions Used 171 laminate H and the 3 x 3 stiffness matrix [A]for balanced symmetric laminates. The function returns a scalar quantity which the desired Poisson's ratio. Gbarry(A,H)-This function calculates the average laminate shear modulus G There are two input arguments to this function-they are the thickness of the laminate H and the 3 x 3 stiffness matrix [A]for balanced symmetric laminates. The function returns a scalar quantity which the desired shear modulus. The following is a listing of the MATLAB source code for these functions: function y Ebarx(A,H) %Ebarx This function returns the average laminate modulus in the x-direction.Its input are two arguments: 名 A -3 x 3 stiffness matrix for balanced symmetric laminates. 名 H-thickness of laminate a inv(A) y=1/(H*a(1,1)); function y Ebary(A,H) %Ebary This function returns the average laminate modulus % in the y-direction.Its input are two arguments: % A -3 x 3 stiffness matrix for balanced symmetric % laminates. H-thickness of laminate a inv(A); y=1/(但*a(2,2); function y NUbarxy(A,H) %NUbarxy This function returns the average laminate Poisson's ratio NUxy.Its input are two arguments: % A-3 x 3 stiffness matrix for balanced symmetric % laminates. H-thickness of laminate a inv(A); y=-a(1,2)/a(1,1); function y NUbaryx(A,H) %NUbaryx This function returns the average laminate 名 Poisson's ratio NUyx.Its input are two arguments: % A-3 x 3 stiffness matrix for balanced symmetric laminates. 名 H-thickness of laminate a inv(A); y=-a(1,2)/a(2,2); function y Gbarxy(A,H) %Gbarxy This function returns the average laminate shear 名 modulus.Its input are two arguments: A -3 x 3 stiffness matrix for balanced symmetric
9.2 MATLAB Functions Used 171 laminate H and the 3 × 3 stiffness matrix [A] for balanced symmetric laminates. The function returns a scalar quantity which the desired Poisson’s ratio. Gbarxy(A, H) – This function calculates the average laminate shear modulus G¯xy. There are two input arguments to this function – they are the thickness of the laminate H and the 3 × 3 stiffness matrix [A] for balanced symmetric laminates. The function returns a scalar quantity which the desired shear modulus. The following is a listing of the MATLAB source code for these functions: function y = Ebarx(A,H) %Ebarx This function returns the average laminate modulus % in the x-direction. Its input are two arguments: % A -3x3 stiffness matrix for balanced symmetric % laminates. % H - thickness of laminate a = inv(A); y = 1/(H*a(1,1)); function y = Ebary(A,H) %Ebary This function returns the average laminate modulus % in the y-direction. Its input are two arguments: % A -3x3 stiffness matrix for balanced symmetric % laminates. % H - thickness of laminate a = inv(A); y = 1/(H*a(2,2)); function y = NUbarxy(A,H) %NUbarxy This function returns the average laminate % Poisson’s ratio NUxy. Its input are two arguments: % A -3x3 stiffness matrix for balanced symmetric % laminates. % H - thickness of laminate a = inv(A); y = -a(1,2)/a(1,1); function y = NUbaryx(A,H) %NUbaryx This function returns the average laminate % Poisson’s ratio NUyx. Its input are two arguments: % A -3x3 stiffness matrix for balanced symmetric % laminates. % H - thickness of laminate a = inv(A); y = -a(1,2)/a(2,2); function y = Gbarxy(A,H) %Gbarxy This function returns the average laminate shear % modulus. Its input are two arguments: % A -3x3 stiffness matrix for balanced symmetric
172 9 Effective Elastic Constants of a Laminate laminates. 名 H-thickness of laminate a inv(A); y=1/(日*a(3,3))； Example 9.1 Show that the effective elastic constants for the laminate can be written in terms of the components A;of the [A]matrix as follows: 瓦=AAA22-A经 A22H (9.10a) A11AA22-A2 ,= AuH (9.10b) 0y= A12 (9.10c A22 Dus A12 (9.10d) A11 Gv=Ace H (9.10e) Solution Starting with (8.11)and(8.12)as follows: N: A11 A12 A12 A22 (9.11) 0 A66 take the inverse of(9.11)to obtain: a11 a12 112 a22 (9.12) 0 0 66 N where A22 a11= (9.13a) A11A22-A72 Au 022= A11A22-A2 (9.13b) A12 a12=A11A2-A (9.13c 1 a66=A66 (9.13d)
172 9 Effective Elastic Constants of a Laminate % laminates. % H - thickness of laminate a = inv(A); y = 1/(H*a(3,3)); Example 9.1 Show that the effective elastic constants for the laminate can be written in terms of the components Aij of the [A] matrix as follows: E¯x = A11AA22 − A2 12 A22H (9.10a) E¯y = A11AA22 − A2 12 A11H (9.10b) ν¯xy = A12 A22 (9.10c) ν¯yx = A12 A11 (9.10d) G¯xy = A66 H (9.10e) Solution Starting with (8.11) and (8.12) as follows: ⎧ ⎪⎨ ⎪⎩ Nx Ny Nxy ⎫ ⎪⎬ ⎪⎭ = ⎡ ⎢ ⎣ A11 A12 0 A12 A22 0 0 0 A66 ⎤ ⎥ ⎦ ⎧ ⎪⎨ ⎪⎩ ε0 x ε0 y γ0 xy ⎫ ⎪⎬ ⎪⎭ (9.11) take the inverse of (9.11) to obtain: ⎧ ⎪⎨ ⎪⎩ ε0 x ε0 y γ0 xy ⎫ ⎪⎬ ⎪⎭ = ⎡ ⎢ ⎣ a11 a12 0 a12 a22 0 0 0 a66 ⎤ ⎥ ⎦ ⎧ ⎪⎨ ⎪⎩ Nx Ny Nxy ⎫ ⎪⎬ ⎪⎭ (9.12) where a11 = A22 A11A22 − A2 12 (9.13a) a22 = A11 A11A22 − A2 12 (9.13b) a12 = A12 A11A22 − A2 12 (9.13c) a66 = 1 A66 (9.13d)
9.2 MATLAB Functions Used 173 Next,substitute(9.13)into (9.8)to obtain the required expressions as follows: 瓦=A1AA22-A经 A22H (9.14a) E,= A11AA22-A12 (9.14b) Au A12 Dry=A22 (9.14c) Dyt A12 A11 (9.14d) ,= (9.14e) MATLAB Example 9.2 Consider a four-layer [0/90s graphite-reinforced polymer composite laminate with the elastic constants as given in Example 2.2.The laminate has total thickness of 0.800 mm.The four layers are of equal thickness.Use MATLAB to determine the five effective elastic constants for this laminate. Solution This example is solved using MATLAB.First,the reduced stiffness matrix [Q]for a typical layer using the MATLAB function ReducedStiffness as follows: EDU>>Q=ReducedStiffness(155.0,12.10,0.248,4.40) Q= 155.7478 3.0153 0 3.0153 12.1584 0 0 0 4.4000 Next,the transformed reduced stiffness matrix [is calculated for each layer using the MATLAB function Qbar as follows: EDU>>Qbar1 Qbar(Q,0) Obar1 155.7478 3.0153 0 3.0153 12.1584 0 0 0 4.4000
9.2 MATLAB Functions Used 173 Next, substitute (9.13) into (9.8) to obtain the required expressions as follows: E¯x = A11AA22 − A2 12 A22H (9.14a) E¯y = A11AA22 − A2 12 A11H (9.14b) ν¯xy = A12 A22 (9.14c) ν¯yx = A12 A11 (9.14d) G¯xy = A66 H (9.14e) MATLAB Example 9.2 Consider a four-layer [0/90]S graphite-reinforced polymer composite laminate with the elastic constants as given in Example 2.2. The laminate has total thickness of 0.800 mm. The four layers are of equal thickness. Use MATLAB to determine the five effective elastic constants for this laminate. Solution This example is solved using MATLAB. First, the reduced stiffness matrix [Q] for a typical layer using the MATLAB function ReducedStiffness as follows: EDU>> Q = ReducedStiffness(155.0, 12.10, 0.248, 4.40) Q = 155.7478 3.0153 0 3.0153 12.1584 0 0 0 4.4000 Next, the transformed reduced stiffness matrix ! Q¯" is calculated for each layer using the MATLAB function Qbar as follows: EDU>> Qbar1 = Qbar(Q, 0) Qbar1 = 155.7478 3.0153 0 3.0153 12.1584 0 0 0 4.4000
174 9 Effective Elastic Constants of a Laminate EDU>>Qbar2 Qbar(Q,90) Qbar2 12.1584 3.0153 -0.0000 3.0153 155.7478 0.0000 -0.0000 0.0000 4.4000 EDU>>Qbar3 Qbar(Q,90) Qbar3 12.1584 3.0153 -0.0000 3.0153 155.7478 0.0000 -0.0000 0.0000 4.4000 EDU>>Qbar4 Qbar(Q,0) Qbar4 155.7478 3.0153 0 3.0153 12.1584 0 0 0 4.4000 Next,the distances zk(=1,2,3,4,5)are calculated as follows: EDU>>z1=-0.400 z1= -0.4000 EDU>>z2=-0.200 Z2= -0.2000 EDU>>z3 =0 z3= 0 EDU>>z4=0.200 z4= 0.2000
174 9 Effective Elastic Constants of a Laminate EDU>> Qbar2 = Qbar(Q, 90) Qbar2 = 12.1584 3.0153 -0.0000 3.0153 155.7478 0.0000 -0.0000 0.0000 4.4000 EDU>> Qbar3 = Qbar(Q, 90) Qbar3 = 12.1584 3.0153 -0.0000 3.0153 155.7478 0.0000 -0.0000 0.0000 4.4000 EDU>> Qbar4 = Qbar(Q, 0) Qbar4 = 155.7478 3.0153 0 3.0153 12.1584 0 0 0 4.4000 Next, the distances zk(k = 1, 2, 3, 4, 5) are calculated as follows: EDU>> z1 = -0.400 z1 = -0.4000 EDU>> z2 = -0.200 z2 = -0.2000 EDU>> z3 = 0 z3 = 0 EDU>> z4 = 0.200 z4 = 0.2000
9.2 MATLAB Functions Used 175 EDU>>z5=0.400 z5= 0.4000 Next,the [A]matrix is calculated using four calls to the MATLAB function Amatriz as follows: EDU>>A=zeros(3,3) A= 0 0 0 0 0 0 0 0 0 EDU>>A Amatrix(A,Qbar1,z1,z2) A= 31.1496 0.6031 0 0.6031 2.4317 0 0 0.8800 EDU>>A Amatrix(A,Qbar2,z2,z3) A= 33.5812 1.2061 -0.0000 1.2061 33.5812 0.0000 -0.0000 0.0000 1.7600 EDU>>A Amatrix(A,Qbar3,z3,z4) A= 36.0129 1.8092 -0.0000 1.8092 64.7308 0.0000 -0.0000 0.0000 2.6400 EDU>>A Amatrix(A,Qbar4,z4,z5) A= 67.1625 2.4122 -0.0000 2.4122 67.1625 0.0000 -0.0000 0.0000 3.5200 Finally,five calls are made to the five MATLAB functions introduced in this chapter to calculate the five effective elastic constants of this laminate
9.2 MATLAB Functions Used 175 EDU>> z5 = 0.400 z5 = 0.4000 Next, the [A] matrix is calculated using four calls to the MATLAB function Amatrix as follows: EDU>> A = zeros(3,3) A = 000 000 000 EDU>> A = Amatrix(A,Qbar1,z1,z2) A = 31.1496 0.6031 0 0.6031 2.4317 0 0 0 0.8800 EDU>> A = Amatrix(A,Qbar2,z2,z3) A = 33.5812 1.2061 -0.0000 1.2061 33.5812 0.0000 -0.0000 0.0000 1.7600 EDU>> A = Amatrix(A,Qbar3,z3,z4) A = 36.0129 1.8092 -0.0000 1.8092 64.7308 0.0000 -0.0000 0.0000 2.6400 EDU>> A = Amatrix(A,Qbar4,z4,z5) A = 67.1625 2.4122 -0.0000 2.4122 67.1625 0.0000 -0.0000 0.0000 3.5200 Finally, five calls are made to the five MATLAB functions introduced in this chapter to calculate the five effective elastic constants of this laminate
176 9 Effective Elastic Constants of a Laminate EDU>>H=0.800 H= 0.8000 EDU>>Ebarx(A,H) ans 83.8448 EDU>>Ebary(A,H) ans 83.8448 EDU>>NUbarxy(A,H) ans 0.0359 EDU>>NUbaryx(A,H) ans 0.0359 EDU>>Gbarxy (A,H) ans 4.4000 MATLAB Example 9.3 Consider a six-layer [+30/0]s graphite-reinforced polymer composite laminate with the elastic constants as given in Example 2.2.The laminate has total thickness of 0.900 mm.The four layers are of equal thickness.Use MATLAB to determine the five effective elastic constants for this laminate. Solution This example is solved using MATLAB.First,the reduced stiffness matrix [Q]for a typical layer using the MATLAB function ReducedStiffness as follows:
176 9 Effective Elastic Constants of a Laminate EDU>> H = 0.800 H = 0.8000 EDU>> Ebarx(A,H) ans = 83.8448 EDU>> Ebary(A,H) ans = 83.8448 EDU>> NUbarxy(A,H) ans = 0.0359 EDU>> NUbaryx(A,H) ans = 0.0359 EDU>> Gbarxy(A,H) ans = 4.4000 MATLAB Example 9.3 Consider a six-layer [±30/0]S graphite-reinforced polymer composite laminate with the elastic constants as given in Example 2.2. The laminate has total thickness of 0.900 mm. The four layers are of equal thickness. Use MATLAB to determine the five effective elastic constants for this laminate. Solution This example is solved using MATLAB. First, the reduced stiffness matrix [Q] for a typical layer using the MATLAB function ReducedStiffness as follows:
9.2 MATLAB Functions Used 177 EDU>>Q=ReducedStiffness(155.0,12.10,0.248,4.40) 0= 155.74783.0153 0 3.015312.1584 0 0 0 4.4000 Next,the transformed reduced stiffness matrix is calculated for each layer using the MATLAB function Qbar as follows: EDU>>Qbar1 Qbar(Q,30) Qbar1 91.1488 31.7170 95.3179 31.7170 19.3541 29.0342 47.6589 14.5171 61.8034 EDU>>Qbar2 Qbar(Q,-30) Qbar2 91.1488 31.7170-95.3179 31.7170 19.3541-29.0342 -47.6589-14.5171 61.8034 EDU>>Qbar3 Qbar(Q,0) Qbar3 155.7478 3.0153 0 3.0153 12.1584 0 0 0 4.4000 EDU>>Qbar4 Qbar(Q,0) Qbar4 155.7478 3.0153 0 3.0153 12.1584 0 0 0 4.4000 EDU>>Qbar5 Qbar(Q,-30) Qbar5 91.1488 31.7170-95.3179 31.7170 19.3541 -29.0342 -47.6589 -14.5171 61.8034
9.2 MATLAB Functions Used 177 EDU>> Q = ReducedStiffness(155.0, 12.10, 0.248, 4.40) Q = 155.7478 3.0153 0 3.0153 12.1584 0 0 0 4.4000 Next, the transformed reduced stiffness matrix ! Q¯" is calculated for each layer using the MATLAB function Qbar as follows: EDU>> Qbar1 = Qbar(Q, 30) Qbar1 = 91.1488 31.7170 95.3179 31.7170 19.3541 29.0342 47.6589 14.5171 61.8034 EDU>> Qbar2 = Qbar(Q, -30) Qbar2 = 91.1488 31.7170 -95.3179 31.7170 19.3541 -29.0342 -47.6589 -14.5171 61.8034 EDU>> Qbar3 = Qbar(Q, 0) Qbar3 = 155.7478 3.0153 0 3.0153 12.1584 0 0 0 4.4000 EDU>> Qbar4 = Qbar(Q, 0) Qbar4 = 155.7478 3.0153 0 3.0153 12.1584 0 0 0 4.4000 EDU>> Qbar5 = Qbar(Q, -30) Qbar5 = 91.1488 31.7170 -95.3179 31.7170 19.3541 -29.0342 -47.6589 -14.5171 61.8034
178 9 Effective Elastic Constants of a Laminate EDU>>Qbar6 Qbar(Q,30) Qbar6 91.1488 31.7170 95.3179 31.7170 19.3541 29.0342 47.6589 14.5171 61.8034 Next,the distances zk(k =1,2,3,4,5,6,7)are calculated as follows: EDU>>z1=-0.450 z1= -0.4500 EDU>>z2=-0.300 z2= -0.3000 EDU>>z3=-0.150 z3= -0.1500 EDU>>z4 =0 z4= 0 EDU>>z5=0.150 z5= 0.1500 EDU>>z6=0.300 z6= 0.3000 EDU>>z7=0.450 z7= 0.4500
178 9 Effective Elastic Constants of a Laminate EDU>> Qbar6 = Qbar(Q, 30) Qbar6 = 91.1488 31.7170 95.3179 31.7170 19.3541 29.0342 47.6589 14.5171 61.8034 Next, the distances zk (k = 1, 2, 3, 4, 5, 6, 7) are calculated as follows: EDU>> z1 = -0.450 z1 = -0.4500 EDU>> z2 = -0.300 z2 = -0.3000 EDU>> z3 = -0.150 z3 = -0.1500 EDU>> z4 = 0 z4 = 0 EDU>> z5 = 0.150 z5 = 0.1500 EDU>> z6 = 0.300 z6 = 0.3000 EDU>> z7 = 0.450 z7 = 0.4500