154 8 Laminate Analysis-Part II H/2 N:= [011(e9+z)+Q12(e9+zk9)+Q16(h2y+zk9)月dz (8.14) -H/2 Expanding (8.14),we obtain: H/2 H/2 H/2 H/2 N:=e2 Qidz+ Quzds+eQ12dz+ Q12zdz -H/2 -H/2 -H/2 -H/2 H/2 H/2 +7v Qio dz+ Q16 zdz (8.15) -H/2 -H/2 Next,we expand the first term of (8.15)as follows: H/2 Qndz= 1dz+…+ ++ 11dz(8.16) -H/2 艺0 k-1 zN-1 Recognizing that Qu is constant within each layer,it can be taken outside the integrals above leading to the following expression: H/2 01dz=Q11(1-z0)+011(2-z1）+…+011(-zk-1） -H/2 +…+Q11(2N-2N-1) (8.17) The above equation can be re-written as follows: H/2 N O1dz=∑O1(k-zk-1)=A1 (8.18) -H/2 k=1 Similarly,we can show that the other five integrals of (8.15)can be written as follows: H/2 N 012dz=02(k-zk-1）=A12 (8.19a) k=1 -H/2 H/2 N Q16d2=∑Q16(2k-zk-1)=A16 (8.19b) -H/2 k=1 H2 Quzd:= 1 ∑O(--)=B (8.19c) H/2 k=
154 8 Laminate Analysis – Part II Nx = H/ 2 −H/2 ! Q¯11 ε 0 x + zκ0 x + Q¯12 ε 0 y + zκ0 y + Q¯16 γ0 xy + zκ0 xy" dz (8.14) Expanding (8.14), we obtain: Nx = ε 0 x H/ 2 −H/2 Q¯11 dz + κ0 x H/ 2 −H/2 Q¯11 zdz + ε 0 y H/ 2 −H/2 Q¯12 dz + κ0 y H/ 2 −H/2 Q¯12 zdz +γ0 xy H/ 2 −H/2 Q¯16 dz + κ0 xy H/ 2 −H/2 Q¯16 zdz (8.15) Next, we expand the first term of (8.15) as follows: H/ 2 −H/2 Q¯11 dz = z1 z0 Q¯11dz + z2 z1 Q¯11dz + ··· + zk zk−1 Q¯11dz + ··· + zN zN−1 Q¯11dz (8.16) Recognizing that Q¯11 is constant within each layer, it can be taken outside the integrals above leading to the following expression: H/ 2 −H/2 Q¯11 dz = Q¯11 (z1 − z0) + Q¯11 (z2 − z1) + ··· + Q¯11 (zk − zk−1) + ··· + Q¯11 (zN − zN−1) (8.17) The above equation can be re-written as follows: H/ 2 −H/2 Q¯11 dz = N k=1 Q¯11 (zk − zk−1) = A11 (8.18) Similarly, we can show that the other five integrals of (8.15) can be written as follows: H/ 2 −H/2 Q¯12 dz = N k=1 Q¯12 (zk − zk−1) = A12 (8.19a) H/ 2 −H/2 Q¯16 dz = N k=1 Q¯16 (zk − zk−1) = A16 (8.19b) H/ 2 −H/2 Q¯11 zdz = 1 2 N k=1 Q¯11 z2 k − z2 k−1 = B11 (8.19c)
8.2 MATLAB Functions Used 155 H/2 -H/2 ah=20保-)= (8.19d) H/2 016dz= Q16(绿-录-1)=B16 (8.19e) 2 -H/2 k=1 Using the remaining two equations of the matrix(7.12),we obtain the general desired expressions as follows: N A=O(2k-2k-1) (8.20) k=1 0(绿-录-1)》 (8.21) k三1 MATLAB Example 8.2 Consider a graphite-reinforced polymer composite laminate with the elastic con- stants as given in Example 2.2.The laminate has total thickness of 0.500 mm and is stacked as a [0/90s laminate.The four layers are of equal thickness.Calculate the [A,[B],and [D]matrices 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: >>Q=ReducedStiffness(155.0,12.10,0.248,4.40) 0= 155.7478 3.0153 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: >Qbar1 Qbar(Q,0) Qbar1 155.7478 3.0153 0 3.0153 12.1584 0 0 4.4000
8.2 MATLAB Functions Used 155 H/ 2 −H/2 Q¯12 zdz = 1 2 N k=1 Q¯12 z2 k − z2 k−1 = B12 (8.19d) H/ 2 −H/2 Q¯16 zdz = 1 2 N k=1 Q¯16 z2 k − z2 k−1 = B16 (8.19e) Using the remaining two equations of the matrix (7.12), we obtain the general desired expressions as follows: Aij = N k=1 Q¯ij (zk − zk−1) (8.20) Bij = 1 2 N k=1 Q¯ij z2 k − z2 k−1 (8.21) MATLAB Example 8.2 Consider a graphite-reinforced polymer composite laminate with the elastic constants as given in Example 2.2. The laminate has total thickness of 0.500 mm and is stacked as a [0/90]S laminate. The four layers are of equal thickness. Calculate the [A], [B], and [D] matrices 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: >> 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: >> Qbar1 = Qbar(Q,0) Qbar1 = 155.7478 3.0153 0 3.0153 12.1584 0 0 0 4.4000