374 Mechanics of Composite Materials,Second Edition E vE 0 12(1-v2) 12(1-v2) [D]= VE E 0 . 12(1-v2) (5.7) 12(1-v2) E 0 0 24(1+v) A laminate is called quasi-isotropic if its extensional stiffness matrix [A] behaves like that of an isotropic material.This implies not only that Au=Azz AA,and but also that these stiffnesses are indepen- 2 dent of the angle of rotation of the laminate.The reason for calling such a laminate quasi-isotropic and not isotropic is that the other stiffness matrices, [B]and [D],may not behave like isotropic materials.Examples of quasi- isotropic laminates include [0/+60],[0/+45/90],and [0/36/72/-36/-72]. Example 5.1 A [0/+60]graphite/epoxy laminate is quasi-isotropic.Find the three stiffness matrices [A],[B],and [D]and show that 1.A1=A2iA6=A6=0;A6=A1A2 2.[Bl≠0，unlike isotropic materials.. 3.[D]matrix is unlike isotropic materials. Use properties of unidirectional graphite/epoxy lamina from Table 2.1.Each lamina has a thickness of 5 mm. Solution From Example 2.6,the reduced stiffness matrix [Q]for the 0graphite/epoxy lamina is 181.8 2.897 0 [Q]= 2.897 10.35 0 (10)Pa. 0 0 7.17 From Equation(2.104),the transformed reduced stiffness matrices for the three plies are 2006 by Taylor Francis Group,LLC
374 Mechanics of Composite Materials, Second Edition . (5.7) A laminate is called quasi-isotropic if its extensional stiffness matrix [A] behaves like that of an isotropic material. This implies not only that A11 = A22, A16= A26 = 0, and , but also that these stiffnesses are independent of the angle of rotation of the laminate. The reason for calling such a laminate quasi-isotropic and not isotropic is that the other stiffness matrices, [B] and [D], may not behave like isotropic materials. Examples of quasiisotropic laminates include [0/±60], [0/±45/90]s, and [0/36/72/–36/–72]. Example 5.1 A [0/±60] graphite/epoxy laminate is quasi-isotropic. Find the three stiffness matrices [A], [B], and [D] and show that 1. . 2. [B] ≠ 0, unlike isotropic materials. 3. [D] matrix is unlike isotropic materials. Use properties of unidirectional graphite/epoxy lamina from Table 2.1. Each lamina has a thickness of 5 mm. Solution From Example 2.6, the reduced stiffness matrix [Q] for the 0° graphite/epoxy lamina is . From Equation (2.104), the transformed reduced stiffness matrices for the three plies are [ ] ( ) ( ) ( ) ( ) D E E E E = − − − − 12 1 12 1 0 12 1 12 1 2 2 2 2 ν ν ν ν ν ν 0 0 0 24 1 3 E h ( ) + ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ν A A A 66 11 12 2 = − A A A A A A A 11 22 16 26 66 11 12 0 2 = = = = − ; ; [ ] . . . . . Q = ( ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ 181 8 2 897 0 2 897 10 35 0 0 0 7 17 109 ) Pa 1343_book.fm Page 374 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC
Failure,Analysis,and Design of Laminates 375 [181.8 2.897 0 [顶= 2.897 10.35 0 (10)Pa, 0 0 7.17 23.65 32.46 20.05 [Ql4o= 32.46 109.4 54.19 (10)Pa, 20.05 54.19 36.74 23.65 32.46 -20.05 [@0= 32.46 109.4 -54.1910)Pa. -20.05 -54.19 36.74 The total thickness of the laminate is h=(0.005)(3)=0.015 m. The midplane is 0.0075 m from the top and bottom of the laminate.Thus, using Equation(4.20), h0=-0.0075m h1=-0.0025m h2=0.0025m h3=0.0075m Using Equation (4.28a)to Equation (4.28c),one can now calculate the stiffness matrices [A],[B],and [D],respectively,as shown in Example 4.2: 1.146 0.3391 0 [A]= 0.3391 1.146 0 (10)Pa-1, 0 0 0.4032 -3.954 0.7391 -0.5013 [B]= 0.7391 2.476 -1.355 (10)Pa-m2, -0.5013 -1.355 0.7391 2006 by Taylor Francis Group,LLC
Failure, Analysis, and Design of Laminates 375 , , The total thickness of the laminate is h = (0.005)(3) = 0.015 m. The midplane is 0.0075 m from the top and bottom of the laminate. Thus, using Equation (4.20), h0 = –0.0075 m h1 = –0.0025 m h2 = 0.0025 m h3 = 0.0075 m Using Equation (4.28a) to Equation (4.28c), one can now calculate the stiffness matrices [A], [B], and [D], respectively, as shown in Example 4.2: , , [ ] . . . . . Q 0 181 8 2 897 0 2 897 10 35 0 0 0 7 17 = ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ( ) 109 Pa [ ] . . . . . . . Q 60 23 65 32 46 20 05 32 46 109 4 54 19 20 05 54 = . . ( ) 19 36 74 109 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ Pa [ ] . . . . . . . Q − = − − − 60 23 65 32 46 20 05 32 46 109 4 54 19 20 05 54 19 36 74 109 − ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ . . ( ) Pa. [ ] . . . . . A = ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ 1 146 0 3391 0 0 3391 1 146 0 0 0 0 4032 ⎥ ⎥ ⎥ ( ) 109 Pa-m [ ] . . . . . . . B = − − − − 3 954 0 7391 0 5013 0 7391 2 476 1 355 0 5013 1 355 0 7391 106 2 − ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ . . ( ) Pa-m 1343_book.fm Page 375 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC
376 Mechanics of Composite Materials,Second Edition 28.07 5.126 -2.507 [D]= 5.126 17.35 -6.774103)Pa-m3. -2.507 6.774 6.328 1.From the extensional stiffness matrix [A], A1=A22=1.146×109Pa-m A16=A26=0 A1-A2=1.146-0.3391×10 2 2 =0.4032×109Pa-m =A66 This behavior is similar to that of an isotropic material.However, a quasi-isotropic laminate should give the same [A]matrix,if a constant angle is added to each of the layers of the laminate.For example,adding 30 to each ply angle of the [0/+60]laminate gives a [30/90/-30]laminate,which has the same [A]matrix as the [0/ ±60]laminate. 2.Unlike isotropic materials,the coupling stiffness matrix [B]of the [0/+60]laminate is nonzero. 3.In an isotropic material, D1=D22, D16=D26=0, and Do=Du-Dg. 2 In this example,unlike isotropic materials,Du D22 because D11=28.07×103Pa-m3 2006 by Taylor Francis Group,LLC
376 Mechanics of Composite Materials, Second Edition . 1. From the extensional stiffness matrix [A], = 0.4032 × 109 Pa-m = A66. This behavior is similar to that of an isotropic material. However, a quasi-isotropic laminate should give the same [A] matrix, if a constant angle is added to each of the layers of the laminate. For example, adding 30° to each ply angle of the [0/±60] laminate gives a [30/90/–30] laminate, which has the same [A] matrix as the [0/ ±60] laminate. 2. Unlike isotropic materials, the coupling stiffness matrix [B] of the [0/±60] laminate is nonzero. 3. In an isotropic material, , , and . In this example, unlike isotropic materials, D11 ≠ D22 because D11 = 28.07 × 103 Pa-m3 [ ] .. . . .. . D = − − − − 28 07 5 126 2 507 5 126 17 35 6 774 2 507 6 774 6 328 103 3 . . ( ) ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ Pa mA A Pa m 11 22 9 == × 1 146 10 . - A A 16 26 = = 0 A A 11 12 9 2 1 146 0 3391 2 10 − = − × . . D D 11 22 = D D 16 26 = = 0 D D D 66 11 12 2 = − 1343_book.fm Page 376 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC
Failure,Analysis,and Design of Laminates 377 D22=17.35×103Pa-m3 D16≠0，D26≠0as D16=-2.507×103Pa-m3 D26=-6.774×103Pa-m3 D1-D2≠D6 2 because D1-D2=28.07×103-5.126×103 2 =11.47×103Pa-m3 D66=6.328×103Pa-m3. One can make a quasi-isotropic laminate by having a laminate with N(N >3)lamina of the same material and thickness,where each lamina is oriented at an angle of 180/N between each other. For example,a three-ply laminate will require the laminae to be oriented at 180/3=60 to each other.Thus,[0/60/-60],[30/90/ -30],and [45/-75/-15]are all quasi-isotropic laminates.One can make the preceding combinations symmetric or repeated to give quasi-isotropic laminates,such as [0/+60],[0/+60]s,and [0/+60]2 laminates.The symmetry of the laminates zeros out the coupling matrix [B]and makes its behavior closer (not same)to that of an isotropic material. Example 5.2 Show that the extensional stiffness matrix for a general N-ply quasi-isotropic laminate is given by U 0 [A]= U u, 0 h (5.8) 0 0 u1-U4 2 2006 by Taylor Francis Group,LLC
Failure, Analysis, and Design of Laminates 377 D22 = 17.35 × 103 Pa-m3 D16 ≠ 0, D26 ≠ 0 as D16 = –2.507 × 103 Pa-m3 D26 = –6.774 × 103 Pa-m3 because = 11.47 × 103 Pa-m3 D66 = 6.328 × 103 Pa-m3. One can make a quasi-isotropic laminate by having a laminate with N (N ≥ 3) lamina of the same material and thickness, where each lamina is oriented at an angle of 180°/N between each other. For example, a three-ply laminate will require the laminae to be oriented at 180°/3 = 60° to each other. Thus, [0/60/–60], [30/90/ –30], and [45/–75/–15] are all quasi-isotropic laminates. One can make the preceding combinations symmetric or repeated to give quasi-isotropic laminates, such as [0/±60]s, [0/±60]s, and [0/±60]2s laminates. The symmetry of the laminates zeros out the coupling matrix [B] and makes its behavior closer (not same) to that of an isotropic material. Example 5.2 Show that the extensional stiffness matrix for a general N-ply quasi-isotropic laminate is given by . (5.8) D D D 11 12 66 2 − ≠ D D 11 12 3 3 2 28 07 10 5 126 10 2 − = . . ×− × [ ] A U U U U U U = h − ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ 1 4 4 1 1 4 0 0 0 0 2 1343_book.fm Page 377 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC
378 Mechanics of Composite Materials,Second Edition where U and U are the stiffness invariants given by Equation(2.132)and h is the thickness of the laminate.Also,find the in-plane engineering stiffness constants of the laminate. Solution From Equation(2.131a),for a general angle ply with angle 0, u=L U2 Cos20 Us Cos40. (5.9) For the kth ply of the quasi-isotropic laminate with an angle0 (Q)=U U2 Cos20g+Us Cos40x (5.10) where 0=5=8=贤-N1,s= 2π From Equation (4.28a), N A1= (5.11) k=1 where f=thickness of kth lamina. Because the thickness of the laminate is h and all laminae are of the same thickness, Nk=1,2,,N, k= (5.12) and,substituting Equation(5.10)in Equation(5.11), A=02u+u,cos29+山，cas46) k=】 (5.13) =hu,+u0∑cos29+u0∑cos40 2006 by Taylor Francis Group,LLC
378 Mechanics of Composite Materials, Second Edition where U1 and U4 are the stiffness invariants given by Equation (2.132) and h is the thickness of the laminate. Also, find the in-plane engineering stiffness constants of the laminate. Solution From Equation (2.131a), for a general angle ply with angle θ, = U1 + U2 Cos2θ + U3 Cos4θ. (5.9) For the kth ply of the quasi-isotropic laminate with an angle θk, = U1 + U2 Cos2θk + U3 Cos4θk, (5.10) where From Equation (4.28a), , (5.11) where tk = thickness of kth lamina. Because the thickness of the laminate is h and all laminae are of the same thickness, (5.12) and, substituting Equation (5.10) in Equation (5.11), (5.13) Q11 ( ) Q11 k θ π θ π θ π θ π 12 1 θ π 2 1 = = …= … = − − = N N k N N N , ,, ,, kN N ( ) , . A tQk k k N 11 11 1 = = ∑ ( ) t h N k = = , , ,............, , k N 1 2 A h N UU U hU U k k k N 11 1 2 3 1 1 =+ + 2 4 = + = ∑( ) Cos Cos θ θ 2 3 1 1 2 4 h N U h N k k N k k N Cos Cos . θ θ + = = ∑ ∑ 1343_book.fm Page 378 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC