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,LLCFailure, 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