408 FINITE ELEMENT ANALYSIS 452 3=0.2mm个 4=0.4mm 03 2=0个 h.=0.8mm 45 Vz1=-0.2mm 20=-0.4mm Figure 9.8:The sublaminate in Example 9.1. Solution.First,we determine the stiffness and compliance matrices.The stiffness matrices [O]and [O]45 for the plies are given by Egs.(3.49)and (3.52)as follows: 148.872.91 0 45.65 36.55 34.79 °= 2.919.71 N 0 [5 36.55 45.65 34.79 10 N m2 0 0 4.55 34.7934.79 38.19 (9.59) The compliance matrix for the 0-degree ply is(Eq.2.224) 6.76 -2.03 -2.03 0 0 0 -2.03 103.63 -62.18 0 0 0 -2.03 -62.18 103.63 0 0 [=°= 0 0 1012m2 0 0 331.61 0 0 N 0 0 0 0 219.78 0 0 0 0 0 0 219.78 (9.60) The compliance matrix for the 45-degree ply is(Eq.2.194,[S]is replaced by [S])as follows: [s5=[T][S[T]1. (9.61) The transformation matrices [T]and [T]are given in Tables 2.15(page 51) and 2.16(page 53).The angle is 45;hence,we have cr=cos 45=s,sin 45= 0.707.The elements of the [S]45 matrix are 81.53 -28.36 -32.10 0 0 -48.447 -28.36 81.53 -32.10 0 0 -48.44 可5= -32.10 -32.10 103.63 0 0 60.15 0 1012m2 0 0 275.69 -55.91 0 N 0 0 0 -55.91 275.69 0 -48.44 -48.44 60.15 0 0 114.44 (9.62)408 FINITE ELEMENT ANALYSIS 452 452 02 02 z0 = 0.4 mm – z3 = 0.2 mm z4 = 0.4 mm z1 = 0.2 mm – z2 = 0 hs = 0.8 mm Figure 9.8: The sublaminate in Example 9.1. Solution. First, we determine the stiffness and compliance matrices. The stiffness matrices [Q] 0 and [Q] 45 for the plies are given by Eqs. (3.49) and (3.52) as follows: [Q] 0 =    148.87 2.91 0 2.91 9.71 0 0 04.55    109 N m2 [Q] 45 =    45.65 36.55 34.79 36.55 45.65 34.79 34.79 34.79 38.19    109 N m2 . (9.59) The compliance matrix for the 0-degree ply is (Eq. 2.224) [S] = [S] 0 =          6.76 −2.03 −2.03 0 0 0 −2.03 103.63 −62.18 0 0 0 −2.03 −62.18 103.63 0 0 0 0 0 0 331.61 0 0 0 0 0 0 219.78 0 0 0 0 0 0 219.78          10−12 m2 N . (9.60) The compliance matrix for the 45-degree ply is (Eq. 2.194, [S ] is replaced by [S]) as follows: [S] 45 = [T
 r ] [S] [T
σ r ] −1 . (9.61) The transformation matrices [T
σ r] and [T
 r] are given in Tables 2.15 (page 51) and 2.16 (page 53). The angle is 45◦; hence, we have cr = cos 45◦ = sr = sin 45◦ = 0.707. The elements of the [S] 45 matrix are [S] 45 =          81.53 −28.36 −32.10 0 0 −48.44 −28.36 81.53 −32.10 0 0 −48.44 −32.10 −32.10 103.63 0 0 60.15 0 0 0 275.69 −55.91 0 000 −55.91 275.69 0 −48.44 −48.44 60.15 0 0 114.44          10−12 m2 N . (9.62)