406 FINITE ELEMENT ANALYSIS (Eq.915) J44 (9.49) Ex J14 J15 J24 J35 (9.50) J64 J65 For a single layer,we have (see Eqs.2.26 and 2.21) -E] (9.51) By combining Eqs.(9.18)and(9.51)and by replacing the integrals by summa- tions,for the sublaminate we obtain =2(a-®网]) (9.52) By comparing Eqs.(9.49)and(9.52),we have [均-元(a--图]) (9.53) For the sublaminate subjected to transverse shear stresses resulting in and with all other strains being zero,Eq.(9.50)becomes 0 J14 J15 0 J2s 0 (9.54) J34 J35 0 J64 J65 To satisfy this equation,the preceding elements of the compliance matrix must be zero as follows: J14 is] 0 J24 J 0 4 J35 0 00 (9.55) J64 0 0406 FINITE ELEMENT ANALYSIS (Eq. 9.15)  γ yz γ xz = J44 J45 J54 J55!τ yz τ xz (9.49)    x  y z γ xy    =      J14 J15 J24 J25 J34 J35 J64 J65       τ yz τ xz . (9.50) For a single layer, we have (see Eqs. 2.26 and 2.21)  γyz γxz = S44 S45 S45 S55!τyz τxz . (9.51) By combining Eqs. (9.18) and (9.51) and by replacing the integrals by summa￾tions, for the sublaminate we obtain  γ yz γ xz = 1 hs * Ks k=1  (zk − zk−1) S44 S45 S45 S55! k τyz τxz . (9.52) By comparing Eqs. (9.49) and (9.52), we have J44 J45 J54 J55! = 1 hs * Ks k=1  (zk − zk−1) S44 S45 S45 S55! k  . (9.53) For the sublaminate subjected to transverse shear stresses resulting in γ yz and γ xz, with all other strains being zero, Eq. (9.50) becomes    0 0 0 0    =      J14 J15 J24 J25 J34 J35 J64 J65       τ yz τ xz . (9.54) To satisfy this equation, the preceding elements of the compliance matrix must be zero as follows:      J14 J15 J24 J25 J34 J35 J64 J65      =      0 0 0 0 0 0 0 0      . (9.55)