404 FINITE ELEMENT ANALYSIS (2.20)give 0:=C33E2 (9.35) C23 (9.36) Txy These equations may be rearranged to yield e= (9.37) 1 ℃3 (9.38) By combining Egs.(9.37)and(9.18),and by replacing the integrals by sum- mations,for the sublaminate we obtain (9.39) (C33)k Equations (9.38)and(9.18)yield 卧 -2 (C3) 0 (9.40) This equation gives the in-plane stresses in the restrained sublaminate Stage 2.We apply the equal and opposite of the stresses,calculated by Eq.(9.40), to the sublaminate.The corresponding strains are obtained from Eqs.(9.28),(9.23), and (9.40).Equations (9.28)and (9.40)result in E2=-[31 J32 J36] =-[J31 Zk Zk-1 (9.41) (C33)k Equations(9.23)and(9.40)give Ex J11 J12 J16 J21 J61 J16 J26 k-k-1 (9.42) 16 s (C3)k404 FINITE ELEMENT ANALYSIS (2.20) give σz = C33z (9.35)    σx σy τxy    =    C13 C23 C63    z. (9.36) These equations may be rearranged to yield z = 1 C33 σz (9.37)    σx σy τxy    =    C13 C23 C63    1 C33 σz. (9.38) By combining Eqs. (9.37) and (9.18), and by replacing the integrals by sum￾mations, for the sublaminate we obtain z = 1 hs * Ks k=1 zk − zk−1 (C33)k  σz. (9.39) Equations (9.38) and (9.18) yield    σ x σ y τ xy    = 1 hs * Ks k=1       C13 C23 C63    k zk − zk−1 (C33)k    σz. (9.40) This equation gives the in-plane stresses in the restrained sublaminate. Stage 2. We apply the equal and opposite of the stresses, calculated by Eq. (9.40), to the sublaminate. The corresponding strains are obtained from Eqs. (9.28), (9.23), and (9.40). Equations (9.28) and (9.40) result in z = −[J31 J32 J36]    σ x σ y τ xy    = −[J31 J32 J36] 1 hs * Ks k=1       C13 C23 C63    k zk − zk−1 (C33)k    σz. (9.41) Equations (9.23) and (9.40) give    x  y γ xy    = −    J11 J12 J16 J21 J22 J26 J61 J62 J66       σ x σ y τ xy    = −    J11 J12 J16 J21 J22 J26 J61 J62 J66    1 hs * Ks k=1       C13 C23 C63    k zk − zk−1 (C33)k    σz. (9.42)