9.4 SUBLAMINATE 405 Stage 3.We combine Egs.(9.39),(9.41),and(9.42)to obtain the strains of a sub- laminate subjected only to the out-of-plane stress o. (9.43) J12 -2k-1 (9.44) (C33)k Comparisons of Eqs.(9.32)and(9.33)with Egs.(9.43)and(9.44)yield J13 J11 J12 J21 J6 J61 、之 (9.45) (- (9.46) SinceTy:=x=0,we have that yyz and y:are zero(Eqs.2.26 and 2.21),and Eq.(9.34)becomes 8-[ (9.47) To satisfy this equation,the preceding elements of the compliance matrix must be zero (9.48) 9.4.3 Step 3.Elements of [due to Out-of-Plane Shear Stresses In this step we determine the elements in the fourth and fifth column of the matrix [] To accomplish this we apply the shear stresses tyz,trz on the sublaminate (Fig.9.7).Since x,,,Txy are zero,the stress-strain relationships are Figure 9.7:Illustration of Step 3.Stress x:and strain Yx on a sublaminate subjected to transverse shear loads (left)and the corresponding average stress Tr and average strain .(right).9.4 SUBLAMINATE 405 Stage 3. We combine Eqs. (9.39), (9.41), and (9.42) to obtain the strains of a sub￾laminate subjected only to the out-of-plane stress σz z =    1 hs * Ks k=1 zk − zk−1 (C33)k  − [J31 J32 J36] 1 hs * Ks k=1       C13 C23 C63    k zk − zk−1 (C33)k       σz (9.43)    x  y γ 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. (9.44) Comparisons of Eqs. (9.32) and (9.33) with Eqs. (9.43) and (9.44) yield    J13 J23 J63    = −    J11 J12 J16 J21 J22 J26 J61 J62 J66    1 hs * Ks k=1       C13 C23 C63    k zk − zk−1 (C33)k    (9.45) J33 = 1 hs * Ks k=1 zk − zk−1 (C33)k  − [J31 J32 J36] 1 hs * Ks k=1       C13 C23 C63    k zk − zk−1 (C33)k    . (9.46) Since τ yz = τ xz = 0, we have that γyz and γxz are zero (Eqs. 2.26 and 2.21), and Eq. (9.34) becomes  0 0  = J43 J53! σ z. (9.47) To satisfy this equation, the preceding elements of the compliance matrix must be zero J43 J53! = 0 0 ! . (9.48) 9.4.3 Step 3. Elements of [J ] due to Out-of-Plane Shear Stresses In this step we determine the elements in the fourth and fifth column of the matrix [J ]. To accomplish this we apply the shear stresses τyz, τxz on the sublaminate (Fig. 9.7). Since σ x, σ y, σ z, τ xy are zero, the stress–strain relationships are τxz γxz τxz γxz Figure 9.7: Illustration of Step 3. Stress τxz and strain γxz on a sublaminate subjected to transverse shear loads (left) and the corresponding average stress τ xz and average strain γ xz (right)