5.2 DEFLECTION OF RECTANGULAR SANDWICH PLATES 181 L=900 mm p 个 L=900 mm 个p SS L=200mm TTTTTTTTT L=200 mm Figure 5.10:The sandwich plates in Example 5.1. uniformly distributed transverse load 500 kN/m2.Calculate the maximum deflec- tion.The core is isotropic (E =2 x 106 kN/m2,ve =0.3). Solution.The tensile and bending stiffnesses are calculated from Table 5.1(page 174)as follows: 430.3465.47 0 [4=2[A川= 65.47 96.34 0 (5.52) 0 m 0 72.02 52.16 7.96 0 [ol=[4+2D'= 7.96 11.71 0 kN.m, (5.53) 0 0 8.76 where [Al and [D]are given in Table 3.7 (page 84)and d=c+t =0.022 m.The shear stiffness matrix is(Eq.5.32) [原-[e-[6 0 ]kN 18615m (5.54) where (see Eq.2.30 and Table 2.10,page 18)Cs-C=769 231 kN/m2, Ec 5=0. We may treat this plate as long when(Eq.4.19) (5.55) In the present problem,Ly/L=4.5 and 3D/D22=4.36.Thus,the preceding condition is satisfied and the long plate expressions may be used.The maximum deflections of the corresponding beam are(Table 7.3,page 332) w= 5 P'LP'L 384E7+ (ss) (5.56) 85 西+ 384E1 (built-in). 8 (5.57)5.2 DEFLECTION OF RECTANGULAR SANDWICH PLATES 181 Lx = 200 mm Ly = 900 mm y x Ly = 900 mm y x ss ss ss ss p p Lx = 200 mm Figure 5.10: The sandwich plates in Example 5.1. uniformly distributed transverse load 500 kN/m2. Calculate the maximum deflec￾tion. The core is isotropic (Ec = 2 × 106 kN/m2, νc = 0.3). Solution. The tensile and bending stiffnesses are calculated from Table 5.1 (page 174) as follows: [A] = 2 [A] t =    430.34 65.47 0 65.47 96.34 0 0 0 72.02    103 kN m (5.52) [D] = 1 2 d2 [A] t + 2 [D] t =    52.16 7.96 0 7.96 11.71 0 0 08.76    kN · m, (5.53) where [A] t and [D] t are given in Table 3.7 (page 84) and d = c + t = 0.022 m. The shear stiffness matrix is (Eq. 5.32) S 11 S 12 S 12 S 22! = d2 c Cc 55 Cc 45 Cc 45 Cc 44 ! = 18 615 0 0 18 615! kN m , (5.54) where (see Eq. 2.30 and Table 2.10, page 18) Cc 55 = Cc 44 = Ec 2(1+νc) = 769 231 kN/m2, Cc 45 = 0. We may treat this plate as long when (Eq. 4.19) Ly Lx > 3 4 , D11 D22 . (5.55) In the present problem, Ly/Lx = 4.5 and 3√4 D11/D22 = 4.36. Thus, the preceding condition is satisfied and the long plate expressions may be used. The maximum deflections of the corresponding beam are (Table 7.3, page 332) w = 5 384 p L4 EI + p L2 8S
(ss) (5.56) w = 1 384 p L4 EI + p L2 8S
(built-in). (5.57)