182 SANDWICH PLATES SS SS Figure 5.11:Rectangular simply supported (ss)sandwich plate subjected to transverse load. SS The maximum deflections of the plate are obtained by replacing EI,S,pby Du,Su1.p(see page 180) 5p位+ (ss) 384D11 (5.58) 8S11 1p+ ǔ=384D18S1 (built-in). (5.59) With the values of Du=52.16kN.m and Su=18 615N,and withL= 0.2 m,the maximum deflections are 元=0.000200+0.000134=0.000334m=0.334mm (ss) (5.60) 0=0.000040+0.000134=0.000174m=0.174mm (built-in). (5.61) 5.2.2 Simply Supported Sandwich Plates-Orthotropic and Symmetrical Layup A simply supported rectangular sandwich plate with dimensions Lx and Ly is subjected to a uniformly distributed load p(Fig.5.11).The layup of the plate is orthotropic (page 176)and symmetrical with respect to the plate's midplane. For a simply supported symmetrical plate subjected to out-of-plane loads only, the in-plane strains in the midplane are zero (see Eq.3.31)as follows: =09=0y8=0. (5.62) Substitution of Eq.(5.62)into the expression of the strain energy(Eq.5.22) gives -me (5.63)182 SANDWICH PLATES x y z L Ly x ss ss ss ss p Figure 5.11: Rectangular simply supported (ss) sandwich plate subjected to transverse load. The maximum deflections of the plate are obtained by replacing EI, S
, p by D11, S 11, p (see page 180) w = 5 384 pL4 x D11 + pL2 x 8S 11 (ss) (5.58) w = 1 384 pL4 x D11 + pL2 x 8S 11 (built-in). (5.59) With the values of D11 = 52.16 kN·m and S 11 = 18 615 kN m , and with Lx = 0.2 m, the maximum deflections are w = 0.000 200 + 0.000 134 = 0.000 334 m = 0.334 mm (ss) (5.60) w = 0.000 040 + 0.000 134 = 0.000 174 m = 0.174 mm (built-in). (5.61) 5.2.2 Simply Supported Sandwich Plates – Orthotropic and Symmetrical Layup A simply supported rectangular sandwich plate with dimensions Lx and Ly is subjected to a uniformly distributed load p (Fig. 5.11). The layup of the plate is orthotropic (page 176) and symmetrical with respect to the plate’s midplane. For a simply supported symmetrical plate subjected to out-of-plane loads only, the in-plane strains in the midplane are zero (see Eq. 3.31) as follows: o x = 0 o y = 0 γ o xy = 0. (5.62) Substitution of Eq. (5.62) into the expression of the strain energy (Eq. 5.22) gives U = 1 2 ) Lx 0 ) Ly 0    {κxκyκxy}[D]    κx κy κxy    + {γxzγyz} S 11 S 12 S 12 S 22!γxz γyz    dydx. (5.63)