4.2 DEFLECTION OF RECTANGULAR PLATES 99 of both long edges of the plate is restricted,as in the top three configurations of Figure 4.3. By substituting Eqs.(4.20),(4.40),and (4.41)into the third and fourth expres- sions of Eq.(4.3),we obtain A6A66 (4.42) We now select a reference plane at a distance o from the midplane (Fig.3.14). The strain components and the stiffnesses referred to this reference plane are identified by the superscript o.(The curvature and the Mr component of the moment are independent of the position of the reference plane and thus do not need to be identified by the superscript e.)For the new reference plane Eq.(4.42) is written as A86 (4.43) Kr The first row of this equation gives B oo=-9e-Kx Ath (4.44) Substitution of Eq.(4.44)into the second row of Eq.(4.43)yields (4.45) In general,the bending moment Mr depends both on ee and K.However, there is a reference plane for which the term in the parentheses in front of ee is zero: B喘- AB=0. (4.46) We recall that the stiffnesses in the midplane(Reference Plane 1)and the new reference plane (Reference Plane 2)are related by(Eq.3.47) A号=A B号=B,-QA (4.47) D号=D-2QBi+Q2A Equations (4.46)and (4.47)give --(a-= (4.48)
4.2 DEFLECTION OF RECTANGULAR PLATES 99 of both long edges of the plate is restricted, as in the top three configurations of Figure 4.3. By substituting Eqs. (4.20), (4.40), and (4.41) into the third and fourth expressions of Eq. (4.3), we obtain 0 Mx = A16 A66 B16 B11 B16 D11! o x γ o xy κx . (4.42) We now select a reference plane at a distance from the midplane (Fig. 3.14). The strain components and the stiffnesses referred to this reference plane are identified by the superscript . (The curvature and the Mx component of the moment are independent of the position of the reference plane and thus do not need to be identified by the superscript .) For the new reference plane Eq. (4.42) is written as 0 Mx = A 16 A 66 B 16 B 11 B 16 D 11 ! o, x γ o, xy κx . (4.43) The first row of this equation gives γ o, xy = −o, x A 16 A 66 − κx B 16 A 66 . (4.44) Substitution of Eq. (4.44) into the second row of Eq. (4.43) yields Mx = B 11 − A 16B 16 A 66 o, x + - D 11 − B2 16 A 66 . κx. (4.45) In general, the bending moment Mx depends both on o, x and κx. However, there is a reference plane for which the term in the parentheses in front of o, x is zero: B 11 − A 16B 16 A 66 = 0. (4.46) We recall that the stiffnesses in the midplane (Reference Plane 1) and the new reference plane (Reference Plane 2) are related by (Eq. 3.47) A i j = Ai j B i j = Bi j − Ai j (4.47) D i j = Di j − 2Bi j + 2Ai j . Equations (4.46) and (4.47) give B11 − A16B16 A66 − A11 − A2 16 A66 = 0. (4.48)
100 THIN PLATES By rearranging this equation,we obtain the position of the reference plane where Eq.(4.46)is satisfied: B1-46B6 A66 0= (4.49) A:-念 For a reference plane at o distance from the midplane,the moment M depends only on Kx as follows: (4.50) Equations(4.50)and (4.24),together with Eq.(4.2),yield the following equi- librium equation for an anisotropic long plate: w°-2=0 long plate dx4 (4.51) unsymmetrical layup, where the symbol is the bending stiffness parameter w=D品 (B%)2 (4.52) A号6 By using Eq.(4.47),may be written as 业=D1-2eB1+e2A1-(86-eAo, (4.53) A66 where e is given by Eq.(4.49). By comparing Eqs.(4.51)and(4.27),we again observe that the equations gov- erning the deflections of long plates and isotropic beams are similar.Therefore, the deflection of a long plate with unsymmetrical layup can be obtained by replac- ing EI/p'by /p in the expression given for the deflection of the corresponding isotropic beam. 4.2.3 Simply Supported Plates-Symmetrical Layup We consider a rectangular plate with dimensions Lr and Ly simply supported along its four edges(Fig.4.7).The layup of the plate is symmetrical and [B]=[O]. The plate is subjected to a uniformly distributed load p. Figure 4.7:Rectangular simply supported (ss)plate subjected to a uniformly dis- tributed transverse load. SS
100 THIN PLATES By rearranging this equation, we obtain the position of the reference plane where Eq. (4.46) is satisfied: = B11 − A16 B16 A66 A11 − A2 16 A66 . (4.49) For a reference plane at distance from the midplane, the moment Mx depends only on κx as follows: Mx = D 11 − B 162 A 66 ! κx. (4.50) Equations (4.50) and (4.24), together with Eq. (4.2), yield the following equilibrium equation for an anisotropic long plate: d4wo dx4 − p = 0 long plate unsymmetrical layup, (4.51) where the symbol is the bending stiffness parameter = D 11 − B 162 A 66 . (4.52) By using Eq. (4.47), may be written as = D11 − 2B11 + 2A11 − (B16 − A16) 2 A66 , (4.53) where is given by Eq. (4.49). By comparing Eqs. (4.51) and (4.27), we again observe that the equations governing the deflections of long plates and isotropic beams are similar. Therefore, the deflection of a long plate with unsymmetrical layup can be obtained by replacing EI/p by /p in the expression given for the deflection of the corresponding isotropic beam. 4.2.3 Simply Supported Plates – Symmetrical Layup We consider a rectangular plate with dimensions Lx and Ly simply supported along its four edges (Fig. 4.7). The layup of the plate is symmetrical and [B] = [0]. The plate is subjected to a uniformly distributed load p. x y z L Ly x p ss ss ss ss Figure 4.7: Rectangular simply supported (ss) plate subjected to a uniformly distributed transverse load.
4.2 DEFLECTION OF RECTANGULAR PLATES 101 Following Whitney,1 we analyze the deflection of this plate by the energy method.For a simply supported plate(symmetrical layup)subjected to out-of- plane loads only,the in-plane strains in the midplane are zero,and Eq.(4.13) simplifies to Du D12 D16 D12 D22 dydx. (4.54) D16 D26 By using the relationships between the curvatures and the deflections given by Eq.(4.2),we obtain u-j[(+(+ 282w° 82w°82w0 82w°282w0 82w°282w° +D6 ax axay 。+D ay axay dydx (4.55) For the applied transverse load p(per unit area)the potential of the external forces is(Eq.2.203) 2 pw)dydx (4.56) We use the Ritz method and select an expression for the deflection that sat- isfies the geometrical boundary conditions.For the simply supported plate under consideration the geometrical boundary conditions require that the deflection be zero along the edges (see Eq.4.8)as follows: x=0 and0≤y≤Ly x=Lr w°=0 and0≤y≤Ly at (4.57) 0≤x≤L and y=0 0≤x≤Lx and y=Ly: The following deflection satisfies these conditions: w°=jsin sin iry inx i=1j=1 Ly (4.58) where I and are the number of terms,chosen arbitrarily,for the summations and wij are constants and are calculated from the principle of stationary potential energy(Eq.2.206)expressed as 8(U+=0. (4.59) Owij awij 11 J.M.Whitney,Structural Analysis of Laminated Anisotropic Plates.Technomic.Lancaster, Pennsylvania,1987,p.133
4.2 DEFLECTION OF RECTANGULAR PLATES 101 Following Whitney,11 we analyze the deflection of this plate by the energy method. For a simply supported plate (symmetrical layup) subjected to out-ofplane loads only, the in-plane strains in the midplane are zero, and Eq. (4.13) simplifies to U = 1 2 ) Lx 0 ) Ly 0 {κx κy κxy} D11 D12 D16 D12 D22 D26 D16 D26 D66 κx κy κxy dydx. (4.54) By using the relationships between the curvatures and the deflections given by Eq. (4.2), we obtain U = 1 2 ) Lx 0 ) Ly 0 / D11∂2wo ∂x2 2 + D22∂2wo ∂y2 2 + D662∂2wo ∂x∂y 2 + 2 D12 ∂2wo ∂x2 ∂2wo ∂y2 + D16 ∂2wo ∂x2 2∂2wo ∂x∂y + D26 ∂2wo ∂y2 2∂2wo ∂x∂y 0 dydx. (4.55) For the applied transverse load p (per unit area) the potential of the external forces is (Eq. 2.203) = − ) Lx 0 ) Ly 0 (pwo)dydx. (4.56) We use the Ritz method and select an expression for the deflection that satisfies the geometrical boundary conditions. For the simply supported plate under consideration the geometrical boundary conditions require that the deflection be zero along the edges (see Eq. 4.8) as follows: wo = 0 at x = 0 and 0 ≤ y ≤ Ly x = Lx and 0 ≤ y ≤ Ly 0 ≤ x ≤ Lx and y = 0 0 ≤ x ≤ Lx and y = Ly . (4.57) The following deflection satisfies these conditions: wo = * I i=1 * J j=1 wi j sin iπx Lx sin jπy Ly , (4.58) where I and J are the number of terms, chosen arbitrarily, for the summations and wi j are constants and are calculated from the principle of stationary potential energy (Eq. 2.206) expressed as ∂πp ∂wi j = ∂ (U + ) ∂wi j = 0. (4.59) 11 J. M. Whitney, Structural Analysis of Laminated Anisotropic Plates. Technomic, Lancaster, Pennsylvania, 1987, p. 133
102 THIN PLATES We now substitute w(from Eq.4.58)into the expressions of U and (Eqs.4.55 and 4.56)and perform the differentiations indicated above.Lengthy but straight- forward algebraic manipulations result in the following system of simultaneous algebraic equations: i,m=1,2,3,..,1 j,n=1,2,3,.,J (4.60) For convenience,we introduce the contracted notation k=(i-1)J+j i=1,2,3,,1 j=1,2,3.,J (4.61) 1=(m-1)J+n m=1,2,3,,1 n=1,2,3,,J (4.62) Equation (4.60)may now be written as I 2ou=n1=1.231x1 (4.63) where Gu(=G)is given in Table 4.1 and,for a uniformly distributed load,pr is if m and n are odd (4.64) if m or n is even Table 4.1.The elements of the matrix [G] Gk=LxLyπ[D(台)+2(D2+2D)()'()2+D2()]s -2L.LrD6[(长)(2)(2)rm'm+(受)（台）()rmra] -2 xLyDa6[()2(2)()rmm+()2()()rmrm] 如-6验 w-含8-升阳 k=0-10J+jj=1.2.3J i=1,2,3.,1 I=(m-1)J+n m=1,2,3.,1 1n=1,2,3,.,J
102 THIN PLATES We now substitute wo (from Eq. 4.58) into the expressions ofU and(Eqs. 4.55 and 4.56) and perform the differentiations indicated above. Lengthy but straightforward algebraic manipulations result in the following system of simultaneous algebraic equations: * I i=1 * J j=1 Gmni jwi j = pmn i, m = 1, 2, 3,..., I j, n = 1, 2, 3,..., J . (4.60) For convenience, we introduce the contracted notation k = (i − 1)J + j i = 1, 2, 3,..., I j = 1, 2, 3,..., J (4.61) l = (m − 1)J + n m = 1, 2, 3,..., I n = 1, 2, 3,..., J . (4.62) Equation (4.60) may now be written as * I×J k=1 Gklwk = pl l = 1, 2, 3,..., I × J, (4.63) where Gkl (= Glk) is given in Table 4.1 and, for a uniformly distributed load, pl is pl = 4pLx Ly π2 mn if m and n are odd 0 if m or n is even . (4.64) Table 4.1. The elements of the matrix [G ] Glk = 1 4 LxLyπ4 " D11 i Lx 4 + 2 D12 + 2D66 i Lx 2 j Ly 2 + D22 j Ly 4# δlk − 2LxLyπ4D16" i Lx 2 m Lx n Ly rimrjn + m Lx 2 i Lx j Ly rmirnj# − 2LxLyπ4D26" j Ly 2 m Lx n Ly rimrjn + n Ly 2 i Lx j Ly rmirnj# δlk = 1 if k = l 0 if k = l ri j = 1 2i i2− j2 1 π if (i − j) is odd 0 if (i − j) is even k = (i − 1)J + j i = 1, 2, 3,... , I j = 1, 2, 3,... , J l = (m − 1)J + n m = 1, 2, 3,... , I n = 1, 2, 3,... , J
4.2 DEFLECTION OF RECTANGULAR PLATES 103 In expanded form,Eq.(4.63)is G11 G12 G1(I×J) 01 G21 Gy P2 (4.65) :: G(IxII G1xJ)(1×J) W(IxJ) P(IxJ) By inverting this equation,we obtain the coefficients wk Wi G11 G12 G1(1×J) P W2 G21 G m (4.66) W(IxJ刀 G(x G1xJ)1xJ)」 P(Ix5) From Egs.(4.2),(3.27),and (4.58)the moments are 总 [D] =[D .(4.67 -名2号coscos号 For an orthotropic plate Di6=D26=0,and Eq.(4.66)becomes 16p Wk=W时= πij[D()+2(D2+2Do(2)'()2+D()] (4.68) where i,j =1,3,5,...(wk=wij =0 when i or j=2,4,6,...). Once the deflections are known,the moments can be calculated by Eq-(4.67) 4.2 Example.A 0.7-m-long and 0.2-m-wide rectangular plate is made of graphite epoxy.The material properties are given in Table 3.6 (page 81).The layup is (+45,/012/445.The 0-degree plies are parallel to the short edge of the plate. The plate is simply supported along all four edges and is subjected to a uniformly distributed transverse load p=50 000 N/m2 (Fig.4.8).Calculate the maximum deflection and the maximum bending moments. Solution.The deflection of the plate is(Eq.4.58) inx:jny (4.69) i=1i=1 Ly
4.2 DEFLECTION OF RECTANGULAR PLATES 103 In expanded form, Eq. (4.63) is G11 G12 ... G1(I×J ) G21 G22 . . . ... G(I×J )1 G(I×J )(I×J ) w1 w2 . . . w(I×J ) = p1 p2 . . . p(I×J ) . (4.65) By inverting this equation, we obtain the coefficients wk w1 w2 . . . w(I×J ) = G11 G12 ... G1(I×J ) G21 G22 . . . ... G(I×J )1 G(I×J )(I×J ) −1 p1 p2 . . . p(I×J ) . (4.66) From Eqs. (4.2), (3.27), and (4.58) the moments are Mx My Mxy = [D] κx κy κxy = [D] 2 I i=1 2 J j=1 wi j iπ Lx 2 sin iπx Lx sin jπy Ly 2 I i=1 2 J j=1 wi j jπ Ly 2 sin iπx Lx sin jπy Ly − 2 I i=1 2 J j=1 wi j2 iπ Lx jπ Ly cos iπx Lx cos jπy Ly . (4.67) For an orthotropic plate D16 = D26 = 0, and Eq. (4.66) becomes wk = wi j = 16p π6i j3 D11 i Lx 4 + 2 D12 + 2D66 i Lx 2 j Ly 2 + D22 j Ly 4 4 , (4.68) where i, j = 1, 3, 5,... (wk = wi j = 0 when i or j = 2, 4, 6,...). Once the deflections are known, the moments can be calculated by Eq. (4.67). 4.2 Example. A 0.7-m-long and 0.2-m-wide rectangular plate is made of graphite epoxy. The material properties are given in Table 3.6 (page 81). The layup is [±45f 2/012/±45f 2]. The 0-degree plies are parallel to the short edge of the plate. The plate is simply supported along all four edges and is subjected to a uniformly distributed transverse load p = 50 000 N/m2 (Fig. 4.8). Calculate the maximum deflection and the maximum bending moments. Solution. The deflection of the plate is (Eq. 4.58) wo = * I i=1 * J j=1 wi j sin iπx Lx sin jπy Ly . (4.69)