Copyrighted Materials 0CpUyPress o CHAPTER FOUR Thin Plates In practice we frequently encounter "thin"plates whose thickness is small com- pared with all other dimensions.Such a plate,undergoing small displacements, may be analyzed with the approximations that the strains vary linearly across the plate,(out-of-plane)shear deformations are negligible,and the out-of-plane normal stress o:and shear stresses tx,tyz are small compared with the in-plane normal ox,oy,and shear try stresses. Under certain conditions,solutions may be obtained for thin plates either by the solution of the differential equations representing equilibrium or by energy methods.Here we demonstrate the use of the first method via the example of long plates and the second method via examples of rectangular plates either with sym- metrical layup or with orthotropic and symmetrical layup.(For orthotropic plates the directions of orthotropy are parallel to the edges of the plate.)We chose these three types of problems because (i)they illustrate the analytical approaches and the use of the relevant equations,(ii)solutions can be obtained without extensive numerical algorithms,and last,but not least,(iii)they are of practical interest. Additionally,and importantly,these problems provide insights that are useful when analyzing plates by numerical methods. Although the specification of orthotropy may seem to be overly restrictive,in fact it does not unduly limit the applicability of the analyses.The reason for this is that plates are often made according to the 10-percent rule,and such plates behave similarly to orthotropic plates.2 Therefore,solutions for orthotropic plates provide good approximations of the deflections,maximum bending moments,buckling loads,and natural frequencies of nonorthotropic plates that have symmetrical layup and are constructed according to the 10-percent rule.The 10-percent rule 1 J.M.Whitney,Structural Analysis of Laminated Anisotropic Plates.Technomic,Lancaster, Pennsylvania,1987. 2 I.Veres and L.P.Kollar,Approximate Analysis of Mid-plane Symmetric Rectangular Composite Plates.Journal of Composite Materials,Vol.36,673-684,2002. 89
CHAPTER FOUR Thin Plates In practice we frequently encounter “thin” plates whose thickness is small compared with all other dimensions. Such a plate, undergoing small displacements, may be analyzed with the approximations that the strains vary linearly across the plate, (out-of-plane) shear deformations are negligible, and the out-of-plane normal stress σz and shear stresses τxz, τyz are small compared with the in-plane normal σx, σy, and shear τxy stresses. Under certain conditions, solutions may be obtained for thin plates either by the solution of the differential equations representing equilibrium or by energy methods.1 Here we demonstrate the use of the first method via the example of long plates and the second method via examples of rectangular plates either with symmetrical layup or with orthotropic and symmetrical layup. (For orthotropic plates the directions of orthotropy are parallel to the edges of the plate.) We chose these three types of problems because (i) they illustrate the analytical approaches and the use of the relevant equations, (ii) solutions can be obtained without extensive numerical algorithms, and last, but not least, (iii) they are of practical interest. Additionally, and importantly, these problems provide insights that are useful when analyzing plates by numerical methods. Although the specification of orthotropy may seem to be overly restrictive, in fact it does not unduly limit the applicability of the analyses. The reason for this is that plates are often made according to the 10-percent rule, and such plates behave similarly to orthotropic plates.2 Therefore, solutions for orthotropic plates provide good approximations of the deflections, maximum bending moments, buckling loads, and natural frequencies of nonorthotropic plates that have symmetrical layup and are constructed according to the 10-percent rule. The 10-percent rule 1 J. M. Whitney, Structural Analysis of Laminated Anisotropic Plates. Technomic, Lancaster, Pennsylvania, 1987. 2 I. Veres and L. P. Koll´ar, Approximate Analysis of Mid-plane Symmetric Rectangular Composite Plates. Journal of Composite Materials, Vol. 36, 673–684, 2002. 89
90 THIN PLATES requires that the plate satisfy the following conditions: The plate is made of unidirectional plies. There are at least three ply orientations. ·The angles between the fibers are at least 15° The number of plies in each fiber direction is at least 10 percent of the total number of plies. Plates conforming to the 10-percent rule have better load bearing capabilities than unidirectional or angle-ply laminates for the following reasons. Unidirectional plies are stiffer and stronger in the 0-degree fiber direction than in the 90-degree direction perpendicular to the fibers.Thus,laminates made of uni- directional plies are ill-suited to carry load in the 90-degree direction.Angle-ply laminates with only two fiber directions do not resist well tensile loads applied along the symmetry axis.Plates made by the 10-percent rule minimize these short- comings. The specification of symmetrical layup is less restrictive than it may appear because the analyses of symmetrical plates(for which([B]=0)can readily be extended to unsymmetrical plates ([B]0)with the use of the reduced bending stiffness [D]*,defined as 3.4.5 [D*=[D]-[BA-[B (4.1) The deflections,maximum bending moments,buckling loads,and natural fre- quencies of unsymmetrical plates can be approximated by replacing [D]by [D]* in the expressions derived for symmetrical plates. 4.1 Governing Equations In this section we summarize the equations used in analyzing thin plates.We employ the x,y,z coordinate system.The origin is at the midplane for plates with symmetrical layup and at a suitably chosen reference plane for plates with unsymmetrical layup. The strains and curvatures of the reference plane(Fig.3.10)are (Egs.3.1,3.8) au° e= avo au°，8v° ax 9= ay y,= ay ax (4.2) 02w° 82w0 282w° Kx三 8x2 ay2 Kxy=一 axay 3 J.M.Whitney,Structural Analysis of Laminated Anisotropic Plates.Technomic,Lancaster. Pennsylvania,1987,p.203. 4 E.Reissner and Y.Stavsky,Bending and Stretching of Certain Types of Heterogeneous Aelotropic Elastic Plates.Journal of Applied Mechanics,Vol.28,402-408,1961. 5 J.E.Ashton,Approximate Solutions for Unsymmetrically Laminated Plates.Joural of Composite Materials,Vol.3,189-191,1969
90 THIN PLATES requires that the plate satisfy the following conditions: The plate is made of unidirectional plies. There are at least three ply orientations. The angles between the fibers are at least 15◦. The number of plies in each fiber direction is at least 10 percent of the total number of plies. Plates conforming to the 10-percent rule have better load bearing capabilities than unidirectional or angle-ply laminates for the following reasons. Unidirectional plies are stiffer and stronger in the 0-degree fiber direction than in the 90-degree direction perpendicular to the fibers. Thus, laminates made of unidirectional plies are ill-suited to carry load in the 90-degree direction. Angle-ply laminates with only two fiber directions do not resist well tensile loads applied along the symmetry axis. Plates made by the 10-percent rule minimize these shortcomings. The specification of symmetrical layup is less restrictive than it may appear because the analyses of symmetrical plates (for which ([B] = 0) can readily be extended to unsymmetrical plates ([B] = 0) with the use of the reduced bending stiffness [D] ∗, defined as 3,4,5 [D] ∗ = [D] − [B][A] −1 [B]. (4.1) The deflections, maximum bending moments, buckling loads, and natural frequencies of unsymmetrical plates can be approximated by replacing [D] by [D] ∗ in the expressions derived for symmetrical plates. 4.1 Governing Equations In this section we summarize the equations used in analyzing thin plates. We employ the x, y, z coordinate system. The origin is at the midplane for plates with symmetrical layup and at a suitably chosen reference plane for plates with unsymmetrical layup. The strains and curvatures of the reference plane (Fig. 3.10) are (Eqs. 3.1, 3.8) o x = ∂uo ∂x o y = ∂vo ∂y γ o xy = ∂uo ∂y + ∂vo ∂x κx = −∂2wo ∂x2 κy = −∂2wo ∂y2 κxy = −2∂2wo ∂x∂y , (4.2) 3 J. M. Whitney, Structural Analysis of Laminated Anisotropic Plates. Technomic, Lancaster, Pennsylvania, 1987, p. 203. 4 E. Reissner and Y. Stavsky, Bending and Stretching of Certain Types of Heterogeneous Aelotropic Elastic Plates. Journal of Applied Mechanics, Vol. 28, 402–408, 1961. 5 J. E. Ashton, Approximate Solutions for Unsymmetrically Laminated Plates. Journal of Composite Materials, Vol. 3, 189–191, 1969
4.1 GOVERNING EQUATIONS 91 Figure 4.1:Forces and loads acting on an element of the plate. where uo and vo are the displacements of the reference plane in the x and y directions,and w is the out-of-plane displacement (deflection)of this plane.The force-strain relationships are (Eq.3.21) [A1 A12 A16 B11 B12 B16 N A2 A 6 B12 Br B26 A6 6 A66 B16 B26 B66 M Bu B12 B16 D11 D12 D16 (4.3) Kx M B12 B22 B26 D12 D22 D26 Ky xy B16 B26 B66 D16 D26 D66 Kxy In the analyses we may employ either the equilibrium equations or the strain energy. The equilibrium equations are5 aNs aNg.=-px ax ay aNy +x」 (4.4) ay aNy=一Py 业+业 ax ay =一P V= OM:aMy ax V= aMy aMey ay ay ax (4.5) where pr,py,and p:are the components of the distributed surface load (per unit area);N,N,and Ny are the in-plane forces (per unit length);Vr and Vy are the transverse shear forces (per unit length);Mr.My and Mry are,respectively,the bending moments and the twist moment(per unit length)(Fig.4.1). 6 S.P.Timoshenko and S.Woinowsky-Krieger,Theory of Plates and Shells.2nd edition.McGraw-Hill, New York,1959,p.80
4.1 GOVERNING EQUATIONS 91 x y x y x y Nx Nxy Nyx Ny Mxy Mx My Myx Vy Vx px py pz Figure 4.1: Forces and loads acting on an element of the plate. where uo and vo are the displacements of the reference plane in the x and y directions, and wo is the out-of-plane displacement (deflection) of this plane. The force–strain relationships are (Eq. 3.21) Nx Ny Nxy Mx My Mxy = A11 A12 A16 B11 B12 B16 A12 A22 A26 B12 B22 B26 A16 A26 A66 B16 B26 B66 B11 B12 B16 D11 D12 D16 B12 B22 B26 D12 D22 D26 B16 B26 B66 D16 D26 D66 o x o y γ o xy κx κy κxy . (4.3) In the analyses we may employ either the equilibrium equations or the strain energy. The equilibrium equations are6 ∂Nx ∂x + ∂Nxy ∂y = −px ∂Ny ∂y + ∂Nxy ∂x = −py (4.4) ∂Vx ∂x + ∂Vy ∂y = −pz Vx = ∂Mx ∂x + ∂Mxy ∂y Vy = ∂My ∂y + ∂Mxy ∂x , (4.5) where px, py, and pz are the components of the distributed surface load (per unit area); Nx, Ny, and Nxy are the in-plane forces (per unit length); Vx and Vy are the transverse shear forces (per unit length); Mx, My and Mxy are, respectively, the bending moments and the twist moment (per unit length) (Fig. 4.1). 6 S. P. Timoshenko and S. Woinowsky-Krieger, Theory of Plates and Shells. 2nd edition. McGraw-Hill, New York, 1959, p. 80
92 THIN PLATES 4.1.1 Boundary Conditions The conditions alongeach edge of the plate must be specified.Boundary conditions for an edge parallel with the y-axis are given below. Along a built-in edge,the deflection wo,the rotation of the edge aw/ax,and the in-plane°，v°displacements are zero: w°=0 aw -=0°=v°=0. (4.6 ax Along a free edge,where no external loads are applied,the bending moment Mr,the replacement shear force?V+a Mry/ax,and the in-plane forces N,Ny are zero: Mx=0 Vx+ 8My=0N:=Nw=0. ay (4.7) Along a simply supported edge,the deflection wo,the bending moment Mr, and the in-plane forces M,Nry are zero: w°=0Mx=0N=Nxy=0. (4.8) When in-plane motions are prevented by the support,the in-plane forces are not zero(N≠0，Nxy≠O)whereas the in-plane displacements are zero: °=0v°=0. (4.9) For an edge parallel with the x-axis,the preceding boundary conditions hold with x and y interchanged. 4.1.2 Strain Energy As we noted previously,solutions to plate problems may be obtained by energy methods that require knowledge of the strain energy.For a linearly elastic material the strain energy is given by Eq.(2.200).Under plane-stress condition the stress components o:,txz,and tyz are zero(Eq.2.121),and the expression for the strain energy simplifies to Ly (oxex +oyey+txyrxy)dzdydx, (4.10) where h and hp are the distances from the reference plane to the plate's surfaces (Fig.3.12).The strain components are (Eq.3.7) (4.11) 7bid,p.84
92 THIN PLATES 4.1.1 Boundary Conditions The conditions along each edge of the plate must be specified. Boundary conditions for an edge parallel with the y-axis are given below. Along a built-in edge, the deflection wo, the rotation of the edge ∂wo/∂x, and the in-plane uo, vo displacements are zero: wo = 0 ∂wo ∂x = 0 uo = vo = 0. (4.6) Along a free edge, where no external loads are applied, the bending moment Mx, the replacement shear force7 Vx + ∂Mxy/∂x, and the in-plane forces Nx, Nxy are zero: Mx = 0 Vx + ∂Mxy ∂y = 0 Nx = Nxy = 0. (4.7) Along a simply supported edge, the deflection wo, the bending moment Mx, and the in-plane forces Nx, Nxy are zero: wo = 0 Mx = 0 Nx = Nxy = 0. (4.8) When in-plane motions are prevented by the support, the in-plane forces are not zero (Nx = 0, Nxy = 0) whereas the in-plane displacements are zero: uo = 0 vo = 0. (4.9) For an edge parallel with the x-axis, the preceding boundary conditions hold with x and y interchanged. 4.1.2 Strain Energy As we noted previously, solutions to plate problems may be obtained by energy methods that require knowledge of the strain energy. For a linearly elastic material the strain energy is given by Eq. (2.200). Under plane-stress condition the stress components σz, τxz, and τyz are zero (Eq. 2.121), and the expression for the strain energy simplifies to U = 1 2 ) Lx 0 ) Ly 0 ) ht −hb (σxx + σyy + τxyγxy) dzdydx, (4.10) where ht and hb are the distances from the reference plane to the plate’s surfaces (Fig. 3.12). The strain components are (Eq. 3.7) x y γxy = o x o y γ o xy + z κx κy κxy . (4.11) 7 Ibid., p. 84
4.2 DEFLECTION OF RECTANGULAR PLATES 93 The stresses and the strains at a point are related by (Eq.3.13) (4.12) By substituting Eqs.(4.11)and (4.12)into Eq.(4.10)and by utilizing the definitions of the [A,[B],[D]matrices (Eq.3.18),we obtain the following expression for the strain energy: 公 「A1 A12 A16 B11 B12 B16 e L 内2 A22 426 B12 B22 B26 1 U= 46 6 A66 B16 B26 B66 dydx. Kx B11 B12 B16 D11 Di2 D16 Kx 0 0 Ky B12 B2 B26 D12 D22 D26 Ky Kxy B16 B26 B66 D16 D26 D66」 Kxy (4.13) The superscript T denotes the transpose of the vector. 4.2 Deflection of Rectangular Plates 4.2.1 Pure Bending and In-Plane Loads We consider an unsupported rectangular plate subjected to pure bending and to in-plane loads(Fig.4.2).The in-plane forces and moments are related to the reference plane's strains and curvatures by Eq.(4.3).Six of the twelve quantities appearing in this equation must be specified as follows: N or e Mx or Kx Ny or e My or Ky (4.14) Nry or esy Mry or Kxy. With six of the quantities chosen(Eq.4.14),the remaining six may be obtained by solving the six simultaneous equations given by Eq.(4.3).Once the curvatures Figure 4.2:Rectangular plate subjected to bending and in-plane loads
4.2 DEFLECTION OF RECTANGULAR PLATES 93 The stresses and the strains at a point are related by (Eq. 3.13) σx σy τxy = [Q] x y γxy . (4.12) By substituting Eqs. (4.11) and (4.12) into Eq. (4.10) and by utilizing the definitions of the [A], [B], [D] matrices (Eq. 3.18), we obtain the following expression for the strain energy: U = 1 2 ) Lx 0 ) Ly 0 o x o y γ o xy κx κy κxy T A11 A12 A16 B11 B12 B16 A12 A22 A26 B12 B22 B26 A16 A26 A66 B16 B26 B66 B11 B12 B16 D11 D12 D16 B12 B22 B26 D12 D22 D26 B16 B26 B66 D16 D26 D66 o x o y γ o xy κx κy κxy dydx. (4.13) The superscript T denotes the transpose of the vector. 4.2 Deflection of Rectangular Plates 4.2.1 Pure Bending and In-Plane Loads We consider an unsupported rectangular plate subjected to pure bending and to in-plane loads (Fig. 4.2). The in-plane forces and moments are related to the reference plane’s strains and curvatures by Eq. (4.3). Six of the twelve quantities appearing in this equation must be specified as follows: Nx or o x Mx or κx Ny or o y My or κy Nxy or o xy Mxy or κxy. (4.14) With six of the quantities chosen (Eq. 4.14), the remaining six may be obtained by solving the six simultaneous equations given by Eq. (4.3). Once the curvatures Nx x Mx Mxy Nxy Nxy My Mxy Ny y z Figure 4.2: Rectangular plate subjected to bending and in-plane loads
94 THIN PLATES A 品 余 Figure 4.3:The different types of supports along the long edges of a transversely loaded long plate. are known,the deflection of the reference surface wo is calculated by using the relationships between the curvatures and the deflection(Eq.4.2)as follows: 82w° 82wo 282w° Kx=一 8x2 Ky=- ay2 Kxy=一 (4.15) The following deflection satisfies these relationships: w=-受-3-受y (4.16 This expression for wo does not include the deflection of the reference plane due to rigid-body motion. 4.2.2 Long Plates We consider a long rectangular plate whose length Ly is large compared with its width L.The long edges may be built-in,simply supported,or free,as shown in Figure 4.3.The plate is subjected to a distributed transverse load p.Neither this load nor the edge supports vary along the longitudinal y direction. When the length of the plate is large compared with its width,away from the short edges the deflected surface may be assumed to be cylindrical(cylindrical deformation,Fig.4.4),and the forces and moments do not vary appreciably along the length.With these approximations the analysis simplifies considerably.Before we undertake the analysis,we establish the length-to-width ratios for which a plate may be considered long. For an isotropic plate the long-plate approximation for the deflection is rea- sonable whens isotropic plate. (4.17) where L,and L are the length and the width of the isotropic plate,respectively. We now establish for orthotropic plates the length-to-width ratios at which the 8bid,p.118
94 THIN PLATES y x Ly Lx Lx p z Figure 4.3: The different types of supports along the long edges of a transversely loaded long plate. are known, the deflection of the reference surface wo is calculated by using the relationships between the curvatures and the deflection (Eq. 4.2) as follows: κx = −∂2wo ∂x2 κy = −∂2wo ∂y2 κxy = −2∂2wo ∂x∂y . (4.15) The following deflection satisfies these relationships: wo = −κx 2 x2 − κy 2 y2 − κxy 2 xy. (4.16) This expression for wo does not include the deflection of the reference plane due to rigid-body motion. 4.2.2 Long Plates We consider a long rectangular plate whose length Ly is large compared with its width Lx. The long edges may be built-in, simply supported, or free, as shown in Figure 4.3. The plate is subjected to a distributed transverse load p. Neither this load nor the edge supports vary along the longitudinal y direction. When the length of the plate is large compared with its width, away from the short edges the deflected surface may be assumed to be cylindrical (cylindrical deformation, Fig. 4.4), and the forces and moments do not vary appreciably along the length. With these approximations the analysis simplifies considerably. Before we undertake the analysis, we establish the length-to-width ratios for which a plate may be considered long. For an isotropic plate the long-plate approximation for the deflection is reasonable when8 L y L x > 3 isotropic plate, (4.17) where L y and L x are the length and the width of the isotropic plate, respectively. We now establish for orthotropic plates the length-to-width ratios at which the 8 Ibid., p. 118.
4.2 DEFLECTION OF RECTANGULAR PLATES 95 Figure 4.4:Cylindrical deformation of a long rectangular plate. long-plate approximation may be applied.To this end we observe that the deflec- tions of an orthotropic plate(with length Ly and width L)and an isotropic plate (with length L,and width L)are similar when (page 109) (4.18) Dit Ly Ly. VDz Thus,from Egs.(4.17)and(4.18)we have that the long-plate approximation is reasonable when the following inequality is satisfied: 2器 orthotropic plate. (4.19) This formula,which is established for orthotropic plates,may also be used as a guide for plates whose layup is not orthotropic. We now proceed with the analysis of long plates in cylindrical bending.The generator of this cylindrical surface is parallel to the longitudinal y-axis of the plate.The curvatures Ky and Kxy of the plate are zero K=0 Kty =0, (4.20) and Kx is (Eq.4.2) 82w° Kx=一 8x2 (4.21) Away from the short edges the forces and moments do not vary along the length of the plate.Thus,from the last of Eq.(4.4)and the first of Eq.(4.5)we have dv d +pz=0 (4.22) dM-V.=0. d (4.23) The load is perpendicular to the surface,and for simplicity we replace p:by p. Thus,by substituting Vr from Eq.(4.23)into Eq.(4.22)we obtain the equilibrium equation d d2+p=0. (4.24) We note that this equation,representing equilibrium,is independent of the material
4.2 DEFLECTION OF RECTANGULAR PLATES 95 a b b a x y Figure 4.4: Cylindrical deformation of a long rectangular plate. long-plate approximation may be applied. To this end we observe that the deflections of an orthotropic plate (with length Ly and width Lx) and an isotropic plate (with length L y and width L x) are similar when (page 109) L x = Lx 4 $ D11 D22 L y = Ly. (4.18) Thus, from Eqs. (4.17) and (4.18) we have that the long-plate approximation is reasonable when the following inequality is satisfied: Ly Lx > 3 4 , D11 D22 orthotropic plate. (4.19) This formula, which is established for orthotropic plates, may also be used as a guide for plates whose layup is not orthotropic. We now proceed with the analysis of long plates in cylindrical bending. The generator of this cylindrical surface is parallel to the longitudinal y-axis of the plate. The curvatures κy and κxy of the plate are zero κy = 0 κxy = 0, (4.20) and κx is (Eq. 4.2) κx = −∂2wo ∂x2 . (4.21) Away from the short edges the forces and moments do not vary along the length of the plate. Thus, from the last of Eq. (4.4) and the first of Eq. (4.5) we have dVx dx + pz = 0 (4.22) dMx dx − Vx = 0. (4.23) The load is perpendicular to the surface, and for simplicity we replace pz by p. Thus, by substituting Vx from Eq. (4.23) into Eq. (4.22) we obtain the equilibrium equation d2Mx dx2 + p = 0. (4.24) We note that this equation, representing equilibrium, is independent of the material.
96 THIN PLATES y L,=700mm SS SS Figure 4.5:The plate in Example 4.1. SS L.=200mm Symmetrical layup.The layup of the plate is symmetrical ([B]=0).We now concern ourselves only with the bending moment M,which,from Eqs.(4.3)and (4.20),is My D11Kx. (4.25) The element Di1 of the matrix [D]is given by Eq.(3.20). By substituting Eq.(4.25)into Eq.(4.24)and by using Eq.(4.21),we obtain the following equilibrium equation for the anisotropic long plate: dw°-p=0 long plate dx+Du (4.26) symmetrical layup. The equation governing the deflection of a transversely loaded isotropic beam 69 dw p' dxE=0 isotropic beam, (4.27) where E is Young's modulus,I is the moment of inertia about the y-axis,and p' is the transverse load per unit length. By comparing Eq.(4.26)and(4.27),we see that the equations describing the deflections of a long plate(symmetrical layup)and an isotropic beam are similar. Consequently,the deflection of a long plate(symmetrical layup)with bending stiffness Du is the same as the deflection of an isotropic beam with bending stiffness EI when the numerical values of the loads are equal (p=p').(Note however that p is per unit area and p'is per unit length.)Thus,the deflection of a long plate with symmetrical layup can be obtained by replacing El/p'by Du/p in the expression given for the deflection of the corresponding isotropic beam. 4.1 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/+4551.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.4.5).Calculate the maximum 9E.P.Popov,Engineering Mechanics of Solids.Prentice-Hall,Englewood Cliffs,New Jersey,1990, p.505. 10 W.D.Pilkey,Formulas for Stresses,Strains,and Structural Matrices.John Wiley Sons,New York, 1994
96 THIN PLATES Lx = 200 mm Ly = 700 mm y x ss ss ss ss p Figure 4.5: The plate in Example 4.1. Symmetrical layup. The layup of the plate is symmetrical ([B] = 0). We now concern ourselves only with the bending moment Mx, which, from Eqs. (4.3) and (4.20), is Mx = D11κx. (4.25) The element D11 of the matrix [D] is given by Eq. (3.20). By substituting Eq. (4.25) into Eq. (4.24) and by using Eq. (4.21), we obtain the following equilibrium equation for the anisotropic long plate: d4wo dx4 − p D11 = 0 long plate symmetrical layup. (4.26) The equation governing the deflection of a transversely loaded isotropic beam is9 d4w dx4 − p EI = 0 isotropic beam, (4.27) where E is Young’s modulus, I is the moment of inertia about the y-axis, and p is the transverse load per unit length. By comparing Eq. (4.26) and (4.27), we see that the equations describing the deflections of a long plate (symmetrical layup) and an isotropic beam are similar. Consequently, the deflection of a long plate (symmetrical layup) with bending stiffness D11 is the same as the deflection of an isotropic beam with bending stiffness EI when the numerical values of the loads are equal (p = p ). (Note however that p is per unit area and p is per unit length.) Thus, the deflection of a long plate with symmetrical layup can be obtained by replacing EI/p by D11/p in the expression10 given for the deflection of the corresponding isotropic beam. 4.1 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. 4.5). Calculate the maximum 9 E. P. Popov, Engineering Mechanics of Solids. Prentice-Hall, Englewood Cliffs, New Jersey, 1990, p. 505. 10 W. D. Pilkey, Formulas for Stresses, Strains, and Structural Matrices. John Wiley & Sons, New York, 1994.
4.2 DEFLECTION OF RECTANGULAR PLATES deflection,the maximum bending moments,and the stresses and strains in each layer. Solution.The bending stiffnesses of the plate are (Table 3.7,page 84)Du= 45.30 N.m and D22=25.26 N.m.We may treat this plate as long when the fol- lowing condition is met (Eq.4.19): Du (4.28) In the present problem the terms in this inequality are Ly/Lx=3.5 and 3D1/D22=3.47.Thus,the preceding condition is satisfied and the long-plate expressions may be used. The maximum deflection of a simply supported beam is(Table 7.3,page 332) 0三 5 PLA 384EI (4.29) The maximum deflection of the plate is obtained by replacing El/p'by Di/p (see page 96).For the plate under consideration Du=45.30N.m and L =0.2 m, and we have 5pL=0.0230m=23.0mm. ǔ=384D1 (4.30) The bending moments are(Eq.3.27) M D11Kx+D12Ky+D16Kty (4.31) My D12Kx D22Ky D26Kxy. (4.32) For a long plate Ky and Kry are zero(Eq.4.20),and Mr and My are M =D11Kx My D12Kx. (4.33) The maximum bending moment M,which arises at Lx/2,is(see Table 7.3, page 332) M=250.00N (4.34) 8 m From Eqs.(4.33)and(4.34)we have (4.35) m From Egs.(4.33)and (4.34)the maximum bending moment My (at L/2)is My=D12Kx D2M,=107.75 N.m (4.36) D11 m 0.431
4.2 DEFLECTION OF RECTANGULAR PLATES 97 deflection, the maximum bending moments, and the stresses and strains in each layer. Solution. The bending stiffnesses of the plate are (Table 3.7, page 84) D11 = 45.30 N · m and D22 = 25.26 N · m. We may treat this plate as long when the following condition is met (Eq. 4.19): Ly Lx > 3 4 , D11 D22 . (4.28) In the present problem the terms in this inequality are Ly/Lx = 3.5 and 3 √4 D11/D22 = 3.47. Thus, the preceding condition is satisfied and the long-plate expressions may be used. The maximum deflection of a simply supported beam is (Table 7.3, page 332) w = 5 384 p L4 EI . (4.29) The maximum deflection of the plate is obtained by replacing EI/p by D11/p (see page 96). For the plate under consideration D11 = 45.30 N · m and Lx = 0.2 m, and we have w = 5 384 pL4 x D11 = 0.0230 m = 23.0 mm. (4.30) The bending moments are (Eq. 3.27) Mx = D11κx + D12κy + D16κxy (4.31) My = D12κx + D22κy + D26κxy. (4.32) For a long plate κy and κxy are zero (Eq. 4.20), and Mx and My are Mx = D11κx My = D12κx. (4.33) The maximum bending moment Mx, which arises at Lx/2, is (see Table 7.3, page 332) Mx = pL2 x 8 = 250.00 N · m m . (4.34) From Eqs. (4.33) and (4.34) we have κx = Mx D11 = 5.52 1 m. (4.35) From Eqs. (4.33) and (4.34) the maximum bending moment My (at Lx/2) is My = D12κx = D12 D11 %&'( 0.431 Mx = 107.75 N · m m . (4.36)
98 THIN PLATES ±45 012 0 .0 0.6 5000 2000g % ±452 Figure 4.6:The nonzero strains and stresses across the thickness of the plate at L/2 in Exam- ple 4.1.The unit of o is 106 N/m2. For the long plate(Ky=Kty =0)the strains at Lx/2 are (Eq.3.7) Ex =Kx=5.52z Ey=0 (4.37) xy=0. The strain distribution at Lr/2 is shown in Figure 4.6.The stresses are calculated by(Eq.3.11) Ox Q11 012 Q16 1 226 (4.38) Q16 Q66 The stiffness matrices for the fabric and for the unidirectional layer are given by Egs.(3.65)and(3.66).The stresses in the top layer (where z=h/2=0.001 m) at Lx/2 are 45.65 251.95 36.55 109×5.52×0.001 201.73 106 N Q16 0 0 (4.39) The stresses in the other layers are calculated similarly.The results are shown in Figure 4.6. Unsymmetrical layup.The layup of the plate is unsymmetrical.One of the long edges must be restrained along the lengthwise direction.With the plate thus restrained,the strain in the longitudinal y direction is zero throughout the plate: e8=0. (4.40) Furthermore,we take the shear force Nry to be zero: Nxy =0. (4.41) Equation(4.41)is valid when one of the long edges of the plate is free to move in the lengthwise y direction.It is only an approximation when the lengthwise motion
98 THIN PLATES z 0 −1 1 0.6 εx z 0 −1 1 500 σx z 0 −1 1 200 σy (%) 012 45f +_ +_ 2 45f 2 Figure 4.6: The nonzero strains and stresses across the thickness of the plate at Lx/2 in Example 4.1. The unit of σ is 106 N/m2. For the long plate (κy = κxy = 0) the strains at Lx/2 are (Eq. 3.7) x = κx z = 5.52z y = 0 (4.37) γxy = 0. The strain distribution at Lx/2 is shown in Figure 4.6. The stresses are calculated by (Eq. 3.11) σx σy τxy = Q11 Q12 Q16 Q12 Q22 Q26 Q16 Q26 Q66 x y γxy . (4.38) The stiffness matrices for the fabric and for the unidirectional layer are given by Eqs. (3.65) and (3.66). The stresses in the top layer (where z = h/2 = 0.001 m) at Lx/2 are σx σy τxy = Q11 Q12 Q16 x = 45.65 36.55 0 109 × 5.52 × 0.001 = 251.95 201.73 0 106 N m2 . (4.39) The stresses in the other layers are calculated similarly. The results are shown in Figure 4.6. Unsymmetrical layup. The layup of the plate is unsymmetrical. One of the long edges must be restrained along the lengthwise direction. With the plate thus restrained, the strain in the longitudinal y direction is zero throughout the plate: o y = 0. (4.40) Furthermore, we take the shear force Nxy to be zero: Nxy = 0. (4.41) Equation (4.41) is valid when one of the long edges of the plate is free to move in the lengthwise y direction. It is only an approximation when the lengthwise motion
