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