104 THIN PLATES L=700 mm Figure 4.8:The plate in Example 4.2. SS L=200 mm The plate is orthotropic,and the bending stiffnesses are (Table 3.7,page 84) D1=45.30N·m,D2=25.26N.m,D12=19.52N.m,D66=20.62Nm.The maximum deflection occurs at the center of the plate,where x =L/2=0.1 m and y=Ly/2=0.35 m.From Eq.(4.68),wij x 103 are Ai 1 2 3 45 6 7 1 24.0389 02.994500.668300.2033 2 0 0 0 0 0 0 0 3 0.1181 00.033000.0148 00.0075 4 0 0 0 0 0 0 0 5 0.0093 00.002900.001500.0009 6 0 0 0 0 0 0 0 0.0017 00.0006 00.0003 00.0002. At the center of the plate the deflection is m号如号 (4.70) 2 We chose to perform the summation up to i=j=7.For i,j=2,4,6,the sine is zero;for i,j=1,5 the sine is unity,and for i,j=3,7 the sine is minus one.The resulting deflection at the midpoint is w°=0.0214m=21.4mm. (4.71) We now assess the length-to-width ratios under which the long-plate approxi- mation is reasonable.To this end,we calculated the maximum deflections of the plate,keeping the width Lx the same while changing the length Ly.In Figure 4.9 we plot the maximum deflections thus calculated versus Ly.In this figure we also included the deflection given by the long-plate approximation(Eq.4.30).The results in this figure show that,in accordance with Eq.(4.19),the long-plate for- mula approximates the deflection well(within 8 percent)when Ly is greater than 3LxD11/D2五=0.694m
104 THIN PLATES Lx = 200 mm Ly = 700 mm y x ss ss ss ss p Figure 4.8: The plate in Example 4.2. The plate is orthotropic, and the bending stiffnesses are (Table 3.7, page 84) D11 = 45.30 N · m, D22 = 25.26 N · m, D12 = 19.52 N · m, D66 = 20.62 N · m). The maximum deflection occurs at the center of the plate, where x = Lx/2 = 0.1 m and y = Ly/2 = 0.35 m. From Eq. (4.68), wi j × 103 are i\ j 1 23456 7 1 24.0389 0 2.9945 0 0.6683 0 0.2033 2 0 00000 0 3 0.1181 0 0.0330 0 0.0148 0 0.0075 4 0 00000 0 5 0.0093 0 0.0029 0 0.0015 0 0.0009 6 0 00000 0 7 0.0017 0 0.0006 0 0.0003 0 0.0002 . At the center of the plate the deflection is wo = * 7 i=1 * 7 j=1 wi j sin iπ 2 sin jπ 2 . (4.70) We chose to perform the summation up to i = j = 7. Fori, j = 2, 4, 6, the sine is zero; for i, j = 1, 5 the sine is unity, and for i, j = 3, 7 the sine is minus one. The resulting deflection at the midpoint is wo = 0.0214 m = 21.4 mm. (4.71) We now assess the length-to-width ratios under which the long-plate approximation is reasonable. To this end, we calculated the maximum deflections of the plate, keeping the width Lx the same while changing the length Ly. In Figure 4.9 we plot the maximum deflections thus calculated versus Ly. In this figure we also included the deflection given by the long-plate approximation (Eq. 4.30). The results in this figure show that, in accordance with Eq. (4.19), the long-plate formula approximates the deflection well (within 8 percent) when Ly is greater than 3Lx √4 D11/D22 = 0.694 m
4.2 DEFLECTION OF RECTANGULAR PLATES 105 20 long-plate approx 10 0.2 0.4 0.60.6940.8 Length,L (m) Figure 4.9:Maximum deflection of the plate in Example 4.2 as a function of the plate length. The bending moments at the center of the plate are(Eq.4.67) M D11D12 M. D2D22 233.941 N.m 102.18J (4.72) m The twist moment at the corner of the plate (x=y=0)is(Eq.4.67) Mxy =-D66 Ly 77 -Ds∑72=-B4s" (4.73) m 4.3 Example.A 0.2-m-long and 0.2-m-wide rectangular plate is made of graphite epoxy unidirectional plies.The material properties are given in Table 3.6(page 81). The layup is [02/452/902/-4521.The plate,simply supported along the four edges (Fig.4.10),is subjected to a uniformly distributed transverse load p =50 000 N/m2. Calculate the maximum deflection and the maximum moments. Solution.The layup of the plate is symmetrical but is not orthotropic.The ben- ding stiffnesses are Du=34.61 N.m,D22 12.34 N.m,D2 =4.58 N.m, =200mm SS D SS ss L:=200 mm Figure 4.10:The plate in Example 4.3
4.2 DEFLECTION OF RECTANGULAR PLATES 105 0 0.2 0.4 0.6 0.694 0.8 1 long-plate approx Deflection, (mm) wo Length, (m) Ly 10 20 Figure 4.9: Maximum deflection of the plate in Example 4.2 as a function of the plate length. The bending moments at the center of the plate are (Eq. 4.67) Mx My = D11 D12 D12 D22 ! 2 7 i=1 2 7 j=1 wi j iπ Lx 2 sin iπ 2 sin jπ 2 2 7 i=1 2 7 j=1 wi j jπ Ly 2 sin iπ 2 sin jπ 2 = 233.94 102.18 N · m m . (4.72) The twist moment at the corner of the plate (x = y = 0) is (Eq. 4.67) Mxy = −D66* 7 i=1 * 7 j=1 wi j2 iπ Lx jπ Ly cos iπx Lx cos jπy Ly = −D66* 7 i=1 * 7 j=1 wi j2 iπ Lx jπ Ly = −113.43 N · m m . (4.73) 4.3 Example. A 0.2-m-long and 0.2-m-wide rectangular plate is made of graphite epoxy unidirectional plies. The material properties are given in Table 3.6 (page 81). The layup is [02/452/902/−452]s. The plate, simply supported along the four edges (Fig. 4.10), is subjected to a uniformly distributed transverse load p = 50 000 N/m2. Calculate the maximum deflection and the maximum moments. Solution. The layup of the plate is symmetrical but is not orthotropic. The bending stiffnesses are D11 = 34.61 N · m, D22 = 12.34 N · m, D12 = 4.58 N · m, Ly = 200 mm y x p Lx = 200 mm ss ss ss ss Figure 4.10: The plate in Example 4.3
106 THIN PLATES Table 4.2.The maximum deflection and the maximum bending and twist moments calculated by the numerical solution and by the orthotropic approximation for the plate in Example 4.3 Mxy Mxy w M My X=y=0 X=Lx.y=0 mm N.m/m numerical 17.60 160.65 64.53 -69.28 36.47 orthotropic 16.93 154.07 63.21 -49.56 49.56 approximation D66=5.14N.m,D6=3.34N.m,D26=3.34N.m(Table3.7,page84).The maximum deflection and the maximum bending and twist moments must be cal- culated from Egs.(4.58),(4.64),(4.66),and(4.67).With the preceding stiffnesses the calculations yield the results given in Table 4.2(first row). The layup follows the 10-percent rule (page 89),and we treat the plate as orthotropic.The deflection of the plate is(Eq.4.58) (4.74) Ly Since the plate is treated as orthotropic Di6=D26=0,and the relevant bend- ing stiffnesses are Du=34.61 N.m,D22 12.34 N.m,D12 =4.58 N.m,D66 5.14N.m(Table 3.7,page 84).The maximum deflection occurs at the center of the plate,where x =Lx/2=0.1 m and y=Ly/2=0.1 m.From Eq.(4.68)wij x 103 are Ai 1 2 3 5 6 7 1 17.3628 00.3409 00.03140 0.0061 2 0 0 0 0 0 0 0 3 0.143900.023800.005200.0014 4 0 0 0 00 0 0 5 0.011900.0030 00.001100.0004 0 00 00 00 1 0.002200.0007 00.000300.0001 At the center of the plate the deflection is 7 w° 2 (4.75) We perform the summation up to i =j=7.For i,j =2,4,6 the sine is zero, for i,j=1,5 the sine is unity,and for i,j=3.7 the sine is minus one.Thus,the resulting deflection at the center of the plate is w°=0.01693m=16.93mm. (4.76)
106 THIN PLATES Table 4.2. The maximum deflection and the maximum bending and twist moments calculated by the numerical solution and by the orthotropic approximation for the plate in Example 4.3 Mxy Mxy wo Mx My x = y = 0 x = L x , y = 0 mm N · m/m numerical 17.60 160.65 64.53 −69.28 36.47 orthotropic 16.93 154.07 63.21 −49.56 49.56 approximation D66 = 5.14 N · m, D16 = 3.34 N · m, D26 = 3.34 N · m (Table 3.7, page 84). The maximum deflection and the maximum bending and twist moments must be calculated from Eqs. (4.58), (4.64), (4.66), and (4.67). With the preceding stiffnesses the calculations yield the results given in Table 4.2 (first row). The layup follows the 10-percent rule (page 89), and we treat the plate as orthotropic. The deflection of the plate is (Eq. 4.58) wo = * I i=1 * J j=1 wi j sin iπx Lx sin jπy Ly . (4.74) Since the plate is treated as orthotropic D16 = D26 = 0, and the relevant bending stiffnesses are D11 = 34.61 N · m, D22 = 12.34 N · m, D12 = 4.58 N · m, D66 = 5.14 N · m (Table 3.7, page 84). The maximum deflection occurs at the center of the plate, where x = Lx/2 = 0.1 m and y = Ly/2 = 0.1 m. From Eq. (4.68) wi j × 103 are i\ j 1 234567 1 17.3628 0 0.3409 0 0.0314 0 0.0061 2 0 000000 3 0.1439 0 0.0238 0 0.0052 0 0.0014 4 0 000000 5 0.0119 0 0.0030 0 0.0011 0 0.0004 6 0 000000 7 0.0022 0 0.0007 0 0.0003 0 0.0001 At the center of the plate the deflection is wo = * 7 i=1 * 7 j=1 wi j sin iπ 2 sin jπ 2 . (4.75) We perform the summation up to i = j = 7. For i, j = 2, 4, 6 the sine is zero, for i, j = 1, 5 the sine is unity, and for i, j = 3, 7 the sine is minus one. Thus, the resulting deflection at the center of the plate is wo = 0.01693 m = 16.93 mm. (4.76)
4.2 DEFLECTION OF RECTANGULAR PLATES 107 The bending moments at the center of the plate are (Eq.4.67) D D12 154.071N.m 63.21 (4.77) m The twist moment at the corner of the plate (x=y=0)is(Eq.4.67) 7 7 M,=-D6∑ in jn inx。jπy 2wi工 cos Lx cos Ly -D6∑2wj 77 i江Π=-49.56Nm. (4.78) m The maximum deflection and the maximum moments thus calculated are in- cluded in Table 4.2(second row).The maximum bending moments and the maxi- mum deflections calculated by the numerical method and the orthotropic approxi- mation are in close agreement,but the maximum twist moments differ significantly. 4.2.4 Plates with Built-In Edges-Orthotropic and Symmetrical Layup We consider a rectangular plate with length L and width Ly built-in along its four edges(Fig.4.11).The layup is orthotropic and symmetrical.The plate is subjected to a uniformly distributed load p. The potential energy of the plate is obtained from Egs.(4.55)and(4.56)by setting Di6 and D26 equal to zero: m=U+2 a()+()》 82w°82w° +D66 (4.79) Figure 4.11:Rectangular plate with built-in edges
4.2 DEFLECTION OF RECTANGULAR PLATES 107 The bending moments at the center of the plate are (Eq. 4.67) Mx My = D11 D12 D12 D22! 2 7 i=1 2 7 j=1 wi j iπ Lx 2 sin iπ 2 sin jπ 2 2 7 i=1 2 7 j=1 wi j jπ Ly 2 sin iπ 2 sin jπ 2 = 154.07 63.21 N · m m . (4.77) The twist moment at the corner of the plate (x = y = 0) is (Eq. 4.67) Mxy = −D66* 7 i=1 * 7 j=1 2wi j iπ Lx jπ Ly cos iπx Lx cos jπy Ly = −D66* 7 i=1 * 7 j=1 2wi j iπ Lx jπ Ly = −49.56N · m m . (4.78) The maximum deflection and the maximum moments thus calculated are included in Table 4.2 (second row). The maximum bending moments and the maximum deflections calculated by the numerical method and the orthotropic approximation are in close agreement, but the maximum twist moments differ significantly. 4.2.4 Plates with Built-In Edges – Orthotropic and Symmetrical Layup We consider a rectangular plate with length Lx and width Ly built-in along its four edges (Fig. 4.11). The layup is orthotropic and symmetrical. The plate is subjected to a uniformly distributed load p. The potential energy of the plate is obtained from Eqs. (4.55) and (4.56) by setting D16 and D26 equal to zero: πp = U + = 1 2 ) Lx 0 ) Ly 0 / D11 ∂2wo ∂x2 2 + D22 ∂2wo ∂y2 2 + D66 2∂2wo ∂x∂y 2 + 2D12 ∂2wo ∂x2 ∂2wo ∂y2 − pwo 0 dydx. (4.79) x y z L Ly x p Figure 4.11: Rectangular plate with built-in edges
108 THIN PLATES The moments at a point x,y are (Eqs.4.2 and 3.27) a2w° 82w0 M.=-Du ax-Dia ay 82w0 82w° My=-Dia a2-Da ay2 (4.80) 282w° Mey =-Do6 axay where wo is the deflection and Du,Di2,D22,D66 are the elements of the stiffness matrix in the x-y coordinate system. The displacements and moments can be calculated when the plate's bending stiffnesses satisfy the following Huber orthotropy relationship12(see Eq.4.153 with K =1): 1 Do=(Du Daz-Du2). (4.81) Although this relationship may not hold exactly,we adopt it for calculating the displacements and the moments.Possible errors introduced by this relationship are discussed on page 111. We introduce the variable =后 (4.82) where a is a constant defined as Du (4.83) Equations(4.79)-(4.83)yield the potential energy and the moments (per unit length)as follows: Lx/a Ly )+)+- D 282w° 2 ax'ay D 82w082w0 +21 D11D22 ax2 ay2 -pwo dydx' (4.84) 82w0 D ay2 82w° 82w° V Dx ay? (4.85) D11D22 2a2w D1D22 ax'ay 12 S.P.Timoshenko and S.Woinowsky-Krieger,Theory of Plates and Shells.2nd edition.McGraw-Hill, New York,1959,p.366
108 THIN PLATES The moments at a point x, y are (Eqs. 4.2 and 3.27) Mx = −D11 ∂2wo ∂x2 − D12 ∂2wo ∂y2 My = −D12 ∂2wo ∂x2 − D22 ∂2wo ∂y2 (4.80) Mxy = −D66 2∂2wo ∂x∂y , where wo is the deflection and D11, D12, D22, D66 are the elements of the stiffness matrix in the x–y coordinate system. The displacements and moments can be calculated when the plate’s bending stiffnesses satisfy the following Huber orthotropy relationship12 (see Eq. 4.153 with K = 1): D66 = 1 2 5 D11D22 − D12 . (4.81) Although this relationship may not hold exactly, we adopt it for calculating the displacements and the moments. Possible errors introduced by this relationship are discussed on page 111. We introduce the variable x = x α , (4.82) where α is a constant defined as α = 4 , D11 D22 . (4.83) Equations (4.79)–(4.83) yield the potential energy and the moments (per unit length) as follows: πp = 1 2 L )x /α 0 ) Ly 0 D22 ∂2wo ∂x2 2 + ∂2wo ∂y2 2 + 1 2 1 − , D2 12 D11D22 2∂2wo ∂x ∂y 2 + 2 , D2 12 D11D22 ∂2wo ∂x2 ∂2wo ∂y2 − pwo dydx (4.84) Mx = α2 −D22 ∂2wo ∂x2 − , D2 12 D11D22 D22 ∂2wo ∂y2 My = − , D2 12 D11D22 D22 ∂2wo ∂x2 − D22 ∂2wo ∂y2 (4.85) Mxy = α −D22 1 2 1 − , D2 12 D11D22 2∂2wo ∂x ∂y . 12 S. P. Timoshenko and S. Woinowsky-Krieger, Theory of Plates and Shells. 2nd edition. McGraw-Hill, New York, 1959, p. 366.