BENDING TO A CYLINDRICAL SURFACE 11 bending moment is several times smaller than the moment gl2/8 which would be obtained if there were no tensile reactions at the ends of the strip. The direct tensile stress o1 and the maximum bending stress o:are now readily expressed in terms of u,g,and the plate constants as follows: Eu 01= S 4uD (10) 6 (11) The maximum stress in the plate is then 0mx=01十02 To show how the curves in Figs.4 and 5 can be used in calculating maximum stresses,let us take a numerical example and assume that a long rectangular steel plate 50 in.wide and in.thick carries a uniformly distributed load g 20 psi.We start with a computation of vUo: E 30.1061 =0.01648 Then,from tables, 10g10(104VUo)=2.217 From the curve A in Fig.4 we find u =3.795,and from Fig.5 we obtain 中0=0.1329. Now,computing stresses by using Eqs.(10)and (11),we find 30·106·3.79521 1=31-0.3910=15,830psi 02=￥·20·104.0.1329=19,930psi 0mx=1十02=35,760psi In calculating the maximum deflection we substitute z =1/2 in Eq.(6) of the deflection curve.In this manner we obtain 5gl 384Dfo(u) (12) cch4-1+号 where fo(u)= 5u4 24 To simplify calculations,values of fo(u)are given by the curve in Fig.5. If there were no tensile reactions at the ends of the strip,the maximum
bending moment is several times smaller than the moment ql2 /S which would be obtained if there were no tensile reactions at the ends of the strip. The direct tensile stress <T\ and the maximum bending stress o-2 are now readily expressed in terms of U1 q, and the plate constants as follows: (10) (H) The maximum stress in the plate is then To show how the curves in Figs. 4 and 5 can be used in calculating maximum stresses, let us take a numerical example and assume that a long rectangular steel plate 50 in. wide and ^- in. thick carries a uniformly distributed load q = 20 psi. We start with a computation of vTTo: Then, from tables, From the curve A in Fig. 4 we find u = 3.795, and from Fig. 5 we obtain ^o = 0.1329. Now, computing stresses by using Eqs. (10) and (11), we find In calculating the maximum deflection we substitute x = 1/2 in Eq. (6) of the deflection curve. In this manner we obtain (12) where To simplify calculations, values of fo(u) are given by the curve in Fig. 5. If there were no tensile reactions at the ends of the strip, the maximum
12 THEORY OF PLATES AND SHELLS deflection would be 5gl4/384D.The effect of the tensile reactions is given by the factor fo(u),which diminishes rapidly with increasing u. Using Fig.5 in the numerical example previously discussed,we find that for u=3.795 the value of fo(u)is 0.145.Substituting this value in Eq.(12),we obtain 0mx=4.74·0.145=0.688in. It is seen from Eg.(8)that the tensile parameter u depends,for a given material of the plate,upon the intensity of the load g and the 60,000 Stresses in steel plotes with simply supported edges 50,000 L/n=220 40,000 L/h200 L/h=80 可 1/h60 /n140 l/h= 2 e30,000 1/h= 00… 80 1/h= 20,000 10,000 Rotio width:thickness =L/h 10 20 30 40 Lood in Ib per sq in. FIG.6 ratio l/h of width to thickness of the plate.From Eqs.(10)and (11) we see that the stresses o and oa are also functions of u,g,and l/h. Therefore,the maximum stress in the plate depends only on the load g and the ratio l/h.This means that we can plot a set of curves giving maximum stress in terms of g,each curve in the set corresponding to a particular value of l/h.Such curves are given in Fig.6.It is seen that because of the presence of tensile forces S,which increase with the load, the maximum stress is not proportional to the load g;and for large values of g this stress does not vary much with the thickness of the plate.By taking the curve marked l/h=100 and assuming g=20 psi,we obtain from the curve the valuecalculated before in the numerical example
deflection would be 5gZ4 /384D. The effect of the tensile reactions is given by the factor /o(w), which diminishes rapidly with increasing u. Using Fig. 5 in the numerical example previously discussed, we find that for u = 3.795 the value of fo(u) is 0.145. Substituting this value in Eq. (12), we obtain «W = 4.74 • 0.145 = 0.688 in. It is seen from Eq. (8) that the tensile parameter u depends, for a given material of the plate, upon the intensity of the load q and the Stresses in steel plates with simply supported edges Stress in Ib per sq in. Ratio width : thickness Lood in Ib per sq in. FIG . 6 ratio l/h of width to thickness of the plate. From Eqs. (10) and (11) we see that the stresses <ri and o-2 are also functions of u, q, and l/h. Therefore, the maximum stress in the plate depends only on the load q and the ratio l/h. This means that we can plot a set of curves giving maximum stress in terms of g, each curve in the set corresponding to a particular value of l/h. Such curves are given in Fig. 6. It is seen that because of the presence of tensile forces S, which increase with the load, the maximum stress is not proportional to the load q; and for large values of q this stress does not vary much with the thickness of the plate. By taking the curve marked l/h = 100 and assuming q — 20 psi, we obtain from the curve the value o^* calculated before in the numerical example
BENDING TO A CYLINDRICAL SURFACE 13 3.Cylindrical Bending of Uniformly Loaded Rectangular Plates with Built-in Edges.We assume that the longitudinal edges of the plate are fixed in such a manner that they cannot rotate.Taking an elemental strip of unit width in the same manner as before(Fig.1)and denoting by Mo the bending moment per unit length acting on the longitudinal edges of the plate,the forces acting on the strip will be as shown in Fig.7. The bending moment at any cross section of the strip is M-器-警-如+， Substituting this expression in Eq.(4),we obtain 器-8-骆+%- (a) The general solution of this equation,using notation(5),will be repre- sented in the following form: C,sinh C,cosh qlix ql2x2 gl4 Mol 8u'D 8uD )-16uD+4w2D (b) Observing that the deflection curve is symmetrical with respect to the middle of the strip,we determine the constants of integration Ci,Ca,and FiG.7 the moment Mo from the following three conditions: 密0 for x=0 andx= (c) w=0 for x =0 Substituting expression (b)for w,we obtain from these conditions ql4 C1=一16uD C2=1 ql4 16uD coth u -器- a coth u = 9l= Au 1拉4() (13) where (w)= 3(u -tanh u) u2 tanh u
3. Cylindrical Bending of Uniformly Loaded Rectangular Plates with Built-in Edges. We assume that the longitudinal edges of the plate are fixed in such a manner that they cannot rotate. Taking an elemental strip of unit width in the same manner as before (Fig. 1) and denoting by Mo the bending moment per unit length acting on the longitudinal edges of the plate, the forces acting on the strip will be as shown in Fig. 7. The bending moment at any cross section of the strip is Substituting this expression in Eq. (4), we obtain (a) The general solution of this equation, using notation (5), will be represented in the following form: (b) Observing that the deflection curve is symmetrical with respect to the middle of the strip, we determine the constants of integration Ci, C2, and FIG. 7 the moment MQ from the following three conditions: (O Substituting expression (b) for w, we obtain from these conditions (13) where
14 THEORY OF PLATES AND SHELLS The deflection w is therefore given by the expression ”=-器Dinh2华+器coth华 gla 9l4 glix 9lx2 、glN 8u*D 8u'D 16uD coth u This can be further simplified and finally put in the following form: w=16uD tanh u (-+ (0-xz 8u*D (14) cosh u For calculating the parameter u we proceed as in the previous article and use Eq.(d)of that article.Substituting in it expression(14)for w and performing the integration,we obtain s0a-( 3 1 1 1 hE 256utanh u256u'sinhu64u384u Substituting S from Eq.(5)and expression(3)for D,the equation for calculating u finally becomes E2hs 81 27 27, 9 (1-ls=16ul tanh u16u sinhiu+4us+8u (15) To simplify the solution of this equation we use the curve in Fig.8,in which the parameter u is taken as abscissa and the ordinates are equal to logio (10vUi),where Ui denotes the right-hand side of Eq.(15). For any given plate we begin by calculating the square root of the left- hand side of Eq.(15),equal to Eh4/[(1-)gl+],which gives us vU1. The quantity logio(10vUi)then gives the ordinate of the curve in Fig.8,and the corresponding abscissa gives the required value of 4. Having u,we can begin calculating the maximum stresses in the plate. The total stress at any point of a cross section of the strip consists of the constant tensile stress i and the bending stress.The maximum bending stress a2 will act at the built-in edges where the bending moment is the largest.Using Eq.(10)to calculate o and Eg.(13)to calculate the bending moment Mo,we obtain Eu 1=31- (16) 6Mo h2 (17) 2 1(u） 0mx■G1十0生 To simplify the calculation of the stress o2,the values of the function i(u)are given by a curve in Fig.5. The maximum deflection is at the middle of the strip and is obtained by
The deflection w is therefore given by the expression This can be further simplified and finally put in the following form: (14) For calculating the parameter u we proceed as in the previous article and use Eq. (d) of that article. Substituting in it expression (14) for w and performing the integration, we obtain Substituting S from Eq. (5) and expression (3) for Z), the equation for calculating u finally becomes (15) To simplify the solution of this equation we use the curve in Fig. 8, in which the parameter u is taken as abscissa and the ordinates are equal to logio (104 Wl) ,where Ui denotes the right-hand side of Eq. (15). For any given plate we begin by calculating the square root of the lefthand side of Eq. (15), equaTto Eh4 /[(1 - v 2 )ql% which gives us VUv The quantity logio (104 V Ui) then gives the ordinate of the curve in Fig. 8, and the corresponding abscissa gives the required value of u. Having u, we can begin calculating the maximum stresses in the plate. The total stress at any point of a cross section of the strip consists of the constant tensile stress (T1 and the bending stress. The maximum bending stress <T2 will act at the built-in edges where the bending moment is the largest. Using Eq. (10) to calculate ax and Eq. (13) to calculate the bending moment M0, we obtain (16) (17) To simplify the calculation of the stress ^2, the values of the function ypiiu) are given by a curve in Fig. 5. The maximum deflection is at the middle of the strip and is obtained by
BENDING TO A CYLINDRICAL SURFACE 15 40 57*019o-2 In+01 607 Curve A:u varies from 0to 4 B:u 4to8 C:u 8 to 12 1.1 2.030 1.0 2.5 Curve B- Curve CCurve A 0.9152.0 0.8 1.5 Log 1oVU (u)for various values of U 0.711.01.0 048 59 26 371 4 82 Value of u FIG.8 substituting x =1/2 in Eq.(14),from which Wmax= gl 384Di(刨 (18) 24/u u where f(u)= 2 sinh u tanh u The function fi(u)is also given by a curve in Fig.5
Curve A: u varies from O+o4 " B: u " » 4 to 8 " C:u' " w 8+012 Curve B Curve C Curve A Log 104Vu,(u) for various values of U Value of U FIG. 8 substituting x = Z/2 in Eq. (14), from which (18) where The function fi(u) is also given by a curve in Fig. 5