BENDING TO A CYLINDRICAL SURFACE 21 the plate toward the other during bending.Tensile forces S in the plate produce contraction of these bars,which results in a displacement A pro- portional to S.*If k is the factor of proportionality depending on the elasticity and cross-sectional area of the bars,we obtain 8=kA or,substituting S 4u2D/12,we obtain 1 Euzh △=无372(1-) and 心+ 31△ Ek u2 =1+1-两 Thus the second factor on the left-hand side of Eqs.(22)and(23)is a constant that can be readily calculated if the dimensions and the elastic properties of the structure are known.Having the magnitude of this factor,the solution of Egs.(22)and (23)can be accomplished in exactly the same manner as used for immovable edges. mm n ny F1G.10 In the general case the second factor on the left-hand side of Eqs.(22) and(23)may depend on the magnitude of the load acting on the struc- ture,and the determination of the parameter u can be accomplished only by the trial-and-error method.This procedure will now be illustrated by an example that is encountered in analyzing stresses in the hull of a ship when it meets a wave.The bottom plates in the hull of a ship are subjected to a uniformly distributed water pressure and also to forces in the plane of the plates due to bending of the hull as a beam.Let b be the width of the ship at a cross section mn (Fig.10)and l be the frame spacing at the bottom.When the hollow of a wave is amidships (Fig. 116),the buoyancy is decreased there and increased at the ends.The effect of this change on the structure is that a sagging bending moment is produced and the normal distance l between the frames at the bottom is increased by a certain amount.To calculate this displacement accu- rately we must consider not only the action of the bending moment M ou the hull but also the effect on this bending of a certain change in The edge support is assumed to be such that A is uniform along the edges
the plate toward the other during bending. Tensile forces S in the plate produce contraction of these bars, which results in a displacement A proportional to S.* Hk is the factor of proportionality depending on the elasticity and cross-sectional area of the bars, we obtain S = M or, substituting S = 4u2D/l2 , we obtain and Thus the second factor on the left-hand side of Eqs. (22) and (23) is a constant that can be readily calculated if the dimensions and the elastic properties of the structure are known. Having the magnitude of this factor, the solution of Eqs. (22) and (23) can be accomplished in exactly the same manner as used for immovable edges. FIG. 10 In the general case the second factor on the left-hand side of Eqs. (22) and (23) may depend on the magnitude of the load acting on the structure, and the determination of the parameter u can be accomplished only by the trial-and-error method. This procedure will now be illustrated by an example that is encountered in analyzing stresses in the hull of a ship when it meets a wave. The bottom plates in the hull of a ship are subjected to a uniformly distributed water pressure and also to forces in the plane of the plates due to bending of the hull as a beam. Let b be the width of the ship at a cross section mn (Fig. 10) and I be the frame spacing at the bottom. When the hollow of a wave is amidships (Fig. 116), the buoyancy is decreased there and increased at the ends. The effect of this change on the structure is that a sagging bending moment is produced and the normal distance I between the frames at the bottom is increased by a certain amount. To calculate this displacement accurately we must consider not only the action of the bending moment M Oii the hull but also the effect on this bending of a certain change in * The edge support is assumed to be such that A is uniform along the edges
22 THEORY OF PLATES AND SHELLS Hogging (a) Sagging (b) Dc.11 tensile forces S distributed along the edges mn and mm of the bottom plate mnmin(Fig.10),which will be considered as a long rectangular plate uniformly loaded by water pressure.Owing to the fact that the plates between the consecutive frames are equally loaded,there M will be no rotation at the longitu- dinal edges of the plates,and they may be considered as absolutely built in along these edges. To determine the value of△， which denotes,as before,the dis- Sb placement of the edge mn toward (a) the edge min,in Fig.10 and which is produced by the hull bending moment M and the tensile reactions S per unit length along the edges mn and mini of the bottom plate,let Centroid A us imagine that the plate mnmin is Centroid A. removed and replaced by uniformly distributed forces S so that the to- tal force along mn and mini is Sb (Fig.12a).We can then say that (b) the displacement a of one frame FG.12 relative to another is due to the bending moment M and to the eccentric load Sb applied to the hull without bottom plating. If A,I,and c are the cross-sectional area,the centroidal moment of inertia,and the distance from the bottom plate to the neutral axis of the
Sagging (b) FIG. 11 tensile forces S distributed along the edges mn and mini of the bottom plate mnmiUi (Fig. 10), which will be considered as a long rectangular plate uniformly loaded by water pressure. Owing to the fact that the plates between the consecutive frames are equally loaded, there will be no rotation at the longitudinal edges of the plates, and they may be considered as absolutely built in along these edges. To determine the value of A, which denotes, as before, the displacement of the edge mn toward the edge mini in Fig. 10 and which is produced by the hull bending moment M and the tensile reactions S per unit length along the edges mn and mini of the bottom plate, let us imagine that the plate mnmini is removed and replaced by uniformly distributed forces S so that the total force along mn and mini is Sb (Fig. 12a). We can then say that the displacement A of one frame relative to another is due to the bending moment M and to the eccentric load Sb applied to the hull without bottom plating. If A7 I, and c are the cross-sectional area, the centroidai moment of inertia, and the distance from the bottom plate to the neutral axis of the Hogging (a) (b) FIG. 12
BENDING TO A CYLINDRICAL SURFACE 23 complete hull section,and if A1,Ti,and ci are the corresponding quanti- ties for the hull section without bottom plates,the latter set of quantities can be derived from the former by the relations A1=A-bh Ac 0= A (b) I1=I-bhc2-A(c1-c)2 The relative displacement AI produced by the eccentrically applied forces Sb is in which the factor 1-2must be introduced if one neglects the lateral strain.The displacement due to the bending moment M is Mel EI Hence the total displacement is 4=41+42=号 [+-] A (d) Substituting in this expression 8=4uD Eu*h 372(1-2) we finally obtain 4= （会+）-器 (d) This quantity must be substituted in Eq.(23)for determining the tensile parameter u. Let us apply this theory to a numerical example.Assume b =54 ft, I=1,668ftA=13.5ft2,c=12.87ft,h=0.75in.=0.0625ft, 45 in.3.75 ft,g 10 psi,M 123,500 ft-tons.From Eqs.(b) we obtain A1=13.5-0.0625·54=10.125ft2 13.51287=17.16ft c1=10.125 I=1,668-559.0-10.125(17.16-12.87)2=922.7ft Substituting these values in expression (d),we calculate A and finally obtain =1.4102-11.48 3Al
complete hull section, and if Ai, Ji, and C\ are the corresponding quantities for the hull section without bottom plates, the latter set of quantities can be derived from the former by the relations (b) The relative displacement Ai produced by the eccentrically applied forces Sb is in which the factor 1 — i>2 must be introduced if one neglects the lateral strain. The displacement due to the bending moment M is Hence the total displacement is (c) Substituting in this expression we finally obtain (d) This quantity must be substituted in Eq. (23) for determining the tensile parameter u. Let us apply this theory to a numerical example. Assume b = 54 ft, J = 1,668 ft4 , A = 13.5 ft2 , c = 12.87 ft, h = 0.75 in. = 0.0625 ft, I = 45 in. = 3.75 ft, q = 10 psi, M = 123,500 ft-tons. From Eqs. (b) we obtain Ai = 13.5 - 0.0625 • 54 = 10.125 ft2 13.5 - 12.87 171Af f C l = 10.125 =17 - 16f t Ji = 1,668 - 559.0 - 10.125(17.16 - 12.87)2 = 922.7 ft4 Substituting these values in expression (d), we calculate A and finally obtain
24 THEORY OF PLATES AND SHELLS Equation(23)then becomes Ehs w2+1.410u2-11.48 q2(1-w2)28 u? =0 1.552Eh4 u2-4.763 or q(1-4V =√U 2u2 Substituting numerical values and taking logarithms of both sides, we obtain u2-4.76 3.597+1l0g10V u2 3=1og1o(104V Using the curve in Fig.8,this equation can be readily solved by the trial-and-error method,and we obtain u=2.187 and,from Fig.5, (u)=0.780.The maximum stress is now calculated by using Eqs. (16)and (17),from which 1-90i43-140时 2=}·10·602.0.780=14,040psi 0mm￥=1+0g=28,640p9i If the bending stress in the plate due to water pressure were neglected and if the bottom plate stress were calculated from the formulaa=Mc/I, we would arrive at a figure of only 13,240 psi. 6.An Approximate Method of Calculating the Parameter u.In calcu- lating the parameter u for plates in which the longitudinal edges do not move in the plane of the plate,we used the equation ”-瓜（ hE (a) which states that the extension of an elemental strip produced by the forces S is equal to the difference between the length of the arc along the deflection curve of the strip and the chord length l.In the particular cases considered in the previous articles,exact expressions for the deflec- tions w were derived,and numerical tables and curves for the right-hand side of Eq.(a)were given.When such tables are not at hand,the solu- tion of the equation becomes complicated,and to simplify the problem recourse should be had to an approximate method.From the discussion of bending of beams it is known:that,in the case of simply supported ends with all lateral loads acting in the same direction,the deflection curve of an elemental strip produced by a combination of a lateral load and an axial tensile force S(Fig.3)can be represented with sufficient See Timoshenko,"Strength of Materials,"part II,3d ed.,p.52,1956
Equation (23) then becomes or Substituting numerical values and taking logarithms of both sides, we obtain Using the curve in Fig. 8, this equation can be readily solved by the trial-and-error method, and we obtain u = 2.187 and, from Fig. 5, ypi(u) = 0.780. The maximum stress is now calculated by using Eqs. (16) and (17), from which 30-106 -4.783 ,. ^ . ai = "3-0.91-60» = 14 ' b °° PS1 a2 = ^ • io • 602 • 0.780 = 14,040 psi o-max = o-i + (T2 = 28,640 psi If the bending stress in the plate due to water pressure were neglected and if the bottom plate stress were calculated from the formula a = Mc/1, we would arrive at a figure of only 13,240 psi. 6. An Approximate Method of Calculating the Parameter u. In calculating the parameter u for plates in which the longitudinal edges do not move in the plane of the plate, we used the equation (a) which states that the extension of an elemental strip produced by the forces S is equal to the difference between the length of the arc along the deflection curve of the strip and the chord length I. In the particular cases considered in the previous articles, exact expressions for the deflections w were derived, and numerical tables and curves for the right-hand side of Eq. (a) were given. When such tables are not at hand, the solution of the equation becomes complicated, and to simplify the problem recourse should be had to an approximate method. From the discussion of bending of beams it is known1 that, in the case of simply supported ends with all lateral loads acting in the same direction, the deflection curve of an elemental strip produced by a combination of a lateral load and an axial tensile force S (Fig. 3) can be represented with sufficient 1 See Timoshenko, "Strength of Materials," part II, 3d ed., p. 52, 1956
BENDING TO A CYLINDRICAL SURFACE 25 accuracy by the equation w=1十a (6) in which wo denotes the deflection at the middle of the strip produced by the lateral load alone,and the quantity a is given by the equation 8S7 a-SorD (c) Thus,a represents the ratio of the axial force S to the Euler critical load for the elemental strip. Substituting expression (b)in Eq.(a)and integrating,we obtain 19）=rw8 hE 4(1+)2 Now,using notation (c)and substituting for D its expression (3),we finally obtain 1+2= 3w6 (24) From this equation the quantity a can be calculated in each particular case,and the parameter u is now determined from the equation u-8Er'a ΓD4-4 (d) To show the application of the approximate Eq.(24)let us take a numerical example.A long rectangular steel plate with simply sup- ported edges and of dimensions l=50 in.and h=in.is loaded with a uniformly distributed load g=20 psi.In such a case 5 gl 00=384D and,after substituting numerical values,Eq.(24)becomes a(1+a)2=269.56 The solution of the equation can be simplified by letting 1+a=x (e) Then x8-x2=269.56 i.e.,the quantity x is such that the difference between its cube and its square has a known value.Thus x can be readily determined from a slide rule or a suitable table,and we find in our case x=6.8109 and a=5.8109
accuracy by the equation w = sin -y- (6) 1 + a I in which W0 denotes the deflection at the middle of the strip produced by the lateral load alone, and the quantity a is given by the equation (c) Thus, a represents the ratio of the axial force S to the Euler critical load for the elemental strip. Substituting expression (6) in Eq. (a) and integrating, we obtain Now, using notation (c) and substituting for D its expression (3), we finally obtain (24) From this equation the quantity a can be calculated in each particular case, and the parameter u is now determined from the equation (d) To show the application of the approximate Eq. (24) let us take a numerical example. A long rectangular steel plate with simply supported edges and of dimensions I = 50 in. and h = i in. is loaded with a uniformly distributed load q = 20 psi. In such a case and, after substituting numerical values, Eq. (24) becomes The solution of the equation can be simplified by letting 1 + a = x (e) Then x 3 - x 2 = 269.56 i.e., the quantity x is such that the difference between its cube and its square has a known value. Thus x can be readily determined from a slide rule or a suitable table, and we find in our case x = 6.8109 and a = 5.8109