16. General theory of culindrical shells

16. General theory of culindrical shells

CHAPTER GENERAL 114 THEORY OF 15 CYLINDRICAL A Circular Cylindrical Shell Loaded SHELLS Symmetrically with Respect to Its Axis In practical applications we frequently encounter problema in which a circular cylindrical shell is submitted to the action of forces distributed symmetrically with respect to the axis of the cylinder The stress distribution in cylindrical boilers submitted to the action of steam pressure, stresses in cylindrical containers having a vertical axis and submitted to internal liquid pressure, and stresses in circular pipes under uniform internal pressure are examples of such problems Fic 235 To establish the equations required for the solution of these problems we consider an element, as shown in Figs 228a and 235, and consider the equations of equilibrium It can be concluded from symmetry that the membrane shearing forces N,, = N,z vanish in this case and that forces N, are constant along the circumference Regarding the transverse shearing forces, it can also be concluded from symmetry that only the forces Q, not vanish Considering the moments acting on the element in Fig 235, we also conclude from symmetry that the twisting moments M,, = M,, vanish and that the bending moments M, are constant along the circumference Under such conditions of symmetry 466 : GENERAL THEORY CF CYLINDRICAL SHELLS 467 three of the six equations of equilibrium of the element are identically satisfied, and we have to consider only the remaining three equations, viz., those obtained by projecting the forces on the x and z axes and by taking the moment of the forces about the y axis Assuming that the external forces consist only of a pressure normal to the surface, these three equations of equilibrium are dN, T- oe: adxdg + N,dxdge dM, adxdy wv " at dy + Zadxdg = Q,adrdge = — (a) The first one indicates that the forces N, are constant,! and we take them If they are different from zero, equal to zero in our further discussion the deformation and stress corresponding to such constant forces can be easily calculated and superposed on stresses and deformations produced by lateral The load remaining following simplified form: dQ: , ly dz tae dM dx can equations two be written ⁄ ¬ Q.=0 in the (0) These two equations contain three unknown quantities: N,, Q:, and Af, To solve the problem we must therefore consider the displacements of points in the middle surface of the shell From symmetry v of the displace- we conclude that the component We thus have to conment in the circumferential direction vanishes sider only the components u and w in the z and z directions, respectively The expressions for the strain components then become fx _ du TT ~~ +0 —~ F (c) Hence, by applying Hooke’s law, we obtain N.= Eh 1g (+9) N, = Hh (e+ =1 v6) = Eh Eh (du +0 (8 ( T?z)= +0 @ | — pv? From the first of these equations it follows that du — dx =v sw a The effect of these forces on bending is neglected in this discussion 468 THEORY OF PLATES AND SHELLS and the second equation gives Ny = - — (e) Considering the bending moments, we conclude from symmetry that there is no change in curvature in the circumferential direction The curvature in the x direction is equal to —d?w/dzx? Using the same equations as for plates, we then obtain M, = »M, _~ _pvw Mu,= | where D= _Ø) =Ð dx? Eh’ 12q — z3 is the flexural rigidity of the shell Returning now to Eqs (b) and eliminating Q, from these equations, we obtain @M.,1y der TgÄy= _ _ from which, by using Eqs (e) and (f), we obtain đ? đ? (Ds) All problems of symmetrical thus reduce to the integration The simplest application of ness of the shell is constant bh _ B= (273) deformation of circular cylindrical shells of Eq (273) this equation is obtained when the thickUnder such conditions Eq (273) becomes d’w , Eh D dnt + Using the notation -— — + aren Eh FD | = Z _ 301 — ah »?) (274) (275) Eq (274) can be represented in the simplified form d3 ˆÖ— | This is the same equation as is obtained for a prismatical bar with a flexural rigidity D, supported by a continuous elastic foundation and submitted to the action of a load of intensity Z.* The general solution of this equation is w = ef(C', cos Bx + C2 sin Bx) + e—6(C; cos Bx + Cy sin Br) + f(x) * See S Timoshenko, “Strength of Materials,’’ part II, 3d ed., p 2, 1956 (277) GENERAL THEORY OF CYLINDRICAL SHELLS 469 in which f(z) is a particular solution of Eq (276), and Ci, , Ca are the constants of integration which must be determined in each particular case from the conditions at the ends of the cylinder Take, as an example, a long circular pipe submitted to the action of bending moments M> and shearing forces Qo, both uniformly distributed along the edge x = (Fig 236) In this case ` there is no pressure Z distributed over the sur- face of the shell, and f(x) = in the general solu- Since the forces applied at the end tion (277) z = produce a local bending which dies out rapidly as the distance x from the loaded end increases, we conclude that the first term on the right-handj side of Eq (277) vanish.’ must Hence, C; = C2 = 0, and we obtain a Mo C ~ C Ly Mo, Q (ø) w = e-6(C3 cos Bx + C4 sin Bz) | Fic 236 The two constants C3; and C, can now be determined from the conditions at the loaded end, which may be written d?w (Maeno = (2) — cons (8) = (8), a aM, _ n (đều oS _ Substituting expression (g) for w, we obtain from these end conditions 1, — gap (Go + BMo) C3 = M Cs = sp , (2) Thus the final expression for w is —Bz w= 581D [8M (sin Bx — cos Bx) — Qo cos Bx] (278) The maximum deflection is obtained at the loaded end, where (w)eno = — gaặp (8Mo + Q)) (279) The negative sign for this deflection results from the fact that w is taken positive toward the axis of the cylinder Observing the fact that the system balanced one and that the length of the from the principle of Saint-Venant; see, “Theory of Elasticity,’’ 2d ed., p 33, The slope at the loaded end is of forces applied at the end of the pipe is a pipe may be increased at will, this follows also for example, Timoshenko and J N Goodier, 1951 470 THEORY OF PLATES AND SHELLS obtained by differentiating expression (278) dw \ (3) ez = 282D [26M COS Bx + Qo(cos Bx This gives + sin Bx) |z—0 = 255D (28M + Qo) (280) Bv introducing the notation g(Bx) (6z) 6(Bx) ¢(Bx) = = = = e-8*(cos e-®*(cos e-® eos e* sin Bx + sin Bz) Bx — sin Bz) Bx | Br (281) the expressions for deflection and its consecutive represented in the following simplified form: derivatives can be w= - ` [Ø8Mof(8z) + Qa0(8z)] Td = 35D [26M o0(6+) + Qu¿(8+)] qd? (283) | d3 ce = = [26M of (82) — Qu(8z)] The numerical values of the functions ¢(@x), (6z), 6(8x), and £(6z) are given in Table 84.'_ The functions @(Øz) and ¥(6z) are represented graphically in Fig 237 It is seen from these curves and from Table 84 -0.4 ¥ -0.2 0.2 0.4 0.6 0.8 1.0 O -~ Fig 237 that the functions defining the bending of the shell approach zero as the quantity Bz becomes large This indicates that the bending produced in the shell is of a local character, as was already mentioned at the beginning when the constants of integration were calculated If the moment M, and the deflection w are found from expressions ‘The figures in this table are taken from the book Berechnung des Eisenbahnoberbaues,” Berlin, 1888 by H Zimmermann, ‘‘ Die GENERAL THEORY OF CYLINDRICAL 471 SHELLS equa(282), the bending moment MM, is obtained from the first of the necestions (f), and the value of the force Ny from Eq (e) Thus all sary information for calculating stresses in the shell can be found 115 Particular Cases of Symmetrical Deformation of Circular CylinBending of a Long Cylindrical Shell by a Load Uniformly drical Shells If the load is far enough Distributed along a Circular Section (Fig 238) (278) can be used for each half of from the ends of the cylinder, solution {F O Z Ip (a) (b) Fic 238 From considerations of symmetry we conclude that the value the shell We thus obtain for the right-hand portion of Qo in this case is —P/2 e bz w= 583D | eu P o(sin Bx — cos Bx) + cos ex | (a) where x is measured from the cross section at which the load is applied use To calculate the moment M, which appears in expression (a) we slope In our case this expression (280), which gives the slope at x = Hence, vanishes because of symmetry and we obtain Substituting becomes (b) Mo = =48 value this (a), expression in the deflection of the shell (283) Ý x g8:D 062) Bx == x7 cos Bt) 881D (ginsin BxBx ++ cos w =— P&™ and by differentiation we find dw Ta —_— d?w = — Ta = 28? —— dx? = P 28 _58D a ; +8 Bx sin e- pba —= — P SS 482D c(z) e-8= (sin Bx — cos Bx) = — qạp ——c-8z 863D e™ cos Bx = P 2D 00+) —— Ý(82) (c) 472 THEORY TaBLE Bx OF 84 TABLE ý PLATES AND OF FUNCTIONS Ử SHELLS 9g, y, 0, AND [ Ệ 0.1 0.2 0.3 0.4 1.0000 0.9907 0,9651 0.9267 0.8784 1.0000 0.8100 0.6398 0.4888 0.3564 1.0000 0.9005 0.8024 0.7077 0.6174 0.0905 0.1627 0.2189 0.2610 0.5 0.6 0.7 0.8 0.9 0.8231 0.7628 0.6997 0.6354 0.5712 0.2415 0.1431 0.0599 —0.00953 —0.0657 0.5323 0.4530 0.3798 0.3131 0.2527 0.2908 0.3099 0.3199 0.3223 _0.3185 1.0 1.1 1.2 1.3 1.4 0.5083 0.4476 0.3899 0.3355 0.2849 —0.1108 —0.1457 —0.1716 —0.1897 —Q.2011 0.1988 0.1510 0.1091 0.0729 0.0419 0.3096 0.2967 0.2807 0.2626 0.2430 1.5 1.6 1.7 1.8 1.9 0.2384 0.1959 0.1576 0.1234 0.0932 —0.2068 —0.2077 —0.2047 —0.1985 —0.1899 0.0158 —0.0059 —0.0235 —0.0376 —0.0484 0.2226 0.2018 0.1812 0.1610 0.1415 2.0 2.1 2.2 2.3 2.4 0.0667 0.0439 0.0244 0.0080 —0.0056 —0.1794 —0.1675 —0.1548 —0.1416 —0.1282 —0.0563 —0.0618 —0.0652 —0.0668 —0.0669 0.1230 0.1057 0.0895 0.0748 0.0613 2.5 2.6 2.7 2.82.9 —0.0166 —0.0254 —0.0320 —0.0369 —0.0403 —0.1149 —0.1019 —0.0895 —0.0777 —0.0666 —0.0658 —0.0636 —0.0608 —0.0573 —0.0534 0.0492 0.0385 0.0287 0.0204 0.0132 3.0 3.1 3.2 3.3 3.4 —0.0423 —0.0431 —0.0431 —0.0422 —0.0408 —0.0563 —0.0469 —0.0383 —0.0306 —0.0237 —0.0493 —0.0450 —0.0407 —0.0364 —0.0525 0.0071 0.0019 —0.0024 —0.0058 —0.0085 3.5 3.6 3.7 3.8 3.9 —0.0389 —0.0366 —0.0341 —0.0314 —0.0286 —0.0177 —0.0124 —0.0079 —0.0040 —0.0008 —0.0283 —0.0245 —0.0210 —0.0177 —0.0147 —0.0106 —0.0121 —0.0131 —0.0137 —0.0140 GENERAL TABLE 84 TABLE Bx ~ 4.0 4.1 4.2 4.3 4.4 —0.0258 —0.0231 —0.0204 —0.0179 —0.0155 4.5 4.6 4.7 4.8 4.9 THEORY OF CYLINDRICAL OF EUNCTIONS Ỷ 473 SHELLS ý, ý, 0, AND ¢ (Continued) Ệ 0.0019 0.0040 0.0057 0.0070 0.0079 —0.0120 —0.0095 —0.0074 —0.0054 —0.0038 —().0139 —0.0136 —Q.0131 —0.0125 —0.0117 —0.0132 —Q.0111 —0.0092 —0.0075 —0.0059 0.0085 0.0089 0.0090 0.0089 0.0087 —0.0023 —0.0011 0.0001 0.0007 0.0014 —0.0108 —0.0100 —0.0091 —0.0082 —0.0073 5.0 51 5.2 5.3 5.4 —0.0046 —0.0033 —0.0023 —0.0014 —0.0006 0.0084 0.0080 0.0075 0.0069 0.0064 0.0019 0.0023 0.0026 0.0028 0.0029 —0.0065 —0.0057 —0.0049 —0.0042 —0.0035 5.5 5.6 5.7 5.8 5.9 0.0000 0.0005 0.0010 0.0013 0.0015 0.0058 0.0052 0.0046 0.0041 0.0036 0.0029 0.0029 0.0028 0.0027 0.0026 —0.0029 —0.0023 —0.0018 —0.0014 —0.0010 6.0 6.1 6.2 6.3 6.4 0.0017 0.0018 0.0019 0.0019 0.0018 0.0031 0.0026 0.0022 0.0018 0.0015 0.0024 0.0022 0.0020 0.0018 0.0017 —0.0007 —0.0004 —0.0002 +0.0001 0.0003 6.5 6.6 6.7 6.8 6.9 7.0 0.0018 0.0017 0.0016 0.0015 0.0014 _ 0.0013 0.0012 0.0009 0.0006 0.0004 0.0002 0.0001 0.0015 0.0013 0.0011 0.0010 0.0008 0.0007 0.0004 0.0005 0.0006 0.0006 0.0006 0.0006 Observing from Eqs (b) and (f) of the preceding article that M, = —D dw dx? Q.= —D dw dx’ we finally obtain the following expressions for the bending moment and shearing force: P (284) 474 THEORY OF PLATES AND SHELLS The results obtained are all graphically represented in Fig 239 It is seen that the maximum deflection is under the load P and that its value P as given by Eq =~ S77 INA we 8@”D rụ) n dw 37” Im 48 Ww p P dx mãn (8X) The maximum a > Ta 2Eh bending moment 1s also under the load and is deter- mined from Eq (284) as P — TM uax — 72 286 The maximum of the absolute value of the shearing force is evidently - YB +] yr (285) = Pa’ 863D Mx= ip vy (AX) a — = ”* (283) is equal _— Qx=-F 0(X) to P/2 The values of all these quantities at a certain distance from the load can be readily obtained by using Table 84 We Fic 239 see from this table and from Fig 239 that all the quantities that determine the bending of the shell are small forz > 7/8 This fact indicates that the bending is of a local character and that a shell of length = 27/8 loaded at the middle will have practically the same maximum deflection and the same maximum stress as a very long shell Having the solution of the problem for the case in which a load is concentrated at a circular cross section, we can readily solve the problem of a load disF€=-~~ L - ý tributed along a certain length of the cylinder ⁄⁄⁄⁄⁄⁄⁄⁄ by applying the principle of superposition As an example let us consider the case of a load of intensity gq uniformly distributed | Al € ole Jag +c ->< b | along a length ¡ of a cylinder (Fig 240) | Assuming that the load is at a considerable KZ} distance from the ends of the cylinder, we can Fig 240 use solution (283) to calculate the deflections | The deflection at a point A produced by an elementary ring load of an intensity! q dé at a distance ~ from A is obtained from expression by substituting gq dé for P and é for x and is (283) jy e~P(cos BE + sin BE) The deflection produced at A by the total load q dé is the load per unit length of circumference distributed over the GENERAL THEORY OF CYLINDRICAL SHELLS 475 length J is then _ | age e-8t(cos BE + sin BE) + I ng ø-?‡(eos 8‡ -+ sin BE) b Cc _ ga’ (2 — e®? cos 8b — e”#° eos 8c) — 9Eh The bending cation of the Cylindrical edges of the moment at a point A can be calculated by similar applimethod of superposition Shell with a Uniform Internal Pressure (Fig 241) If the shell are free, the internal pressure p produces only a hoop stress Oo; = i a and the radius of the cylinder increases by the amount _ 2%, _ par o =F > Eh (ở) If the ends of the shell are built in, as shown in Fig 241la, they cannot move out, and local bending occurs at the edges If the length J of the Mo, Ferrers 4s —É At ⁄ 11111 11111 ⁄ (a) ⁄ a HTTTHTTTTE song sai 22 2gNhi ¡: |LIILLIIILLI (b) Fig 241 shell is sufficiently large, we can use solution bending, the moment M> the and shearing (278) to investigate this force Qo being determined from the conditions that the deflection and the slope along the built-in According to these conditions, Eqs edge x = O (Fig 241a) vanish (279) and (280) of the press article become sip (28M, + Qo) = where is given by Eq (da) Solving for My and Qo, we obtain 3S Q, = —463Dâ = — Ễ ® M, = 262D6= 3g? (287) ... between The magnitude of the forces per unit length of each ring and the pipe 450 THEORY OF PLATES AND SHELLS the circumference of the tube will be denoted by P The magnitude of P will now be determined... part II, 3d ed., p 2, 1956 (277) GENERAL THEORY OF CYLINDRICAL SHELLS 469 in which f(z) is a particular solution of Eq (276), and Ci, , Ca are the constants of integration which must be determined... 1888 by H Zimmermann, ‘‘ Die GENERAL THEORY OF CYLINDRICAL 471 SHELLS equa(282), the bending moment MM, is obtained from the first of the necestions (f), and the value of the force Ny from Eq (e)