16. General theory of culindrical shells

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GENERAL THEORY OF CYLINDRICAL SHELLS

114 A Circular Cylindrical Shell Loaded 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, do 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

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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- at dy oe: adxdg + N,dxdge + Zadxdg = 0 (a) dM, wv 0 adxdy — Q,adrdge = 0

The first one indicates that the forces N, are constant,! and we take them equal to zero in our further discussion If they are different from zero, the deformation and stress corresponding to such constant forces can be easily calculated and superposed on stresses and deformations produced by lateral load The remaining two equations can be written in the following simplified form: dQ: , ly 7 dz tae ⁄ dM ¬ dx Q.=0 (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

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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 _Ø) Mu,= =Ð dx? | Eh’ where D= 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= 4 from which, by using Eqs (e) and (f), we obtain đ? đ? bh -— — (Ds) + aren (273)

All problems of symmetrical deformation of circular cylindrical shells thus reduce to the integration of Eq (273)

The simplest application of this equation is obtained when the thickness of the shell is constant Under such conditions Eq (273) becomes d’w , Eh | D dnt + 3 = Z (274) Using the notation _ Eh _ 301 — »?) B= FD ah (275) Eq (274) can be represented in the simplified form d3 ˆÖ— 9 |

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) (277)

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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 = 0 (Fig 236) In this case

there is no pressure Z distributed over the sur- `

face of the shell, and f(x) = 0 in the general solu- a Mo

tion (277) Since the forces applied at the end C | z = 0 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) must vanish.’ C Ly Mo, Hence, C; = C2 = 0, and we obtain Q 0

w = e-6(C3 cos Bx + C4 sin Bz) (ø) 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 (2) — aM, _ n (đều _ oS cons (8) = 0 (8), a Substituting expression (g) for w, we obtain from these end conditions 1, M ,

C3 = — gap (Go + BMo) 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

1

(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 The slope at the loaded end is

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obtained by differentiating expression (278) This gives dw \ ez (3) = 282D [26M COS Bx + Qo(cos Bx + sin Bx) |z—0 1 = 255D (28M + Qo) (280) Bv introducing the notation g(Bx) = e-8*(cos Bx + sin Bz) (6z) = e-®*(cos Bx — sin Bz) 6(Bx) = e-đ eos Bx | Â(Bx) = e* sin Br (281) the expressions for deflection and its consecutive derivatives can be represented in the following simplified form: w= - ` [Ø8Mof(8z) + Qa0(8z)] d 1 T = 35D [26M o0(6+) + Qu¿(8+)] (283) qd? 1 | 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 0.2 0.4 0.6 0.8 1.0 O ¥ 1 2 3 4 5 -~ 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

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(282), the bending moment MM, is obtained from the first of the equations (f), and the value of the force Ny from Eq (e) Thus all necessary information for calculating stresses in the shell can be found

115 Particular Cases of Symmetrical Deformation of Circular Cylin- drical Shells Bending of a Long Cylindrical Shell by a Load Uniformly Distributed along a Circular Section (Fig 238) If the load is far enough from the ends of the cylinder, solution (278) can be used for each half of {F Z O Ip (a) (b) Fic 238

the shell From considerations of symmetry we conclude that the value of Qo in this case is —P/2 We thus obtain for the right-hand portion

e bz P

w= 583D | eu o(sin Bx — cos Bx) + 5 cos ex | (a) where x is measured from the cross section at which the load is applied To calculate the moment M, which appears in expression (a) we use expression (280), which gives the slope at x = 0 In our case this slope vanishes because of symmetry Hence,

and we obtain

Mo = = 48 (b)

Substituting this value in expression (a), the deflection of the shell becomes

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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 as given by Eq (283) is P 0 =~ S77 Ww = a = Pa’ p ”* 863D 2Eh

INA we 8@”D rụ) The maximum bending moment 1s

n dw P also under the load and is deter-

37” dx mãn (8X) mined from Eq (284) as Im 7 P (285) 48 > Ta — TM uax — 72 286

Mx= ip vy (AX) The maximum of the absolute value

a of the shearing force is evidently

- YB equal to P/2 The values of all

— +] _— these quantities at a certain dis-

yr Qx=-F 0(X) tance from the load can be readily obtained by using Table 84 We 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 1 = 27/8 loaded at the middle will have practi- cally 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 con- centrated at a circular cross section, we can

Fic 239

readily solve the problem of a load dis- F€=-~~ L - ý tributed along a certain length of the cylinder ⁄⁄⁄⁄⁄⁄⁄⁄ 3 by applying the principle of superposition Al € ole Jag As an example let us consider the case of a +c ->< b load of intensity gq uniformly distributed | |

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 (283) by substituting gq dé for P and é for x and is

jy e~P(cos BE + sin BE)

The deflection produced at A by the total load distributed over the

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length J is then

b Cc

_ | age e-8t(cos BE + sin BE) + I ng ø-?‡(eos 8‡ -+ sin BE)

_ ga’

— 9Eh (2 — e®? cos 8b — e”#° eos 8c)

The bending moment at a point A can be calculated by similar application of the method of superposition

Cylindrical Shell with a Uniform Internal Pressure (Fig 241) If the edges of the shell are free, the internal pressure p produces only a hoop stress a Oo; = i 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 a HTTTHTTTTE 5

4s Nhi song sai At - —É 2g 22 - ⁄ : 1111111111 ¡ |LIILLIIILLI ⁄ (a) ⁄ (b) Fig 241

shell is sufficiently large, we can use solution (278) to investigate this bending, the moment M> and the shearing force Qo being determined from the conditions that the deflection and the slope along the built-in edge x = O (Fig 241a) vanish According to these conditions, Eqs (279) and (280) of the press article become

sip (28M, + Qo) = 0 where 6 is given by Eq (da)

Solving for My and Qo, we obtain

3S

M, = 262D6 = 3g? Q, = —463Dâ = — Ễ (287)

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We thus obtain a positive bending moment and a negative shearing force acting as shown in Fig 24la Substituting these values in expressions (282), the deflection and the bending moment at any distance from the end can be readily calculated using Table 84

If, instead of built-in edges, we have simply supported edges as shown in Fig 241b, the deflection and the bending moment M, vanish along the edge My, = 0, and we obtain, by using Eq (279),

_Qạọ = —26°Đầ

By substituting these values in solution (278) the deflection at any distance from the end can be calculated

It was assumed in the preceding discussion that the length of the shell is large If this is not the case, the bending at one end cannot be considered as independent of the conditions at the other end, and recourse must be had to the general solution (277), which contains four constants of integration The particular solution of Eq (276) for the case of uniform load (Z = —p) is —p/484D = —pa?/Eh The general solution (277) can then be put in the following form by the introduction of hyperbolic functions in place of the exponential functions:

2

40 = — ĐT + Cy; sin Bx sinh Bz + C2 sin Bx cosh Bx

+ C; cos Bx sinh Bx + C, cos Bx cosh 6z (e) If the origin of coordinates is taken at the middle of the cylinder, as shown in Fig 241b, expression (e) must be an even function of x Hence

C2 = C3; = 0 (f)

The constants C; and C, must now be selected so as to satisfy the conditions at the ends If the ends are simply supported, the deflection and the bending moment //, must vanish at the ends, and we obtain d?w | (w)z—1/2 = 0 () = 0 (9) Substituting expression (e) in these relations and remembering that C = C3 = 0, we find pa’

Fh + CC, sin a sinh a + Cy, cos a cosh a = 0 (h)

Ci cos a cosh a — Cy, sin a sinh a = 0

where, for the sake of simplicity,

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From these equations we obtain

C, = 2 sin a sinh a _ pa 2sina sinha

1 Eh sin? a sinh? a + cos? a cosh?a Eh cos 2a + cosh 2a ()

C, = pa? cos a cosh a _ pa? 2 cos a cosh a J

4 ~ Eh sin? a sinh? a + cos? a cosh? a Eh cos 2a + cosh 2a

Substituting the values (7) and (f) of the constants in expression (e) and observing from expression (275) that Eh 64œ1+] we obtain _ pl — 2sinasinhe HH — 64 Det ( cos 2a + cosh 2a sin Bx sinh Bx _ 2 cos a cosh a cos 2a + cosh 2a

cos Bx cosh az) (1)

In each particular case, if the dimensions of the shell are known, the quantity a, which is dimensionless, can be calculated by means of notation (7) and Eq (275) By substituting this value in expression (Ð the deflection of the shell at any point can be found

For the middle of the shell, substituting 2 = 0 in expression (1), we obtain

_ pl _ 2 cosa cosh a

(00)z—o = 64 Da‘ (1 cos 2a + cosh 53) (m) When the shell is long, a becomes large, the second term in the parentheses of expression (m) becomes small, and the deflection approaches the value (d) calculated for the case of free ends This indicates that in the case of long shells the effect of the end supports upon the deflection at the middle is negligible Taking another extreme case, viz., the case when a is very small, we can show by expanding the trigonometric and hyperbolic functions in power series that the expression in parentheses in Eq (m) approaches the value 5a‘/6 and that the deflection (J) approaches that for a uniformly loaded and simply supported beam of length / and flexural rigidity D

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At the middle of the shell this moment is

_ pl? sin a sinh a (0)

4a? cos 2a + cosh 2a 5

It is seen that for large values of a, that is, for long shells, this moment becomes negligibly small and the middle portion is, for all practical pur- poses, under the action of merely the hoop stresses pa/h

The case of a cylinder with built-in edges (Fig 241a) can be treated in a similar manner Going directly to the final result,! we find that the bending moment M, acting along the built-in edge is (M,)z—o — _ P sinh 2œ —sin 2a ? 1o = s8snh 2z Lan Đz— 2g? X;(22) (288) where (2a) = sinh 2a — sin 2a a X2 sinh 2œ + sin 2a

In the case of long shells, a is large, the factor x2(2a) in expression (288) approaches unity, and the value of the moment approaches that given by the first of the expressions (287) For shorter shells the value of the factor x2(2a) in (288) can be taken from Table 85 TABLE 85 2a xi(2ø) x2(2a) xa(2ø) 0.2 5.000 0.0068 0.100 0.4 2.502 0.0268 0.200 0.6 1.674 0.0601 0.300 0.8 1.267 0.1065 0.400 1.0 1.033 0.1670 0.500 1.2 0.890 0.2370 0.596 1.4 0.803 0.3170 0.689 1.6 0.755 0.4080 0.775 1.8 0.735 0.5050 0.855 2.0 0.738 0.6000 0.925 2.5 0:802 0.8220 1.045 3.0 0.893 0.9770 1.090 — 3.5 0.966 1.0500 1.085 4.0 1.005 1.0580 1.050 4.5 1.017 1.0400 1.027 5.0 1.017 1.0300 1.008

Cylindrical Shell Bent by Forces and Moments Distributed along the Edges In the preceding section this problem was discussed assuming

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that the shell is long and that each end can be treated independently In the case of shorter shells both ends must be considered simultaneously by using solution (e) with four constants of integration Proceeding as in the previous cases, the following results can be obtained For the case of bending by uniformly distributed shearing forces Qo (Fig 242a), the deflection and the slope at the ends are

(w) _ —_ 2Quổa! cosh 2œ + cos 2x _ _ 2Quổa? (2ø)

UJz=0,c=l CC Eh sinh 2a + sin 2a Eh O° (289)

dw _ 4 2Q,62a? sinh 2a — sin 2a _ + 2Qo8°a? (2ø)

“\đr/ cac © Eh sinh2a+sin2a ~~ Eh xe

In the case of bending by the moments AZo (Fig 2426), we obtain (w) — — 2M 6?a? sinh 20 — sin 2a 2M 6?’ (2ø)

— Eh sinh 2a + sm 2x - Eh X?°“Z (290)

dw —+ 4Mo83ø? cosh 2œ — oos 2x _ „ 4Äa8°a” (2ø) |

AX /;—0.x=i oT Eh sinh 2a + sin2a —- 7m

In the case of long shells, the factors xi, x2, and x3 in expressions (289) and (290) are close to unity, and the results coincide with those given by Qo Qo M M { yf ` + z h* x ' ~ MM ' Z Am — > to =——— L > -|}——- - 2d - - QoŸỶ ÝQo Mọ Mo (a) (b) Fic 242

expressions (279) and (280) To simplify the calculations for shorter shells, the values of functions x1, x2, and x; are given in Table 85

Using solutions (289) and (290), the stresses in a long pipe reinforced by equidistant rings (Fig 243) and submitted to the action of uniform internal pressure p can be readily discussed |

Assume first that there are norings Then, under the action of internal pressure, hoop stresses o, = pa/h will be produced, and the radius of the pipe will increase by the amount |

_ pa’

° = Tih

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the circumference of the tube will be denoted by P The magnitude of P will now be determined from the condition that the forces P produce a deflection of the pipe under the ring equal to the expansion 6 created by the internal pressure p In calculating this deflection we observe that a portion of the tube between two adjacent rings may be considered as the shell shown in Fig 242a and b In this case Qo = —3P, and the magnitude of the bending moment M> under a ring is determined from the condition that dw/dzx = 0 at that point

eon] nan |e Hence from Eqs (289) and (290) we find

! , I 1 t lỊ ! , — PB?a* x:(2œ ) + 4M cos a? X3(2a) = 0

re "Ti h I | 4 ‘7 from which Eh

—— Fig 243 M, = Px:@a) 4Bx3(2a) (p)

If the distance | between the rings is large,! the quantity 2a = “ va X⁄3(1 1— y?) — z?)

is also large, the functions x2(2«) and x3;(2a) approach unity, and the moment M, approaches the value (286) For calculating the force P entering in Eq (p) the expressions for deflections as given in Eqs (289) and (290) must be used These expressions give

“Be x¡(2a) — PBa? x3(2a) _ 5 = pa? xi 2Eh x;(2a) - Eh or Pg |x@ _Ä aS | _ ohh _, (291) For large values of 2a this reduces to PBa? 2Eh

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forces P produce in the ring a tensile force Pa and that the corresponding increase of the inner radius of the ring 1s’

Pa?

ôi — TT

where A is the cross-sectional area of the ring To take this extension into account we substitute 6 — 6; for 6 in Eq (291) and obtain

PB |xi@œ _ | -p-— (293)

From this equation, P can be readily obtained by using Table 85, and the moment found by substituting p — (Ph/A) for p in Eq (292)

If the pressure p acts not only on the cylindrical shell but also on the ends, longitudinal forces - pe Ni = 5 are produced in the shell The extension of the radius of the cylinder is then pa’ 1 and the quantity p(1 — 4v) must be substituted for p in Kas (292) and (293) |

Equations (293) and (291) can also be used in the case of external uniform pressure provided the compressive stresses in the ring and in the shell are far enough from the critical stresses at which buckling may occur.2 This case is of practical importance in the design of submarines and has been discussed by several authors.’

116 Pressure Vessels The method illustrated by the examples of the preceding article can also be applied in the analysis of stresses in cylindrical vessels submitted to the action of internal pressure.* In discussing the ‘““membrane theory” it was repeatedly indicated that this theory fails to represent the true stresses in those portions of a shell close to the

1It is assumed that the cross-sectional dimensions of the ring are small in comparison with the radius a

2 Buckling of rings and cylindrical shells is discussed in 8 Timoshenko, ‘‘Theory of Elastic Stability,” 1936

2 See paper by K von Sanden and K Giinther, ‘‘Werft und Reederei,'' vol 1, 1920, pp 163-168, 189-198, 216-221, and vol 2, 1921, pp 505-510

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edges, since the edge conditions usually cannot be completely satisfied by considering only membrane stresses A similar condition in which the membrane theory is inadequate is found in cylindrical pressure vessels at the joints between the cylindrical portion and the ends of the vessel At these joints the membrane stresses are usually accompanied by local bending stresses which are distributed symmetrically with respect to the axis of the cylinder These local stresses can be calculated by using

solution (278) of Art 114 : :

Let us begin with the simple case of a cylindrical vessel with hemispherical ends (Fig 244).! At a sufficient distance from the joints mn O Mo Mo, SKE X m mạ - DY x - (“| % AQ , 1) y 8 | n Dp ET 7 (a) | Fic 244

and m,n; the membrane theory is accurate enough and gives for the cylindrical portion of radius a

N, = © Ni = pa (a)

where p denotes the internal pressure

For the spherical ends this theory gives a uniform tensile force

_ pa

N= %5 | (b)

The extension of the radius of the cylindrical shell under the action of the forces (a) is _ par(, _» | and the extension of the radius of the spherical ends is _ pa’ do = Eh (1 — v) (d)

Comparing expressions (c) and (d), it can be concluded that if we consider only membrane stresses we obtain a discontinuity at the joints as represented in Fig 244b This indicates that at the joint there must act

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shearing forces Qo and bending moments Mo uniformly distributed along the circumference and of such magnitudes as to eliminate this discontinuity ‘he stresses produced by these forces are sometimes called discontinuity stresses

In calculating the quantities Qo and Mo we assume that the bending is of a local character so that solution (278) can be applied with sufficient accuracy in discussing the bending of the cylindrical portion The investigation of the bending of the spherical ends represents a more complicated problem which will be fully discussed in Chap 16 Here we obtain an approximate solution of the problem by assuming that the bending is of importance only in the zone of the spherical shell close to the joint and that this zone can be treated as a portion of a long cylindrical shell! of radius a If the thickness of the spherical and the cylindrical portion of the vessel is the same, the forces Qo produce equal rotations of the edges of both portions at the joint (Fig 244b) This indicates that M, vanishes and that Qo alone is sufficient to eliminate the discontinuity The magnitude of Qo is now determined from the condition that

the sum of the numerical values of the deflections of the edges of the two

parts must be equal to the difference 6; — 62 of the radial expansions furnished by the membrane theory Using Eq (27 9) for the deflections, we obtain BD ` ` 2Eh from which, by using notation (275), _ Pa*8'D _ p | Go = “DEh ~ 86 (e)

Having obtained this value of the force Qo, the deflection and the bending moment M, can be calculated at any point by using formulas (282), which give?

Qo

- _pử?_ _ &a

Substituting expression (e) for Qo and expression (275) for 8 in the formula for M,, we obtain

M = ~ 831 —-) anp _e(gr) (f)

1 Meissner, in the above-mentioned paper, showed that the error in the magnitude of the bending stresses as calculated from such an approximate solution is small for thin hemispherical shells and is smaller than 1 per cent if a/h > 30

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This moment attains its numerical maximum at the distance x = 7/48, at which point the derivative of the moment is zero, as can be seen from the fourth of the equations (282)

Combining the maximum bending stress produced by M, with the membrane stress, we find

_ ap , 3 ap T\ _—_ ap

This stress which acts at the outer surface of the cylindrical shell is about 30 per cent larger than the membrane stress acting in the axial direction In calculating stresses in the circumferential direction in addition to the membrane stress pa/h, the hoop stress caused by the deflection w as well as the bending stress produced by the moment M, = »M, must be considered In this way we obtain at the outer surface of the cylindrical shell

_ op _ Hw _ bv _ 3P _i - Ov

Taking v = 0.3 and using Table 84, we find

(o:)max = 1.03292 h at Bx = 1.85 (h)

Since the membrane stress is smaller in the ends than in the cylinder sides, the maximum stress in the spherical ends is always smaller than the calculated stress (h) Thus the latter stress is the determining factor in the design

of the vessel |

The same method of calculating discontinuity stresses can be applied in the case of ends having the form of an ellipsoid of revolution ‘The membrane stresses in this case are obtained from expressions (263) and (264) (see page 440) At the joint mn which represents the equator of the ellipsoid (Fig 245), the stresses in the direction of the meridian and in the equatorial direction are, respectively, o\ 6 = = (1-55) (2) The extension of the radius of the equator is pi 2 yg) = PU we oo = By (06 — Yee) = (1 5p 3

Substituting this quantity for 62 in the previous calculation of the shearing force Qo, we find

pa’ a2

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and, instead of Eq (e), we obtain

p a

Qo = 88 b

It is seen that the shearing force Qo in the case of ellipsoidal ends is larger than in the case of hemispherical ends in the ratio a?/b? The discontinuity stresses will evidently increase in the same proportion For example, taking a/b = 2, we obtain, from expressions (g) and (h), oap LÍ ap z)max ——” 2 172 — (cz) 5y vs ai (7) - 2h ()„„„ = 1.128 + | Again, (ơ;)„;„ 1s the largest stress and 1s consequently the determining factor in design.?

117 Cylindrical Tanks with Uniform Wall Thickness If a tank is submitted to the action of a liquid pressure, as shown in Fig 246, the stresses in the wall can be analyzed by using Eq (276) Substituting in this equation Z= —y(d — 2) (a) where y is the weight per unit volume of the liquid, we obtain d4w _ ¥(d — =) eg | dg + *P% = D Oo Fea: A particular solution of this equation is — d— 2 w= — oa _ _ ¥( _ ue (c) lf = SẼ sa Ax '

This expression represents the radial expansion L Qo L

of a cylindrical shell with free edgesunderthe ⁄⁄⁄ ⁄ OMS “ Sah

IG

action of hoop stresses Substituting expres-

sion (c) in place of f(z) in expression (277), we obtain for the complete solution of Eq (6)

w = e*(Ci cos Bx + C2sin Bx) + e~**(C; cos Bx + Cysin Br) — vd —

In most practical cases the wall thickness h is small in comparison with both the radius a and the depth d of the tank, and we may consider the shell as infinitely long The constants Ci and C2 are then equal to zero,

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and we obtain

w = e *=(C; cos Bx + Cy sin Br) — ve — Da (d) The constants C; and C, can now be obtained from the conditions at the bottom of the tank Assuming that the lower edge of the wall is built into an absolutely rigid foundation, the boundary conditions are oq (0)„-ạ = Œy — TT =0 (2) = | —8Œz7?“(eo Bx + sin Bx) —B+z — QC] ya" —= — ya" = + BC,e—§*(cos Bx — sin Bx) + Eh L B(C, — Cs) + ih 0 From these equations we obtain _ yard _t C= f= m (4 3) Expression (d) then becomes w= _ [2-2 — e | d cos ee + (a — 2) sin áz from which, by using the notation of Eqs (281), we obtain _ _ ya'd — ¬ " = =0 1~g— 808) — (1 — an) (42) | (e

From this expression the deflection at any point can be readily calculated by the use of Table 84 The force N, in the circumferential direction is then

Ny = ~ 2M = yaa|1 —§ — 98a) = (1= ag)t@5)| Ơ

From the second derivative of expression (e) we obtain the bending moment

M, = —D Fe = PES — (a) + (1 - 3) s65)

_ Fay | He + (1 x) 062) |_ vyadh — „ưa _

Having expressions (f) and (g), the maximum stress at any point can readily be calculated in each particular case The bending moment has its maximum value at the bottom, where it is equal to

_vy._ _ 1) yadh

(Me)oow = Mo = (1 ~ 33) tee —

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The same result can be obtained by using solutions (279) and (280) (pages 469, 470) Assuming that the lower edge of the shell is entirely free, we obtain from expression (c)

_ _ yard đu \ _ ya?

(01)z—p = — Tp S )., ~ Eh (2)

To eliminate this displacement and rotation of the edge and thus satisfy the edge conditions at the bottom of the tank, a shearing force Qo and bending moment M,> must be applied as indicated in Fig 246 The magnitude of each of these quantities is obtained by equating expressions (279) and (280) to expressions (z) taken -with reversed signs ‘This gives

1 _ vad

— gap (Mo + Qo) = + “ER

1 _ ya?

585p (28Mo + 6u) = Eh

From these equations we again obtain expression (/) for Mo, whereas for the shearing force we find!

_ vyadh | — 1

đo Vid — ») (28 3) (J)

Taking, as an example, a = 30 ft, d = 26 ft, h = 141n.,, y = 0.03613 lb per n.3,

and » = 0.25, we find 6 = 0.01824 in.~! and fd = 5.691 For such a value of @d our

assumption that the shell is infinitely long results in an accurate value for the moment and the shearing force, and we obtain from expressions (h) and (9) Ke = 2a - >| Mẹ = 13,960 in.-]b perin Qo = —563.6 lb per in- ——— -_—_—

In the construction of steel tanks, metallic sheets of ”! n several different thicknesses are very often used as

shown in Fig 247 Applying the particular solution mM, ny (c) to each portion of uniform thickness, we find that

the differences in thickness give rise to discontinuities TROT r7" in the đisplacement ¡ along the joints mn and mini d b

These discontinuities, together with the displace- Fic, 247 ments at the bottom ab, can be removed by apply-

ing moments and shearing forces Assuming that the vertical dimension of each portion is sufficiently large to justify the application of the formulas for an infinitely large shell, we calculate the discontinuity moments and shearing forces as before by using Eqs (279) and (280) and applying at each joint the two conditions that the adjacent portions of the shell have equal deflections and a common tangent If the use of formulas (279) and (280) derived for an infinitely long shell cannot be justified, the general solution containing four constants of integration must be applied to each portion of the tank The determination of the constants under such conditions becomes much more complicated, since the fact that each joint cannot be treated mee nnn Q n= >

Trang 23

independently necessitates the solution of a system of simultaneous equations This

problem can be solved by approximate methods.}

118 Cylindrical Tanks with Nonuniform Wall Thickness In the case of tanks of nonuniform wall thickness the solution of the problem requires the integration of Eq (273), considering the flexural rigidity D and the thickness hf as no longer constant but as functions of x We have thus to deal with a linear differential equation of fourth order with variable coefficients As an example, let us consider the case when the thickness of the wall is a linear function of the coordinate z.* Taking’ the origin of the coordinates as shown in Fig 248, we have for the thickness of the wall and for the flexural rigidity the expressions ba3 = ax ° D = ———— 3 12(1 — »2)” 6 and Eq (273) becomes d? đ?ụ 121 — ?? 121 — »? — a 30 4 ( Vv ) +? = — ( v )yŒ Xo) (b) dx? dz? aq? Ea’ | The particular solution of this equation is 1 2 O — _ Yr — Xo ——^ KK | m Ea 2 €)

->‡-œ a ! %9 This solution represents the radial expansion of a shell x i with free edges under the internal pressure +(# — zo)- pa + Asa result of the displacement (c) a certain amount of

— — TT — TỊ i ¡ bending of the generatrices of the cylinder occurs : i: The corresponding bending moment is N N ở ụ ~ d uM = D d?w, _ +œ?a®+o (đ) : : dz? 6(1 — v?) \ N

Z 2227277772 2 x This moment is independent of z and is in all practical

¬ 2Q - cases of such small magnitude that its action can

usually be neglected

Fie 248 To obtain the complete solution of Eq (b) we have to add to the particular solution (c) the solution of the homogeneous equation

d? d?w 12(1 — pv?)

5 (2) on

1 An approximate method of solving this problem was given by C Runge, Z Math Phystk, vol 51, p 254, 1904 This method was applied by K Girkmann in a design of a large welded tank; see Stahlbau, vol 4, p 25, 1931

Trang 24

vhich, upon division by z, can be also written

1d (= = 4 2A”), 9 ()

x ax? dz? o?a?

Che solution of this equation of the fourth order can be reduced to that of two equa- ions of the second order! if we observe that 1 d? d?w 1d dild dw = [73 ) = —-— fg? — | —-— [ 2? — x dx? dz? 2 dz dx | x dz dz “or simplification we introduce the following symbols: 1d dw L(w) =~ = (= =) (f) of = 12(1 — v?) G) œ?q? Equation (e) then becomes ¬ | L{L(w)] + ptw = 0 (h) ind can be rewritten in one of the two following forms: L{L(w) + ip?u] — ?p?[L(0) + ip?w] = 0 L{L(w) — ip%o] + ip*{L(w) — ip*w] = 0 6) where ? = 4V —] We see that Eq (h) is satisfied by the solutions of the second-order equations L(w) + ip?w = 0 (9) L(w) — ip?w = 0 (k) Assuming that ` W1 = gi + U2 We = gs + ty (1) are the two linearly independent solutions of Eq (7), it can be seen that Ws = Ø1 — 1? and Ws = @š — tựa (m) are the solutions of Eq (k) All four solutions (1) and (m) together then represent

the complete system of independent solutions of Eq (h) By using the sums and the differences of solutions (J) and (m), the general solution of Eq (h) can be represented in the following form:

+ = Chợọi + Cove + Cavs + Cigs (n)

in which Ci, ., Cs are arbitrary constants Thus the problem reduces to the determination of four functions ¢1, , g4, which can all be obtained if the com-

plete solution of either Eq (j) or Eq (k&) is known

Taking Eq (j) and substituting for L(w) its meaning (f), we obtain đo _ đu

— — rn? =

#a T27 10010 0 (o)

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By introducing new variables n=%2 Vic ¢=wvz | (p) Eq (0) becomes d2 d oo tag tt Dr =0 (r) We take as a solution of this equation the power series fi = đọ + đi + gan? + - - ° (s)

Substituting this series in Eq (r) and equating the coefficients of each power of 7 to zero, we obtain the following relation between the coefficients of series (s):

(n? — 1)dn + GQn-2 = O (2)

Applying this equation to the first two coefficients and taking a_1 = a_2 = 0, we find

that ao = 0 and that a; can be taken equal to any arbitrary constant Calculating

the further coefficients by means of Eq (t), we find that series (s) is

-ot|,-@4_~". 7 c | =Œ"

trơn 2:42 198:4:6 2:4:600:8 ` |- cn )

where /¡() is the Bessel function of the first kind and of the first order For our further discussion it is advantageous to use the relation

d „3 „ 6 dJo

J 1(7) = — — đn 1 — — 22 Ta 4)? ——— — —————————— (2:4- 6)? + eo ee | = —— dn (v)

in which the series in brackets, denoted by Jo, is the Bessel function of the first kind

and of zero order Substituting the expression 2p X⁄ iz for n [see notation (p)] in

the series representing Jo(n) and collecting the real and the imaginary terms, we obtain Jo(n) = vip Vx) + i2(2p X⁄3) (w) where 2 Vz 4 2 X⁄+ 8 tip Vz) =1 — C2 V2, Oe Ve) (2 - 4)? (2-4-6 8)? ` _ _ (294) /2(2; A/2 - — (ÊP V2)? „ (2p V2)° _ _— Œp V2)” ee 22 (2-4-6)? (2-4-6-8- 10)? The solution (u) then gives ° tì = —CllW1(2p V2) + 122p V⁄2)] (a’) where y, and y, denote the derivatives of the functions (294) with respect to the argument 2p X⁄ +

The second integral of Eq (r) is of a more complicated form Without derivation it can be represented in the form

Trang 29

TABLE 86 TABLE OF THE y(x) FuNctTions (Continued) + ýa(z) ⁄4(z) ae) + .@) dx 4.00 —0.0014 +0.0230 —0.0152 —0.0200 4.10 —0.0028 +0.0211 —0.0127 —0.0193 4.20 —0.0039 +0.0192 —0.0104 —0.0185 4.30 —0.0049 +0.0174 —0.0085 —0.0177 4.40 —0.0056 +0.0156 —0.0065 —0.0168 4.50 —0.0062 +0.0140 —0.0049 —0.0158 4.60 —0.0066 +0.0125 —Q.0035 —0.0148 4.70 —0.0069 +0.0110 —0.0023 —0.0138 4.80 —0.0071 -++0 0097 —0.0012 —0.0129 4.90 —0.0071 +0.0085 —0.0003 —0.0119 5.00 —0.0071 +0.0073 +0.0005 —0.0109 5.10 —0.0070 +0.0063 +0.0012 —0.0100 5.20 —0.0069 +0.0053 +0.0017 —0.0091 5.30 —0.0067 +0.0044 +0.0022 —0.0083 5.40 —0.0065 + 0.0037 +0.0025 —0.0075 5.50 —0.0062 +0.0029 +0.0028 —0.0067 5.60 —0.0059 + 0.0023 +0.0030 —0.0060 5.70 —0.0056 +0.0017 +0.0032 —0.0053 5.80 —0.0053 +0.0012 +0.0033 —0.0047 5.90 —0.0049 +0 0008 +0.0033 —0.0041 6.00 —0.0046 +0.0004 +0.0033 —0.0036 Having solutions (a’) and (b’) of Eq (r), we conclude that the gencral solution (n) of Eq (e) is 1 = a = levi Jz) + Cw (2p V2) + Cw (2 Wz) ụ + Cw (2p Vz)} (e)

Numerical values of the functions yi, , ¥4 and their first derivatives are given

in Table 86.1 A graphical representation of the functions yj, , ¥4 i8 given in Fig 249 It is seen that the values of these functions increase or decrease rapidly as the distance from the end increases This indicates that in calculating the constants of integration in solution (c’) we can very often proceed as we did with functions (281), z.e., by considering the cylinder as an infinitely long one and using at each edge only two of the four constants in solution (c’)

1This table was calculated by F Schleicher; see ‘“Kreisplatten auf elastischer

Unterlage,’’ Berlin, 1926 The well-known Kelvin functions may be used in place of the functions ¥, to which they relate as follows: ¥i(z) = ber z, ¥2(x) = — bei z,

ý:(z) = —(2/z) kel z, ¿ = —(2/z) kerz For more accurate tables of the functions

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92 Úz;, Ủa” 40+ 02+ 30† W vq 20+ o.1 + 10+ 5 of LES Si 2 tot, of LAST ESE “10+ -20+ ve -0.1 > -30+ -40+ -0.24 Fic 249

In applying the general theory to particular cases, the calculation of the consecutive arivatives of +ø is simplified if we use the following relations: vi (£) = z(‡) — cứi( //() = —w() — (9 1 (d’) v3 (é) = ý4(£) — ;2(Ð Wa (&) = —4(Ệ) — AAG ‘here the symbol £ is used in place of 2p V⁄z From expression (c’) we then obtain Ne=- “Tụ =——— ^= 4⁄2ICw/ (£) + Cz:(£) + Cz#;(£@ + C#„(@] — (e) 8 =S—— [ớilga(b) — 20/(Đ] — Ơlga(E) + 2/40] & 2+ X⁄z | + Œ:[gW.(8) — 2;()] — Caléva(d) + 2K Œ) đ?ụ› Ea? f.= -Do7 Ra we X⁄z{ŒI()3;(£) — 4(0:(£@) + 8/:(] — C:[(£)*#;(£) — 4(‡)ý:(‡) — Sự;(£)] -++ C;[(£)?⁄,(@) — 4(£)#4(£) + Sự;(£)] — C,[(£)*;(£) — 4(£)v:(#) — 8,(Ð]] (g) dM, Eat === Hư X⁄z{Cila(@ + 2/:(Ð] + Œ:[‡Uz(#) — 2:(] + Œ:[tUz(£) + 2/,(Ð] + C,lt,() — 29;(]) ()

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condi-tions The values of the functions ¥, , ¥4 and their derivatives are to be taken

from Table 86 if 2p ~/z <6 For larger values of the argument, the following

asymptotic expressions are sufficiently accurate: 1 — et/V2 cog ( = — = Ệ L8 V⁄2z‡ c 3 1 Ệ LÍ y(t) =Z — =1 án cóc Tà : X⁄2zx£ 3⁄2 8 vilé) = AC) ~ ct/V⁄2 cos (S, + 3 2È 8 v(t) ¥ — ft et/V2 sin (củ; 42 V⁄2zt | 4/2 8 2 2 (296) 2 Ệ V4) Nà 3 (5 + 3 2 Ệ ee) we — „| =e-t/V⁄2 —— 0) Va í mm (Yr ) 2 Ệ ery ow „| e—t/V2 ` - “49 V3 ° ¬ (Wi )

As an example, consider a cylindrical tank of the same general dimensions as that used in the preceding article (page 487), and assume that the thickness of the wali varies

from 14 in at the bottom to 34 in at the top In such a case the distance of the

origin of the coordinates (Fig 248) from the bottom of the tank is

d +20 = $d = 416 in

Hencé, (2p V⁄2);~z,+a = 21.45 Forsuch a large value of the argument, the functions vi, , wand their first derivatives can be replaced by their asymptotic expressions (296) The deflection and the slope at the bottom of the tank corresponding to the particular solution (c) are

on _ ya? TT + Lo zeroed = _-_T8_ #ọ —_ ()

Ha (to +d)? -

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‘alculating the values of functions ¥, ¥2 and their derivatives from the asymptotic

yrmulas (296) and substituting the resulting values in Eqs (j’), we obtain ya? 1 C,; = —269 — N Ea a/d + zo ya? 1 ba /d + Xo rhere | N = (e~t/V2 ⁄2z‡)t~s.ss

ubstituting these values of the constants in expression (g’) we find for the bending

10oment at the bottom

Mạ = 13,900 lb-in per in

n the same manner, by using expression (h’), we find the magnitude of the shearing

2rce at the bottom of the tank as

Qo = 527 |b per in

‘hese results do not differ much from the values obtained earlier for a tank with niform wall thickness (page 487)

119 Thermal Stresses in Cylindrical Shells Uniform Temperature Jistribution If a cylindrical shell with free edges undergoes a uniform emperature change, no thermal stresses will be produced But if the dges are supported or clamped, free expansion of the shell is prevented, nd local bending stresses are set up at the edges Knowing the thermal xpansion of a shell when the edges are free, the values of the reactive aoments and forces at the edges for any kind of symmetrical support an be readily obtained by using Eqs (279) and (280), as was done in he cases shown in Fig 241

Temperature Gradient in the Radial Direction Assume that t; and te re the uniform temperatures of the cylindrical wall at the inside and the utside surfaces, respectively, and that the variation of the temperature hrough the thickness is linear In such a case, at points at a large dis- ance from the ends of the shell, there will be no bending, and the stresses an be calculated by using Eq (51), which was derived for clamped plates see page 50) Thus the stresses at the outer and the inner surfaces are

Batty — ts)

Ce =O = FOG) (a)

vyhere the upper sign refers to the outer surface, indicating that a tensile tress will act on this surface if t1 > te

Trang 33

in this case we observe that at the edge the stresses (a) result in uniformly distributed moments Mo (Fig 250a) of the amount

Batty —_ t2)h?

12(1 — ) (6)

My = —-

To obtain a free edge, moments of the same magnitude but opposite in direction must be superposed (lig 250b) Hence the stresses at a free edge are obtained by superposing upon the stresses (a) the stresses produced by the moments — Mp (Fig 250b) These latter stresses can be readily calculated by using solution (278) From this solution it follows that Fatty — t;)h? _ vE a(t: — te)h? (M,)z-o = “TẤT =) Bia» t6) Eh My _ Eha(t, — te) 1 — py? œ 28D 2 /3(1 — ») vì e

It is seen that at the free edge the maximum thermal stress acts in the

circumferential direction and is obtained by adding to the stress (a) the stresses produced by the moment M, and the force N, Assuming that

ti > te, we thus obtain

(oy) max = ` = (1 —y+ v5") (e)

For v = 0.3 this stress is about 25 per cent greater than the stress (a) calculated at points at a large distance from the ends We can therefore conclude that if a crack will occur in a brittle (M¿)z—o = v(M,);—o = (Ny) z=0 —= — = (U)z—o =

C x material such as glass due to a temperature

Mo difference ¡ — t2, it will start at the edge and + (a) will proceed in the axial direction Ina similar oes —x manner the stresses can also be calculated

“Mo P| in cases in which the edges are clamped or

2 (b) supported.!

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This expansion can be eliminated and the shell can be brought to its initial diameter by applying an external pressure of an intensity Z such that 2 Tự = œgF'(+) which gives Eh Z = —— F(a) (f)

A load of this intensity entirely arrests the thermal expansion of the shell and produces in it only circumferential stresses having a magnitude

Z

oo = — T = — HaF (x) (g)

To obtain the total thermal stresses, we must superpose on the stresses (q) the stresses that will be produced in the shell

by a load of the intensity —Z This latter load must be applied in order to make the lateral surface of the shell free from the external load given by Eq (f) The stresses produced in the shell by the load —Z are obtained by the integration of the differential equation (276), which in this case becomes d4w Kha KT Mol [Mo — + 484w = — —— F(z) (h) 4 — dx Da " (to-t,) (b)

As an example of the application of this Mx equation let us consider a long cylinder, as `:

) _— _— shown in lig 25la, and assume that the part

of the cylinder to the right of the cross section (c) mn has a constant temperature to, whereas that Fig 251

to the left side has a temperature that decreases linearly to a temperature t; at the end x = 6 according to the relation

_ (to — f1)

t = to 5

The temperature change at a point in this portion is thus

F(a) =t— t= — 5 We | (2)

Substituting this expression for the temperature change in Eq (h), we find that the particular solution of that equation is

Trang 35

The displacement corresponding to this particular solution is shown in Fig 251b, which indicates that there is at the section mn an angle of discontinuity of the magnitude

ad

= +> (to — t1) | (k)

To remove this diseontinulty the moments Mạ must be applied Since the stress o, corresponding to the particular solution (7) cancels the stresses (g), we conclude that the stresses produced by the moments My are the total thermal stresses resulting from the above-described decrease in temperature If the distances of the cross section mn from the ends of the cylinder are large, the magnitude of the moment M > can be obtained at once from Eq (280) by substituting

to obtain!

Substituting for 6 its value from expression (275) and taking v = 0.3, we find that the maximum thermal stress is

(02) mac = cm 0 _ 9, 353 5 TC /ah(e — bì (m)

It was assumed in this calculation that the length b to the end of the cylinder is large If this is not the case, a correction to the moment (I) must be calculated as follows In an infinitely long shell the moment M yroduces at the distance x = b a moment and a shearing force (Fig 251c)? that are given by the general solution (282) as ©

2

M, = —D&Y = Moe(8))

d®w in)

Q = —DTY = —26Mat(6b)

Since at the distance x = b we have a free edge, it is necessary to apply there a moment and a force of the magnitude

—M, = —Mog(6b) —Qz = 26M (6b) (0)

in order to eliminate the forces (n) (Fig 251c)

11f t9 — t; is positive, as was assumed in the derivation, Mo is negative and thus has the direction shown in Fig 2516

Trang 36

The moment produced by the forces (0) at the cross section mn gives the desired correction AM which is to be applied to the moment (1) Its value can be obtained from the third of the equations (282) if we substitute init —M (6b) for My and —28M of (6b)* for Qo These substitutions give

d’w —_

AM = —D—] = —Midle(6b)}? — 2M of (8b)]? (p)

As a numerical example, consider a cast-iron cylinder having the following dimensions:a = 942 in., h = 12 in., b = 44in.; a = 101-107’, = 14 - 108 psi, to —_ ty = 180°C The formula (m) then gives Ømax = 7,720 psi | (q) In calculating the correction (p), we have _ pPa-”) 1 _ a= ae ang GB) Bb = 1.50 and, from Table 84, y(6b) = 0.238 t(@) = 0.223 Hence, from Eq (p), AM = —M,(0.238? + 2 - 0.2232) = —0.156M

This indicates that the above-calculated maximum stress (¢) must be diminished by 15.6 per cent to obtain the correct maximum value of the thermal stress

The method shown here for the calculation of thermal stresses in the case of a linear temperature gradient (z) can also be easily applied in cases in which F(z) has other than a linear form |

120 Inextensional Deformation of a Circular Cylindrical Shell.? If

the ends of a thin circular cylindrical shell are free and the loading is not symmetrical with respect to the axis of the cylinder, the deformation con- sists principally in bending In such cases the magnitude of deflection can be obtained with sufficient accuracy by neglecting entirely the strain in the middle surface of the shell An example of such a loading condition is shown in Fig 252 The shortening of the vertical diameter along which the forces P act can be found with good accuracy by considering only the bending of the shell and assuming that the middle surface is inextensible

Let us first consider the limitations to which the components of dis-

placement are subject if the deformation of a cylindrical shell is to be inextensional Taking an element in the middle surface of the shell at a point O and directing the coordinate axes as shown in Fig 253, we shall

* The opposite sign to that in expression (0) is used here, since Eqs 282 are derived for the direction of the x axis opposite to that shown in Fig 251a

Trang 37

denote by u, v, and w the components in the 2, y, and z directions of the displacement of the point O The strain in the z direction is then

đu

Ox (a)

€y —

Tn calculating the strain in the circumferential direction we use Eq (a) (Art 108, page 446) Thus,

Gap a ©)

The shearing strain in the middle surface can be expressed by Ou Ov

Yzo = ade ax (c)

which is the same as in the case of small deflections of plates except that a dy takes the place of dy The condition that the deformation is inexten- fae ———- + / ~ N ¡ @ : 7Ì |_ - - 2a - _—†' | Po ——-~t* t0 : / _—Y T = L ⁄ Pk-c >, P | \ ' Fic 252 Fig 253

sional then requires that the three strain components in the middle surface must vanish; 7.e., Ov 2 0 OU Ov | (đ) These requirements are satisfied if we take the displacements in the following form: 1M) 0 U› =Œ > (dn COS NE — A, Sin ng) ?+= Ì we (e) wi = —a > n(an Sin ng + ay cos nv) n=]

Trang 38

cross sections of the shell deform identically On these displacements we can superpose displacements two of which vary along the length of the cylinder and which are given by the following series: 0 U2 = —a » = (b, sin ng + bi cos nv) n=1 Ùạ = 2 » (b, cos ny — bj sin ng) (f) n=1 We = —2 » n(b, sin ne + bf cos ne) n=1

It can be readily proved by substitution in Eqs (d) that these expressions also satisfy the conditions of inextensibility Thus the general expressions for displacements in inextensional deformation of a cylindrical shell are

w= tạ Cua 0 =0, 4+9 wWw=wW+U2 (9)

In calculating the inextensional deformations of a cylindrical shell under the action of a given system of forces, it is advantageous to use the energy method To establish the required expression for the strain energy of bending of the shell, we begin with

the calculation of the changes of curvature of ds

the middle surface of the shell The change a eo

of curvature in the direction of the generatrix dạ” |

| \

is equal to zero, since, as can be seen from \

2xpressions (e) and (f), the generatrices re- ds 2D n, dw, d?w ds

° ° /

main straight The change of curvature of n \ ds dsế the circumference is obtained by comparing Fic 254 the curvature of an element mn of the circum-

Trang 39

Hence the change in curvature is 2 de + 2 as Xe (a — w) dg adp a? 0¢? By using the second of the equations (d) we can also write 1 fov ö? Xe = a? & + mì (h) The bending moment producing this change in curvature is D fav , dw Me — É + me)

and the corresponding strain energy of bending per unit area can be calcu- fated as in the discussion of plates (see page 46) and is equal to

D TS [ “—.L— Ì =— fav ở? \? D —— ở? \? )

2a‘ & + ae) 2a (u + 5) ớ,

In addition to bending, there will be a twist of each element such as that shown at point O in Fig 253 In calculating this twist we note that during deformation an element of a generatrix rotates! through an angle equal to —dw/dzx about the y axis and through an angle equal to dv/dz about the z axis Considering a similar element of a generatrix at a circumferential distance a dy from the first one, we see that its rotation about the y axis, as a result of the displacement w, is ow 0ö? an dp an” (J) The rotation of the same element in the plane tangent to the shell is Ov ở (2) 2z + ay dep

Because of the central angle dy between the two elements, the latter rotation has a component with respect to the y axis equal to?

— —dy (k)

From results (j) and (k) we conclude that the total angle of twist between the two elements under consideration is |

0” dv

—Xz„0 dọ = — (52 + =) dp

Trang 40

and that the amount of strain energy per unit area due to twist is (see page 47)

D(A — v) ( d?w g0\?

a? ( ant =) °

Adding together expressions (7) and (1) and integrating over the surface of the shell, the total strain energy of a cylindrical shell undergoing an inextensional deformation is found to be

07*w s{ 9” đu\Ÿ

= ao |] {Ge + sợ) +90 — 0n (ng + áp) |4 dm

Substituting for w and v their expressions (g) and integrating, we find for a cylinder of a length 21 (Fig 252) the following expression for strain energy: V=7r7Dl » m—U [ns |a*œ; +a?) += = 12(b2 + b2) | n=2 + 2(1 — z)a*(b‡ + ut (297) This expression does not contain a term with n = 1, since the corresponding displacements v1 = a(ai cos » — aj; sin ¢) (m) Wi = —a(aisin g + a; cos ¢) ’

represent the displacement of the circle in its plane as a rigid body The vertical and horizontal components of this displacement are found by substituting ¢ = 7/2 in expressions (m) to obtain

(V1) p=2/2 = — aaj (Wi) y=n/2 = —Ga,

Such a displacement does not contribute to the strain energy

The same conclusion can also be made regarding the displacements represented by the terms with n = 1 in expressions (f)

Let us now apply expression (297) for the strain energy to the calculation of the deformation produced in a cylindrical shell by two equal and opposite forces P acting along a diameter at a distance c from the middle! (Fig 252) These forces produce work only during radial displacements w of their points of application, 7.e., at the points x =c, ¢ = 0, and g =z Also, since the terms with coefficients a, and ba in the expressions for w; and wz [see Eqs (e) and (f)] vanish at these points, only terms with coefficients a, and 6, will enter in the expression for deformation By using the 1 The case of a cylindrical shell reinforced by elastic rings with two opposite forces acting along a diameter of every ring was discussed by R S Levy, J Appl Mechanics,

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