16. General theory of culindrical shells

16. General theory of culindrical shells

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 sub- mitted 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 ele- ment in Fig 235, we also conclude from symmetry that the twisting moments M,, = M,, vanish and that the bending moments M, are con- stant along the circumference Under such conditions of symmetry

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)

by lateral load The remaining two equations can be written in the following simplified form:

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 we conclude that the component v of the displace- ment in the circumferential direction vanishes We thus have to con- sider only the components u and w in the z and z directions, respectively The expressions for the strain components then become

and the second equation gives

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 equa- tions as for plates, we then obtain

is the flexural rigidity of the shell

Returning now to Eqs (b) and eliminating Q, from these equations,

of this equation is

w = ef(C', cos Bx + C2 sin Bx)

+ e—6(C; cos Bx + Cy sin Br) + f(x) (277)

* See S Timoshenko, “Strength of Materials,’’ part II, 3d ed., p 2, 1956

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

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

1 Observing the fact that the system of forces applied at the end of the pipe is a balanced one and that the length of the pipe may be increased at will, this follows also from the principle of Saint-Venant; see, for example, 8 Timoshenko and J N Goodier,

“Theory of Elasticity,’’ 2d ed., p 33, 1951

obtained by differentiating expression (278) This gives

-0.4 -0.2

0 0.2 0.4 0.6

0.8 1.0

If the moment M, and the deflection w are found from expressions

‘The figures in this table are taken from the book by H Zimmermann, ‘‘ Die Berechnung des Eisenbahnoberbaues,” Berlin, 1888

(282), the bending moment MM, is obtained from the first of the equa- tions (f), and the value of the force Ny from Eq (e) Thus all neces- sary 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

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

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,

TaBLE 84 TABLE OF FUNCTIONS 9g, y, 0, AND [

TABLE 84 TABLE OF EUNCTIONS ý, ý, 0, AND ¢ (Continued)

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

n dw P also under the load and is deter-

37” dx mãn (8X) mined from Eq (284) as

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

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

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 | |

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

1 q dé is the load per unit length of circumference

length J is then

_ | 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 appli- cation 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

Mo, Ferrers a HTTTHTTTTE 5

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

Solving for My and Qo, we obtain

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

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 dis- tance 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 con- sidered 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 uni- form load (Z = —p) is —p/484D = —pa?/Eh The general solution (277) can then be put in the following form by the introduction of hyper- bolic 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

The constants C; and C, must now be selected so as to satisfy the con- ditions at the ends If the ends are simply supported, the deflection and the bending moment //, must vanish at the ends, and we obtain

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,

From these equations we obtain

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

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

(00)z—o = 64 Da‘ (1 cos 2a + cosh 53) (m) When the shell is long, a becomes large, the second term in the paren- theses 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

Differentiating expression (J) twice and multiplying it by D, the bend- ing moment is found as

At the middle of the shell this moment is

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

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

1 Both cases are discussed in detail by I G Boobnov in his ‘Theory of Structure

of Ships,”’ vol 2, p 368, St Petersburg, 1913 Also included are numerical tables which simplify the calculations of moments and deflections

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)

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

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

Now, taking the rings into consideration and assuming that they are abso- lutely rigid, we conclude that reactive forces will be produced between each ring and the pipe The magnitude of the forces per unit length of

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 mag- nitude 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

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?

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

116 Pressure Vessels The method illustrated by the examples of the preceding article can also be applied in the analysis of stresses in cylindri- cal 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 com- parison 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

4See also M Esslinger, ‘‘Statische Berechnung von Kesselbéden,’’ Berlin, 1952;

G Salet and J Barthelemy, Bull Assoc Tech Maritime Aeronaut., vol 44, p 505, 1945; J L Maulbetsch and M Hetényi, ASCH Design Data, no 1, 1944, and F Schultz-Grunow, Ingr.-Arch., vol 4, p 545, 1933; N L Svensson, J Appl Mechanics, vol 25, p 89, 1958

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

where p denotes the internal pressure

For the spherical ends this theory gives a uniform tensile force

1 This case was discussed by E Meissner, Schweiz Bauzig., voi 86, p 1, 1925

shearing forces Qo and bending moments Mo uniformly distributed along the circumference and of such magnitudes as to eliminate this discon- tinuity ‘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 cylindri- cal 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 discon- tinuity 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,

2 Note that the direction of Qo in Fig 244 is opposite to the direction in Fig 236

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

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 con- sidered In this way we obtain at the outer surface of the cylindrical shell

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

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,

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),

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 solu- tion 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,

1 More detail regarding stresses in boilers with ellipsoidal ends can be found in the book by Héhn, “Uber die Festigkeit der gewélbten Béden und der Zylinderschale,”’ Zurich, 1927 Also included are the results of experimental investigations of dis- continuity stresses which are in a good agreement with the approximate solution _ See also Schultz-Grunow, loc cit

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

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)

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

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

— gap (Mo + Qo) = + “ER

585p (28Mo + 6u) = Eh From these equations we again obtain expression (/) for Mo, whereas for the shearing force we find!

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

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

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

->‡-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

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

¬ 2Q -yÌ 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

* H Reissner, Beton u Eisen, vol 7, p 150, 1908; see also W Fliigge, “Statik und Dynamik der Schalen,” 2d ed., p 167, Berlin, 1957 For tanks slightly deviating from the cylindrical form see K Federhofer, Osterr Bauzettschrift, vol 6, p 149, 1951; and for tanks with thickness varying in accordance with a quadratic law, see Feder- hofer, Osterr Ingr.-Arch., vol 6, p 43, 1952 A parameter method, akin to that explained i in Art 40, has been used by H Faure, Proc Ninth Congr Appl Mech Brussels, vol 6, p 297, 1957 Many data regarding the design of water tanks are found in W 8 Gray, “Reinforced Concrete Reservoirs and Tanks,” London, 1954, and in V Lewe, “Handbuch fiir Eisenbetonbau,” vol 9, Berlin, 1934

vhich, upon division by z, can be also written

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:

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

By introducing new variables

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

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)

The second integral of Eq (r) is of a more complicated form Without derivation

it can be represented in the form

te = C''lWj;(2s Vz) + 2 X⁄2)] | (b')

TABLE 86 TABLE OF THE (rz) FUNCTIONS

TABLE 86 TasLe oF THE (xz) FuNcTIons (Continued)

TABLE 86 TABLE OF THE /(Z) FuNcTIONS (Continued)

TABLE 86 TABLE OF THE y(x) FuNctTions (Continued)

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

By means of these formulas the deflections and the stresses can be calculated at

ny point, provided the constants Ci, , C, are determined from the edge

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:

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

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

yrmulas (296) and substituting the resulting values in Eqs (j’), we obtain

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

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

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

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

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

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

Temperature Gradient in the Axial Direction

If the temperature is constant through the thickness of the wall but varies along the length of the cylinder, the pro- blem can be easily reduced to the solution of Eq (274).? Let ¢ = F(x) be the increase of the temperature of the shell from a certain uniform initial temperature Assuming that the shell is divided into infinitely thin rings

by planes perpendicular to the z axis and denoting the radius of the shell by

a, the radial expansion of the rings due to the temperature change is aaF' (x)