axisymmetric constriction of a circular cylinder

AXISYMMETRIC CONSTRICTION OF A CIRCULAR CYLINDER UNDER UNIFORM AXIAL COMPRESSION ByDOO-SUNG LEE

[Received 1 February 1994. Revise 30 March 1994] SUMMARY The compression of an elastic circular cylinder under a uniform load is considered when radial expansion is prevented over a central portion of the curved surface by a rigid constraint concentric with the cylinder. From the mixed boundary conditions on the outer surface, dual series equations are obtained which are subsequently reduced to a Fredholm integral equation of the second kind amenable to numerical solution. An alternative derivation of this integral is also presented. The numerical solution of this integral equation is obtained, and the physical quantities of interest are calculated. 1. Introduction

in this paper is a boundary-value problem of mixed type in elastostatics which arises in the theoretical determination of the stress in an elastic circular cylinder. Recently, the author (1) has considered the problem of shrink fit in an elastic half-space having a cylindrical cavity the analysis of which can be extended to the axisymmetric contact problem involving a cylinder of finite length discussed in the present paper. This sort of problem naturally arises in the case when a circular cylinder is compressed by axial loads while the radial expansion is prevented over a central region of the curved surface by a rigid constraint. In reality this resembles the situation when a rivet is expanded into a circular hole in a relatively rigid plate as illustrated in Fig. I. Investigations into the symmetrical deformations of a circular cylinder have appeared in the past. Notably Vaughan and Allwood (2) obtained dual series equations in the course of studying the constriction of a cylinder, much the same problem as ours. They limited their investigations to relatively short cylinders, and obtained an approximate solution. Tranter and Craggs (3) treated the discontinuities occurring in the boundary values over the surface of a circular cylinder by the use of integral transforms. On the other hand Yaw and Cakmak (4) obtained the stress and displacement fields for a band applied rigidly to a hollow cylinder. Sneddon (5) has also examined the case of an infinite cylinder loaded over half of its surface by a constant pressure. In more recent years, contact problems concerning circular cylinders have been considered by various authors (6 to 9) using different methods including Filippova and Chebkov (10) who considered the contact problem for a O F CONCERN

[Q. Jl Mccfa. ippL M i l h , VoL 48, PL I. 1995]

( Oxford UnJmsty Press I99S

Downloaded from http://qjmam.oxfordjournals.org/ at Seoul National University on December 30, 2015

(Department of Mathematics, Konkuk University, 93-1 Mojin-Dong, Sungdong-Gu, Seoul, Korea)

90

DOO-SUNG LEE

t •M-

Y//////////////A FIG. 1. Constriction of the cylinder under axial compression

prestressed finite cylinder when the lateral surface is compressed uniformly by forces but its end faces are free of loads. The technique employed in this paper is to express the components of the displacement vector and stress tensor in terms of suitable harmonic and biharmonic functions which satisfy the equations of elastic equilibrium. The boundary conditions of mixed type lead to dual series equations which are subsequently reduced to a Fredholm integral equation of the second kind. In section 3, an alternative derivation of this integral equation is presented. In section 4, a numerical solution of this integral equation is obtained and finally stress-intensity factors are tabulated for various values of the length and radius of the cylinder. Other quantities of physical interest are also presented. For special functions, we use the notation of Erdelyi el al. (11), whose work is also a readily available source for the properties of special functions. 2. Formulation of the problem and derivation of the integral equation We take the length and the radius of the cylinder to be 2d and a, respectively, and use the cylindrical polar coordinates (p, 0, z) with z-axis as the axis of the cylinder. The notation employed for the components of the displacement vector and the stress tensor are those used in (12). Consider a circular cylinder confined in a hole of a rigid plate and compressed by the uniform pressure p per unit area at its ends. We assume that the contact surface of the cylinder with the hole whose axial distance is 2d — 2c is prevented from any radial expansion. Then the boundary conditions for the present problem can be mathematically stated as follows. On the curved surface p = a


i

CONSTRICTION OF A CIRCULAR CYLINDER

91

the boundary conditions are -p)dP.

(2.19)

It can be easily shown (15) that if 7,(/a) = 0 then H ( / ? > A) =

~/jIT;.2 ' • ( ^ M ) ^ )

(2.20)

and

V 0(fr>y f V/

14

).

(2.2i)

Jo

On putting (2.20) and (2.21) into (2.18), we obtain after simplification that IX—d + sinh IX—d

8 2!,

Pm®*

,~


If we multiply this equation by p and integrate it, we obtain the following equation:


95

Substituting the values of the coefficients aH from (2.13) into (2.22) we find that on interchanging the order of integration and summation

8a'

sinh Xmd

-

g(t)F(t,XJdt,

(2.23)

where (2.24)

(Pi +1 2 ) 1 Further,

'

2

(2.25)

dX '

where (2.26)

«+ To evaluate the infinite series in (2.26), we consider the integral

1

el"J0(pzt) ^—^ cosec nz dz, c, z2P2 + A2

where ft = n/d, and the contour consists of the real axis indented at — N,..., — 1, 0, 1 , . . . , N, together with the large semi-circular arc in the upper half-plane shown in Fig. 2.

-2

-1

0

t

2

FIG. 2. The contour C,

N

Rez


PIMP.*)

R 2

96

D O O - S U N G LEE

We can easily show that -

J0(pnt)

_d['-

J0(xt)

A

_

1 ,d

2)^1

0

sinh Ad

(

)

Making use of the result (11)

(2-28)

we see that ip(t, /.) can be written in the form .nz + / n (sinh knz + >.nz cosh ).nz)} n= 1

= - , " If we invert this Abel-type integral equation, and

• (3.5)

make use of the


+ P £ K3l{b«

98

DOO-SUNG LEE

relations f' cos £z dz

j

'zsinhAzdz

and (2.17), we find that y(t) is related to dn, cn and /„ through the equation -

-

^(U)^U

Jo

„ is given by

2 2 2 a Jo ^0 (C + ^ )

If we substitute A(c,) from (3.4) in (3.8) we find that ,/Jdt,

(3.9)

a Jo

where F,(i,/.) = Now

^2

2 2 = max(f, u).

(4.5)

The singularity of K(t/u) at u = f can be removed as follows. If we use the formula (16)

where E(k) is the complete elliptic integral of the second kind, we can write the


Jo sinhrfy


103

integral involving singular terms in the integral equation as

bY65~\t The infinite series of (2.35) does converge easily (not even with 40 terms), and is thus unsuitable for numerical evaluation. Thus we try to evaluate it in an alternative way. We decompose h3(t, '/.„) as /i3(f, /.„) = sinh /.„ J[/i,(/. n f) - /i 2 (/.f)] , where 2 sinh 2 d/.n

1- e

2iA

"

and h2(u)

= / 0 (u) + H/,(M) - L 0 (M) - ML_ ,(M);

(4.7)

then the infinite series W3(u, t) can be put as

a „= i 2/ n J + sinh 2/.nd x [hx(/.nt)hx(/.nu)

- hx(/.nt)h2(/.nu) -

hl(/.nu)h2(/.nt)

(4.8)

].

Because of the factor e~1?-d, the first three terms in (4.8) are very rapidly convergent but the last term is not. The last term can be put in an alternative form, by integrating the function

f

(4.9)

Jo around the contour shown in Fig. 4, and in (4.9) the function S(z) is defined as S(z)=

2: sinh2 zcl 2zd + sinh 2zd

.

(4.10)


The integration is performed by the trapezoidal rule replacing the differentials by their finite differences. For E(k), the following approximate formula is available within 01 per cent error (17):

104

DOO-SUNG LEE Imr

FICJ. 4. The contour C3

It can be easily shown that n Z, 2/.. sinh2/.-* 2 - z2y

n

106

n 0-3

LEE

TABLE 2. The variation with d, a, and c of S.I.F. ( — k/p) (d = 3") c/d 0-4 0-1 0-2 0-6 0-8 a/d 1-5 2-0 30 1-5 2-0 3-0

0-0323 0-0375 0-0423 0-0500 0-0580 0-0666

1-0

a = n d = 7t c/d = 0-5 n = 0-3

0-3547 0-3828 0-4149 0-5344 0-5803 0-6335

0-1445 0-1475 01499 0-2208 0-2663 0-2310

0-5040 0-5869 0-6992 0-7567 0-8847 10660

ft,. 0 —d—-

0-9 0-8297 0-9503 1-2112 1-2409 I -4248 1-8225

0-6836 0-8349 10579 10225 1-2518 1-5934

-z Pp ^ >

«."-1IHJ "

•

•

0-5

1

1

1

0-2

0-4

0-6

0-8

10

this paper,

(2)

z Fici. 5a. Surface radial displacement versus z/c;

2-0 ~-

a= n d =n v

c/d

= 0-5 = 0-3

1-0 -

1

0-2

1

1

z-c d-c

CMS

FIG. 5b. Surface stress versus (z — c)/(d — c)\

0-8

this paper,

1-0

(2)


0-4

DOO-SUNG

107


o = 4-5 —sr

17-0-4 1-7

* * s

s

™~~\

•z

—«

H

r~/^

K

0-8

^~~"—-—Z>^ c/d-04

04

^ ^ - ^

•

c/d-0-2 ^~"""~

i

0-2

i

0-8

04

1-0

FIG. 6a. Surface radial displacement versus z/c

2-5 -

\

a = 4-5 d = 30

n = 0-4 eld" 0-7 2-0 -

c/d=0-6 / 1-5 -

cld=0A

\

^

1-0 -

c/rf=0-2

— 0-2

04

—~

06

0-8

z-c d-c FIG. 6b. Surface stress versus (r — c)/(d — c)

1-0


c/