The torsion of a circular cylinder containing a symmetric array of edge cracks

Inr. J. Engng Sci., 1972, Vol. 10, pp. 80 1-8 12. Pergamon Press.

THE TORSION CONTAINING

Printed in Gnat Britain

OF A CIRCULAR CYLINDER A SYMMETRIC ARRAY OF EDGE CRACKS

J. TWEED Department of Mathematics, The University of Glasgow, Scotland and

D. P.ROOKJZ TheRoyal Aircraft Establishment, Famborough, Hams, England Abstract- In this paper the authors make use of a recent advance in the theory of Mellin transforms to find the stress intensity factors and crack energy of a symmetric array of edge cracks in a circular cylinder under torsion. Two special cases are considered in detail and the results are given in graphical form. 1. INTRODUCTION

IN RECENT years, the problem of determining the distribution of stress in the vicinity of a crack in a bar under torsion has been considered by a number of authors [ l-61 all of whom have made use of conformal mapping techniques and some of whose results appear to be in conflict. It is the object of this paper to show first of all that problems of this type can sometimes be solved by means of transform techniques and secondly to try to clear up at least one of the anomalies at present in the literature. We do this by considering the following problem: The cylinder 0 =S r s b, 0 G 6 s 27r, 0 s z =SL contains a symmetric array of edge cracks which are defined by the relations 0 < bc s r d b, 8 = 2k?r/n, k = 0, 1, . . ., n - 1, (Fig. 1). The end z = 0 is fixed in the &plane and the other end is acted upon by a couple whose moment T lies along the z-axis; the stress intensity factors and the work done on opening the cracks are calculated. 2. REDUCTION

OF THE PROBLEM

TO AN INTEGRAL

EQUATION

Before we can calculate the quantities of interest we require to find a solution of the equations of elasticity for the region occupied by the cylinder which is such that (i) the crack surfaces and the lateral surfaces of the cylinder are stress free, (ii) on each end of the cylinder the normal stress and the tangential stress in the r-direction is zero

(iii) the resultant force on each end of the cylinder is zero and the resultant moment just balances the applied torque. In order to do this we will assume that the couple acting on the end z= L produces in the cylinder a twist of (Yper unit length and that the displacement field thus set up is of the form u=O,u=arzandw=cucp(r,6) 801

(2.1)

802

J. TWEED

and D. P. ROOKE

Fig. 1.

where U, D and w are the displacements in the r, 6 and z-directions respectively. It is well known[7] that the stresses corresponding to such a displacement field are given by the relations

(2.2) where p is the shear modulus, 8 (r, 8) is a solution of the partial differential equation (2.3) a is related to the applied torque T and the torsional rigidity D through the equations cy= T/D

(2.4)

where (2.51 A being a cross section of the cylinder. By symmetry therefore, it follows that the problem reduces to that of finding a function cp(r, 8) which is a solution of the equation (2.3) in the sector 0 < r < b, 0 < 8 < p/n and which satisfies the conditions

?!ie p’ ar and $ are all finite at r =

0,

(2.6)

6pp(r,7rfn) = 0,O S r si b,

(2.7)

C$

(2.8)

(6, 8) = 0,O C 29 G r/n, rp(r,O)=O,Oerrbc

(2.9)

803

Torsion of a circuhr cylinder

and

$gr,O)=-

(2.10)

r2, be < r < b.

In order to find a suitable representation for cp(r, 8) we shall begin by superimposing the solutions of problems 1 and 2 below. Problem 1. Find a solution @)(r, 8) of the equation (2.3) in the sector 0 < r < b, 0 < 6 < r/n, which is such that (i) (p(l),F

and $$

are finite at r = 0

and (ii)

@‘(r, a/n) = (p”‘(r,O) = 0,O s r s b.

On applying the method of separation of variables solution of this problem may be written in the form @)(r,

$9) = g

it is easily shown that the

akrkn sin (knzl)).

(2.11)

k=l

Problem 2. Find a solution qt2’(rT 8) of the equation (2.3) for the region 0 < r < 01, 0 < 6 < ?rln which is such that (i) q(2) is 0 (r) at r = 0 and is bounded at infmity and (ii) CpC2’(r, r/n) = 0,O S r < 9 By utilising the properties of the Mellin transform [ 131 it is not difficult to show that in this case the,required solution is given by

d2’(r96) =9.X-’

[

s-lA(s)

sin (6-?r/n)s sin

(Ts,n)

;

r

1

(2.12)

where W1 is the inverse Mellin transform and - 1 < Re (s) < 0. It should now be clear that the function cp(r, 6) = %Jtnz-l S-IA(S) [

+

2 akrkn sin

sin (6- 7r/n)s sin (Ts,n)

;r19

(2.13)

(kn#) )- 1 < &?(s) < 0,

k=l

which is obtained by superimposing the results (2.11) and (2.12) is a solution of the equation (2.3) in the sector 0 < r < b, 0 < 6 < r/n, which automatically satisfies the conditions (2.6) and (2.7). To complete the solution we now have to choose the function A (s) and the sequence {&}T in such a way that the remaining boundary conditions are satisfied.

804

J. TWEED and D. P. ROOKE

If we apply the condition (2.8) and make use of the result

we find that

1

A(s) k=l

0 < 8 < ?r/n, and hence that ak=*!IR-’ ,rrbkn Similarly, on applying the conditions the duaI integral equations

where-l

< Re(s)

,k=l,2,3

Cot (m/n);

,....

(2.14)

(2.9) and (2.10) we find that A(s) has to satisfy

r] = 0,

W’[s-lA(s); m-‘[A(s)

1

/$;b

OSrSbc bc < r < b

r] =-?-F(r),

(2.15)

< Oand m F(r) = I: knakrkn.

(2.16)

k=l

If we now assume that A (s ) can be written in the form A(s) = rkp(t”)P+“-‘dt

(2.17)

we find, on making use of the result %JP[s-~; r] = --H(r-

l),Re(s)

< 0,

that !IP[s-~A(s);

rl =

0, -

I

0 G r G be,

r bc P-+(P)

dt, bc < r < b.

so that the first of the equations (2.15) is satisfied automatically. Similarly, on substituting from (2.17) into the right hand side of the second of the dual equations and making use of the result YJP’[cot (7rs/n);r]

=nr”[7r(l-P)]-‘,-1

C Re(s)

we find that 2JP[A(s)

cot (7rs/n); r] =F

n b P-$(P) s bc P-P

dt

< 0,

Torsion of a circular cylinder so

805

that this equation will be satisfied if

?+F(r) r~

,

bc -C r < b.

(2.18)

But, from (2.14) and (2.16) we see that

= -;[,

P-$(P)

{5

(rt/b’)*“}dt

k=l

so that (2.18) now takes the form -1 * ntn-‘p(P) ?r I bc P-P

dt _I

* t+lp(P) ?r I be tin-PP

dt

(2.19) =-rzmn,bc

< r < b.

If we now let q(t) = p(b’V), y = P, P = bnp and then make the change of variables P = b?, we find that (2.19) reduces to

=-

(2.20) b2--np(2--n)/n,

.,, <

p <

1.

Also, if we put u = p-l we find that (2.20) implies that

(2.21) =

__pau-(n+2)ln,

1

<

().

pm

<

and on grouping these equations together and making use of the fact that

I’ ‘d’) yl--Pr

dT

=

I

1

T-2dT-1)

&

7-P

we discover that -1 IT

T’ h(7) ypdT

IY

= g(p), 7 < P < y-l,

(2.22)

J. TWEED and D. P. ROOKE

806

where we have written

h(7) = h(T) = q(7),

y
h.2(7) =-&J(7-1),

(2.23)

g1 (p) = - byP-n)‘n, y < p < 1 g,(p) = - b2-np++~‘n, 1 < p < y-1..

g(p) =

3. THE SOLUTION

OF THE INTEGRAL

(2.24)

EQUATION

The integral equation (2.22) is well known and Tricomi[8] tion is given by A(Y) y‘-’ y_.rdy)dy

has shown that its solu-

1

(3.1)

where A(T) = { (y-l -7) (~-y)}l’~ and C is an arbitrary constant. In order to determine C we note that hz (T) = - ~-~h, (7-l), g(y) = y-‘g(y-‘) proceed as follows : For 1 < T < y-l we have 1 b(T)

=-T-2h(T-1)

1

A(Y) y‘-’ --g(y)dy

= A(7)r

and

and therefore, for y < r < 1 hl(7) =

1 ‘-’ A(&dZ) d( 1 ‘-’

-l

Y Butby(3.l)weseethat,fory

A(Y)

(3.2)

y ,_.-,g(y)dy

~~T~A(T-‘)

5-7


‘-’ A(y)&) Y

Y-7

dy

(3.3)

1

and together with (3.2) this implies that C = 0 and hence that h(7)

4. THE STRESS

-=-A(*)

INTENSITY

1

1 ~

y-’ A(Y)

y

FACTOR

(3.4)

,_.g(y)dy

AND

THE CRACK

In this section we shall be concerned with the calculation factor K and the total crack energy W which are given by K=limiJ

[2(bc-r)]“2u*,,(r,o),

ENERGY

of the stress intensity (4.1)

and W= n/.xx&w(r,O)dr respectively.

(4.2)

Torsion of a circular cylinder

807

Since a*fi(r, 0) =-

P Q(r,O) r [ a?9

+rZ

1

b

_ panP_’ IT

1

bc

weseethat K = limitCMlr”-I {2(bc-r)}1/2 P&c_ IT X

[Ib bc

=

nt”-‘p(P) P-P

E.f bfl-ly(n-1)/n

I”

dt_

nf"-'p(P) tin-t-v”

bc

limit

{ 2b

l/n _

(y

dt

p’/“)}

I

l/2 I:’

(4.3)

2

&

WV-

where h(r) is given by (3.4). But ‘-’ Am y-7

y

J

y-’ A(y)g(y)

dy

dy

P-Y

‘-’ A(y)g(y) y_p

y

dyvOsp<~

and therefore it follows that K =

pab”-’

2bf2”-l)”

1’2 V’ A (y )g(y )

T --[n(l-?)I

I,

y-y

dy

(4.4)

+2,

b

I

w(r,O)dr

be

808

J. TWEED and D. P. ROOKE

and therefore since q(r,O) = ,iC t”-‘p(t”)dt, bc < r < b we see that 1 DIP

=

q

_

,@+2

xfl-1( I

C

1 -X2)q(Xn) d_X

where by (3.4)

4w

=

1.zch

$gJ I’Ak”) [A+& c

It follows that

1

(4.5)

d+(z”){&+j-+}~*

(4.6)

D/p=;

b4 ?rz ‘T-I [

where Z=

n2

lxn-‘(l---f) C AW) I

If we now express (Yin terms of I and substitute the result into (4.4) we see that the stress intensity factor K is given by the formula 2n( 1 -c”)

K = b5’2[ti:2-I]

“’

I c(l+c”)

(4.7) Similarly, on making use of the fact that W=np$[Crp(r,O)dr we see that in terms of Zthe crack energy is given by ?TT2 I P. w = 2pb4 [7?/2-Z12 5. THE SPECIAL

CASE

(4.8) n= 1.

From (4.6), (4.7) and (4.8) we see that in this case

=

T(1+c)[2(1--)~1’2{(3~-_2c+3)tan-lcl/2-3(1-c)cl’2}

4b5’2c5’2[7?/2 -I]

7

(5.1)

and 7TT21 w = 2pb4[1T2/2-I]2’

(5.2)

809

Torsion of a circular cylinder

where Z is given by

V-3)

+ 2 tan-’ cl’3 1_c_[6c+5(1+c2)](1-c)3 cl/2 24~ I _[5(l+cz)“-4c1](1-c)3 64c3 + I x

I

(1--~)~(1+C)~~-~~~,~_7(~~~;)~_(1+C+3~~,~-C)” 8~3 (l+a

I

tan-lcl’2_(1

-3772. c1/2

C

I

I

It should be noted that the expression we have just obtained for K does not agree with that given by Sih [ l] who appears to be in error, for as is easily verified, as c tends

to zero the expression given above tends to 482112T/[5(97? -64)b3j2] which is a well known result [6] but the stress factor given by Sih [ l] and repeated in Paris and Sih [ 151 tends to infinity and this is clearly wrong. 6. THE SPECIAL

CASE n = 2

In this case we see that

T

1 1

l-cz 1’2[(l+C2)2tan-‘c-c(l--cP)], = b5’3C3’2[7r3/2-z] 1+c2

(6.1)

and W is expressed in tenus of Zthrough (4.8) where Zis given by the equation

= 2tan-‘c +

(l-332-g/2 2C3

Lllzsvd.

lONo.9--E

2(1-C3)_(1-C3)3+(1+C3)2tan-lc+ cl C 2C3 *

(1-c4)3tan_lc 4c4

1 (6.2)

810

J. TWEED 7. LIMITING

and D. P. ROOKE CASES

AND

RESULTS

When the crack length ‘u’ is very small (a 6 b, i.e. c - 1) the stress field in the vicinity of the crack approximates to that due to an edge crack in a semi-infinite sheet subject to a constant shear stress at infinity (a,* = 7Csay). The shear stress 7, is equal to the stress at r = b in an untracked bar subject to a torque T and is given by 2T 7c=-3 rb

(7.1)

The stress intensity factor K,, for an edge crack of length ‘u’ in such a stress field is given by [I61 Ko = T&l'* (7.2) and the energy W. by *a*

7rr wo=*.

(7.3)

From (4.7) and (4.8) we see, after a little analysis, that (7.4) and na*T* rpbs

limJtw=-

(7.5)

and these results are in agreement with (7.1), (7.2) and (7.3). It follows that for n 2 1 h;Tt K = ,&P* = &

(7.6)

and li$

02

0.3

W = nWo.

0.4

0.5

06

(7.7)

0.7

c/b

Fig. 2. The variationof KaSi2/T with a/b.

06

0.9

I.0

811

Torsion of a circular cylinder

n=I

I

1

@I

I

0.2

I

0.3

I

0.4

I

0.5

I

,

0.6

0.7

I

o-6

I

OS

I

I.0

a/b Fig. 3. The variation of W/B’, with n/b.

The limit of K as the crack approaches the centre of the cylinder is given in section fiveforn= l.Forn > lwefindthat lil$ K = 0.

(7.8)

For the special cases considered in sections five and six above the stress intensity factor K and the total crack energy W have been computed for several values of the ratio a/b. The results are given in graphical form in Figs. 2 and 3 above which show respectively how Ka5j2/T and W/W, vary with a/b. REFERENCES [II G. C. SIH,J. appl. Mech. 30,419 (1963). PI L. A. WIGGLESWORTH, Proc. Lond. math. Sot. 47.20 (1940). Proc. R. Sot. A170,365 (1939). r31 L. A. WIGGLESWORTH, TN-58-88 (Air Research and Development Command TN-4) (1958). [41 P. M. QUINLAN,AFOSR [51 E. A. SHIRYAEV, P.M.M. 20,555 (1956). 161 A. C. STEVENSON, Phil. Trans. R. Sot. A237,161(1939). Mathematical Theory of Elusticify. McGraw-Hill (1956). 171 I. S. SOKOLNIKOFF, PI F. A. TRICOMI, Q.JI. Marh. 2,199 (1951). A. ERDELYI, W. MAGNUS, F. OBERHETTINGER and F. G. TRICOMI, Tables [91 A. ERDIkYI, of Integral Transforms, Vol. 1. McGraw-Hill (1954). Integral Transforms and Operarional Calculus. Pergamon [lOI V. A. DITKIN and A. P. PRODNIKOV, Press (1965). Introduction to the Theory of Fourier Integrals. Oxford University Press (1937). 1111 E. TITCHMARSH, iI21 I. N. SNEDDON, Fourier Transforms. McGraw-Hill (1951). iI31 C. J. TRANTER, Integral Transforms in Mathematical Physics. Wiley (1956). and I. N. RYZHIK, Tables of Integruls, Series and Proucts. Academic Press 1141 I. S. GRADSHTEYN (1965). [151 P. C. PARIS and G. C. SIH,A.S.T.M. Sp. Tech. Pub. 181,30 (1965). Crack Problems in the Classical Theory of Elasticity. [I61 I. N. SNEDDON and M. LOWENGRUB, Wiley (1969). (Received 1 October 197 1) Resume-Dam cet article, les auteurs utilisent de ticents progrks darts la theorie des transformations de Mellin, pour trouver les facteurs d’intensitk de contrainte et l’energie de fissure dun r&au symetrique de fissures de hard dans un cylindre circulaire sous l’effet dune torsion. Deux cas particuliers sont consider& en d&ail et les rksultats sont donn6.s sous forme graphique. ZmnfmungIn dieser Arheit verwenden die Verfasser einen jiingst in der The&e Mellin’scher Transforme gemachten Fortschritt, urn die Spannungsintensititsfaktoren und die Rissemergie einer synunetrischen Reihe von Kantenrissen in einem Kreiiylinder unter Torsion zu flnden. Zwei SpezialfXlle werden eingehend untersucht und die Resultate werden in graphischer Form gegeben.

812

J. TWEED

and D. P. ROOKE

Sommario-In questa memoria gli AA. fanno uso di un recente progress0 nella teoria delle trasformazioni di Mellin per scoprire i fattori d’intensiti delle sollecitazioni e dell’energia d’incrinatura di un fascia simmetrico d’incrinature di bordo (marginali) in un cilindro circolare sotto tensione. Si esaminano nei particolari due casi speciali e i risultati sono presentati in forma grafica. Akrpaw WieHTbI

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