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The stress intensity factors of a radial crack in a finite rotating elastic disc
Inr. J. Engng Sci., 1972, Vol. 10, pp. 709-714.
Pergamon Press.
Printed in Great Britain
THE STRESS INTENSITY FACTORS OF A RADIAL CRACK IN A FINITE ROTATING ELASTIC DISC D. P. ROOKE The Royal Aircraft Establishment,
Famborough,
Hams., England
and J. TWEED The University of Glasgow, Scotland Abstract-The elastic problem of a radial crack in a finite rotating elastic disc is reduced to the solution of a Fredholm equation. From this solution the stress intensity factor at both tips and the crack formation energy are derived. 1. INTRODUCTION IT HAS been shown by Tweed et al.[ l] that the stress intensity factors and the strain energy for a radial crack, subject to an arbitrary internal pressure, in a finite disc can be expressed in terms of a function which is the solution of a Fredholm equation. In that paper [ I] the plane strain case of a constant internal pressure was considered. We will here consider the pressure which gives results appropriate to the case of a crack in a uniformly rotating disc. 2. BASIC
EQUATIONS
Consider a disc of unit radius containing a crack defined, in plane polar word coordinates, by 6 = 0, 0 < a G r s b < 1 which is subject to an internal pressure f(r). The stress intensity factors, K, and Kb, and the crack energy, W, under plane strain conditions are given by the following: for the tip at r = G
K, = &$a);
)
for the tip at r = b and
,=_2(h*)
(2)
E
where q is Poisson’s ratio and E is Young’s modulus. P(t) is the solution of the integral equation. (3) with s(f)
=;
,-bq(b-y)(y-n)J.(y) a
y--I
709
dy
(4)
710
D. P. ROOKE
and J. TWEED
and
where J=i(nib--2r) -J=* 0
J =fi-2 t
(6) 1
(7)
t
p(a+b-2ubp)
~~~--up)(l-~~)+
2(1 -pflVL/(l
(1 -ptj*
-ap)
(1 -bp)
(9)
and J =a -3
t
V’(l-ap)(l-bp)+ (1 -pt)”
p(2-ppr)(afb-2abp) 2(1 -pt)*V(l -ap) (1 -bpj
$(a---b)” ~8(l-ppr)[(1-ap)(l-bp)]R~~~ The function a=SrSb,
P(t)
(10)
is closely related to the crack shape function
3. ROTATlNG
o(r,O) since [ 11, for
DISC
For the problem of a crack in a disc, of density po, rotating at an angular velocity w the pressure takes the form
f(4 =Po[“f;td -~.fi~dl
(12)
where
f,(r) = 1
and
j;(r)
= rZ/R2
where R is the radius of the disc (numerically equal to unity). If we let P{(t), i = 1,2 be defined by (13)
711
Stress intensity factors of a radial crack
where
Ib
&(t) =+
,I
a-YHY-41;(y) Y--r
dy
(14)
then it follows from (3) and (4) that
(13
-~Pz(t)l.
P(t) =Po[P,(t) From (1) we have
-apzc41
(16)
[P,(b) -aPz(b)l.
(17)
K, = POJ ---[PI(Q) &) and b=-Poj/&y
If K. is the stress intensity factor of a Griffith crack of length (b - a) which is opened by a constant pressure po, then clearly we may write KU’
K’2’
K”’
&
_=LL_aLL Ko
Ko
K’2’
K,dL_&_
and
K
Ko
0
(18) Ko
Ko
where K;’ Ko
2
and
-b-iP,(n)
(19)
From (2) and (12) the energy can be written as
1
dr’
(20)
Since, from (1 l), crack closure at Y= a implies
P(t)
Vi@-t)
dt=O
(t-u)
(21)
we have,
(r-at”/3W
,=_2(1-7%‘:
E
a-f)
[P,(t)__P,(t)Idt
(f-a)
(22)
Let IV0be the energy of a Griffith crack of length (b - a) which is opened by a constant pressurep,, then w
0
= 7r(l -?q)&(b-N)
4E
(23)
712
D. P. ROOKE
and J. TWEED
and
w -=-
1
a
L
wow,-ffwq
W”
w3+7
a*w,
I
(24
where tP, (t) dt V(b-t) (t-a) tP2 (t) dt V’(b-t) (f-a)
(25) t3Pl(t) dt V/(&t) (t-a) W A=WO
8 ?T(b-a)2z?2
t3Pz(t) dt kqb-t) (t-a)
In order to evaluate K,, Kb, and W we require S,(t) and S,(t) from (14). s,(r) which is the same function as S(t), for a constant pressure, in [l] is given by
s,(t) = Elementary
a+b-2t 2
(26)
.
integration of (14) for i = 2 gives S,(t)
=& f
(a+b-2t)
4. NUMERICAL
The integral equations formula, i.e. (13) becomes
+v
(a+b+2t)].
PROCEDURE
(13) are solved using the Gauss-Chebyshev
P,(tj) -z
i
(27)
M(tjyfk.)P<(th_)=
S,(tj)
quadrature
(28)
k=1
fori= 1,2andj,k= 1,2 . . . . n where IZis the order of the Gauss-Chebyshev mation. The arguments tj are given by tj=-
u+b+b-a -COS 2 2
Solution of the set of simultaneous From ( 13) it follows that
n(2j- 1) 2n .
linear equations
approxi-
(29)
(28) yields P&j) for i = 1 and 2.
713
Stress intensity factors of a radial crack
and (30)
Thus K,l&, and KJ& can now be calculated using (18) and {19). To the same approximation the energy terms are obtained as:
and
To calculate W8 and W, R is put equal to unity in the above equations.
Since the stress intensity factors and the crack energy are physical quantities, the way in which they vary with a and b must be independent of the mathematical technique used to find them. It follows that we may dispense with the restriction 0 < a < B -=z1 which was made at the beginning of [I] only to faciiitate the use of the Mehin transform, By using the procedure outiined above, the authors have constructed a comprehensive 3.0 I-
K,/K K,IK
n
0
----
t
2*Q
llES
Vol. 10, No. 8 - E
$-.._a=-o-9
2.5
i
714
D. P. ROOKE
I -0.8
I -b.6
I -0.4
I -0.2
and J. TWEED
I
I
DO
02
I 0.4
I 0.6
I 0.8
I IO
b
Fig. 2. The variation of WIW, with a and b for the case in which Poisson’s ratio is 0.3.
set of tables which give the values of K~‘/K,, Kf’/&, i = 1,2 and W ‘j’/W,,, j = 1,2,3,4 for values of a and b satisfying the condition - 1 < a < b < 1. These tables, which are accurate to four signi~~~t figures may be obtained from either author. Figures 1 and 2 show the variation of I&/&, KblKo and W/W, with a and b for the special case in which Poisson’s ratio is 0.3. REFERENCES [l] [2] [3] [4]
J. TWEED, S. C. DAS and D. P. ROOKE, Inf. J. Engng Sci. 10,323 (1972). S. TIMOSHENKO and J. N. GOODIER, Theory ofElasticity. McGraw-Hill (1965). M. ABRAMOWITZ and I. A. STEGUN, Handbook of Mathematical Functions. Dover (196.5). 1. S. GRADSHTEYN and I. M. RYZHIK, Tables of Integrals, Series and Products. Academic Press (1965). (Received
1 October 197 1)
Resume-Le probleme classique d’une fissure radiale dans un disque 6lastique fini en rotation est ramene a la r&olution d’une equation de Fredholm. A partir de cette solution, le facteur d’intensite de contrainte a chaque extremiti? et i’energie de formation de la fissure sont deduits. Zusa~~~ungDas elastische Problem eines Radialrisses in einer endlichen sich drehenden Scheibe wird auf die Lijsung einer Fredholm’schen Gleichung reduziert. Fur diese LSsung werden der Spannungsintensit;itsfaktor an beiden Spitzen und die Rissbildungsenergie abgeleiter. Sommario-I1 problema elastic0 di un’incrinatura radiale in un disco elastic0 finito in rotazione B ridotto atla soluzione di un’equazione di Fredholm. Da questa soluzione il fattore d’intensim deile solleci~ioni adentrambe~epuntee~ener~adel~fo~~ione del~inc~n~uraso~o~s~ttivamentec~colate. A~cT~~KTHpo6neMa 06 YIQJY~OCTE~AJISIpanHanbIioii ~peuuiIib1B KOHeSHOM, BpamaIoIIIeMca, yIIpyroM AUCKe CBOAHTCX K PeIIIeHHIO YpaBHeHHR @pW'OJlbMa. Ha OCHOBe 3TOrO pfXPi%iB5lBbIBOAXTCR KOJ@WUtIkSITbIHHTCHCHBHOCTA HElIlpSfXCeH%i5ty o6enx KO~I~OBEI3HeprHIO 06pa30BaIWI TPUUaHbI.
Pergamon Press.
Printed in Great Britain
THE STRESS INTENSITY FACTORS OF A RADIAL CRACK IN A FINITE ROTATING ELASTIC DISC D. P. ROOKE The Royal Aircraft Establishment,
Famborough,
Hams., England
and J. TWEED The University of Glasgow, Scotland Abstract-The elastic problem of a radial crack in a finite rotating elastic disc is reduced to the solution of a Fredholm equation. From this solution the stress intensity factor at both tips and the crack formation energy are derived. 1. INTRODUCTION IT HAS been shown by Tweed et al.[ l] that the stress intensity factors and the strain energy for a radial crack, subject to an arbitrary internal pressure, in a finite disc can be expressed in terms of a function which is the solution of a Fredholm equation. In that paper [ I] the plane strain case of a constant internal pressure was considered. We will here consider the pressure which gives results appropriate to the case of a crack in a uniformly rotating disc. 2. BASIC
EQUATIONS
Consider a disc of unit radius containing a crack defined, in plane polar word coordinates, by 6 = 0, 0 < a G r s b < 1 which is subject to an internal pressure f(r). The stress intensity factors, K, and Kb, and the crack energy, W, under plane strain conditions are given by the following: for the tip at r = G
K, = &$a);
)
for the tip at r = b and
,=_2(h*)
(2)
E
where q is Poisson’s ratio and E is Young’s modulus. P(t) is the solution of the integral equation. (3) with s(f)
=;
,-bq(b-y)(y-n)J.(y) a
y--I
709
dy
(4)
710
D. P. ROOKE
and J. TWEED
and
where J=i(nib--2r) -J=* 0
J =fi-2 t
(6) 1
(7)
t
p(a+b-2ubp)
~~~--up)(l-~~)+
2(1 -pflVL/(l
(1 -ptj*
-ap)
(1 -bp)
(9)
and J =a -3
t
V’(l-ap)(l-bp)+ (1 -pt)”
p(2-ppr)(afb-2abp) 2(1 -pt)*V(l -ap) (1 -bpj
$(a---b)” ~8(l-ppr)[(1-ap)(l-bp)]R~~~ The function a=SrSb,
P(t)
(10)
is closely related to the crack shape function
3. ROTATlNG
o(r,O) since [ 11, for
DISC
For the problem of a crack in a disc, of density po, rotating at an angular velocity w the pressure takes the form
f(4 =Po[“f;td -~.fi~dl
(12)
where
f,(r) = 1
and
j;(r)
= rZ/R2
where R is the radius of the disc (numerically equal to unity). If we let P{(t), i = 1,2 be defined by (13)
711
Stress intensity factors of a radial crack
where
Ib
&(t) =+
,I
a-YHY-41;(y) Y--r
dy
(14)
then it follows from (3) and (4) that
(13
-~Pz(t)l.
P(t) =Po[P,(t) From (1) we have
-apzc41
(16)
[P,(b) -aPz(b)l.
(17)
K, = POJ ---[PI(Q) &) and b=-Poj/&y
If K. is the stress intensity factor of a Griffith crack of length (b - a) which is opened by a constant pressure po, then clearly we may write KU’
K’2’
K”’
&
_=LL_aLL Ko
Ko
K’2’
K,dL_&_
and
K
Ko
0
(18) Ko
Ko
where K;’ Ko
2
and
-b-iP,(n)
(19)
From (2) and (12) the energy can be written as
1
dr’
(20)
Since, from (1 l), crack closure at Y= a implies
P(t)
Vi@-t)
dt=O
(t-u)
(21)
we have,
(r-at”/3W
,=_2(1-7%‘:
E
a-f)
[P,(t)__P,(t)Idt
(f-a)
(22)
Let IV0be the energy of a Griffith crack of length (b - a) which is opened by a constant pressurep,, then w
0
= 7r(l -?q)&(b-N)
4E
(23)
712
D. P. ROOKE
and J. TWEED
and
w -=-
1
a
L
wow,-ffwq
W”
w3+7
a*w,
I
(24
where tP, (t) dt V(b-t) (t-a) tP2 (t) dt V’(b-t) (f-a)
(25) t3Pl(t) dt V/(&t) (t-a) W A=WO
8 ?T(b-a)2z?2
t3Pz(t) dt kqb-t) (t-a)
In order to evaluate K,, Kb, and W we require S,(t) and S,(t) from (14). s,(r) which is the same function as S(t), for a constant pressure, in [l] is given by
s,(t) = Elementary
a+b-2t 2
(26)
.
integration of (14) for i = 2 gives S,(t)
=& f
(a+b-2t)
4. NUMERICAL
The integral equations formula, i.e. (13) becomes
+v
(a+b+2t)].
PROCEDURE
(13) are solved using the Gauss-Chebyshev
P,(tj) -z
i
(27)
M(tjyfk.)P<(th_)=
S,(tj)
quadrature
(28)
k=1
fori= 1,2andj,k= 1,2 . . . . n where IZis the order of the Gauss-Chebyshev mation. The arguments tj are given by tj=-
u+b+b-a -COS 2 2
Solution of the set of simultaneous From ( 13) it follows that
n(2j- 1) 2n .
linear equations
approxi-
(29)
(28) yields P&j) for i = 1 and 2.
713
Stress intensity factors of a radial crack
and (30)
Thus K,l&, and KJ& can now be calculated using (18) and {19). To the same approximation the energy terms are obtained as:
and
To calculate W8 and W, R is put equal to unity in the above equations.
Since the stress intensity factors and the crack energy are physical quantities, the way in which they vary with a and b must be independent of the mathematical technique used to find them. It follows that we may dispense with the restriction 0 < a < B -=z1 which was made at the beginning of [I] only to faciiitate the use of the Mehin transform, By using the procedure outiined above, the authors have constructed a comprehensive 3.0 I-
K,/K K,IK
n
0
----
t
2*Q
llES
Vol. 10, No. 8 - E
$-.._a=-o-9
2.5
i
714
D. P. ROOKE
I -0.8
I -b.6
I -0.4
I -0.2
and J. TWEED
I
I
DO
02
I 0.4
I 0.6
I 0.8
I IO
b
Fig. 2. The variation of WIW, with a and b for the case in which Poisson’s ratio is 0.3.
set of tables which give the values of K~‘/K,, Kf’/&, i = 1,2 and W ‘j’/W,,, j = 1,2,3,4 for values of a and b satisfying the condition - 1 < a < b < 1. These tables, which are accurate to four signi~~~t figures may be obtained from either author. Figures 1 and 2 show the variation of I&/&, KblKo and W/W, with a and b for the special case in which Poisson’s ratio is 0.3. REFERENCES [l] [2] [3] [4]
J. TWEED, S. C. DAS and D. P. ROOKE, Inf. J. Engng Sci. 10,323 (1972). S. TIMOSHENKO and J. N. GOODIER, Theory ofElasticity. McGraw-Hill (1965). M. ABRAMOWITZ and I. A. STEGUN, Handbook of Mathematical Functions. Dover (196.5). 1. S. GRADSHTEYN and I. M. RYZHIK, Tables of Integrals, Series and Products. Academic Press (1965). (Received
1 October 197 1)
Resume-Le probleme classique d’une fissure radiale dans un disque 6lastique fini en rotation est ramene a la r&olution d’une equation de Fredholm. A partir de cette solution, le facteur d’intensite de contrainte a chaque extremiti? et i’energie de formation de la fissure sont deduits. Zusa~~~ungDas elastische Problem eines Radialrisses in einer endlichen sich drehenden Scheibe wird auf die Lijsung einer Fredholm’schen Gleichung reduziert. Fur diese LSsung werden der Spannungsintensit;itsfaktor an beiden Spitzen und die Rissbildungsenergie abgeleiter. Sommario-I1 problema elastic0 di un’incrinatura radiale in un disco elastic0 finito in rotazione B ridotto atla soluzione di un’equazione di Fredholm. Da questa soluzione il fattore d’intensim deile solleci~ioni adentrambe~epuntee~ener~adel~fo~~ione del~inc~n~uraso~o~s~ttivamentec~colate. A~cT~~KTHpo6neMa 06 YIQJY~OCTE~AJISIpanHanbIioii ~peuuiIib1B KOHeSHOM, BpamaIoIIIeMca, yIIpyroM AUCKe CBOAHTCX K PeIIIeHHIO YpaBHeHHR @pW'OJlbMa. Ha OCHOBe 3TOrO pfXPi%iB5lBbIBOAXTCR KOJ@WUtIkSITbIHHTCHCHBHOCTA HElIlpSfXCeH%i5ty o6enx KO~I~OBEI3HeprHIO 06pa30BaIWI TPUUaHbI.