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Dugdale model for strain hardening materials
0013-7944/92 $5.00 + 0.00 Q 1992 Pergamon Press pk.
Engine&tg Frachue MechanicsVol. 41, No. 6, pp. 843-871, 1992 F’rintcd in Great Britain.
DUGDALE
MODEL FOR STRAIN MATERIALS
HARDENING
X. G. CHEN,? X. R. WU and M. G. YAN Institute of Aeronautical Materials, Beijing 100095, P.R.C. Abstract-A Dugdale model for hardening materials under plane stress conditions is proposed. Crack tip opening displacements (CTODs) of different geometries, compact, center crack tension, double edge crack tension and three point bend specimens, have been calculated by means of the weight function method. The results have been compared with the J-integral from EPRI’s handbook [y. Kumar, M. D. German and C. F. Shih, EPRI NP-1931, Project 1237-1, Technical Report (1981)] for strain hardening exponent n = 3,5,7,10,13,16,20 and two different RambergOsgood coefficients. The results indicate that the proposed model could well correlate CIODs from the Dugdale model and the J-integral, thus it is an alternative tool to take account of the strain hardening effect in engineering applications.
1. INTRODUCTION RECENT development of elastic-plastic fracture mechanics has revealed that the crack tip opening displacement (CTOD) method has great potential in engineering applications of fracture mechanics. Calculating CTOD by the Dugdale model is of great interest to us in many ways: the Dugdale model is simple yet very powerful. The successful application of the Dugdale model in a fatigue closure model and in an elastic-plastic analysis[l-31 indicates that the model has great potential. In the present paper, a Dugdale model which includes the strain hardening effects of the materials is presented. In contrast to previous papers[l-61, the present paper systematically examines the influences of different parameters of materials, different geometries and the different stress distributions of the plastic zone. The stress distribution given by Hoffman and Seeger in ref. [4] (in which only the case of a center crack in an infinite plate under remote tension was considered) has been adopted in the present paper, and CTODs have been calculated by means of the proposed Dugdale model for the following combination cases:
materials with strain hardening exponent: n = 3,5,7, 10, 13, 16,20; coefficient a in Ramberg-Osgood stress-strain relation: a = 3/7, 1; geometry: center crack finite plate subjected to tension (CCT) compact tension specimen C(T) three point bending specimen (TPB) double edge notch finite plate subjected to tension (DECT). Also, systematic comparison 2. DUGDALE
has been made with the J-integral
from the EPRI’s handbook[7].
MODEL OF STRAIN HARDENING FOR AN INFINITE CCT PLATE
MATERIALS
First, let us briefly reveal the concept of the Dugdale model[6] (Fig. 1). Under the plane stress condition, there is a plastic zone on the extension line of the crack in front of the crack tip. The length of the plastic zone is p and its width is denoted as t. Dugdale[6] assumed that the plastic zone could be considered as closed crack surfaces on which the tensile yield stress co is applied. On the new assumed crack tip (which is also the boundary of the elastic and plastic regions), there should be no singularity for stress, thus the stress indensity factor caused by the a, and external applied force should be zero, i.e. &(a + P) = &,(a + P), (1) tPresent address: Department of Civil Engineering, University of Maryland at College Park, MD 20742, U.S.A. 843
844
X. G. CHEN et al.
00
P
(b)
t
11-1 Y
X
ZL
1_
’
a
W
CCT
DECT
CT
(d)
P’z
b
a w
W
c
TPB Fig. 1. (a) Center cracked panel subjected to remote tension (CCT). (b) Double edge cracked panel subjected to remote tension (DECT). (c) Compact tension specimen. (d) Three point bending specimen (TPB). (e) Single edge notched specimen subjected to remote tension (SENT).
Dugdale model for strain hardening materials
845
where the stress intensity factor (SIF) induced by external force is &(a + p); the SIF induced by a, is &,,(a + p). In many cases, a,, was taken as the yield stress; however, it is often taken as the flow stress to account for the strain hardening effect. There are several papers discussing the Dugdale model for strain hardening materials. In one of them, a very interesting result was given by Hoffman and Seeger[4]. Theoretically speaking, the Dugdale model described above can only work when the materials are perfectly plastic, though by taking a, as the flow stress one could take into account some strain hardening effects. If a modified Dugdale model is used which could take n into account, a new stress distribution on the plastic zone p(x) (instead of constant stress distribution uo) should be defined to replace a,. The question is: if we use the exact stress distribution as p(x) on a plastic zone (which could be obtained by FEM to certain accuracy) to replace tensile yield stress, can we get a satisfactory CTOD from the Dugdale model? The work of Hoffman and Seeger pointed out that: even if the exact stress distribution is taken as p(x), the calculated CTOD still has large discrepancies from the expected CTOD. Thus p(x) should be well defined to obtain an accurate CTOD value. Yet, there are no clearly specified roles to guide one as to how to chose p(x). Since there exists a relation for the J-integral and CTOD J = const CTOD we could use this condition as a judgement for p(x). In ref.[4], Hoffman and Seeger took p(x) as follows: P(x)=ao
I/n+
I
( >; &
(3)
comparatively good results were given in their paper. n is strain hardening exponent and a is initial crack length; refer to Fig. 2 for the coordinate system. In the present paper, we adopted the same p(x) as eq. (3); more general results are given in this paper for the infinite plate case. For a center crack in an infinite plate subjected to a remote tension (CCT), the stress intensity factor and crack opening displacement (COD) of the plate could be obtained by means of a Green function (or weight function) method. The weight function of the center cracked plate is given by I, . :
m(x, a) = 2 J(
(a2 _ x2)-‘/2 >
and the SIF is expressed by K,(d) = whered=a+p.
‘p(x)+, sa
Letx=a+p(l-<)=d-p<;
d) dx,
(5)
thenp(x)couldbeexpressedas
p(cg = uo(l - <)-“@+I),
(6)
thus K,(d) could be written in another form: (7) Finally, we have
K, 00
KM) =-=,&)G(~) CO
and the plastic zone in front of the crack tip could be expressed as
X. G. CHEN et al.
846
where
G
0 $
+.
(10)
In the above equation, r(t) is the gamma function; F(ol, 8, y, z) is the super-geometry G(p/2d) could also be expressed as the following series:
function[8].
(11) where
b,=
1
(12) where B(a, /3) is the beta function. 2.1. Special cases of plastic zone representation In the following, we discuss the presentation of the plastic zone under several special conditions. (1) Small scale yielding (S.S.Y.). n-co Under such a condition
k@) 2
p=s (4 II
00
*
(13)
Suppose that the external load is uniform remote tension, then (14)
and
which is the S.S.Y. plastic zone equation from the Dugdale model. (2) Small scale yielding (S.S.Y.). p/2d x 0 For small scale yielding, F and G can be written in a very simple form:
F(a,B,y,O) = 1 G(0)
=,/n
(16)
Dugdale model for strain hardening materials
847
so that the plastic zone can be simplified as
1 > I
;+-$ 2 ( ). p=2( > r-n ( > r
1 KM) 2 a,
n+l
[
If eq. (14) holds, then
r
(
i+& n
1 &(a) 2
p=z(
00
)
0 n+l
[
2
2
r (r
!+A!( 2 n+l
2
rn
( n+l
,I
(17)
*
(18)
> > Ii
(3) Perfectly plastic materials. n --t cc For n = co, the closed form solution of function G can be obtained:
(19) Thus the plastic zone is given by
If eq. (14) holds, then the general crack tip plastic zone expression from the Dugdale model can be obtained from eq. (20):
=sec (2%>110
p
1.
(21)
Special cases of CTOD Next, we calculate the CTOD from the Dugdale model. In the Dugdale model, the CTOD be expressed as 6, = 6, - 8, = 2v*,
(22)
where VD is the crack tip displacement. From eq. (22) and using the weight function method (refs [9-l l]), V. can be further given by
(23) where m(a, x) is the weight function of the SIF, and
ao=
x
1 a
x2a x
(24)
X. G. CHEN et al.
848
After a complicated derivation (see Appendix l), 8, can be written in the following form: a,(x) = f
dK,(Om(t, s4
x) d<
d d =&P(r)Z(r)
+
s a
I (I
where
(25)
-$
s
PO) = NJ d(t)
If eq. (3) holds, then from eq. (26) P(t)
= p’ln+’
(!ig >
_ u)n/n+‘60*
(t
(27)
Thus P(a) = 0. Also, since Z(d) = 0, 8, can be simplified as
s d
II
P(t) -At2-x2
(28)
In the following, we will discuss several special cases of CTOD. (1) n-+c~ and x = a As the first case, we consider the CTOD of perfect plastic material. From Appendix we know that
1,
(29)
.
If the external loading is uniform remote tension, then &(a)
= %
(d2 - u~)‘/~;
(30)
from eq. (22) we can get the famous CTOD expression *,=sln(sec(;z)).
(31)
(2)n+cc andx#u For the COD being not at the crack tip (which is useful in the fatigue crack closure model), a closed form solution of the COD of perfectly plastic material can also be obtained from Appendix 1, i.e. S,(x) =
2
,/(d2 - a’) + ,/(d2 -x2)
,/(d2 - x2)
-?$bjl”
,/(d2-u2)-&d2-x2)
for x < a; and &d2 - a’) + &d2 - x2) 6i’(x)
=
!$
J(d2
-
x2){
arccos(
;)
u&d2
-
+
- x2) + x&d2
(d2-x2)-x&d’-a’)
In
J(d2
- x2)
_
a2)
_
&2
_
x2)
(33)
Dugdale model for strain hardening materials
849
for x > a. If we assume that cos(8)
;=
(34) 3 = cos(4) then eqs (29), (32), and (33) can be further simplified as d,(x) = $
[0 sin(B) - co@)ln(se@))J
(35)
for x = a;
6,(x) =
sin(d) - 4
cos
sin@)+ SW 1
(e)ln
sin(e)
_
sin(d)
+
4cos(4W
‘) +‘l} [2si~~~~~~~f~~
forx
80sin(4)-4cos(Qln I
sin@) + sin(4) / ; +4cos($)ln sin(d) - sin(4)
(37)
for x > a. (3) Small scale yielding. p/2d x 0 Under the S.S.Y. condition the following approximation
equation exists: (38)
where we introduce the following transform: x = d - ps, and r = d - pt in eq. (28). (A) At crack tip x =a, s = 1: 40, n + 1 Jp dP(a)=Zn m
,/(d2 - a’)G(O).
(39)
G(0) could be replaced by plastic zone p and K, from eqs (14) and (9). Thus we have
w
=- ;r+/(d2-a2). (B) At x #a, s # 1: 1
1 :+n
‘2’2
n+l’s
i .
(41)
3. DUGDALE MODEL FOR STRAIN HARDENING MATERIALS lN FINITE CRACKED PLATES For a finite cracked plate, we still use the stress distribution p(x) defined in eq. (3), and we will use the same method as described in Section 2 to get the CTOD. Before we go into detail about the calculation of the CTOD, let us discuss the weight function method. 3.1. Two-dimensional weight function method Weight function is a very efficient way of calculating SIFs and crack face opening displacement of an elastic cracked body. The key point in application of the weight function method is to derive the specific weight function at a specific condition. The weight functions for 2D crack problems have been given by many researchers[9,10,12]. The effective range of these weight functions is
850
X. G. CHEN et al.
usually at a/w < 0.6, which is enough for the elastic fracture problem; however, it is far from enough for the elasti*plastic fracture problem we consider here, because the initial crack length is usually taken as a/w = 0.4-0.5. Thus a rather wide range of functions is needed for this purpose. (1) Wide range weight function method for center crack[l 1, 131 In ref. [l l] the authors gave a wide range weight function which has an effective range a/w GO.85 It has the following form: (421 where
(43)
and
G(a/W)=;(@
-$,(+))
Z&,is a reference SIF corresponding to CT,while 0 is the reference stress. At almost the same time, Fett et a1.[13] gave another weight function for a center cracked plate. With reference to the method given in [13], rearranging the weight function in the form of ref. [12], the rearranged Fett’s weight function is written as (46) where
(47)
Dugdale model for strain hardening materials
and
851
3%
0
G-f+
4J2
HA 0W
(48)
9
where f0 is defined as before. It has been proved that the weight function given in ref. [l l] and ,the weight functions given in eqs (46) and (42) have the same accuracy. (2) Wide range weight function for edge cracked plate In ref. [13], Fett et al. also give the weight function for edge crack problems; it is valid for the range a/W < 0.75. After rearranging it according to the form of ref. [9], the weight function for an edge crack is exactly the same as eq. (46) except for G, H and 9.
G
0 +
0’ =35$$,_,dh d; 0
(49)
3.2. Dugdale model for strain hardening materials in a finite cracked plate As we said before, we still use the Hoffman and Seeger stress distribution assumption when we deal with the finite cracked plate problem. We use Fett’s weight function to calculate the SIF and COD. Since the weight functions for center and edge cracks have the same form we can analyze them in a general way. To calculate the CTOD, insert eq. (3) and eq. (46) into eq. (5), and integrate the integral; we could first get the SIF which is a function of the variable crack length a.
where B(a, /I) is the beta function[8]; it could be expressed in gamma functions: Jj(a
/j) 9
=
r(a)r(B)
(51)
W + 8) ’
p* is the plastic zone in front of the crack tip for crack length a. We keep it constant during the next integration for the CTOD, while p = d - a is a variable during the following integration because of the change of crack length a. To calculate the COD, the following integration must be performed: d J$(P*,
s cl0
0m@,
0
dt.
(52)
The reader will argue that the p* will also change with crack length a; this is true, however we obtain better results if we keep p* constant. There are two points to note: the tirst is the determination of the plastic zone; the second is the implementation of the singularity on the integral. Because of the complex stress distribution p(x), the plastic zone could not easily be expressed in a closed from. Yet we know that for a given stress ratio, a/u,,, the length of the plastic zone
X. G. CHEN et al.
852
could be determined, where Q is the remote reference stress. Therefore we could use a trial and error or iteration method to numerically solve for p. From eq. (1) we solve the stress ratio as a function of the plastic zone R(p)
;=w
(53)
and then solve for p from eq. (53). Equation (51) could only be solved numerically. Notice the integrand is singular at 5 = x. Although accuracy will increase somewhat by using more Gaussian integrating points, the drawback of this approach is that it is time consuming and with the increase of the Gaussian points, the accuracy will saturate. In the present paper, the linearized method is adopted toimplement the singular integral. The singular integral can be separated into two parts: integrand =
T(a) m’
where T(x) is a non-singular integral. Thus the integral can be written as (55) where 6 is a chosen small number. In the above equation, the second term is a non-singularity integral, thus it can be treated as an ordinary integral; the singularity comes from the first term. The integral range varies from (x, x + c). In such a small region, the function T(x) could be considered as a linear function, thus we can have following form: T(r) = T(x) + B(r -x),
(56)
where x + cx > 5 > x, B = T’(x). By doing so, the first term in eq. (55) can be rewritten as
s
JO+‘)T(x) + /.?(r -x)
X
J
d< = 20(x)&x)
+ ; /?@x)~‘~.
(57)
If x = 0, the only change needed is to change from x + ex to x + Ed.The c can be neither too large nor too small. If L is large, the T(x) may not be linear in the region (x,x + ox); while if L is too small, the effect of singularity will come into the second term. By trial and error, we chose c = 0.01 as a small number. 4. APPLICATION OF STRAIN HARDENING DUGDALE MODEL AND COMPARISON WITH THE J-INTEGRAL FROM EPRI[7l In this section, we apply the model we developed above to different cracked geometries; CTODs of the following geometries are calculated and compared with the J of EPRI: (a) (b) (c) (d)
center crack tension plate (CCT) compact tension specimen (CT) double edge cracked tension plate (DECT) three point bending bar (TPB).
The strain hardening exponent n is taken as n = 3,5,7,10,13,16,20 in order to compare with EPRI’s work. The Ramberg-Osgood stress-strain relation is taken for the same reason. The coefficient c1is taken as o! = 3/7 and 1. The main reason to choose a = 3/7 is based on the following consideration. From the three parameter stress-strain relation (58) or
Dugdale model for strain hardening materials
853
and in many cases, ~r,,~/4,,,x 1, we could get a = 3/7. The second case, a = 1, is basically from the MIL-HBDK-5D[14], where the stress-strain relation is expressed as n E--e=0.002 -!?- ) w ( GO.2 > or in the form of the Ramberg-Osgood stress-strain relation tY 0.002 -=--, 6.2
knowing that eo,2= 0.002, we have a = 1. According to the EPRI[‘TJ handbook, specimen is
Co.2
0
” (61)
(
60.2
>
the evaluation
equation
of the J-integral
for a
J = JeW + J,,@, n),
(62)
where
Jeb-4 =E
K:Gd
(63)
In the above equations, b = 2 for plane stress, 6 for plane strain. P is the unit thickness load carried by the specimen; POis the unit thickness limit load carried by the specimen. 4.1. Center crack panel subjected to remote tension (CCT) For a CCT specimen, the plastic J-integral is given as Jp = aaoco *($).h,(~,n)*(~-t’,
where b = W - a, and PO= 2boo for plane stress and 4ba,/J3 for plane strain. The h, (a/W, n) can be found in the table of the EPRI handbook[7]. For convenience of calculation, we fit these h, into polynomial form whenever it is possible.
for a CCT specimen, the aj coefficients for the plane stress condition are given in Table 1. Table 1.Coefficients of h, function, center crack 3
n
i
5.218172
1 2 3 4 5 6
- 11.39764 15.31509 - 12.91599 5.107191 0 0
Eft
0.8
4
QJ
4
0
7
5 6.514183 -20.71263 41.36186 - 57.27338 47.56169 - 16.54982 0
8.36303 -41.88701 114.9541 - 166.6059 93.4695 22.79248 - 30.58944
1.7
1.5
tEf: maximum error in polynomial fitting.
16
%
4
7.511297 - 30.88694 78.1464 - 126.2213 112.78 -40.7358 0
0.9
13
10
8.373961 -45.06873 128.7712 - 186.456 92.79952 45.27715 -43.35869 3
dLI 7.880014 -42.6278 121.6382 - 173.8297 76.12039 60.35604 -49.28308 4.9
20
4 7.32982 -41.92877 127.3355 - 195.9416 104.0834 48.6895 -49.51298
X. G. CHEN et al.
854
To compare the J and CTOD from our model, the SIF equation for the CCT specimen is needed
where (67) The reference SIFf,
for the weight function calculation is taken as f(a/W).
4.2. Compact tension (CT) and three point bending (TPB) specimens For CT and TPB specimens, the plastic J-integrals
have the same form, which is given as
J,=ro,r..b.h,(~,n).(~~+“.
(68)
The difference comes from the expression of limit load P,, and the h, function. (1) Compact tension specimen (CT) P,,, the unit thickness limit load, is 1.455qba, for plane strain, l.O7l~ba, for plane stress. The parameter q is expressed as 059)
? =[(;>‘+2(3+2lln_[(3+1].
To calculate the elastic J, the SIF for the CT specimen under pin-loading is needed. Such a result can be found from the recent work by Mall and Newman[lS]:
K,(a) = [+$0.886+464(;)-
l3.32(;~+
14.72($,.,(;y)&
(70)
W with accuracy 0.3%. The reference SIF f.needed for the generation of the weight function is usually taken as the SIF of the specimen under uniform loading on the crack face. Such a solution could also be found from [15]. However, the SIF expression of the CT specimen under uniform loading on the crack surface in [15] is quite complicated, and we fitted it into a polynomial as below to make it easy to use.
.
(71)
Table 2. Coefficients of h, function, compact tension n
3
5
7
10
13
i 0 1 :
4 1.001198 7.121677 -38.37178 65.97446
CLI 1A07833 2.069074 -25.01254 52.09884
% 1.266582 2.5158 -29.86555 67.40027
QJ 1A00438 -0.023668 - 65.41306 23.82534
4 1.509037 -2.682314 - 49.83837 13.58019
4 5 6
-28.73521 -26.71126 20.64867
-24.12447 -26.7087 21.27885
-41.83951 -21.24579 22.98825
-47.51505 -18.71816 24.76117
-40.70305 - 13.09316 19.85806
Eft
0.8
0.7
tEf: maximum error in polynomial fitting.
1
1.2
1.1
16
20
4 2.017384 - 7.982558 34.07126 1.918588
4 1.382444 -4.024957 -44.16989 7.979807
-41.86072 -3.472516 16.62728 0.5
-43.37432 - 7.554049 18.66797 1.3
Dugdale model for strain hardening materials
855
The fitted eq. (70) has a maximum of 0.83% error from the result of Mall and Newman[lS] in the range of 0.2
plane strain
0.5363 *a0 . $
plane stress,
PO=
(72)
I where L is the span of the specimen. For the case we are concerned with, a standard specimen, 2L/W = 4. The SIF of a standard TPB specimen for elastic J calculation can be found from the SIF handbook[l6]. From the discrete data in [16], we can have a polynomial fit: 5 * + 1169*002 (W) 5 &(a) = --jjPJW -0.04933 + 63.4269 (W> 5 - 368 *4393 (W) - 1690.447 (;>‘+
980.2072 (;y),
(73)
while the reference SIFf, could be taken as the SIF of a single edge crack panel under uniform tension, because they have the same geometry. The problem we have in this case is that we cannot find the&, for L/W = 4. As an approximation, we take thef, of an infinite long plate instead. This approximation will produce some error if L/W = 2, and little error when L/W > 3. The h, function for the TPB was not fitted due to the difficulty of polynomial fitting. The result of this is that, during the CTOD calculation, the initial crack length a~/ W can only be chosen as exactly the same as the crack length given in [7], i.e. so/W = l/8, l/4,3/8, l/2,5/8,6/8,7/8. (3) Double edge cracked panel subjected to tension (DECT) Plastic J is given in [7] as p n+l
J,=at~~~~~b -h,
*
,
F
0
0
(74)
where a0 W
plane strain
PO=
(75) plane stress.
Is 3 The SIF K,(a) for elastic J calculation is expressed as K,(a) =foJ(na),
(76)
where 1.122-0.561
(77)
The above equation has the accuracy of 0.5% for any a/ W[16). The reference SIF fo for weight function generation is taken as f(a/W). The h, (u/W, n) functions are fitted in the same way as for the CCT specimen. Table 3 shows the a coefficients for DECT. (4) Single edge cracked panel subjected to tension (SECT) Plastic J is given as (78)
856
X. G. CWEN et al. Table 3. Coefficients n
3
j 0
t(, 0.2689648
: 3 4 5 6
- 19.2348 8.158046 17.34996 -5.261816 0 0
Eft
0.3
ofh, function,
5
double edge cracked sp&mens 10
7
4 0.797909
crt 1.321107
-24.88236 7.258186 29.64258 - 11.90544 0 0
-40.91061 7.109418 79.71513 - 69.08565 22.84852 0
0.6
13
GLI 2.766505 -25.14265 -2.124447 77.85988 -85.68216 33.4283 0
0.1
4 5.021851 -21.97722 39.35135 -24.82852 -4.59675 7.938473 0 0.5
0.5
16
20
QI 7.612527 -43.70007 99.53776 -88.67284 -5.413372 29.10341 0 1.4
TEE maximum error in polynomial fitting.
where P&J=
1.455$~_~, plane strain I 1.072~~~* plane stress.
(79)
01
(80)
tf is expressed as
2
‘1 =
1+
[
;
I/Z
-;.
The SIF for elastic J can be found in [16]: &@I =“&/W, where f(a/FV) is
The reference SIF, fO, can be chosen as f(u/ W). There are problems with the hi function; as a/W approaches 0.25, the value of h, increases dramatically, which leads to the difficulty of polynomial fitting. After careful examination of the origin of hl values in refs [7, 17-191, the authors find that there are big differences in the values of h, of SECT in different papers, especially for the plane strain problem. After noticing the nonsmooth (o~illation) problem of h, and h3, Parks et af. proposed a consistency check method for the hi function[l7]. Another method, penalty method, was also proposed by Shih and Needleman in the finite element calculation in order to deal with the full plastic problem of the cracked body[l8, 191. The results in [18, 191 show that, for 1
where d,, is a function of tq/E and hardening exponent n. d,, weakly depends on Q/E.
Dugdale model for strain hardening materials
857
Here, we use 2V, to stand for the CTOD from the Dugdale model which distinguishes the true CTOD 6,. Along this line, first, we examine the ratio of .I/( V,a,) given the initial crack length a/W; then we discuss the crack size effects on this ratio; and last we will show the results of shape effects on the ratio. 5.1. Stress ratio, crack size and geometry efects Given the crack length a/W, and changing the stress ratio a/a,, , we can see that the ratios of J and (J/ VDao)are relatively constant for n > 3 when a = 3/7. For highly strain hardening materials, like n = 3, the ratio of J and V. is highly stress state dependent, for a = 3/7. Figures 2 through 9 show the crack size and stress ratio effects on J/(oo V,) for n = 5 and 16. There are further similar figures of different geometries for n = 3, 7, 10, 13 and 20 which are not shown here.
Fig. 2. Comparison of J and CTOD calculated from the Dugdale model for strain hardening materials, stress ratio and crack size effects on J/(u~ V,), CCT, o( = 317, n = 5.
“I
I
I
Fig. 3. Comparison of J and CTOD calculated from the Dugdale model for strain hardening materials, stress ratio and crack size effects on J/(0, VD), DECT, a = 3/7, n = 5.
858
X. G. CHEN ef al.
Fig. 4. Comparison of J and CTOD calculated from the Dugdale model for strain hardening materials, stress ratio and crack size effects on J/(uO V,), CT, a = 3/7, n = 5.
6
6
4
2
u
Q u
16
47
00
Fig. 5. Comparison of J and CTOD calculated from the Dugdale model for strain hardening materials, stress ratio and crack size effects on J/(u, V,),TPB, a = 3/7, n = 5.
Dugdale model for strain hardening materials
1
VD00 IU
0
6
*
P
u
0
P
7
6
0
10
x70”
0
a0
Fig. 6. Comparison of J and CTOD calculated from the Dugdale model for strain hardening materials, stress ratio and crack size effects on J/(a, V,), CCT, u = 3/7, n = 16.
J ‘D
‘0
-a
Fig. 7. Comparison of J and CLOD calculated from the Dugdale model for strain hardening materials, stress ratio and crack size effects on J/(u, V,), DECT, a = 3/7, n = 16.
859
860
X. G. CHEN ef al.
Fig, 8. Comparison of J and CTOD calcuIated from the Dugdale model for strain hardening materials, stress ratio and crack size effects on J/(q, V,), CT, a = 3/7, n = 16.
Fig. 9. Comparison of J and CTOD caiculated from the Dugdale model for strain hardening materials, stress ratio and crack size effects on J&T,, V,,), TPB, a = 3/7, n = 16.
Dugdale model for strain hardening materials
861
For a CCT specimen, ct = 3/7 and n = 5, the value of J/(a,, V,) is slightly stress ratio and crack size dependent. The increasing tendency of J/(crOVD) with a/a0 is observed. For a less strain hardening material, n = 16, we find much less stress ratio dependency of J/(a,, V,), though there is still some crack size dependence. There is almost no crack size effect and very slight stress state dependence in the DECT of n = 5, u = 317. The same is found for n = 16, a = 3/7. Although the DECT and CCT are both tension type specimens, the behaviors of them are not expected to be the same. For a CT specimen, the curves of J/(0,, V,) for a = 3/7, n = 5 and n = 16 are quite similar, while the average value of J/(oO V,) for n = 16, a = 3/7 is higher than that of n = 5. The TPB specimens behave differently from the CT specimens. With a small stress ratio, the J/(0,, VD) is
-
0 OO
Fig. 10. Comparison of J and CTOD calculated from the Dugdale model for strain hardening materials, geometry effect on J/(u, V,), plotted against stress ratio, a = 3/7, n = 5.
,!!!L RRlF-
DmYvTAw
n7aFvLTm
a--
3 7
Fig. 11. Comparison of J and CTOD calculated from the Dugdale model for strain hardening materials, geometry effect on J/(u,,V&, plotted against 6, a = 3/7, n = 5.
862
X. G, CHIP? et al.
almost constant for different crack sizes, while for large stress ratio, it increases dramatically due to the large plastic deformation. Now, if we consider the differences due to the crack length and types of geometries, we will find that, given the same stress ratio, the plastic zone in front of the crack tip will be diEerent for diRerent crack lengths and different specimens. Thus, for the same-stress status, some specimens have only a very small plastic zone, while the others may have a very big one. It is reasonable to compare the J and CTOD on the same plastic deformation bases by defining a plastic deformation measure. Let us define a new parameter c3
Fig. 12. Comparison of J and CTOD calculated from the Dugdale model for strain hardening materials, geometry effect on .!/(a0 V,), plotted against stress ratio, a = 3/i’, n = 16.
Fig. 13. Comparison of J and CTOD calculated from the Dugdale model for strah hrdening geometry effect on J/(u*V,), plotted against cS, u = 3/7, n = 16.
materials,
Dugdale model for strain hardening materials
863
where b = (W - a) is the ligament of the specimen. Given a certain plastic zone size, 8 is the measure of the plastic zone size with respect to the ligament. p/d = cJb/ W. Plotting the .I/( V,q,) against 6, the results for four types of geometries are shown in Figs 11 and 13, and the same comparisons in terms of stress ratio are given in Figs 10 and 12. Little geometry dependence is found in Pigs 10 through 13. Similar results are shown for a = 1 in Figs 14 and 15. It is found that when z = 1, great size effects will affect the calculated CTOD when a material is highly strain hardening. This is due to the fact the model does not take account of the parameter cc. The summary of these computations is given in Tables 4 through IO.
db W Fig. 14. Comparison of J and CTOD calculated from the Dugdale model for strain hardening materials, geometry effect on J/(u,,Vo), plotted against stress ratio, a = 1, n = 5.
I
a=1
Fig. 15. Comparison of J and CTOD calculated from the Dugdale model for strain hardening materials, geometry effect on $/(a, v,), plotted agRinSt 6, K = 1, n = 16.
X. G. CHEN et al.
Dugdale model for strain hardening matehls
865
w
0.3 < w < 0.80
3
(%)
0.3 < cl < 0.55
2.9338
11.6 8.6 7.4 7.6 8.4
2.8729 2.8351 2.8514 2.%11 3.1484 11.1 19.1
Es (%)
JI(~,cy)
CCT 5
0.3 < w < 0.55
2.6211
6.5 5.1 4.7 2.5 3
2.5243 2.5147 2.5486 2.6659 2.8519
5
5.5 17.8
9.8 12.5 13.6 16.5
Es (%)
0.3 < o < 0.65
2.%98
2.884 2.9252 3.042 3.0278
J/w/,)
0.2
2.4883
2.4718 2.4685 2.4952 2.5176
JI(V,oJ
13.4 17.2
Es (%)
JI(~,ey)
7
2 15.3
1.4 4.6 11.95 13.9
Es (%)
< 0.65
7
0.1
2.2122
2.2521 2.21 2.1825 2.2tMl
J/(V,u,)
3.2 9.7
8.6 7.8 6.6 4.9
Es (%)
<0.60
10
Table 9. Compact tension, CT (a = 1)
due to the different crack length a/W. due to stress ratio. due to both crack length a/W and stress ratio.
4.6 18.1
16.9 13.7 10.4 9.8
tech maximum error of I/( V&) $Es maximum error ofJ/(V&J #Et: maximum error of J/( V&J
range
EfTective
Average J/(~,e,)
4.2953
4.2834 4.1985 4.3929 4.3063
0.1125 0.250 0.375 ::ZZ
w (%) EtQ (%)
J/(~fPy)
n
a/W
(“Q
17 13.4 11.2 10.1 9.7
6.3 17
lx
0.3 < 0 < 0.55
3.9815
3.9565 3.8352 3.7869 3.854 4.025
J/woe,)
3
tEc: maximum error of J/( V&J due to the different crack length a/W. #ES maximum error of J/( V,F,) due to stress ratio. §Et: maximum error of J/( V&.) due to both crack length a/W and stress ratio.
-ge
EKective
I
J/(~LP”)
Average
Ect (%) Et0 (%)
0.125 0.250 0.375 O.u)o 0.600
o/W
n
13
5.8 19.6
11.6 14.5 12.4 14.2
Es (%)
0.05 < w < 0.65
2.0832
2.1502 2.094 2.0558 2.0328
Jl(V,cy)
0.3 < UJ< 0.55
3.6118
12.7 10.5 9.3 8.6 7.1
3.8193 3.7077 3.6136 3.517 3.4016 12.3 23.8
Es (%)
J/(v,ey)
3
Table 8. Center crack, CCT, and double edge crack, DECT (a = 1) 5
7.7 19
6.5 19.9
12.6 18.5 15.3 13.5
Es (%)
0.05 < w < 0.65
2.0243
2.1015 2.0391 1.9836 1.9728
16
12 7.9 6.4 6.4 6.6
Es (%)
0.3 < o < 0.80
2.7646
2.8626 2.8126 2.7697 2.719 2.692
JI(~,ey)
Jwae,)
DECT
20
10.3 19.8
5.6 10.2 15.3 12.6
Es (%)
0.05 < 0 < 0.55
2.0019
2.1049 2.0433 1.9505 1.9017
JMvLloy)
g
2
9
x
(%I
0.3 < o < 0.80
Bffective range
5
46.3
26.7
27.7 20.5
9.8 18.3 29.9
Es (%)
0.3 < o < 0.65
3.0013
3.2776 2.9837
2.8949 2.5872 3.263
JlWD~y)
I
26.2
6.8
24.5 6.7
2.4 8 24.5
Es (%I
0.2 < w < 0.65
2.5868
2.6827 2.5115
2.5157 2.5429 2.6811
JWDfJJ
tEc: maximum error of J/( V,F,) due to the different crack length a/W. $Ek maximum error of J/( V&J due to stress ratio. #Et: maximum error of J/( V&J due to both crack length a/W and stress ratio.
4.5304
JWney)
55.7
Et5 (%) Average
14.8 7
18.3 20.7 19.4
w
12.3
4.6289 4.4207
0.500 0.600
3
(%)
4.228 4.6249 4.4795
6.125 0.250 0.375
w
JWD~,)
n
a/W
10
16.4
6.5
13.4 1
1.6 2.6 12.8
Es (%)
0.1 < w < 0.60
2.241
2.2811 2.1699
2.311 2.2544 2.3069
JWdJJ
13
10.7
9.2
6.9 2.7
2.5 6.8 6.4
Es (%)
0.05 < o < 0.65
2.1322
2.105 2.0354
2.2221 2.1513 2.1467
Jl(V,ey)
Table 10. Three. ooint bendina_ snecimen. TBB (a . = 1), _ 16
14.4
9.1
9.5 8
2.1 8.8 7.4
Es (%)
0.05 < o < 0.65
2.1262
2.0137 1.9903
2.1718 2.0946 2.0638
JWney)
20
20.1
9.9
16 9.6
2.8 9.8 10.7
EL?(%)
0.3 < 0 < 0.55
2.1024
I .9367 1.9453
2.1293 2.0508 2.OtMl
J/(V,cy)
9 w
p
k
F
3.
B
;
y$
X. G. CHEN et al.
868
5.2. The limitations of the present model
There are two limitations on the Dugdale model presented here. (1) The accuracy evaluation method of the model. Since the Dugdale model is a simplified model, there are no ways to judge its accuracy by itself. Other methods, such as experimental method or finite element method, should be used to evaluate the model. (2) The present Dugdale model does not take the Ramberg-Osgood coefficient c1into account; choosing an a different from 3/7 or 1 may cause different results. Further study is needed to improve this model. 6. CONCLUSION A Dugdale model for strain hardening materials has been presented. The weight function method has been used to calculate the CTOD by this model. It is shown that the present model gives quite promising results after comparison with the J-integral solution from EPRI’s work. Since this model simplifies the elastic-plastic fracture problem into an elastic fracture problem, it is very easy to calculate the CTOD with this model, which makes the engineering application easier than before. The results are only valid under the plane stress condition, and more work should be done for the plane strain problems. Acknowledgement-The authors are grateful for the support from the Institute of Aeronautical Materials, Beijing, on this project when the first author worked in the Institute during 1985 to 1989.
REFERENCES G. S. Wang, B. Palmberg and A. F. Blom, The crack propagation in the plate subjected to spectrum loads. Technical Note, The Aeronautical Research Institute of Sweden (1988). J. C..Newman, Jr., ASTM STP 748, 53-84 (1981). J. C. Newman. Jr.. ASTM STP 8%. 139-166 (1985). M. Hoffman and T. Seeger, Dugdale solutions for strain hardening materials, in The Crack Tip Opening Displacemenf in Elastic-Plastic Fracture Mechanics, pp. 57-77, Proceedings of the Workshop on the CTOD Methodology, Geesthacht, Germany (1985). J. C. Newman, Jr., Engng Fracture Mech. 1, 137-154 (1968). D. S. Dugdale, J. Mech. Pkys. Soli& 8, 100-104 (1960). V. Kumar, M. D. German and C. F. Shih, An engineering approach for elastic-plastic fracture analysis. EPRI NP-1931, Project 1237-1, Technical Report (1981). Z. X. Wang and D. R. Gu, General Theory of Special Functions. Science Press, Beijing (1979) (in Chinese). X. R. Wu, Engng Fracture Meek. 20, 35-49 (1984). X. R. Wu and J. Carlsson, J. Mech. Phys. Sohds 31, 485-497 (1983). X. R. Wu and X. G. Chen, Engng Fracture Mech. 33, 877-886 (1989). H. J. Petroski and J. D. Achenback, Engng Fracture Mech. 10;257-266 (1978). T. Fett, C. Matteck and D. Munx, Engng Fracture Mech. 27, 697-715 (1987). MIL-HDBK-SD (1986). S. Mall and J. C. Newman, Jr., ASTM STP 868, 113-128 (1985). Chinese Aerospace Establishment (CAE), Stress Inrensity Factor Handbook. Science Press, Beijing (1981) (in Chinese). D. M. Parks, V. Kumar and C. F. Shih, ASTM STP 803, 1-307-I-383 (1983). C. F. Shih and A. Needleman, J. appl. Mech. 51, 48-56 (1984). C. F. Shih and A. Needleman, J. appl. Mech. 51, 56-64 (1984). J. W. Hutchinson, A. Needleman and C. F. Shih, in Fracture Meckanics (Edited by N. Perrone et al.), pp. 515-528. University Press of Virginia, Charlottesville, VA (cited from[ll]). W. Sauter and V. Kumar, Section 3.2 of Combined Fifth and Sixth Semiannual Report to EPRI, General Electric Company Schenectady, NY, General Electric Report, SRD-82-048 (1982) (cited from[l7]). C. F. Shih, J. Mech. Phys. Solids 29, 305-326 (1981).
APPENDIX 1. CRACK FACE DISPLACEMENT CALCULATION OF THE CENTER CRACK IN AN INFINITE PLATE BY DUGDALE MODEL FOR STRAIN HARDENING MATERIALS CTOD could be written as 6,=;
:K,(S)m(S.x)dC.
(AlI
s where
ho=
1 x
a
x>a x
(A2)
Dugdale model for strain hardening materials
869
The weight function m(a, x) is 8iven as i (cZ _ x2)-‘/2, J( > so that the SIF is expressed in terms of weight function and crack face stress m(a, x) 5 2
(A3)
(A4) where p(t) is the stress distribution on the crack surface. Substituting eqs (A4) and (A3) into (Al), we have p(r)e(r2
- x~)-“~({~ - r2)“2 dr d<.
(AS)
Referring to Fig. Al, we change the sequence of the integration, and we can obtain p(r)t(t2
- x~)-“~(<~ - r2)-“2dC
646)
Notice that I({, t) =
{(r2 - x~)-I’~(~~- ,2)-“2 d{ (A7)
Thus we can simplify the integral into the following form: (A8) where
Z,(r) =
C-d
or,t)
t-t
.
(A9)
(1) If x < (I, then d, = II, and the first term in eq. (A8) equals zero. Equation (A8) becomes Sp=;
dp(r)Z,(r) dr.
s (I
(Al’3
Sometimes it is helpful to change eq. (AlO) into the following form, integrating it by parts:
(All)
d
________-
___...--._____ -
4
-I----
---_-es
d
Fig. Al. Integration area.
X. G. CHEN et al.
870 where z,(f)
=
l”
Jv* +x3 f (d2- r3 (d’+xq+(d2-13
P(t) = If
J
p(t) dt.
zJto=
t/il+1
( > A
009
(I -
and the crack face ~spla~ment
ar”+‘u,
W3)
W4)
becomes a,=&~Q’ZW$+/($$)dr.
tAl8)
In this particular case, with load p(t) expressed as eq. (A13), the transform from eq. (AIO) to eq. (A15) changes the l/(n + 1) order of singularity in the inte8ral to the -l/2 order of singularity in the integral, which is much easier to in@rate. (2) If x > u, then 4 = x, and it follows that Z,(f) = Z,(r). Equation (A8) has the form
Equation (Al6) has the same expression as eq. (AIO), so that eq. (AIO) is in fact valid for both x
(A18)
where S, is the crack face displacement produced by the remote tensile stress, which is given as G,+/(d2-u’)
IA19
from eq. (1). and knowing that the stress intensity factor due to crack face loading and the remote tension in the plate with a center crack is KJd, a) = 2eo/(~)arccos(~)
K&4 = a,/Wh
WW
we can express the CTOD in the following way by substituting the eqs (A20), (1) and (A19), (A17) into (A18):
WU In this case, the material is still perfectly plastic; we calculate the crack face displacement which is not at the crack tip. From (AU), the integration could be implemented and we obtain
6422) where Z2 is
z-
J
d
’-
2t dt
II (r2 - x2),/(d2 - t2)
,/(d* + a2) + (d2 -x2) =~lnc+/(dz+.2)-(d2-x2)
(~23)
Dugdale model for strain hardening materials
871
and Z, is d
dr
I
W4)
z1 = II (r* - XqJ(d2 - rq. This integration has two solutions according to two different cases. (A) x
6425) (B) x > a dr (12- X2)&2 - x3
‘G&q”
aJ(&+x2)+xJ(d2-u2) a&P-x2)-x (CP--2)
(Received 7 March 1991)
.
(A261
Engine&tg Frachue MechanicsVol. 41, No. 6, pp. 843-871, 1992 F’rintcd in Great Britain.
DUGDALE
MODEL FOR STRAIN MATERIALS
HARDENING
X. G. CHEN,? X. R. WU and M. G. YAN Institute of Aeronautical Materials, Beijing 100095, P.R.C. Abstract-A Dugdale model for hardening materials under plane stress conditions is proposed. Crack tip opening displacements (CTODs) of different geometries, compact, center crack tension, double edge crack tension and three point bend specimens, have been calculated by means of the weight function method. The results have been compared with the J-integral from EPRI’s handbook [y. Kumar, M. D. German and C. F. Shih, EPRI NP-1931, Project 1237-1, Technical Report (1981)] for strain hardening exponent n = 3,5,7,10,13,16,20 and two different RambergOsgood coefficients. The results indicate that the proposed model could well correlate CIODs from the Dugdale model and the J-integral, thus it is an alternative tool to take account of the strain hardening effect in engineering applications.
1. INTRODUCTION RECENT development of elastic-plastic fracture mechanics has revealed that the crack tip opening displacement (CTOD) method has great potential in engineering applications of fracture mechanics. Calculating CTOD by the Dugdale model is of great interest to us in many ways: the Dugdale model is simple yet very powerful. The successful application of the Dugdale model in a fatigue closure model and in an elastic-plastic analysis[l-31 indicates that the model has great potential. In the present paper, a Dugdale model which includes the strain hardening effects of the materials is presented. In contrast to previous papers[l-61, the present paper systematically examines the influences of different parameters of materials, different geometries and the different stress distributions of the plastic zone. The stress distribution given by Hoffman and Seeger in ref. [4] (in which only the case of a center crack in an infinite plate under remote tension was considered) has been adopted in the present paper, and CTODs have been calculated by means of the proposed Dugdale model for the following combination cases:
materials with strain hardening exponent: n = 3,5,7, 10, 13, 16,20; coefficient a in Ramberg-Osgood stress-strain relation: a = 3/7, 1; geometry: center crack finite plate subjected to tension (CCT) compact tension specimen C(T) three point bending specimen (TPB) double edge notch finite plate subjected to tension (DECT). Also, systematic comparison 2. DUGDALE
has been made with the J-integral
from the EPRI’s handbook[7].
MODEL OF STRAIN HARDENING FOR AN INFINITE CCT PLATE
MATERIALS
First, let us briefly reveal the concept of the Dugdale model[6] (Fig. 1). Under the plane stress condition, there is a plastic zone on the extension line of the crack in front of the crack tip. The length of the plastic zone is p and its width is denoted as t. Dugdale[6] assumed that the plastic zone could be considered as closed crack surfaces on which the tensile yield stress co is applied. On the new assumed crack tip (which is also the boundary of the elastic and plastic regions), there should be no singularity for stress, thus the stress indensity factor caused by the a, and external applied force should be zero, i.e. &(a + P) = &,(a + P), (1) tPresent address: Department of Civil Engineering, University of Maryland at College Park, MD 20742, U.S.A. 843
844
X. G. CHEN et al.
00
P
(b)
t
11-1 Y
X
ZL
1_
’
a
W
CCT
DECT
CT
(d)
P’z
b
a w
W
c
TPB Fig. 1. (a) Center cracked panel subjected to remote tension (CCT). (b) Double edge cracked panel subjected to remote tension (DECT). (c) Compact tension specimen. (d) Three point bending specimen (TPB). (e) Single edge notched specimen subjected to remote tension (SENT).
Dugdale model for strain hardening materials
845
where the stress intensity factor (SIF) induced by external force is &(a + p); the SIF induced by a, is &,,(a + p). In many cases, a,, was taken as the yield stress; however, it is often taken as the flow stress to account for the strain hardening effect. There are several papers discussing the Dugdale model for strain hardening materials. In one of them, a very interesting result was given by Hoffman and Seeger[4]. Theoretically speaking, the Dugdale model described above can only work when the materials are perfectly plastic, though by taking a, as the flow stress one could take into account some strain hardening effects. If a modified Dugdale model is used which could take n into account, a new stress distribution on the plastic zone p(x) (instead of constant stress distribution uo) should be defined to replace a,. The question is: if we use the exact stress distribution as p(x) on a plastic zone (which could be obtained by FEM to certain accuracy) to replace tensile yield stress, can we get a satisfactory CTOD from the Dugdale model? The work of Hoffman and Seeger pointed out that: even if the exact stress distribution is taken as p(x), the calculated CTOD still has large discrepancies from the expected CTOD. Thus p(x) should be well defined to obtain an accurate CTOD value. Yet, there are no clearly specified roles to guide one as to how to chose p(x). Since there exists a relation for the J-integral and CTOD J = const CTOD we could use this condition as a judgement for p(x). In ref.[4], Hoffman and Seeger took p(x) as follows: P(x)=ao
I/n+
I
( >; &
(3)
comparatively good results were given in their paper. n is strain hardening exponent and a is initial crack length; refer to Fig. 2 for the coordinate system. In the present paper, we adopted the same p(x) as eq. (3); more general results are given in this paper for the infinite plate case. For a center crack in an infinite plate subjected to a remote tension (CCT), the stress intensity factor and crack opening displacement (COD) of the plate could be obtained by means of a Green function (or weight function) method. The weight function of the center cracked plate is given by I, . :
m(x, a) = 2 J(
(a2 _ x2)-‘/2 >
and the SIF is expressed by K,(d) = whered=a+p.
‘p(x)+, sa
Letx=a+p(l-<)=d-p<;
d) dx,
(5)
thenp(x)couldbeexpressedas
p(cg = uo(l - <)-“@+I),
(6)
thus K,(d) could be written in another form: (7) Finally, we have
K, 00
KM) =-=,&)G(~) CO
and the plastic zone in front of the crack tip could be expressed as
X. G. CHEN et al.
846
where
G
0 $
+.
(10)
In the above equation, r(t) is the gamma function; F(ol, 8, y, z) is the super-geometry G(p/2d) could also be expressed as the following series:
function[8].
(11) where
b,=
1
(12) where B(a, /3) is the beta function. 2.1. Special cases of plastic zone representation In the following, we discuss the presentation of the plastic zone under several special conditions. (1) Small scale yielding (S.S.Y.). n-co Under such a condition
k@) 2
p=s (4 II
00
*
(13)
Suppose that the external load is uniform remote tension, then (14)
and
which is the S.S.Y. plastic zone equation from the Dugdale model. (2) Small scale yielding (S.S.Y.). p/2d x 0 For small scale yielding, F and G can be written in a very simple form:
F(a,B,y,O) = 1 G(0)
=,/n
(16)
Dugdale model for strain hardening materials
847
so that the plastic zone can be simplified as
1 > I
;+-$ 2 ( ). p=2( > r-n ( > r
1 KM) 2 a,
n+l
[
If eq. (14) holds, then
r
(
i+& n
1 &(a) 2
p=z(
00
)
0 n+l
[
2
2
r (r
!+A!( 2 n+l
2
rn
( n+l
,I
(17)
*
(18)
> > Ii
(3) Perfectly plastic materials. n --t cc For n = co, the closed form solution of function G can be obtained:
(19) Thus the plastic zone is given by
If eq. (14) holds, then the general crack tip plastic zone expression from the Dugdale model can be obtained from eq. (20):
=sec (2%>110
p
1.
(21)
Special cases of CTOD Next, we calculate the CTOD from the Dugdale model. In the Dugdale model, the CTOD be expressed as 6, = 6, - 8, = 2v*,
(22)
where VD is the crack tip displacement. From eq. (22) and using the weight function method (refs [9-l l]), V. can be further given by
(23) where m(a, x) is the weight function of the SIF, and
ao=
x
1 a
x2a x
(24)
X. G. CHEN et al.
848
After a complicated derivation (see Appendix l), 8, can be written in the following form: a,(x) = f
dK,(Om(t, s4
x) d<
d d =&P(r)Z(r)
+
s a
I (I
where
(25)
-$
s
PO) = NJ d(t)
If eq. (3) holds, then from eq. (26) P(t)
= p’ln+’
(!ig >
_ u)n/n+‘60*
(t
(27)
Thus P(a) = 0. Also, since Z(d) = 0, 8, can be simplified as
s d
II
P(t) -At2-x2
(28)
In the following, we will discuss several special cases of CTOD. (1) n-+c~ and x = a As the first case, we consider the CTOD of perfect plastic material. From Appendix we know that
1,
(29)
.
If the external loading is uniform remote tension, then &(a)
= %
(d2 - u~)‘/~;
(30)
from eq. (22) we can get the famous CTOD expression *,=sln(sec(;z)).
(31)
(2)n+cc andx#u For the COD being not at the crack tip (which is useful in the fatigue crack closure model), a closed form solution of the COD of perfectly plastic material can also be obtained from Appendix 1, i.e. S,(x) =
2
,/(d2 - a’) + ,/(d2 -x2)
,/(d2 - x2)
-?$bjl”
,/(d2-u2)-&d2-x2)
for x < a; and &d2 - a’) + &d2 - x2) 6i’(x)
=
!$
J(d2
-
x2){
arccos(
;)
u&d2
-
+
- x2) + x&d2
(d2-x2)-x&d’-a’)
In
J(d2
- x2)
_
a2)
_
&2
_
x2)
(33)
Dugdale model for strain hardening materials
849
for x > a. If we assume that cos(8)
;=
(34) 3 = cos(4) then eqs (29), (32), and (33) can be further simplified as d,(x) = $
[0 sin(B) - co@)ln(se@))J
(35)
for x = a;
6,(x) =
sin(d) - 4
cos
sin@)+ SW 1
(e)ln
sin(e)
_
sin(d)
+
4cos(4W
‘) +‘l} [2si~~~~~~~f~~
forx
80sin(4)-4cos(Qln I
sin@) + sin(4) / ; +4cos($)ln sin(d) - sin(4)
(37)
for x > a. (3) Small scale yielding. p/2d x 0 Under the S.S.Y. condition the following approximation
equation exists: (38)
where we introduce the following transform: x = d - ps, and r = d - pt in eq. (28). (A) At crack tip x =a, s = 1: 40, n + 1 Jp dP(a)=Zn m
,/(d2 - a’)G(O).
(39)
G(0) could be replaced by plastic zone p and K, from eqs (14) and (9). Thus we have
w
=- ;r+/(d2-a2). (B) At x #a, s # 1: 1
1 :+n
‘2’2
n+l’s
i .
(41)
3. DUGDALE MODEL FOR STRAIN HARDENING MATERIALS lN FINITE CRACKED PLATES For a finite cracked plate, we still use the stress distribution p(x) defined in eq. (3), and we will use the same method as described in Section 2 to get the CTOD. Before we go into detail about the calculation of the CTOD, let us discuss the weight function method. 3.1. Two-dimensional weight function method Weight function is a very efficient way of calculating SIFs and crack face opening displacement of an elastic cracked body. The key point in application of the weight function method is to derive the specific weight function at a specific condition. The weight functions for 2D crack problems have been given by many researchers[9,10,12]. The effective range of these weight functions is
850
X. G. CHEN et al.
usually at a/w < 0.6, which is enough for the elastic fracture problem; however, it is far from enough for the elasti*plastic fracture problem we consider here, because the initial crack length is usually taken as a/w = 0.4-0.5. Thus a rather wide range of functions is needed for this purpose. (1) Wide range weight function method for center crack[l 1, 131 In ref. [l l] the authors gave a wide range weight function which has an effective range a/w GO.85 It has the following form: (421 where
(43)
and
G(a/W)=;(@
-$,(+))
Z&,is a reference SIF corresponding to CT,while 0 is the reference stress. At almost the same time, Fett et a1.[13] gave another weight function for a center cracked plate. With reference to the method given in [13], rearranging the weight function in the form of ref. [12], the rearranged Fett’s weight function is written as (46) where
(47)
Dugdale model for strain hardening materials
and
851
3%
0
G-f+
4J2
HA 0W
(48)
9
where f0 is defined as before. It has been proved that the weight function given in ref. [l l] and ,the weight functions given in eqs (46) and (42) have the same accuracy. (2) Wide range weight function for edge cracked plate In ref. [13], Fett et al. also give the weight function for edge crack problems; it is valid for the range a/W < 0.75. After rearranging it according to the form of ref. [9], the weight function for an edge crack is exactly the same as eq. (46) except for G, H and 9.
G
0 +
0’ =35$$,_,dh d; 0
(49)
3.2. Dugdale model for strain hardening materials in a finite cracked plate As we said before, we still use the Hoffman and Seeger stress distribution assumption when we deal with the finite cracked plate problem. We use Fett’s weight function to calculate the SIF and COD. Since the weight functions for center and edge cracks have the same form we can analyze them in a general way. To calculate the CTOD, insert eq. (3) and eq. (46) into eq. (5), and integrate the integral; we could first get the SIF which is a function of the variable crack length a.
where B(a, /I) is the beta function[8]; it could be expressed in gamma functions: Jj(a
/j) 9
=
r(a)r(B)
(51)
W + 8) ’
p* is the plastic zone in front of the crack tip for crack length a. We keep it constant during the next integration for the CTOD, while p = d - a is a variable during the following integration because of the change of crack length a. To calculate the COD, the following integration must be performed: d J$(P*,
s cl0
0m@,
0
dt.
(52)
The reader will argue that the p* will also change with crack length a; this is true, however we obtain better results if we keep p* constant. There are two points to note: the tirst is the determination of the plastic zone; the second is the implementation of the singularity on the integral. Because of the complex stress distribution p(x), the plastic zone could not easily be expressed in a closed from. Yet we know that for a given stress ratio, a/u,,, the length of the plastic zone
X. G. CHEN et al.
852
could be determined, where Q is the remote reference stress. Therefore we could use a trial and error or iteration method to numerically solve for p. From eq. (1) we solve the stress ratio as a function of the plastic zone R(p)
;=w
(53)
and then solve for p from eq. (53). Equation (51) could only be solved numerically. Notice the integrand is singular at 5 = x. Although accuracy will increase somewhat by using more Gaussian integrating points, the drawback of this approach is that it is time consuming and with the increase of the Gaussian points, the accuracy will saturate. In the present paper, the linearized method is adopted toimplement the singular integral. The singular integral can be separated into two parts: integrand =
T(a) m’
where T(x) is a non-singular integral. Thus the integral can be written as (55) where 6 is a chosen small number. In the above equation, the second term is a non-singularity integral, thus it can be treated as an ordinary integral; the singularity comes from the first term. The integral range varies from (x, x + c). In such a small region, the function T(x) could be considered as a linear function, thus we can have following form: T(r) = T(x) + B(r -x),
(56)
where x + cx > 5 > x, B = T’(x). By doing so, the first term in eq. (55) can be rewritten as
s
JO+‘)T(x) + /.?(r -x)
X
J
d< = 20(x)&x)
+ ; /?@x)~‘~.
(57)
If x = 0, the only change needed is to change from x + ex to x + Ed.The c can be neither too large nor too small. If L is large, the T(x) may not be linear in the region (x,x + ox); while if L is too small, the effect of singularity will come into the second term. By trial and error, we chose c = 0.01 as a small number. 4. APPLICATION OF STRAIN HARDENING DUGDALE MODEL AND COMPARISON WITH THE J-INTEGRAL FROM EPRI[7l In this section, we apply the model we developed above to different cracked geometries; CTODs of the following geometries are calculated and compared with the J of EPRI: (a) (b) (c) (d)
center crack tension plate (CCT) compact tension specimen (CT) double edge cracked tension plate (DECT) three point bending bar (TPB).
The strain hardening exponent n is taken as n = 3,5,7,10,13,16,20 in order to compare with EPRI’s work. The Ramberg-Osgood stress-strain relation is taken for the same reason. The coefficient c1is taken as o! = 3/7 and 1. The main reason to choose a = 3/7 is based on the following consideration. From the three parameter stress-strain relation (58) or
Dugdale model for strain hardening materials
853
and in many cases, ~r,,~/4,,,x 1, we could get a = 3/7. The second case, a = 1, is basically from the MIL-HBDK-5D[14], where the stress-strain relation is expressed as n E--e=0.002 -!?- ) w ( GO.2 > or in the form of the Ramberg-Osgood stress-strain relation tY 0.002 -=--, 6.2
knowing that eo,2= 0.002, we have a = 1. According to the EPRI[‘TJ handbook, specimen is
Co.2
0
” (61)
(
60.2
>
the evaluation
equation
of the J-integral
for a
J = JeW + J,,@, n),
(62)
where
Jeb-4 =E
K:Gd
(63)
In the above equations, b = 2 for plane stress, 6 for plane strain. P is the unit thickness load carried by the specimen; POis the unit thickness limit load carried by the specimen. 4.1. Center crack panel subjected to remote tension (CCT) For a CCT specimen, the plastic J-integral is given as Jp = aaoco *($).h,(~,n)*(~-t’,
where b = W - a, and PO= 2boo for plane stress and 4ba,/J3 for plane strain. The h, (a/W, n) can be found in the table of the EPRI handbook[7]. For convenience of calculation, we fit these h, into polynomial form whenever it is possible.
for a CCT specimen, the aj coefficients for the plane stress condition are given in Table 1. Table 1.Coefficients of h, function, center crack 3
n
i
5.218172
1 2 3 4 5 6
- 11.39764 15.31509 - 12.91599 5.107191 0 0
Eft
0.8
4
QJ
4
0
7
5 6.514183 -20.71263 41.36186 - 57.27338 47.56169 - 16.54982 0
8.36303 -41.88701 114.9541 - 166.6059 93.4695 22.79248 - 30.58944
1.7
1.5
tEf: maximum error in polynomial fitting.
16
%
4
7.511297 - 30.88694 78.1464 - 126.2213 112.78 -40.7358 0
0.9
13
10
8.373961 -45.06873 128.7712 - 186.456 92.79952 45.27715 -43.35869 3
dLI 7.880014 -42.6278 121.6382 - 173.8297 76.12039 60.35604 -49.28308 4.9
20
4 7.32982 -41.92877 127.3355 - 195.9416 104.0834 48.6895 -49.51298
X. G. CHEN et al.
854
To compare the J and CTOD from our model, the SIF equation for the CCT specimen is needed
where (67) The reference SIFf,
for the weight function calculation is taken as f(a/W).
4.2. Compact tension (CT) and three point bending (TPB) specimens For CT and TPB specimens, the plastic J-integrals
have the same form, which is given as
J,=ro,r..b.h,(~,n).(~~+“.
(68)
The difference comes from the expression of limit load P,, and the h, function. (1) Compact tension specimen (CT) P,,, the unit thickness limit load, is 1.455qba, for plane strain, l.O7l~ba, for plane stress. The parameter q is expressed as 059)
? =[(;>‘+2(3+2lln_[(3+1].
To calculate the elastic J, the SIF for the CT specimen under pin-loading is needed. Such a result can be found from the recent work by Mall and Newman[lS]:
K,(a) = [+$0.886+464(;)-
l3.32(;~+
14.72($,.,(;y)&
(70)
W with accuracy 0.3%. The reference SIF f.needed for the generation of the weight function is usually taken as the SIF of the specimen under uniform loading on the crack face. Such a solution could also be found from [15]. However, the SIF expression of the CT specimen under uniform loading on the crack surface in [15] is quite complicated, and we fitted it into a polynomial as below to make it easy to use.
.
(71)
Table 2. Coefficients of h, function, compact tension n
3
5
7
10
13
i 0 1 :
4 1.001198 7.121677 -38.37178 65.97446
CLI 1A07833 2.069074 -25.01254 52.09884
% 1.266582 2.5158 -29.86555 67.40027
QJ 1A00438 -0.023668 - 65.41306 23.82534
4 1.509037 -2.682314 - 49.83837 13.58019
4 5 6
-28.73521 -26.71126 20.64867
-24.12447 -26.7087 21.27885
-41.83951 -21.24579 22.98825
-47.51505 -18.71816 24.76117
-40.70305 - 13.09316 19.85806
Eft
0.8
0.7
tEf: maximum error in polynomial fitting.
1
1.2
1.1
16
20
4 2.017384 - 7.982558 34.07126 1.918588
4 1.382444 -4.024957 -44.16989 7.979807
-41.86072 -3.472516 16.62728 0.5
-43.37432 - 7.554049 18.66797 1.3
Dugdale model for strain hardening materials
855
The fitted eq. (70) has a maximum of 0.83% error from the result of Mall and Newman[lS] in the range of 0.2
plane strain
0.5363 *a0 . $
plane stress,
PO=
(72)
I where L is the span of the specimen. For the case we are concerned with, a standard specimen, 2L/W = 4. The SIF of a standard TPB specimen for elastic J calculation can be found from the SIF handbook[l6]. From the discrete data in [16], we can have a polynomial fit: 5 * + 1169*002 (W) 5 &(a) = --jjPJW -0.04933 + 63.4269 (W> 5 - 368 *4393 (W) - 1690.447 (;>‘+
980.2072 (;y),
(73)
while the reference SIFf, could be taken as the SIF of a single edge crack panel under uniform tension, because they have the same geometry. The problem we have in this case is that we cannot find the&, for L/W = 4. As an approximation, we take thef, of an infinite long plate instead. This approximation will produce some error if L/W = 2, and little error when L/W > 3. The h, function for the TPB was not fitted due to the difficulty of polynomial fitting. The result of this is that, during the CTOD calculation, the initial crack length a~/ W can only be chosen as exactly the same as the crack length given in [7], i.e. so/W = l/8, l/4,3/8, l/2,5/8,6/8,7/8. (3) Double edge cracked panel subjected to tension (DECT) Plastic J is given in [7] as p n+l
J,=at~~~~~b -h,
*
,
F
0
0
(74)
where a0 W
plane strain
PO=
(75) plane stress.
Is 3 The SIF K,(a) for elastic J calculation is expressed as K,(a) =foJ(na),
(76)
where 1.122-0.561
(77)
The above equation has the accuracy of 0.5% for any a/ W[16). The reference SIF fo for weight function generation is taken as f(a/W). The h, (u/W, n) functions are fitted in the same way as for the CCT specimen. Table 3 shows the a coefficients for DECT. (4) Single edge cracked panel subjected to tension (SECT) Plastic J is given as (78)
856
X. G. CWEN et al. Table 3. Coefficients n
3
j 0
t(, 0.2689648
: 3 4 5 6
- 19.2348 8.158046 17.34996 -5.261816 0 0
Eft
0.3
ofh, function,
5
double edge cracked sp&mens 10
7
4 0.797909
crt 1.321107
-24.88236 7.258186 29.64258 - 11.90544 0 0
-40.91061 7.109418 79.71513 - 69.08565 22.84852 0
0.6
13
GLI 2.766505 -25.14265 -2.124447 77.85988 -85.68216 33.4283 0
0.1
4 5.021851 -21.97722 39.35135 -24.82852 -4.59675 7.938473 0 0.5
0.5
16
20
QI 7.612527 -43.70007 99.53776 -88.67284 -5.413372 29.10341 0 1.4
TEE maximum error in polynomial fitting.
where P&J=
1.455$~_~, plane strain I 1.072~~~* plane stress.
(79)
01
(80)
tf is expressed as
2
‘1 =
1+
[
;
I/Z
-;.
The SIF for elastic J can be found in [16]: &@I =“&/W, where f(a/FV) is
The reference SIF, fO, can be chosen as f(u/ W). There are problems with the hi function; as a/W approaches 0.25, the value of h, increases dramatically, which leads to the difficulty of polynomial fitting. After careful examination of the origin of hl values in refs [7, 17-191, the authors find that there are big differences in the values of h, of SECT in different papers, especially for the plane strain problem. After noticing the nonsmooth (o~illation) problem of h, and h3, Parks et af. proposed a consistency check method for the hi function[l7]. Another method, penalty method, was also proposed by Shih and Needleman in the finite element calculation in order to deal with the full plastic problem of the cracked body[l8, 191. The results in [18, 191 show that, for 1
where d,, is a function of tq/E and hardening exponent n. d,, weakly depends on Q/E.
Dugdale model for strain hardening materials
857
Here, we use 2V, to stand for the CTOD from the Dugdale model which distinguishes the true CTOD 6,. Along this line, first, we examine the ratio of .I/( V,a,) given the initial crack length a/W; then we discuss the crack size effects on this ratio; and last we will show the results of shape effects on the ratio. 5.1. Stress ratio, crack size and geometry efects Given the crack length a/W, and changing the stress ratio a/a,, , we can see that the ratios of J and (J/ VDao)are relatively constant for n > 3 when a = 3/7. For highly strain hardening materials, like n = 3, the ratio of J and V. is highly stress state dependent, for a = 3/7. Figures 2 through 9 show the crack size and stress ratio effects on J/(oo V,) for n = 5 and 16. There are further similar figures of different geometries for n = 3, 7, 10, 13 and 20 which are not shown here.
Fig. 2. Comparison of J and CTOD calculated from the Dugdale model for strain hardening materials, stress ratio and crack size effects on J/(u~ V,), CCT, o( = 317, n = 5.
“I
I
I
Fig. 3. Comparison of J and CTOD calculated from the Dugdale model for strain hardening materials, stress ratio and crack size effects on J/(0, VD), DECT, a = 3/7, n = 5.
858
X. G. CHEN ef al.
Fig. 4. Comparison of J and CTOD calculated from the Dugdale model for strain hardening materials, stress ratio and crack size effects on J/(uO V,), CT, a = 3/7, n = 5.
6
6
4
2
u
Q u
16
47
00
Fig. 5. Comparison of J and CTOD calculated from the Dugdale model for strain hardening materials, stress ratio and crack size effects on J/(u, V,),TPB, a = 3/7, n = 5.
Dugdale model for strain hardening materials
1
VD00 IU
0
6
*
P
u
0
P
7
6
0
10
x70”
0
a0
Fig. 6. Comparison of J and CTOD calculated from the Dugdale model for strain hardening materials, stress ratio and crack size effects on J/(a, V,), CCT, u = 3/7, n = 16.
J ‘D
‘0
-a
Fig. 7. Comparison of J and CLOD calculated from the Dugdale model for strain hardening materials, stress ratio and crack size effects on J/(u, V,), DECT, a = 3/7, n = 16.
859
860
X. G. CHEN ef al.
Fig, 8. Comparison of J and CTOD calcuIated from the Dugdale model for strain hardening materials, stress ratio and crack size effects on J/(q, V,), CT, a = 3/7, n = 16.
Fig. 9. Comparison of J and CTOD caiculated from the Dugdale model for strain hardening materials, stress ratio and crack size effects on J&T,, V,,), TPB, a = 3/7, n = 16.
Dugdale model for strain hardening materials
861
For a CCT specimen, ct = 3/7 and n = 5, the value of J/(a,, V,) is slightly stress ratio and crack size dependent. The increasing tendency of J/(crOVD) with a/a0 is observed. For a less strain hardening material, n = 16, we find much less stress ratio dependency of J/(a,, V,), though there is still some crack size dependence. There is almost no crack size effect and very slight stress state dependence in the DECT of n = 5, u = 317. The same is found for n = 16, a = 3/7. Although the DECT and CCT are both tension type specimens, the behaviors of them are not expected to be the same. For a CT specimen, the curves of J/(0,, V,) for a = 3/7, n = 5 and n = 16 are quite similar, while the average value of J/(oO V,) for n = 16, a = 3/7 is higher than that of n = 5. The TPB specimens behave differently from the CT specimens. With a small stress ratio, the J/(0,, VD) is
-
0 OO
Fig. 10. Comparison of J and CTOD calculated from the Dugdale model for strain hardening materials, geometry effect on J/(u, V,), plotted against stress ratio, a = 3/7, n = 5.
,!!!L RRlF-
DmYvTAw
n7aFvLTm
a--
3 7
Fig. 11. Comparison of J and CTOD calculated from the Dugdale model for strain hardening materials, geometry effect on J/(u,,V&, plotted against 6, a = 3/7, n = 5.
862
X. G, CHIP? et al.
almost constant for different crack sizes, while for large stress ratio, it increases dramatically due to the large plastic deformation. Now, if we consider the differences due to the crack length and types of geometries, we will find that, given the same stress ratio, the plastic zone in front of the crack tip will be diEerent for diRerent crack lengths and different specimens. Thus, for the same-stress status, some specimens have only a very small plastic zone, while the others may have a very big one. It is reasonable to compare the J and CTOD on the same plastic deformation bases by defining a plastic deformation measure. Let us define a new parameter c3
Fig. 12. Comparison of J and CTOD calculated from the Dugdale model for strain hardening materials, geometry effect on .!/(a0 V,), plotted against stress ratio, a = 3/i’, n = 16.
Fig. 13. Comparison of J and CTOD calculated from the Dugdale model for strah hrdening geometry effect on J/(u*V,), plotted against cS, u = 3/7, n = 16.
materials,
Dugdale model for strain hardening materials
863
where b = (W - a) is the ligament of the specimen. Given a certain plastic zone size, 8 is the measure of the plastic zone size with respect to the ligament. p/d = cJb/ W. Plotting the .I/( V,q,) against 6, the results for four types of geometries are shown in Figs 11 and 13, and the same comparisons in terms of stress ratio are given in Figs 10 and 12. Little geometry dependence is found in Pigs 10 through 13. Similar results are shown for a = 1 in Figs 14 and 15. It is found that when z = 1, great size effects will affect the calculated CTOD when a material is highly strain hardening. This is due to the fact the model does not take account of the parameter cc. The summary of these computations is given in Tables 4 through IO.
db W Fig. 14. Comparison of J and CTOD calculated from the Dugdale model for strain hardening materials, geometry effect on J/(u,,Vo), plotted against stress ratio, a = 1, n = 5.
I
a=1
Fig. 15. Comparison of J and CTOD calculated from the Dugdale model for strain hardening materials, geometry effect on $/(a, v,), plotted agRinSt 6, K = 1, n = 16.
X. G. CHEN et al.
Dugdale model for strain hardening matehls
865
w
0.3 < w < 0.80
3
(%)
0.3 < cl < 0.55
2.9338
11.6 8.6 7.4 7.6 8.4
2.8729 2.8351 2.8514 2.%11 3.1484 11.1 19.1
Es (%)
JI(~,cy)
CCT 5
0.3 < w < 0.55
2.6211
6.5 5.1 4.7 2.5 3
2.5243 2.5147 2.5486 2.6659 2.8519
5
5.5 17.8
9.8 12.5 13.6 16.5
Es (%)
0.3 < o < 0.65
2.%98
2.884 2.9252 3.042 3.0278
J/w/,)
0.2
2.4883
2.4718 2.4685 2.4952 2.5176
JI(V,oJ
13.4 17.2
Es (%)
JI(~,ey)
7
2 15.3
1.4 4.6 11.95 13.9
Es (%)
< 0.65
7
0.1
2.2122
2.2521 2.21 2.1825 2.2tMl
J/(V,u,)
3.2 9.7
8.6 7.8 6.6 4.9
Es (%)
<0.60
10
Table 9. Compact tension, CT (a = 1)
due to the different crack length a/W. due to stress ratio. due to both crack length a/W and stress ratio.
4.6 18.1
16.9 13.7 10.4 9.8
tech maximum error of I/( V&) $Es maximum error ofJ/(V&J #Et: maximum error of J/( V&J
range
EfTective
Average J/(~,e,)
4.2953
4.2834 4.1985 4.3929 4.3063
0.1125 0.250 0.375 ::ZZ
w (%) EtQ (%)
J/(~fPy)
n
a/W
(“Q
17 13.4 11.2 10.1 9.7
6.3 17
lx
0.3 < 0 < 0.55
3.9815
3.9565 3.8352 3.7869 3.854 4.025
J/woe,)
3
tEc: maximum error of J/( V&J due to the different crack length a/W. #ES maximum error of J/( V,F,) due to stress ratio. §Et: maximum error of J/( V&.) due to both crack length a/W and stress ratio.
-ge
EKective
I
J/(~LP”)
Average
Ect (%) Et0 (%)
0.125 0.250 0.375 O.u)o 0.600
o/W
n
13
5.8 19.6
11.6 14.5 12.4 14.2
Es (%)
0.05 < w < 0.65
2.0832
2.1502 2.094 2.0558 2.0328
Jl(V,cy)
0.3 < UJ< 0.55
3.6118
12.7 10.5 9.3 8.6 7.1
3.8193 3.7077 3.6136 3.517 3.4016 12.3 23.8
Es (%)
J/(v,ey)
3
Table 8. Center crack, CCT, and double edge crack, DECT (a = 1) 5
7.7 19
6.5 19.9
12.6 18.5 15.3 13.5
Es (%)
0.05 < w < 0.65
2.0243
2.1015 2.0391 1.9836 1.9728
16
12 7.9 6.4 6.4 6.6
Es (%)
0.3 < o < 0.80
2.7646
2.8626 2.8126 2.7697 2.719 2.692
JI(~,ey)
Jwae,)
DECT
20
10.3 19.8
5.6 10.2 15.3 12.6
Es (%)
0.05 < 0 < 0.55
2.0019
2.1049 2.0433 1.9505 1.9017
JMvLloy)
g
2
9
x
(%I
0.3 < o < 0.80
Bffective range
5
46.3
26.7
27.7 20.5
9.8 18.3 29.9
Es (%)
0.3 < o < 0.65
3.0013
3.2776 2.9837
2.8949 2.5872 3.263
JlWD~y)
I
26.2
6.8
24.5 6.7
2.4 8 24.5
Es (%I
0.2 < w < 0.65
2.5868
2.6827 2.5115
2.5157 2.5429 2.6811
JWDfJJ
tEc: maximum error of J/( V,F,) due to the different crack length a/W. $Ek maximum error of J/( V&J due to stress ratio. #Et: maximum error of J/( V&J due to both crack length a/W and stress ratio.
4.5304
JWney)
55.7
Et5 (%) Average
14.8 7
18.3 20.7 19.4
w
12.3
4.6289 4.4207
0.500 0.600
3
(%)
4.228 4.6249 4.4795
6.125 0.250 0.375
w
JWD~,)
n
a/W
10
16.4
6.5
13.4 1
1.6 2.6 12.8
Es (%)
0.1 < w < 0.60
2.241
2.2811 2.1699
2.311 2.2544 2.3069
JWdJJ
13
10.7
9.2
6.9 2.7
2.5 6.8 6.4
Es (%)
0.05 < o < 0.65
2.1322
2.105 2.0354
2.2221 2.1513 2.1467
Jl(V,ey)
Table 10. Three. ooint bendina_ snecimen. TBB (a . = 1), _ 16
14.4
9.1
9.5 8
2.1 8.8 7.4
Es (%)
0.05 < o < 0.65
2.1262
2.0137 1.9903
2.1718 2.0946 2.0638
JWney)
20
20.1
9.9
16 9.6
2.8 9.8 10.7
EL?(%)
0.3 < 0 < 0.55
2.1024
I .9367 1.9453
2.1293 2.0508 2.OtMl
J/(V,cy)
9 w
p
k
F
3.
B
;
y$
X. G. CHEN et al.
868
5.2. The limitations of the present model
There are two limitations on the Dugdale model presented here. (1) The accuracy evaluation method of the model. Since the Dugdale model is a simplified model, there are no ways to judge its accuracy by itself. Other methods, such as experimental method or finite element method, should be used to evaluate the model. (2) The present Dugdale model does not take the Ramberg-Osgood coefficient c1into account; choosing an a different from 3/7 or 1 may cause different results. Further study is needed to improve this model. 6. CONCLUSION A Dugdale model for strain hardening materials has been presented. The weight function method has been used to calculate the CTOD by this model. It is shown that the present model gives quite promising results after comparison with the J-integral solution from EPRI’s work. Since this model simplifies the elastic-plastic fracture problem into an elastic fracture problem, it is very easy to calculate the CTOD with this model, which makes the engineering application easier than before. The results are only valid under the plane stress condition, and more work should be done for the plane strain problems. Acknowledgement-The authors are grateful for the support from the Institute of Aeronautical Materials, Beijing, on this project when the first author worked in the Institute during 1985 to 1989.
REFERENCES G. S. Wang, B. Palmberg and A. F. Blom, The crack propagation in the plate subjected to spectrum loads. Technical Note, The Aeronautical Research Institute of Sweden (1988). J. C..Newman, Jr., ASTM STP 748, 53-84 (1981). J. C. Newman. Jr.. ASTM STP 8%. 139-166 (1985). M. Hoffman and T. Seeger, Dugdale solutions for strain hardening materials, in The Crack Tip Opening Displacemenf in Elastic-Plastic Fracture Mechanics, pp. 57-77, Proceedings of the Workshop on the CTOD Methodology, Geesthacht, Germany (1985). J. C. Newman, Jr., Engng Fracture Mech. 1, 137-154 (1968). D. S. Dugdale, J. Mech. Pkys. Soli& 8, 100-104 (1960). V. Kumar, M. D. German and C. F. Shih, An engineering approach for elastic-plastic fracture analysis. EPRI NP-1931, Project 1237-1, Technical Report (1981). Z. X. Wang and D. R. Gu, General Theory of Special Functions. Science Press, Beijing (1979) (in Chinese). X. R. Wu, Engng Fracture Meek. 20, 35-49 (1984). X. R. Wu and J. Carlsson, J. Mech. Phys. Sohds 31, 485-497 (1983). X. R. Wu and X. G. Chen, Engng Fracture Mech. 33, 877-886 (1989). H. J. Petroski and J. D. Achenback, Engng Fracture Mech. 10;257-266 (1978). T. Fett, C. Matteck and D. Munx, Engng Fracture Mech. 27, 697-715 (1987). MIL-HDBK-SD (1986). S. Mall and J. C. Newman, Jr., ASTM STP 868, 113-128 (1985). Chinese Aerospace Establishment (CAE), Stress Inrensity Factor Handbook. Science Press, Beijing (1981) (in Chinese). D. M. Parks, V. Kumar and C. F. Shih, ASTM STP 803, 1-307-I-383 (1983). C. F. Shih and A. Needleman, J. appl. Mech. 51, 48-56 (1984). C. F. Shih and A. Needleman, J. appl. Mech. 51, 56-64 (1984). J. W. Hutchinson, A. Needleman and C. F. Shih, in Fracture Meckanics (Edited by N. Perrone et al.), pp. 515-528. University Press of Virginia, Charlottesville, VA (cited from[ll]). W. Sauter and V. Kumar, Section 3.2 of Combined Fifth and Sixth Semiannual Report to EPRI, General Electric Company Schenectady, NY, General Electric Report, SRD-82-048 (1982) (cited from[l7]). C. F. Shih, J. Mech. Phys. Solids 29, 305-326 (1981).
APPENDIX 1. CRACK FACE DISPLACEMENT CALCULATION OF THE CENTER CRACK IN AN INFINITE PLATE BY DUGDALE MODEL FOR STRAIN HARDENING MATERIALS CTOD could be written as 6,=;
:K,(S)m(S.x)dC.
(AlI
s where
ho=
1 x
a
x>a x
(A2)
Dugdale model for strain hardening materials
869
The weight function m(a, x) is 8iven as i (cZ _ x2)-‘/2, J( > so that the SIF is expressed in terms of weight function and crack face stress m(a, x) 5 2
(A3)
(A4) where p(t) is the stress distribution on the crack surface. Substituting eqs (A4) and (A3) into (Al), we have p(r)e(r2
- x~)-“~({~ - r2)“2 dr d<.
(AS)
Referring to Fig. Al, we change the sequence of the integration, and we can obtain p(r)t(t2
- x~)-“~(<~ - r2)-“2dC
646)
Notice that I({, t) =
{(r2 - x~)-I’~(~~- ,2)-“2 d{ (A7)
Thus we can simplify the integral into the following form: (A8) where
Z,(r) =
C-d
or,t)
t-t
.
(A9)
(1) If x < (I, then d, = II, and the first term in eq. (A8) equals zero. Equation (A8) becomes Sp=;
dp(r)Z,(r) dr.
s (I
(Al’3
Sometimes it is helpful to change eq. (AlO) into the following form, integrating it by parts:
(All)
d
________-
___...--._____ -
4
-I----
---_-es
d
Fig. Al. Integration area.
X. G. CHEN et al.
870 where z,(f)
=
l”
Jv* +x3 f (d2- r3 (d’+xq+(d2-13
P(t) = If
J
p(t) dt.
zJto=
t/il+1
( > A
009
(I -
and the crack face ~spla~ment
ar”+‘u,
W3)
W4)
becomes a,=&~Q’ZW$+/($$)dr.
tAl8)
In this particular case, with load p(t) expressed as eq. (A13), the transform from eq. (AIO) to eq. (A15) changes the l/(n + 1) order of singularity in the inte8ral to the -l/2 order of singularity in the integral, which is much easier to in@rate. (2) If x > u, then 4 = x, and it follows that Z,(f) = Z,(r). Equation (A8) has the form
Equation (Al6) has the same expression as eq. (AIO), so that eq. (AIO) is in fact valid for both x
(A18)
where S, is the crack face displacement produced by the remote tensile stress, which is given as G,+/(d2-u’)
IA19
from eq. (1). and knowing that the stress intensity factor due to crack face loading and the remote tension in the plate with a center crack is KJd, a) = 2eo/(~)arccos(~)
K&4 = a,/Wh
WW
we can express the CTOD in the following way by substituting the eqs (A20), (1) and (A19), (A17) into (A18):
WU In this case, the material is still perfectly plastic; we calculate the crack face displacement which is not at the crack tip. From (AU), the integration could be implemented and we obtain
6422) where Z2 is
z-
J
d
’-
2t dt
II (r2 - x2),/(d2 - t2)
,/(d* + a2) + (d2 -x2) =~lnc+/(dz+.2)-(d2-x2)
(~23)
Dugdale model for strain hardening materials
871
and Z, is d
dr
I
W4)
z1 = II (r* - XqJ(d2 - rq. This integration has two solutions according to two different cases. (A) x
6425) (B) x > a dr (12- X2)&2 - x3
‘G&q”
aJ(&+x2)+xJ(d2-u2) a&P-x2)-x (CP--2)
(Received 7 March 1991)
.
(A261