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Crack opening displacement for two unequal straight cracks with coalesced plastic zones – A modified Dugdale model
Applied Mathematical Modelling 35 (2011) 3788–3796
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Crack opening displacement for two unequal straight cracks with coalesced plastic zones – A modified Dugdale model R.R. Bhargava a, S. Hasan b,⇑ a b
Department of Mathematics, Indian Institute of Technology Roorkee, Roorkee 247 667, India Department of Mathematics, Jamia Millia Islamia (Central University), New Delhi 110 025, India
a r t i c l e
i n f o
Article history: Received 22 December 2009 Received in revised form 16 January 2011 Accepted 2 February 2011 Available online 23 February 2011 Keywords: Coalesced plastic zones Crack opening displacement Dugdale model
a b s t r a c t The problem investigated is of an infinite plate weakened by two collinear unequal hairline straight quasi-static cracks. Uniform constant tension is applied at infinity in a direction perpendicular to the rims of the cracks. Consequently the rims of the cracks open in Mode I type deformation. The tension at infinity is increased to the limit such that the plastic zones developed at the two adjacent interior tips of cracks get coalesced. To arrest the crack from further opening normal cohesive variable stress distribution is applied on the rims of the plastic zones. Closed form analytic expressions are obtained for load bearing capacity and crack opening displacement (COD). An illustrative case is discussed to study the behavior of load bearing capacity and crack opening displacement with respect to affecting parameters viz. crack length, plastic zone length and inter crack distance between the two cracks. Results obtained are reported graphically and analyzed. Ó 2011 Elsevier Inc. All rights reserved.
1. Introduction Dugdale [1] proposed a closure model for a slit weakening a plate. The slit was opened in Mode I type deformation. The plastic zones developed due to the remotely applied tension were closed by distributing yield point stress over their rims. The approached further modified [2] when the rims of developed plastic zones were subjected to variable stress distribution to arrest the crack from further opening. Dugdale model solution was obtained [3] for the case when an infinite plate was weakened by two collinear straight cracks. Papers [4,5] modified the Dugdale model solution for two collinear equal straight cracks by closing the plastic zones developed by distributing linearly and quadratically varying stress over their rims. The model was further modified [6] for the case when applied load was characterized by the specified remote strain. Enriched finite element penalty functions method has been used in [7] for modeling interface crack with contact zones. An analytic solution for two equal length collinear strip yield cracks weakening an unbounded plate has been obtained in [8] using complex variable technique developed in [9]. The problem was further extended [10] to the strip yield analysis for two collinear unequal straight cracks weakening an infinite sheet. Mixed mode Dugdale–Barenblatt model has been established for a semi-infinite crack in an ideally elastic–plastic thin plate loaded by a pair of self-equilibrating concentrated forces at the crack rims in [11]. The problem investigated in the present paper is of an infinite elastic perfectly-plastic plate weakened by two unequal collinear straight quasi-static cracks. The rims of the crack open in Mode I type deformation by remotely applied uniform constant unidirectional tension consequently the rims of the cracks open in self-similar fashion forming a plastic zone ahead of each tip of the cracks. The tension at infinity is increased to the limit such that plastic zones developed at two adjacent ⇑ Corresponding author. E-mail addresses: [email protected] (R.R. Bhargava), [email protected], [email protected] (S. Hasan). 0307-904X/$ - see front matter Ó 2011 Elsevier Inc. All rights reserved. doi:10.1016/j.apm.2011.02.018
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interior tips of the two cracks get coalesced. Rims of the plastic zones, in turn, are subjected to quadratically varying normal cohesive stress distribution arresting the crack from further opening. The analytic expressions are obtained for the crack opening displacement at each tip of the cracks. A qualitative study has been presented for the COD at interior and exterior tips of the cracks in graphical form. Also load required to arrest the crack opening when the size of the plastic zone increases is computed and plotted. 2. Mathematical formulation According to the complex variable technique [9], the stress components Pij (i, j = x, y) may be expressed in terms of two complex potential functions U(z) and X(z) as
h i Pxx þ Pyy ¼ 2 UðzÞ þ UðzÞ ; Pyy iPxy
ð1Þ
¼ UðzÞ þ XðzÞ ðz zÞU0 ðzÞ;
ð2Þ
where bar over the function denotes its complex conjugate. Prime after the function denotes its differentiation with respect to argument. Consider an infinite homogeneous, isotropic plate which occupies xy-plane, and is cut along n collinear straight cracks Li S (i = 1, 2, 3 . . . n) lying on x-axis. Union of these cracks is denoted by L ¼ ni¼1 Li . If infinite boundary of the plate be kept unloaded and the rims of L be subjected to the stress distribution P yy ; P xy then using Eq. (2) following two Hilbert problems are obtained.
Uþ ðtÞ þ X ðtÞ ¼ Pþyy iP þxy ;
þ
U ðtÞ þ X ðtÞ ¼
Pyy
iP xy ;
ð3Þ on L;
ð4Þ
0
under the assumption Lim fyU ðt þ iyÞg ¼ 0. y!0
Superscripts + and denote the limiting value of the function when any point t on any of the cracks, other than end points, is approached from the positive y-plane (y > 0) and the negative y-plane (y < 0), respectively. Adding and subtracting Eqs. (3) and (4) following two Hilbert problems are obtained
fUðtÞ þ XðtÞgþ þ fUðtÞ þ XðtÞg ¼ 2pðtÞ;
ð5Þ
þ fUðtÞ XðtÞg fUðtÞ XðtÞg ¼ 2qðtÞ;
ð6Þ
2pðtÞ ¼ Pþyy þ Pyy iðP þxy þ Pxy Þ;
ð7Þ
where
2qðtÞ ¼
ðPþyy
Pyy Þ
iðPþxy
Pxy Þ:
ð8Þ
Solution of the Eq. (5) in absence of forces at infinity may be written as
UðzÞ þ XðzÞ ¼
1 piXðzÞ
Z L
X þ ðtÞpðtÞ 2Pn ðzÞ dt þ : tz XðzÞ
ð9Þ
And the solution of Eq. (6) may be written as
UðzÞ XðzÞ ¼
1 piXðzÞ
Z L
qðtÞ dt C0 ; tz
ð10Þ
where
Pn ðzÞ ¼ C 0 zn þ C 1 zn1 þ þ C n ;
ð11Þ
and
ðXðzÞ ¼
n Y
ðz ak Þ1=2 ðz bk Þ1=2 :
ð12Þ
k¼1
Constants Ci (i = 1, 2, 3 . . . n) are determined by boundary condition at infinity and single-valuedness of displacements on crack Lk(k = 1, 2, 3 . . . n). The end points of crack Lk are denoted by ak, bk, (k = 1, 2 . . . n). 3. Statement of the problem An infinite elastic perfectly-plastic plate, occupying xoy-plane, is weakened by two collinear unequal straight cracks L1 and L2. The crack L1 lies from [d1, 0] to [c1, 0] and L2 lies along [b1, 0] to [a1, 0] on the x-axis. At the infinite boundary of the plate uniform constant tension, ryy = r1 is applied which opens the rims of the cracks L1 and L2 in Mode I type
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deformations. Consequently the plastic zones developed ahead of the tips d1, c1, b1 and a1 of the cracks L1 and L2. The tension at boundary is increased to the limit such that plastic zones developed at the two adjacent interior tips c1 and b1 (of the cracks L1 and L2) get coalesced. This coalesced plastic zone from [c1, 0] to [b1, 0] is denoted by C1. The plastic zone from [d, 0] to [d1, 0] is denoted by C3 and C2 denotes the plastic zone [a1, 0] to [a, 0]. Each rim of the plastic zones C1, C2 and C3 is subjected to normal stress distribution Pyy ¼ t2 rye ; Pxy ¼ 0 to arrest the crack from further opening. Entire configuration is schematically depicted in Fig. 1. 4. Solution of the problem The problem stated in Section 3 is solved by superimposition of the solution of two auxiliary problems contributing towards the stress singularity at the tip of the cracks. These two problems are appropriately derived from the above problem. These problems are termed Problem I and Problem II. 4.1. Problem I An infinite elastic perfectly-plastic plate is weakened by a single crack C occupying the interval [d, 0] to [a, 0] on x-axis {formed of the union of C3, L1, C1, L2 and C2 which are defined in Section 3}. Boundary conditions of the problem are (i) The rims of the crack C are stress free. (ii) Infinite boundary of the plate is subjected to uniform stress distribution ryy = r1. (iii) Displacements are single-valued around the rims of crack C. Using boundary condition (i) and Eqs. (5)–(8) following two Hilbert problems are obtained
n
n o þ UI ðtÞ þ XI ðtÞ ¼ 0; n oþ n o UI ðtÞ XI ðtÞ UI ðtÞ XI ðtÞ ¼ 0:
UI ðtÞ þ XI ðtÞ
oþ
ð13Þ ð14Þ
Superscript I denotes that the function refer to Problem I. Solution of the Eqs. (13) and (14) using boundary condition (i) and (ii) and Eqs. (9)–(12) may be written as
UðzÞ þ XðzÞ ¼
2ðC 0 z þ C 1 Þ ; XðzÞ
UðzÞ XðzÞ ¼
r1 2
:
Using boundary condition (iii) the desired complex potential function for problem-I may be written as
r
UI ðzÞ ¼ pffiffiffiffiffiffiffiffiffiffiffi1pffiffiffiffiffiffiffiffiffiffiffi z 2 za zd
aþd r1 : 2 4
ð15Þ
4.2. Problem II An infinite elastic perfectly-plastic plate is weakened by a straight crack C as defined in Problem I. The boundary conditions of the problem are
Fig. 1. Configuration of the problem.
R.R. Bhargava, S. Hasan / Applied Mathematical Modelling 35 (2011) 3788–3796
(i) (ii) (iii) (iv)
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No stresses are acting at infinite boundary of the plate. 2 Rims of the Ci(i = 1, 2, 3), segment of the crack C, are subjected to stress distribution P yy ¼ t rye ; P xy ¼ 0. Rims of the Li(i = 1, 2), segment of the crack C, are stress free. Displacements are single-valued along the rims of crack C.
Boundary conditions (ii) and (iii) and Eqs. (2)–(8) give following two Hilbert problems
n
n o þ UII ðtÞ þ XII ðtÞ ¼ 2t2 rye ; n oþ n o UII ðtÞ XII ðtÞ UII ðtÞ XII ðtÞ ¼ 0;
UII ðtÞ þ XII ðtÞ
oþ
ð16Þ ð17Þ
S on C ¼ 3i¼1 Ci . Superscript II denote that quantities refer to Problem II. Desired complex potential UII(z) is written, solving Eqs. (16) and (17), as
UII ðzÞ ¼ where J ¼
rye
pffiffiffiffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffiffiffiffi ðJ zf2 Þ 2p z a z d
2f 1 f2 ðaþdÞf12 f0
rye ðf1 þ zf0 Þ pffiffiffiffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffiffiffiffi z2 rye za zdþ FðzÞ; 2p 2pi
ð18Þ
þ ðf3 þ 2adf1 ða þ dÞf0 Þ
fi ¼ Ri ðd1 Þ Ri ðdÞ þ Ri ðb1 Þ Ri ðc1 Þ þ Ri ðaÞ Ri ða1 Þ;
i ¼ 0; 1; 2; 3;
! ! t2 5ða þ dÞt 5ða þ dÞ2 2ad pffiffiffiffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffiffiffiffi ða þ dÞ 5ða þ dÞ2 1 2t þ a þ d ; at td þ 3ad sin 12 3 4 ad 3 8 4
R3 ðtÞ ¼
! t 3ða þ dÞ pffiffiffiffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffiffiffiffi 3ða þ dÞ2 ad 1 2t þ a þ d R2 ðtÞ ¼ at td sin ; 2 4 2 ad 8 pffiffiffiffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffiffiffiffi a þ d 1 2t þ a þ d sin ; R1 ðtÞ ¼ a t t d 2 ad
R0 ðtÞ ¼ sin
1
2t þ a þ d ; ad
2
3 1 1 ðazÞðd1 dÞþðad1 ÞðzdÞ ðazÞðb1 dÞþðab1 ÞðzdÞ pffiffiffiffiffiffipffiffiffiffiffiffipffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffi tanh pffiffiffiffiffiffipffiffiffiffiffiffipffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffi 7 6 tanh 2 az zd ad1 d1 d 2 az zd ab1 b1 d 7 6 7: FðzÞ ¼ 6 4 5 1 ðazÞðc1 dÞþðac1 ÞðzdÞ 1 ðazÞða1 dÞþðaa1 ÞðzdÞ þtanh pffiffiffiffiffiffipffiffiffiffiffiffipffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffi þ tanh pffiffiffiffiffiffipffiffiffiffiffiffipffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffi 2 az zd ac1
c1 d
2 az zd aa1
ð19Þ
a1 d
5. Plastic zone length Plastic zone length developed at each exterior tip of two cracks is obtained using the condition that the stresses remain finite through out the plate.
UI ðzÞ þ UII ðzÞ–1: Consequently following two conditions are obtained at the outer tip of the bigger crack i.e. a
r1 a2 rye
¼
2ðJ af2 Þ
pðd aÞa2
;
ð20Þ
and at the outer tip of the smaller crack i.e. d
r1 2ðJ df2 Þ ¼ : 2 rye pða dÞd2
d
ð21Þ
Values of a and d are determined from above equations for prescribed t2rr1 ; a1 ; b1 ; c1 ; d1 . ye The plastic zone length at the tip a1 and d1 are calculated from ja1 aj and jd d1j, respectively. 6. Crack opening displacement Crack opening displacement (COD) at each of the tip is calculated by the method given in [2].The parting of the crack face is given by
D¼
Z 4 Im Zdz ðfor plane stress caseÞ E
ð22Þ
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where Westergaard stress function, Z = Non-singular term of 2U(z), E denotes Young’s modulus and rye/E = .01. The crack opening displacement (COD) at each of the tip is given by. dt ¼ lim D; t denotes the crack tip. z!t COD at the crack tip a1
2
da1
3 g 0 ða1 Þ þ 13 g 2 ða1 Þ G1 g 1 ða1 Þ þ aþd g 0 ða1 Þ f0 þ G2 g 3 ða1 Þ þ G3 g 4 ða1 Þ 2 7 6 3 3 7 6 a dÞþðad1 Þða1 dÞ 1 Þðd p1ffiffiffiffiffiffiffi ffipffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffi 7 6 þ 31 d31 tanh1 pðaa ffiffiffiffiffiffiffi ffi 7 6 2 aa1 a1 d ad1 d1 d 7 6 4rye 6 7; 3 3 ¼ a1 1 b1 ðaa1 Þðb1 dÞþðab1 Þða1 dÞ 7 6 pE 6 þ 3 3 tanh pffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffi 7 2 aa1 a1 d ab1 b1 d 7 6 7 6 5 4 a31 c31 1 ðaa1 Þðc1 dÞþðac1 Þða1 dÞ 3 3 tanh pffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffi a d ac c d 2 aa 1
1
1
1
COD at the crack tip b1
2
db1
3 g 0 ðb1 Þ þ 13 g 2 ðb1 Þ G1 g 1 ðb1 Þ þ aþd g 0 ðb1 Þ f0 þ G2 g 3 ðb1 Þ þ G3 g 4 ðb1 Þ 2 7 6 3 3 7 6 ðab1 Þðd1 dÞþðad1 Þðb1 dÞ ffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffi 7 6 þ b31 d31 tanh1 p 7 6 2 ab1 b1 d ad1 d1 d 7 4rye 6 7; 6 3 3 ¼ c 1 b ðab1 Þðc1 dÞþðac1 Þðb1 dÞ 7 1 1 pE 6 p ffiffiffiffiffiffiffi ffi p ffiffiffiffiffiffiffiffi p ffiffiffiffiffiffiffi ffi pffiffiffiffiffiffiffiffi 7 6 3 3 tanh 2 ab1 b1 d ac1 c1 d 7 6 7 6 5 4 a31 b31 1 ðab1 Þða1 dÞþðaa1 Þðb1 dÞ pffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffi 3 3 tanh 2
ab1
b1 d aa1
a1 d
COD at the crack tip c1
2
3 g 0 ðc1 Þ þ 13 g 2 ðc1 Þ G1 g 1 ðc1 Þ þ aþd g 0 ðc1 Þ f0 þ G2 g 3 ðc1 Þ þ G3 g 4 ðc1 Þ 2 7 6 3 3 7 6 c dÞþðad1 Þðc1 dÞ 1 Þðd p1ffiffiffiffiffiffiffi ffipffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffi 7 6 þ 31 d31 tanh1 pðac ffiffiffiffiffiffiffi ffi 7 6 2 ac1 c1 d ad1 d1 d 7 4rye 6 7; 6 c 3 3 dc 1 ¼ b 1 ðac Þðb dÞþðab Þðc dÞ 7 6 1 p1ffiffiffiffiffiffiffiffipffiffiffiffiffiffiffi 1 ffip 1 ffiffiffiffiffiffiffiffi pE 6 þ 31 31 tanh pffiffiffiffiffiffiffiffi 7 2 ac1 c1 d ab1 b1 d 7 6 7 6 5 4 3 3 c1 a1 1 ðac1 Þða1 dÞþðaa1 Þðc1 dÞ 3 3 tanh pffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffi 2 ac1
c1 d aa1
a1 d
COD at the crack tip d1
2
dd1
3 g 0 ðd1 Þ þ 13 g 2 ðd1 Þ G1 g 1 ðd1 Þ þ aþd g 0 ðd1 Þ f0 þ G2 g 3 ðd1 Þ þ G3 g 4 ðd1 Þ 2 7 6 3 3 7 6 ðad1 Þðb1 dÞþðab1 Þðd1 dÞ ffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffi 7 6 þ d31 b31 tanh1 p 7 6 2 ad1 d1 d ab1 b1 d 7 4rye 6 7; 6 3 c3 ¼ 1 ðad1 Þðc1 dÞþðac1 Þðd1 dÞ d 7 pE 6 pffiffiffiffiffiffipffiffiffiffiffiffipffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffi 7 6 31 31 tanh az zd ac c d 2 1 1 7 6 7 6 5 4 a31 d31 1 ðad1 Þða1 dÞþðaa1 Þðd1 dÞ pffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffi 3 3 tanh 2
ad1
d1 d aa1
a1 d
where
pffiffiffiffiffiffiffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffiffiffiffiffiffiffi a d1 d1 d þ a b1 b1 d a c1 c1 d a a1 a1 d; pffiffiffiffiffiffiffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3G2 ¼ d1 a d1 d1 d þ b1 a b1 b1 d c1 a c1 c1 d a1 a a1 a1 d; pffiffiffiffiffiffiffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 3G3 ¼ d1 a d1 d1 d þ b1 a b1 b1 d c21 a c1 c1 d a21 a a1 a1 d; " # ð2z a dÞ pffiffiffiffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffiffiffiffi ða dÞ2 1 2z þ a þ d az zd g 0 ðzÞ ¼ ; sin 4 ad 8 ( ) qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 1 ð2z a dÞða þ dÞ ða þ dÞða dÞ2 1 a þ d 2t 2 ða zÞðz dÞ ; g 1 ðzÞ ¼ ðða zÞðz dÞÞ þ sin 3 8 ad 16 ! z 3ða þ dÞ pffiffiffiffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffiffiffiffi 3ða þ dÞ2 ad 1 2z þ a þ d az zd sin ; g 2 ðzÞ ¼ 2 4 2 ad 8 pffiffiffiffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffiffiffiffi a þ d 1 2z þ a þ d 1 2z þ a þ d g 3 ðzÞ ¼ a z z d ; g 4 ðzÞ ¼ sin : sin 2 ad ad
G1 ¼
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7. Case study A qualitative study is presented for studying the behavior of crack opening and its arrest possibilities with respect to the parameter affecting the crack arrest. Load ratio used to denote the ratio of the applied at infinity to the load applied on the rims of the developed plastic zone. Crack ratio (CR) refers the ratio of the length of bigger crack to that of smaller crack. Fig. 2 depicts the variation of load ratio vs plastic zone length to crack length ratio at the outer tip of the bigger crack. For each crack ratio, the requirement of load to arrest the crack opening increases as plastic zone length is increased. It is interesting to note if the two cracks are equal in length, the load required to arrest the crack opening is the maximum. As the size of one crack is increased vis-à-vis the other crack then fewer load is required to arrest the crack opening. This is because the presence of the bigger crack itself affects the opening of the smaller one. Also the size of the plastic zone becomes smaller and smaller as one of the cracks becomes bigger as compare to other one, as expected. The same variation of load ratio (load applied at infinity to load prescribed at plastic zone) versus the plastic zone length to crack ratio is plotted at the exterior tip of the smaller crack in Fig. 3. In this case also it is observed that as the plastic zone
0.45 CR=1
σ∞ t σ ye
CR=2
2
CR=3
0.225
CR=4 CR=5
CR = length of bigger crack to length of smaller crack ratio 0 0.00
0.31
0.62
Plastic zone length/ crack length
CR=1
CR=2
CR=3
1
CR=4
CR=5
Fig. 2. Variation of load ratio with plastic zone length at the outer tip of bigger crack (say a) for inner plastic zone = .02.
σ∞ t σ ye 2
0.5
CR = length of bigger crack to length of smaller crack Ratio 0 0.00
0.31
0.62
Plastic zone length/ crack length Fig. 3. Variation of load ratio with plastic zone length at the outer tip of smaller crack (say d) for inner plastic zone = .02.
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size increases the load ratio required to arrest the opening of the crack also increases. But as the crack ratio increases the behavior here is opposite to that in Fig. 2. as the crack ratio is increased more load is required for arrest. Figs. 4–7 show the crack opening displacement with respect to plastic zone to crack length ratio: at the outer tip d1 of smaller crack, at the inner tip c1of smaller crack, at the inner tip b1 of bigger crack and at the outer tip a1 of bigger crack, respectively. Crack opening displacement is least at the outer tip of the smaller crack and also it may be noted that the variation at this tip is opposite that in case of others three crack tips. The crack ratio, CR, is increased from 1 to 5. The variations of COD at the inner tips of the smaller and bigger cracks are identical. They make a perfect opening for coalesce which is assumed to happen too. This proves the correctness of the analysis. It is to be noted in both the Figs. 5 and 6 that although COD increases with crack ratio but the increase is not much. Individually for each crack ratio the COD increases when plastic zone length to crack length ratio is increased. Fig. 7 depicts the variation of COD at the outer tip of the bigger crack. The variation in COD is almost linear but steep up. The crack starts opening for larger values of plastic zone to crack length ratio and for higher values of crack ratio.
0.15
COD / smaller crack length
CR = length of bigger crack to length of smaller crack ratio
CR=1
CR=2
0.075
CR=3
CR=4
CR=5
0 0.3
0
0.6
Plastic zone length/ crack length Fig. 4. Variation of crack opening displacement with plastic zone length at the outer tip of smaller crack (d1).
1
CR=2 CR=3 CR=1 CR=4
COD/ smaller crack length
CR=5
0.5
CR = length of bigger crack to length of smaller crack ratio
0 0
0.3
0.6
Plastic zone length/ crack length Fig. 5. Variation of crack opening displacement with plastic zone length at the inner tip of smaller crack (c1).
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CR=2
1
CR=3
CR=1
CR=4
COD/ smaller crack length
CR=5
0.5
CR = length of bigger crack to length of smaller crack ratio
0 0
0.3
0.6
Plastic zone length/ crack length Fig. 6. Variation of crack opening displacement with plastic zone length at the inner tip of bigger crack (b1).
CR=5
1
CR=4
CR=3
COD/ smaller crack length
CR=2
CR=1
0.5
CR = length of bigger crack to length of smaller crack ratio 0 0
0.3
0.6
Plastic zone length/ crack length Fig. 7. Variation of crack opening displacement with plastic zone length at the outer tip of bigger crack (a1).
8. Conclusions Following are the conclusions: The problem of two unequal collinear straight cracks with a coalesced plastic zone has been investigated using complex variable technique. Analytic expressions are obtained to compute plastic zones developed at exterior tips of the smaller and bigger cracks and also crack opening displacements (COD). Variation of the load ratio at the exterior tips of smaller and bigger crack with respect to increasing crack ratio shows opposite behavior. This is because when bigger crack is much bigger then the smaller one, the smaller crack comes much in distress.
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The crack opening displacement plotted at the four tips of the two cracks also confirms that the interior tip of the crack COD has a coalescing behavior. And the behavior of the COD at the exterior tip of the smaller crack has same increasing trend. But as the crack ratio is increased, CR, it is noted that the variation of COD at the interior tip of the smaller crack is opposite to all other tips of the two cracks. The model is valid for small scale yielding.
Acknowledgements Authors are grateful to Prof. R. D. Bhargava [Senior Professor and Head (retd.), Dept. of Mathematics, Indian Institute of Technology Bombay, Mumbai] for his valuable suggestions during the course of this work. The authors are grateful to the referees and the editor for their suggestions, which improved the understandability of the paper. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11]
D.S. Dugdale, Yielding of sheets containing slits, J. Mech. Phys. Solids 8 (1960) 100–104. L.P. Harrop, Application of a modified Dugdale model to the K vs. COD relation, Eng. Fract. Mech. 10 (1978) 807–816. P.S. Theocaris, Dugdale model for two collinear unequal cracks, Eng. Fract. Mech. 18 (1983) 545–559. R.R. Bhargava, S.C. Agrawal, Cohesive quadratic load distribution arresting crack opening – the Dugdale Approach, J. Theor. Appl. Mech. Booklet 4, Bulgaria, (1996). R.R. Bhargava, S.C. Agrawal, Modified Dugdale model for two collinear cracks with coalesced interior plastic zone, in: 19th International Congress of Theoretical and Applied Mechanics, Kyoto, Japan, August 25–31, (1996). C.H. Wang, L.R.F. Rose, G.X. Chen, A strain-based Dugdale model for cracks under generally yielding conditions, Int. J. Fract. 113 (2002) 77–104. S. Beretta, M. Carboni, A Strip-Yield algorithm for the analysis of closure evaluation near the crack tip, Eng. Fract. Mech. 72 (2005) 1222–1237. R.A. Collins, D.J. Cartwright, An analytical solution for two equal-length collinear strip yield cracks, Eng. Fract. Mech. 68 (7) (2001) 915–924. N.I. Muskhelisvili, Some Basic Problems of the Mathematical Theory of Elasticity, Groningen P. Noordhoff Ltd., 1953. English translation. Nishimura Toshihiko, Strip yield analysis of two collinear unequal cracks in an infinite sheet, Eng. Fract. Mech. 69 (2002) 1173–1191. Li Jia, A Dugdale–Barenblatt model for a plane stress semi-infinite crack under mixed mode concentrated forces, Int. J. Fract. 88 (1997) 153–166.
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Crack opening displacement for two unequal straight cracks with coalesced plastic zones – A modified Dugdale model R.R. Bhargava a, S. Hasan b,⇑ a b
Department of Mathematics, Indian Institute of Technology Roorkee, Roorkee 247 667, India Department of Mathematics, Jamia Millia Islamia (Central University), New Delhi 110 025, India
a r t i c l e
i n f o
Article history: Received 22 December 2009 Received in revised form 16 January 2011 Accepted 2 February 2011 Available online 23 February 2011 Keywords: Coalesced plastic zones Crack opening displacement Dugdale model
a b s t r a c t The problem investigated is of an infinite plate weakened by two collinear unequal hairline straight quasi-static cracks. Uniform constant tension is applied at infinity in a direction perpendicular to the rims of the cracks. Consequently the rims of the cracks open in Mode I type deformation. The tension at infinity is increased to the limit such that the plastic zones developed at the two adjacent interior tips of cracks get coalesced. To arrest the crack from further opening normal cohesive variable stress distribution is applied on the rims of the plastic zones. Closed form analytic expressions are obtained for load bearing capacity and crack opening displacement (COD). An illustrative case is discussed to study the behavior of load bearing capacity and crack opening displacement with respect to affecting parameters viz. crack length, plastic zone length and inter crack distance between the two cracks. Results obtained are reported graphically and analyzed. Ó 2011 Elsevier Inc. All rights reserved.
1. Introduction Dugdale [1] proposed a closure model for a slit weakening a plate. The slit was opened in Mode I type deformation. The plastic zones developed due to the remotely applied tension were closed by distributing yield point stress over their rims. The approached further modified [2] when the rims of developed plastic zones were subjected to variable stress distribution to arrest the crack from further opening. Dugdale model solution was obtained [3] for the case when an infinite plate was weakened by two collinear straight cracks. Papers [4,5] modified the Dugdale model solution for two collinear equal straight cracks by closing the plastic zones developed by distributing linearly and quadratically varying stress over their rims. The model was further modified [6] for the case when applied load was characterized by the specified remote strain. Enriched finite element penalty functions method has been used in [7] for modeling interface crack with contact zones. An analytic solution for two equal length collinear strip yield cracks weakening an unbounded plate has been obtained in [8] using complex variable technique developed in [9]. The problem was further extended [10] to the strip yield analysis for two collinear unequal straight cracks weakening an infinite sheet. Mixed mode Dugdale–Barenblatt model has been established for a semi-infinite crack in an ideally elastic–plastic thin plate loaded by a pair of self-equilibrating concentrated forces at the crack rims in [11]. The problem investigated in the present paper is of an infinite elastic perfectly-plastic plate weakened by two unequal collinear straight quasi-static cracks. The rims of the crack open in Mode I type deformation by remotely applied uniform constant unidirectional tension consequently the rims of the cracks open in self-similar fashion forming a plastic zone ahead of each tip of the cracks. The tension at infinity is increased to the limit such that plastic zones developed at two adjacent ⇑ Corresponding author. E-mail addresses: [email protected] (R.R. Bhargava), [email protected], [email protected] (S. Hasan). 0307-904X/$ - see front matter Ó 2011 Elsevier Inc. All rights reserved. doi:10.1016/j.apm.2011.02.018
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interior tips of the two cracks get coalesced. Rims of the plastic zones, in turn, are subjected to quadratically varying normal cohesive stress distribution arresting the crack from further opening. The analytic expressions are obtained for the crack opening displacement at each tip of the cracks. A qualitative study has been presented for the COD at interior and exterior tips of the cracks in graphical form. Also load required to arrest the crack opening when the size of the plastic zone increases is computed and plotted. 2. Mathematical formulation According to the complex variable technique [9], the stress components Pij (i, j = x, y) may be expressed in terms of two complex potential functions U(z) and X(z) as
h i Pxx þ Pyy ¼ 2 UðzÞ þ UðzÞ ; Pyy iPxy
ð1Þ
¼ UðzÞ þ XðzÞ ðz zÞU0 ðzÞ;
ð2Þ
where bar over the function denotes its complex conjugate. Prime after the function denotes its differentiation with respect to argument. Consider an infinite homogeneous, isotropic plate which occupies xy-plane, and is cut along n collinear straight cracks Li S (i = 1, 2, 3 . . . n) lying on x-axis. Union of these cracks is denoted by L ¼ ni¼1 Li . If infinite boundary of the plate be kept unloaded and the rims of L be subjected to the stress distribution P yy ; P xy then using Eq. (2) following two Hilbert problems are obtained.
Uþ ðtÞ þ X ðtÞ ¼ Pþyy iP þxy ;
þ
U ðtÞ þ X ðtÞ ¼
Pyy
iP xy ;
ð3Þ on L;
ð4Þ
0
under the assumption Lim fyU ðt þ iyÞg ¼ 0. y!0
Superscripts + and denote the limiting value of the function when any point t on any of the cracks, other than end points, is approached from the positive y-plane (y > 0) and the negative y-plane (y < 0), respectively. Adding and subtracting Eqs. (3) and (4) following two Hilbert problems are obtained
fUðtÞ þ XðtÞgþ þ fUðtÞ þ XðtÞg ¼ 2pðtÞ;
ð5Þ
þ fUðtÞ XðtÞg fUðtÞ XðtÞg ¼ 2qðtÞ;
ð6Þ
2pðtÞ ¼ Pþyy þ Pyy iðP þxy þ Pxy Þ;
ð7Þ
where
2qðtÞ ¼
ðPþyy
Pyy Þ
iðPþxy
Pxy Þ:
ð8Þ
Solution of the Eq. (5) in absence of forces at infinity may be written as
UðzÞ þ XðzÞ ¼
1 piXðzÞ
Z L
X þ ðtÞpðtÞ 2Pn ðzÞ dt þ : tz XðzÞ
ð9Þ
And the solution of Eq. (6) may be written as
UðzÞ XðzÞ ¼
1 piXðzÞ
Z L
qðtÞ dt C0 ; tz
ð10Þ
where
Pn ðzÞ ¼ C 0 zn þ C 1 zn1 þ þ C n ;
ð11Þ
and
ðXðzÞ ¼
n Y
ðz ak Þ1=2 ðz bk Þ1=2 :
ð12Þ
k¼1
Constants Ci (i = 1, 2, 3 . . . n) are determined by boundary condition at infinity and single-valuedness of displacements on crack Lk(k = 1, 2, 3 . . . n). The end points of crack Lk are denoted by ak, bk, (k = 1, 2 . . . n). 3. Statement of the problem An infinite elastic perfectly-plastic plate, occupying xoy-plane, is weakened by two collinear unequal straight cracks L1 and L2. The crack L1 lies from [d1, 0] to [c1, 0] and L2 lies along [b1, 0] to [a1, 0] on the x-axis. At the infinite boundary of the plate uniform constant tension, ryy = r1 is applied which opens the rims of the cracks L1 and L2 in Mode I type
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deformations. Consequently the plastic zones developed ahead of the tips d1, c1, b1 and a1 of the cracks L1 and L2. The tension at boundary is increased to the limit such that plastic zones developed at the two adjacent interior tips c1 and b1 (of the cracks L1 and L2) get coalesced. This coalesced plastic zone from [c1, 0] to [b1, 0] is denoted by C1. The plastic zone from [d, 0] to [d1, 0] is denoted by C3 and C2 denotes the plastic zone [a1, 0] to [a, 0]. Each rim of the plastic zones C1, C2 and C3 is subjected to normal stress distribution Pyy ¼ t2 rye ; Pxy ¼ 0 to arrest the crack from further opening. Entire configuration is schematically depicted in Fig. 1. 4. Solution of the problem The problem stated in Section 3 is solved by superimposition of the solution of two auxiliary problems contributing towards the stress singularity at the tip of the cracks. These two problems are appropriately derived from the above problem. These problems are termed Problem I and Problem II. 4.1. Problem I An infinite elastic perfectly-plastic plate is weakened by a single crack C occupying the interval [d, 0] to [a, 0] on x-axis {formed of the union of C3, L1, C1, L2 and C2 which are defined in Section 3}. Boundary conditions of the problem are (i) The rims of the crack C are stress free. (ii) Infinite boundary of the plate is subjected to uniform stress distribution ryy = r1. (iii) Displacements are single-valued around the rims of crack C. Using boundary condition (i) and Eqs. (5)–(8) following two Hilbert problems are obtained
n
n o þ UI ðtÞ þ XI ðtÞ ¼ 0; n oþ n o UI ðtÞ XI ðtÞ UI ðtÞ XI ðtÞ ¼ 0:
UI ðtÞ þ XI ðtÞ
oþ
ð13Þ ð14Þ
Superscript I denotes that the function refer to Problem I. Solution of the Eqs. (13) and (14) using boundary condition (i) and (ii) and Eqs. (9)–(12) may be written as
UðzÞ þ XðzÞ ¼
2ðC 0 z þ C 1 Þ ; XðzÞ
UðzÞ XðzÞ ¼
r1 2
:
Using boundary condition (iii) the desired complex potential function for problem-I may be written as
r
UI ðzÞ ¼ pffiffiffiffiffiffiffiffiffiffiffi1pffiffiffiffiffiffiffiffiffiffiffi z 2 za zd
aþd r1 : 2 4
ð15Þ
4.2. Problem II An infinite elastic perfectly-plastic plate is weakened by a straight crack C as defined in Problem I. The boundary conditions of the problem are
Fig. 1. Configuration of the problem.
R.R. Bhargava, S. Hasan / Applied Mathematical Modelling 35 (2011) 3788–3796
(i) (ii) (iii) (iv)
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No stresses are acting at infinite boundary of the plate. 2 Rims of the Ci(i = 1, 2, 3), segment of the crack C, are subjected to stress distribution P yy ¼ t rye ; P xy ¼ 0. Rims of the Li(i = 1, 2), segment of the crack C, are stress free. Displacements are single-valued along the rims of crack C.
Boundary conditions (ii) and (iii) and Eqs. (2)–(8) give following two Hilbert problems
n
n o þ UII ðtÞ þ XII ðtÞ ¼ 2t2 rye ; n oþ n o UII ðtÞ XII ðtÞ UII ðtÞ XII ðtÞ ¼ 0;
UII ðtÞ þ XII ðtÞ
oþ
ð16Þ ð17Þ
S on C ¼ 3i¼1 Ci . Superscript II denote that quantities refer to Problem II. Desired complex potential UII(z) is written, solving Eqs. (16) and (17), as
UII ðzÞ ¼ where J ¼
rye
pffiffiffiffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffiffiffiffi ðJ zf2 Þ 2p z a z d
2f 1 f2 ðaþdÞf12 f0
rye ðf1 þ zf0 Þ pffiffiffiffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffiffiffiffi z2 rye za zdþ FðzÞ; 2p 2pi
ð18Þ
þ ðf3 þ 2adf1 ða þ dÞf0 Þ
fi ¼ Ri ðd1 Þ Ri ðdÞ þ Ri ðb1 Þ Ri ðc1 Þ þ Ri ðaÞ Ri ða1 Þ;
i ¼ 0; 1; 2; 3;
! ! t2 5ða þ dÞt 5ða þ dÞ2 2ad pffiffiffiffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffiffiffiffi ða þ dÞ 5ða þ dÞ2 1 2t þ a þ d ; at td þ 3ad sin 12 3 4 ad 3 8 4
R3 ðtÞ ¼
! t 3ða þ dÞ pffiffiffiffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffiffiffiffi 3ða þ dÞ2 ad 1 2t þ a þ d R2 ðtÞ ¼ at td sin ; 2 4 2 ad 8 pffiffiffiffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffiffiffiffi a þ d 1 2t þ a þ d sin ; R1 ðtÞ ¼ a t t d 2 ad
R0 ðtÞ ¼ sin
1
2t þ a þ d ; ad
2
3 1 1 ðazÞðd1 dÞþðad1 ÞðzdÞ ðazÞðb1 dÞþðab1 ÞðzdÞ pffiffiffiffiffiffipffiffiffiffiffiffipffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffi tanh pffiffiffiffiffiffipffiffiffiffiffiffipffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffi 7 6 tanh 2 az zd ad1 d1 d 2 az zd ab1 b1 d 7 6 7: FðzÞ ¼ 6 4 5 1 ðazÞðc1 dÞþðac1 ÞðzdÞ 1 ðazÞða1 dÞþðaa1 ÞðzdÞ þtanh pffiffiffiffiffiffipffiffiffiffiffiffipffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffi þ tanh pffiffiffiffiffiffipffiffiffiffiffiffipffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffi 2 az zd ac1
c1 d
2 az zd aa1
ð19Þ
a1 d
5. Plastic zone length Plastic zone length developed at each exterior tip of two cracks is obtained using the condition that the stresses remain finite through out the plate.
UI ðzÞ þ UII ðzÞ–1: Consequently following two conditions are obtained at the outer tip of the bigger crack i.e. a
r1 a2 rye
¼
2ðJ af2 Þ
pðd aÞa2
;
ð20Þ
and at the outer tip of the smaller crack i.e. d
r1 2ðJ df2 Þ ¼ : 2 rye pða dÞd2
d
ð21Þ
Values of a and d are determined from above equations for prescribed t2rr1 ; a1 ; b1 ; c1 ; d1 . ye The plastic zone length at the tip a1 and d1 are calculated from ja1 aj and jd d1j, respectively. 6. Crack opening displacement Crack opening displacement (COD) at each of the tip is calculated by the method given in [2].The parting of the crack face is given by
D¼
Z 4 Im Zdz ðfor plane stress caseÞ E
ð22Þ
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where Westergaard stress function, Z = Non-singular term of 2U(z), E denotes Young’s modulus and rye/E = .01. The crack opening displacement (COD) at each of the tip is given by. dt ¼ lim D; t denotes the crack tip. z!t COD at the crack tip a1
2
da1
3 g 0 ða1 Þ þ 13 g 2 ða1 Þ G1 g 1 ða1 Þ þ aþd g 0 ða1 Þ f0 þ G2 g 3 ða1 Þ þ G3 g 4 ða1 Þ 2 7 6 3 3 7 6 a dÞþðad1 Þða1 dÞ 1 Þðd p1ffiffiffiffiffiffiffi ffipffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffi 7 6 þ 31 d31 tanh1 pðaa ffiffiffiffiffiffiffi ffi 7 6 2 aa1 a1 d ad1 d1 d 7 6 4rye 6 7; 3 3 ¼ a1 1 b1 ðaa1 Þðb1 dÞþðab1 Þða1 dÞ 7 6 pE 6 þ 3 3 tanh pffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffi 7 2 aa1 a1 d ab1 b1 d 7 6 7 6 5 4 a31 c31 1 ðaa1 Þðc1 dÞþðac1 Þða1 dÞ 3 3 tanh pffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffi a d ac c d 2 aa 1
1
1
1
COD at the crack tip b1
2
db1
3 g 0 ðb1 Þ þ 13 g 2 ðb1 Þ G1 g 1 ðb1 Þ þ aþd g 0 ðb1 Þ f0 þ G2 g 3 ðb1 Þ þ G3 g 4 ðb1 Þ 2 7 6 3 3 7 6 ðab1 Þðd1 dÞþðad1 Þðb1 dÞ ffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffi 7 6 þ b31 d31 tanh1 p 7 6 2 ab1 b1 d ad1 d1 d 7 4rye 6 7; 6 3 3 ¼ c 1 b ðab1 Þðc1 dÞþðac1 Þðb1 dÞ 7 1 1 pE 6 p ffiffiffiffiffiffiffi ffi p ffiffiffiffiffiffiffiffi p ffiffiffiffiffiffiffi ffi pffiffiffiffiffiffiffiffi 7 6 3 3 tanh 2 ab1 b1 d ac1 c1 d 7 6 7 6 5 4 a31 b31 1 ðab1 Þða1 dÞþðaa1 Þðb1 dÞ pffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffi 3 3 tanh 2
ab1
b1 d aa1
a1 d
COD at the crack tip c1
2
3 g 0 ðc1 Þ þ 13 g 2 ðc1 Þ G1 g 1 ðc1 Þ þ aþd g 0 ðc1 Þ f0 þ G2 g 3 ðc1 Þ þ G3 g 4 ðc1 Þ 2 7 6 3 3 7 6 c dÞþðad1 Þðc1 dÞ 1 Þðd p1ffiffiffiffiffiffiffi ffipffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffi 7 6 þ 31 d31 tanh1 pðac ffiffiffiffiffiffiffi ffi 7 6 2 ac1 c1 d ad1 d1 d 7 4rye 6 7; 6 c 3 3 dc 1 ¼ b 1 ðac Þðb dÞþðab Þðc dÞ 7 6 1 p1ffiffiffiffiffiffiffiffipffiffiffiffiffiffiffi 1 ffip 1 ffiffiffiffiffiffiffiffi pE 6 þ 31 31 tanh pffiffiffiffiffiffiffiffi 7 2 ac1 c1 d ab1 b1 d 7 6 7 6 5 4 3 3 c1 a1 1 ðac1 Þða1 dÞþðaa1 Þðc1 dÞ 3 3 tanh pffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffi 2 ac1
c1 d aa1
a1 d
COD at the crack tip d1
2
dd1
3 g 0 ðd1 Þ þ 13 g 2 ðd1 Þ G1 g 1 ðd1 Þ þ aþd g 0 ðd1 Þ f0 þ G2 g 3 ðd1 Þ þ G3 g 4 ðd1 Þ 2 7 6 3 3 7 6 ðad1 Þðb1 dÞþðab1 Þðd1 dÞ ffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffi 7 6 þ d31 b31 tanh1 p 7 6 2 ad1 d1 d ab1 b1 d 7 4rye 6 7; 6 3 c3 ¼ 1 ðad1 Þðc1 dÞþðac1 Þðd1 dÞ d 7 pE 6 pffiffiffiffiffiffipffiffiffiffiffiffipffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffi 7 6 31 31 tanh az zd ac c d 2 1 1 7 6 7 6 5 4 a31 d31 1 ðad1 Þða1 dÞþðaa1 Þðd1 dÞ pffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffi 3 3 tanh 2
ad1
d1 d aa1
a1 d
where
pffiffiffiffiffiffiffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffiffiffiffiffiffiffi a d1 d1 d þ a b1 b1 d a c1 c1 d a a1 a1 d; pffiffiffiffiffiffiffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3G2 ¼ d1 a d1 d1 d þ b1 a b1 b1 d c1 a c1 c1 d a1 a a1 a1 d; pffiffiffiffiffiffiffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 3G3 ¼ d1 a d1 d1 d þ b1 a b1 b1 d c21 a c1 c1 d a21 a a1 a1 d; " # ð2z a dÞ pffiffiffiffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffiffiffiffi ða dÞ2 1 2z þ a þ d az zd g 0 ðzÞ ¼ ; sin 4 ad 8 ( ) qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 1 ð2z a dÞða þ dÞ ða þ dÞða dÞ2 1 a þ d 2t 2 ða zÞðz dÞ ; g 1 ðzÞ ¼ ðða zÞðz dÞÞ þ sin 3 8 ad 16 ! z 3ða þ dÞ pffiffiffiffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffiffiffiffi 3ða þ dÞ2 ad 1 2z þ a þ d az zd sin ; g 2 ðzÞ ¼ 2 4 2 ad 8 pffiffiffiffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffiffiffiffi a þ d 1 2z þ a þ d 1 2z þ a þ d g 3 ðzÞ ¼ a z z d ; g 4 ðzÞ ¼ sin : sin 2 ad ad
G1 ¼
R.R. Bhargava, S. Hasan / Applied Mathematical Modelling 35 (2011) 3788–3796
3793
7. Case study A qualitative study is presented for studying the behavior of crack opening and its arrest possibilities with respect to the parameter affecting the crack arrest. Load ratio used to denote the ratio of the applied at infinity to the load applied on the rims of the developed plastic zone. Crack ratio (CR) refers the ratio of the length of bigger crack to that of smaller crack. Fig. 2 depicts the variation of load ratio vs plastic zone length to crack length ratio at the outer tip of the bigger crack. For each crack ratio, the requirement of load to arrest the crack opening increases as plastic zone length is increased. It is interesting to note if the two cracks are equal in length, the load required to arrest the crack opening is the maximum. As the size of one crack is increased vis-à-vis the other crack then fewer load is required to arrest the crack opening. This is because the presence of the bigger crack itself affects the opening of the smaller one. Also the size of the plastic zone becomes smaller and smaller as one of the cracks becomes bigger as compare to other one, as expected. The same variation of load ratio (load applied at infinity to load prescribed at plastic zone) versus the plastic zone length to crack ratio is plotted at the exterior tip of the smaller crack in Fig. 3. In this case also it is observed that as the plastic zone
0.45 CR=1
σ∞ t σ ye
CR=2
2
CR=3
0.225
CR=4 CR=5
CR = length of bigger crack to length of smaller crack ratio 0 0.00
0.31
0.62
Plastic zone length/ crack length
CR=1
CR=2
CR=3
1
CR=4
CR=5
Fig. 2. Variation of load ratio with plastic zone length at the outer tip of bigger crack (say a) for inner plastic zone = .02.
σ∞ t σ ye 2
0.5
CR = length of bigger crack to length of smaller crack Ratio 0 0.00
0.31
0.62
Plastic zone length/ crack length Fig. 3. Variation of load ratio with plastic zone length at the outer tip of smaller crack (say d) for inner plastic zone = .02.
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size increases the load ratio required to arrest the opening of the crack also increases. But as the crack ratio increases the behavior here is opposite to that in Fig. 2. as the crack ratio is increased more load is required for arrest. Figs. 4–7 show the crack opening displacement with respect to plastic zone to crack length ratio: at the outer tip d1 of smaller crack, at the inner tip c1of smaller crack, at the inner tip b1 of bigger crack and at the outer tip a1 of bigger crack, respectively. Crack opening displacement is least at the outer tip of the smaller crack and also it may be noted that the variation at this tip is opposite that in case of others three crack tips. The crack ratio, CR, is increased from 1 to 5. The variations of COD at the inner tips of the smaller and bigger cracks are identical. They make a perfect opening for coalesce which is assumed to happen too. This proves the correctness of the analysis. It is to be noted in both the Figs. 5 and 6 that although COD increases with crack ratio but the increase is not much. Individually for each crack ratio the COD increases when plastic zone length to crack length ratio is increased. Fig. 7 depicts the variation of COD at the outer tip of the bigger crack. The variation in COD is almost linear but steep up. The crack starts opening for larger values of plastic zone to crack length ratio and for higher values of crack ratio.
0.15
COD / smaller crack length
CR = length of bigger crack to length of smaller crack ratio
CR=1
CR=2
0.075
CR=3
CR=4
CR=5
0 0.3
0
0.6
Plastic zone length/ crack length Fig. 4. Variation of crack opening displacement with plastic zone length at the outer tip of smaller crack (d1).
1
CR=2 CR=3 CR=1 CR=4
COD/ smaller crack length
CR=5
0.5
CR = length of bigger crack to length of smaller crack ratio
0 0
0.3
0.6
Plastic zone length/ crack length Fig. 5. Variation of crack opening displacement with plastic zone length at the inner tip of smaller crack (c1).
R.R. Bhargava, S. Hasan / Applied Mathematical Modelling 35 (2011) 3788–3796
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CR=2
1
CR=3
CR=1
CR=4
COD/ smaller crack length
CR=5
0.5
CR = length of bigger crack to length of smaller crack ratio
0 0
0.3
0.6
Plastic zone length/ crack length Fig. 6. Variation of crack opening displacement with plastic zone length at the inner tip of bigger crack (b1).
CR=5
1
CR=4
CR=3
COD/ smaller crack length
CR=2
CR=1
0.5
CR = length of bigger crack to length of smaller crack ratio 0 0
0.3
0.6
Plastic zone length/ crack length Fig. 7. Variation of crack opening displacement with plastic zone length at the outer tip of bigger crack (a1).
8. Conclusions Following are the conclusions: The problem of two unequal collinear straight cracks with a coalesced plastic zone has been investigated using complex variable technique. Analytic expressions are obtained to compute plastic zones developed at exterior tips of the smaller and bigger cracks and also crack opening displacements (COD). Variation of the load ratio at the exterior tips of smaller and bigger crack with respect to increasing crack ratio shows opposite behavior. This is because when bigger crack is much bigger then the smaller one, the smaller crack comes much in distress.
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The crack opening displacement plotted at the four tips of the two cracks also confirms that the interior tip of the crack COD has a coalescing behavior. And the behavior of the COD at the exterior tip of the smaller crack has same increasing trend. But as the crack ratio is increased, CR, it is noted that the variation of COD at the interior tip of the smaller crack is opposite to all other tips of the two cracks. The model is valid for small scale yielding.
Acknowledgements Authors are grateful to Prof. R. D. Bhargava [Senior Professor and Head (retd.), Dept. of Mathematics, Indian Institute of Technology Bombay, Mumbai] for his valuable suggestions during the course of this work. The authors are grateful to the referees and the editor for their suggestions, which improved the understandability of the paper. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11]
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