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Cracks emanating from circular hole or square hole in rectangular plate in tension
Engineering Fracture Mechanics 73 (2006) 1743–1754 www.elsevier.com/locate/engfracmech
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Cracks emanating from circular hole or square hole in rectangular plate in tension Xiangqiao Yan Research Laboratory on Composite Materials, Harbin Institute of Technology, Harbin 150001, PR China Received 10 January 2006; received in revised form 1 February 2006; accepted 1 February 2006 Available online 15 March 2006
Abstract A numerical analysis of cracks emanating from a circular hole (Fig. 1) or a square hole (Fig. 2) in rectangular plate in tension is performed by means of the displacement discontinuity method with crack-tip elements (a boundary element method) presented recently by the author. Detail solutions of the stress intensity factors (SIFs) of the two plane elastic crack problems are given, which can reveal the effect of geometric parameters of the cracked bodies on the SIFs. By comparing the SIFs of the two crack problems with those of the center crack in rectangular plate in tension (Fig. 3), in addition, an effect of the circular hole or the square hole on the SIFs of the center crack is discussed in detail. The numerical results reported here also illustrate that the boundary element method is simple, yet accurate for calculating the SIFs of complex crack problems in finite plate. Ó 2006 Published by Elsevier Ltd. Keywords: Stress intensity factor; Boundary element; Crack-tip element; Displacement discontinuity
1. Introduction Due to the stress concentration effect around the hole, cracks are likely to initiate at the hole under the action of fatigue loading. Consequently, a number of papers dealing with hole edge crack problems are available. Bowie [1] gave solutions of a circular hole with a single edge crack and a pair of symmetrical edge cracks in a plate under tension. Newman [2] by means of the boundary collocation method, and Murakami [3] by using the body force method performed analysis of the tension problem for an elliptical hole with symmetrical edge cracks. Tweed and Rooke [4] used the Mellin transform technique to make analysis of biaxial tensions for a branching crack emanating a circle hole. Isida et al. [5] made an analysis of a slant crack emanating from an elliptical hole under uniaxial tension and shear at infinity by using the body force method. In this paper, a numerical analysis of cracks emanating from a circular hole (Fig. 1) or a square hole (Fig. 2) in rectangular plate in tension is performed by means of the displacement discontinuity method with crack-tip
E-mail address: [email protected] 0013-7944/$ - see front matter Ó 2006 Published by Elsevier Ltd. doi:10.1016/j.engfracmech.2006.02.003
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σ
R 2H a
a 2W σ
Fig. 1. A fair of cracks emanating from a circular hole in rectangular plate in tension.
σ
c
c 2c
2H
a
a 2W σ
Fig. 2. A fair of cracks emanating from a square hole in rectangular plate in tension.
σ
H
2a
H
2W σ Fig. 3. Schematic of a center cracked rectangular plate in tension.
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elements (a boundary element method) presented recently by the author [6]. For the crack problem shown in Fig. 1, the documented Ref. [7] reported the detailed solutions to the SIFs when H/W is very large (larger than 2)(at this time, the effect of H/W on the SIFs is negligible). For cracks emanating from a square hole in infinite plate in tension, the documented Ref. [7] reported also the solutions to the SIFs. But to the author’s knowledge, detail solutions to the SIFs of the two plane elastic problems shown in Figs. 1 and 2 have not been obtained when H/W is smaller than 2. In this paper, specifically, detail solutions of the SIFs of the two plane elastic crack problems are given, which can reveal the effect of geometric parameters of the cracked bodies on the SIFs. By comparing the SIFs of the two crack problems with those of the center crack in rectangular plate in tension (Fig. 3), in addition, an effect of the circular hole or the square hole on the SIFs of the center crack is discussed in detail. The numerical results reported here illustrate also that the boundary element method is simple, yet accurate for calculating the SIFs of complex crack problems in finite plate. 2. Brief description of the numerical approach In this section, the displacement discontinuity method with crack-tip elements presented by the author [6] is described briefly. It consists of the constant displacement discontinuity element presented by Crouch and Starfield [8] and the crack-tip displacement discontinuity elements due to the author. 2.1. Brief introduction of constant displacement discontinuity method [8] The displacement discontinuity Di is defined as the difference in displacement between the two sides of the segment [8]: Dx ¼ ux ðx; 0 Þ ux ðx; 0þ Þ; Dy ¼ uy ðx; 0 Þ uy ðx; 0þ Þ.
ð1Þ
The solution to the subject problem is given by Crouch and Starfield [8]. The displacements and stresses can be written as ux ¼ Dx ½2ð1 mÞF 3 ðx; yÞ yF 5 ðx; yÞ þ Dy ½ð1 2mÞF 2 ðx; yÞ yF 4 ðx; yÞ; uy ¼ Dx ½ð1 2mÞF 2 ðx; yÞ yF 4 ðx; yÞ þ Dy ½2ð1 mÞF 3 ðx; yÞ yF 5 ðx; yÞ;
ð2Þ
and rxx ¼ 2GDx ½2F 4 ðx; yÞ þ yF 6 ðx; yÞ þ 2GDy ½F 5 ðx; yÞ þ yF 7 ðx; yÞ; ryy ¼ 2GDx ½yF 6 ðx; yÞ þ 2GDy ½F 5 ðx; yÞ yF 7 ðx; yÞ; rxy ¼ 2GDx ½F 5 ðx; yÞ þ yF 7 ðx; yÞ þ 2GDy ½yF 6 ðx; yÞ.
ð3Þ
G and v in these equations are shear modulus and Poisson’s ratio, respectively. Functions F2 through F7 are described in Ref. [8]. Eqs. (2) and (3) are used by Crouch and Starfield [8] to set up a constant displacement discontinuity boundary element method. 2.2. Crack-tip displacement discontinuity elements By using Eqs. (2) and (3), recently, the author [6] presented crack-tip displacement discontinuity elements, which can be classified as the left and the right crack-tip displacement discontinuity elements to deal with crack problems in general plane elasticity. The following gives basic formulas of the left crack-tip displacement discontinuity element. For the left crack-tip displacement discontinuity element, its displacement discontinuity functions are chosen as 1 1 aþn 2 aþn 2 Dx ¼ H s ; Dy ¼ H n ; ð4Þ a a
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where Hs and Hn are the tangential and normal displacement discontinuity quantities at the center of the element, respectively. Here, it is noted that the element has the same unknowns as the two-dimensional constant displacement discontinuity element. But it can be seen that the displacement discontinuity functions defined in (4) can model the displacement fields around the crack tip. The stress field determined by the displacement discontinuity functions (4) possesses r1/2 singularity around the crack tip. Based on Eqs. (2) and (3), the displacements and stresses at a point (x, y)) due to the left crack-tip displacement discontinuity element can be obtained, ux ¼ H s ½2ð1 mÞB3 ðx; yÞ yB5 ðx; yÞ þ H n ½ð1 2mÞB2 ðx; yÞ yB4 ðx; yÞ; uy ¼ H s ½ð1 2mÞB2 ðx; yÞ yB4 ðx; yÞ þ H n ½2ð1 mÞB3 ðx; yÞ yB5 ðx; yÞ;
ð5Þ
and rxx ¼ 2GH s ½2B4 ðx; yÞ þ yB6 ðx; yÞ þ 2GH n ½B5 ðx; yÞ þ yB7 ðx; yÞ; ryy ¼ 2GH s ½yB6 ðx; yÞ þ 2GH n ½B5 ðx; yÞ yB7 ðx; yÞ; rxy ¼ 2GH s ½B5 ðx; yÞ þ yB7 ðx; yÞ þ 2GH n ½yB6 ðx; yÞ;
ð6Þ
where functionsB2 through B7 are described in Ref. [6]. 2.3. Implementation of the present numerical approach Crouch and Starfield [8] used Eqs. (2) and (3) to set up constant displacement discontinuity boundary element method. Similarly, we can use Eqs. (5) and (6) to set up boundary element equations associated with the crack-tip elements. The constant displacement discontinuity element together with the crack-tip elements is combined easily to form a very effective numerical approach for calculating the SIFs of general plane cracks. In the boundary element implementation, the left or the right crack-tip element is placed locally at the corresponding left or right crack tip on top of the constant displacement discontinuity elements that cover the entire crack surface and the other boundaries. The method is called a displacement discontinuity method with cracktip elements. 3. Computational formulas of the stress intensity factors and a test example The objective of many analyses of linear elastic crack problems is to obtain the SIFs KI and KII. Based on the displacement field around the crack tip, the following formulas exist pffiffiffiffiffiffi pffiffiffiffiffiffi G 2p G 2p 0:5 limfDy ðrÞ=r g; K II ¼ limfDx ðrÞ=r0:5 g; KI ¼ ð7Þ 4ð1 mÞ r!0 4ð1 mÞ r!0 where Dy(r) and Dx(r) are the normal and shear components of displacement discontinuity at a distance r from the crack tip(s). For the practical purpose, the limits in Eq. (7) can be approximated by simply evaluating the expression for a fixed value of r, small in relation to the size of the crack. By means of the crack-tip displacement discontinuity functions defined in Eq. (4), thus, the approximate formulas of the stress intensity factors KI and KII can be obtained by letting r in Eq. (7) be a, a half length of crack-tip element. pffiffiffiffiffiffi pffiffiffiffiffiffi 2pGH n 2pGH s pffiffiffi ; K II ¼ pffiffiffi . KI ¼ ð8Þ 4ð1 mÞ a 4ð1 mÞ a To illustrate the accuracy of the numerical approach for calculating the SIFs of a complex plane crack problem in finite body, a test example is given here. A center cracked rectangular plate in tension (see Fig. 3) is analyzed by means of the boundary element method. In this analysis, the symmetric conditions of the crack body are used. The following cases are considered a=W ¼ 0:1; 0:2; 0:3; 0:4; 0:5; 0:6; 0:7; H =W ¼ 0:4; 0:5; 0:6; 0:7; 0:8; 0:9; 1:0; 1:5; 2:0;
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Table 1 Normalized SIFs of center cracked rectangular plate in tension a/W
0.1 0.2 0.3 0.4 0.5 0.6 0.7
H/W 0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.5
2.0
1.0613 1.2454 1.5078 1.8302 2.2320 2.7838 3.6246
1.0379 1.1650 1.3594 1.6156 1.9510 2.4046 3.0179
1.0252 1.1219 1.2753 1.4856 1.7605 2.1121 2.5410
1.0186 1.0958 1.2201 1.3918 1.6117 1.8797 2.1935
1.0145 1.0775 1.1789 1.3180 1.4935 1.7040 1.9547
1.0116 1.0638 1.1468 1.2606 1.4034 1.5770 1.7953
1.0101 1.0531 1.1223 1.2167 1.3371 1.4885 1.6906
1.0053 1.0302 1.0699 1.1280 1.2110 1.3199 1.5176
1.0039 1.0269 1.0629 1.1170 1.1964 1.3150 1.5004
Table 2 Normalized SIFs of center cracked rectangular plate in tension reported in Ref. [7] a/W
0.1 0.2 0.3 0.4 0.5 0.6 0.7
H/W 0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.5
1
1.069 1.256 1.520 1.843 2.247 2.806 3.67
1.046 1.175 1.371 1.629 1.967 2.424 3.04
1.033 1.130 1.285 1.497 1.773 2.123 2.55
1.026 1.103 1.228 1.400 1.619 1.883 2.19
1.021 1.083 1.184 1.323 1.496 1.702 1.94
1.017 1.067 1.150 1.262 1.403 1.572 1.78
1.014 1.055 1.123 1.216 1.334 1.481 1.68
1.007 1.029 1.066 1.122 1.203 1.32 –
1.006 1.026 1.058 1.109 1.187 1.303 1.488
Regarding discretization, the number of elements on a half of the crack is 20 and the other boundaries are discretized according to the limitationpthat ffiffiffiffiffiffi all boundary elements have approximately the same length [6]. The calculated SIFs normalized by r pa are given in Table 1. For the comparison purpose, Table 2 lists the numerical results reported in Ref. [7]. It is found from Tables 1 and 2 that the present numerical results are in excellent agreement with those reported in Ref. [7]. 4. Numerical results and discussion In this section, the plane elastic crack problems shown in Figs. 1 and 2 are analyzed by using the displacement discontinuity method with crack-tip elements. 4.1. Cracks emanating from a circular hole in rectangular plate in tension For the purpose of illustrating the accuracy of the numerical results obtained in this subsection, first the following two cases are analyzed for the plane elastic crack problem shown in Fig. 1, W =R ¼ H =R ¼ 10;
a=R ¼ 1:2
W =R ¼ H =R ¼ 10;
a=R ¼ 7.
and Obviously, the former can be regarded approximately as cracks emanating from a circular hole in an infinite plate in tension. While the latter can be regarded approximately as a center crack in a finite square plate in tension, i.e. the effect of the circular hole on the SIFs can be neglected. Regarding discretization, the number of boundary elements on a quarter of the circular hole is 50 and the other boundaries are discretized according to the limitation pffiffiffiffiffiffithat all boundary elements have approximately the same length [6]. The calculated SIFs normalized by r pa for the former and the latter are given, respectively, in Tables 3 and 4. For the comparison purpose, Tables 3 and 4 also list the numerical results reported in Ref. [7]. From Tables 3 and 4, it is found that the numerical results are in very good agreement with those reported in Ref. [7].
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Table 3 Comparison of normalized SIF
a/R = 1.2
Finite plate
Infinite plate [7]
1.0140
0.9851
Table 4 Comparison of normalized SIF
a/R = 7(a/W = 0.7)
Present
Ref [7]
1.7148
1.68
Secondly, we study in detail the plane elastic crack problem shown in Fig. 1. The following cases are considered: a=R ¼ 1:02; 1:04; 1:06; 1:08; 1:1; 1:15; 1:2; 1:5; 2:0; H =W ¼ 0:5; 0:6; 0:7; 0:8; 0:9; 1:0; 1:5; 2:0; a=W ¼ 0:2; 0:3; 0:4; 0:5; 0:6; 0:7. Regarding discretization, the number of boundary elements on a quarter of the circular hole shown in Fig. 1 is varied with a/Rand given in Table 5 and the other boundaries are discretized according to the limitation that pffiffiffiffiffiffi all boundary elements have approximately the same length. The calculated SIFs normalized by r pa are given in Tables 6–11. Now we compare the SIFs listed in Tables 1 and 6–11 to study the effect of the circular hole on the SIFs of the center crack. For this, we denote a/R, a/W and H/W by ar, aw and Hw, respectively, and denote the SIFs listed in Tables 1 and 6–11 by KIcc and K 0Icc , respectively, which can be expressed as K Icc ¼ K Icc ðH w Þ and K 0Icc ¼ K 0Icc ðar ; H w Þ for a given aw. From Tables 1 and 6–11, we find: (1) With increase of ar, K 0Icc fast increases monotonously regardless of the size of aw, and when ar reaches a certain value arc, K 0Icc equals KIcc. Generally, arc depends on Hw for a given aw. For aw = 0.3, for example, arc when Hw = 0.5 and 2.0 is about 1.12 and 1.19, respectively;
Table 5 Variation of number of elements on a quarter of the circular hole shown in Fig. 2 with a/R a/R
Number of elements
1.02
1.04
1.06
1.08
1.1
1.15
1.2
1.5
2.0
250
200
167
175
160
133
100
100
50
Table 6 Normalized SIFs of cracks emanating from a circular hole in rectangular plate in tension (a/W = 0.2) a/R
1.02 1.04 1.06 1.08 1.1 1.15 1.2 1.5 2.0
H/W 0.5
0.6
0.7
0.8
0.9
1.0
0.5638 0.7631 0.8913 0.9822 1.0503 1.1596 1.2212 1.2963 1.2627
0.5362 0.7263 0.8485 0.9357 1.0009 1.1058 1.1650 1.2379 1.2074
0.5201 0.7048 0.8238 0.9088 0.9723 1.0749 1.1326 1.2037 1.1738
0.5083 0.6890 0.8058 0.8893 0.9517 1.0527 1.1093 1.1795 1.1506
0.4982 0.6758 0.7907 0.8730 0.9344 1.0342 1.0903 1.1598 1.1321
0.4894 0.6642 0.7776 0.8586 0.9195 1.0182 1.0740 1.1434 1.1167
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Table 7 Normalized SIFs of cracks emanating from a circular hole in rectangular plate in tension (a/W = 0.3) a/R
1.02 1.04 1.06 1.08 1.1 1.15 1.2 1.5 2.0
H/W 0.5
0.6
0.7
0.8
0.9
1.0
1.5
2.0
0.7197 0.9692 1.1272 1.2376 1.3197 1.4493 1.5206 1.5977 1.5380
0.6680 0.8991 1.0454 1.1477 1.2236 1.3434 1.4088 1.4786 1.4239
0.6353 0.8554 0.9943 1.0921 1.1639 1.2778 1.3398 1.4034 1.3512
0.6084 0.8194 0.9531 1.0472 1.1166 1.2262 1.2857 1.3458 1.2966
0.5836 0.7870 0.9161 1.0073 1.0745 1.1810 1.2393 1.2982 1.2520
0.5614 0.7578 0.8832 0.9719 1.0374 1.1417 1.1990 1.2580 1.2155
0.5022 0.6807 0.7963 0.8781 0.9399 1.0391 1.0952 1.1603 1.1315
0.4902 0.6653 0.7795 0.8595 0.9212 1.0196 1.0761 1.1424 1.1177
Table 8 Normalized SIFs of cracks emanating from a circular hole in rectangular plate in tension (a/W = 0.4) a/R
1.02 1.04 1.06 1.08 1.1 1.15 1.2 1.5 2.0
H/W 0.5
0.6
0.7
0.8
0.9
1.0
1.5
2.0
0.9642 1.2922 1.4950 1.6325 1.7348 1.8924 1.9765 2.0493 1.9444
0.8913 1.1897 1.3735 1.4982 1.5897 1.7298 1.8030 1.8574 1.7565
0.8376 1.1168 1.2883 1.4045 1.4895 1.6183 1.6846 1.7242 1.6253
0.7849 1.0469 1.2083 1.3180 1.3977 1.5188 1.5806 1.6129 1.5193
0.7321 0.9784 1.1305 1.2349 1.3102 1.4255 1.4848 1.5161 1.4320
0.6839 0.9161 1.0605 1.1599 1.2323 1.3433 1.4013 1.4359 1.3622
0.5635 0.7599 0.8850 0.9726 1.0378 1.1410 1.1975 1.2515 1.2095
0.5430 0.7336 0.8557 0.9408 1.0052 1.1072 1.1640 1.2223 1.1870
Table 9 Normalized SIFs of cracks emanating from a circular hole in rectangular plate in tension (a/W = 0.5) a/R
1.02 1.04 1.06 1.08 1.1 1.15 1.2 1.5 2.0
H/W 0.5
0.6
0.7
0.8
0.9
1.0
1.5
2.0
1.3305 1.8697 2.1461 2.3034 2.4334 2.6202 2.7130 2.7502 2.5642
1.2969 1.7119 1.9574 2.1105 2.2249 2.3883 2.4673 2.4727 2.2739
1.2134 1.5953 1.8202 1.9612 2.0645 2.2085 2.2741 2.2448 2.0434
1.1108 1.4604 1.6666 1.7973 1.8910 2.0215 2.0786 2.0345 1.8483
0.9981 1.3168 1.5054 1.6270 1.7135 1.8349 1.8893 1.8509 1.6894
0.8959 1.1851 1.3593 1.4727 1.5542 1.6705 1.7244 1.7019 1.5674
0.6601 0.8836 1.0230 1.1185 1.1887 1.2962 1.3525 1.3859 1.3225
0.6258 0.8395 0.9739 1.0661 1.1347 1.2412 1.2980 1.3411 1.2895
(2) With continuous increase of ar, K 0Icc increases slowly and reaches its maximum K 0Iccm at some value arm; (3) Continuous increase of ar leads to K 0Icc to decrease slowly and K 0Icc approaches KIcc when ar is large enough. Fig. 4 shows the normalized SIFs of cracks emanating from a circular hole in infinite plate in tension (Data in Fig. 4 are from Ref. [7, p. 240]), from which it can be seen that arc and arm are approximately equal to 1.25 and 1.50, respectively. Fig. 5 gives the normalized SIFs of cracks emanating from a circular hole in rectangular plate in tension as aw = 0.4, Hw = 0.5, from which it is found that arc and arm are approximately equal to 1.078
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Table 10 Normalized SIFs of cracks emanating from a circular hole in rectangular plate in tension (a/W = 0.6) a/R
1.02 1.04 1.06 1.08 1.1 1.15 1.2 1.5 2.0
H/W 0.6
0.7
0.8
0.9
1.0
1.5
2.0
2.0121 2.7902 3.1451 3.3150 3.4585 3.6377 3.7076 3.5663 3.1396
1.9728 2.5592 2.8781 3.0503 3.1772 3.3267 3.3752 3.1503 2.7054
1.7724 2.2861 2.5688 2.7237 2.8339 2.9577 2.9887 2.7356 2.3377
1.5221 1.9659 2.2158 2.3533 2.4541 2.5665 2.5951 2.3769 2.0546
1.2886 1.6742 1.8941 2.0227 2.1151 2.2259 2.2610 2.1019 1.8514
0.8130 1.0775 1.2380 1.3449 1.4219 1.5350 1.5898 1.5841 1.4856
0.7565 1.0055 1.1586 1.2610 1.3361 1.4480 1.5050 1.5186 1.4412
Table 11 Normalized SIFs of cracks emanating from a circular hole in rectangular plate in tension (a/W = 0.7) a/R
1.02 1.04 1.06 1.08 1.1 1.15 1.2 1.5 2.0
H/W 0.7
0.8
0.9
1.0
1.5
2.0
3.4981 4.8000 5.3240 5.4848 5.6563 5.7862 5.7663 4.9424 3.8223
3.3016 4.2044 4.6400 4.7955 4.9248 4.9790 4.9060 4.0108 3.0863
2.6998 3.4156 3.7702 3.9096 4.0106 4.0495 3.9809 3.2555 2.5789
2.0987 2.6592 2.9594 3.0934 3.1941 3.2602 3.2300 2.7286 2.2495
1.0654 1.3932 1.5885 1.7120 1.8008 1.9204 1.9692 1.8872 1.7306
0.9741 1.2786 1.4618 1.5803 1.6659 1.7856 1.8391 1.7956 1.6745
1.1 1.0
Normalized SIFs
0.9 0.8 0.7 0.6 0.5 0.4 0.3 1.0
1.5
2.0
2.5
3.0
3.5
4.0
a/R
Fig. 4. Normalized SIFs of cracks emanating from a circular hole in infinite plate in tension.
and 1.50, respectively. From Tables 6–11 and Fig. 4, arc and arm can be determined approximately for given aw and Hw and it is found that, as aw and Hw vary, arc changes from 1.0 to 1.25 but arm almost is a constant, 1.5. After introducing the parameters arc and arm, it can be concluded from Tables 6–11 that: (1) when ar is less than arc, the circular hole has a shielding effect on the crack, and the closer ar is to 1.0 the stronger the effect is; (2) when ar is larger than arc, the circular hole has an amplifying effect on the SIFs of the crack and the amplifying effect is the most obvious at arm.
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2.2
Normalized SIFs
2.0
1.8
1.6
1.4
1.2
1.0
0.8 1
2
3
4
5
6
7
a/R
Fig. 5. Normalized SIFs of cracks emanating from a circular hole in rectangular plate in tension (a/W = 0.4, H/W = 0.5).
4.2. Cracks emanating from a square hole in rectangular plate in tension In this subsection, we study the plane elastic crack problem shown in Fig. 2. The following cases are considered: a=c ¼ 1:02; 1:04; 1:06; 1:08; 1:1; 1:15; 1:2; 1:5; 2:0; 4:0; 7:0; H =W ¼ 0:5; 0:6; 0:7; 0:8; 0:9; 1:0; 1:5; 2:0; a=W ¼ 0:2; 0:3; 0:4; 0:5; 0:6; 0:7. Regarding discretization, the number of boundary elements on a quarter of the square hole shown in Fig. 2 is varied with a/c and given in Table 12 and the other boundaries are discretized according to the limitation that pffiffiffiffiffiffi all boundary elements have approximately the same length. The calculated SIFs normalized by r pa are given in Tables 13–18. Table 12 Variation of number of elements on a quarter of the square hole shown in Fig. 2 with a/c a/c
Number of elements
1.02
1.04
1.06
1.08
1.1
1.15
1.2
1.5
2.0
4.0
7.0
250
200
167
175
160
133
100
100
50
50
50
Table 13 Normalized SIFs of cracks emanating from a square hole in rectangular plate in tension (a/W = 0.2) a/c
1.02 1.04 1.06 1.08 1.1 1.15 1.2 1.5 2.0
H/W 0.5
0.6
0.7
0.8
0.9
1.0
1.2576 1.2882 1.3020 1.3086 1.3124 1.3140 1.3108 1.2779 1.2361
1.1995 1.2285 1.2413 1.2477 1.2512 1.2529 1.2499 1.2204 1.1832
1.1660 1.1944 1.2068 1.2130 1.2164 1.2177 1.2149 1.1863 1.1509
1.1422 1.1698 1.1823 1.1883 1.1917 1.1931 1.1902 1.1626 1.1287
1.1224 1.1498 1.1620 1.1680 1.1713 1.1729 1.1701 1.1434 1.1109
1.1054 1.1324 1.1448 1.1505 1.1541 1.1558 1.1533 1.1276 1.0963
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As done in the subsection above, we compare the SIFs listed in Tables 1 and 13–18 to study the effect of the square hole on the SIFs of the center crack. For this, we denote a/c, a/W and H/W by ac, aw and Hw, respectively, and denote the SIFs listed in Tables 1 13–18 by KIcc and KIsc, respectively, which can be expressed as KIcc = KIcc (Hw) and KIsc = KIsc (ac, Hw) for a given aw. From Tables 1 and 13–18, we find: (1) For any aw, KIsc(ac, Hw) > KIcc(Hw). Only when ac is large enough KIsc (ac, Hw) approaches KIcc(Hw). For aw = 0.5 and Hw = 0.5, for example, KIsc at ac = 7 is 1.9985 (see Fig. 7) which almost equals KIcc(Hw) (= 1.9510). (2) For any given aw there is acm at which KIsc(ac, Hw) has a maximum value. Generally, acm depends on aw and Hw. For aw = 0.5, for example, acm when Hw = 0.5 and 1.0 is 1.1 and 1.06.
Table 14 Normalized SIFs of cracks emanating from a square hole in rectangular plate in tension (a/W = 0.3) a/c
1.02 1.04 1.06 1.08 1.1 1.15 1.2 1.5 2.0
H/W 0.5
0.6
0.7
0.8
0.9
1.0
1.5
2.0
1.5711 1.6084 1.6253 1.6319 1.6367 1.6347 1.6267 1.5690 1.4940
1.4527 1.4863 1.5006 1.5069 1.5103 1.5086 1.5010 1.4499 1.3857
1.5711 1.6084 1.6253 1.6319 1.6367 1.6347 1.4237 1.3748 1.3161
1.3245 1.3543 1.3665 1.3721 1.3743 1.3720 1.3651 1.3177 1.2640
1.2763 1.3052 1.3168 1.3222 1.3242 1.3222 1.3157 1.2714 1.2217
1.2345 1.2625 1.2742 1.2795 1.2817 1.2799 1.2743 1.2329 1.1875
1.1243 1.1517 1.1643 1.1693 1.1730 1.1740 1.1714 1.1421 1.1100
1.1016 1.1293 1.1431 1.1475 1.1523 1.1541 1.1529 1.1259 1.0975
Table 15 Normalized SIFs of cracks emanating from a square hole in rectangular plate in tension (a/W = 0.4) a/c
1.02 1.04 1.06 1.08 1.1 1.15 1.2 1.5 2.0
H/W 0.5
0.6
0.7
0.8
0.9
1.0
1.5
2.0
2.0390 2.0880 2.1096 2.1192 2.1236 2.1187 2.1026 2.0037 1.8676
1.8617 1.9028 1.9197 1.9251 1.9279 1.9192 1.9030 1.8078 1.6903
1.7381 1.7743 1.7877 1.7914 1.7925 1.7818 1.7652 1.6717 1.5656
1.6303 1.6634 1.6747 1.6779 1.6775 1.6667 1.6503 1.5614 1.4657
1.5311 1.5622 1.5725 1.5756 1.5750 1.5647 1.5498 1.4688 1.3846
1.4440 1.4738 1.4840 1.4873 1.4868 1.4782 1.4655 1.3934 1.3205
1.2272 1.2554 1.2670 1.2715 1.2737 1.2718 1.2664 1.2245 1.1821
1.1893 1.2177 1.2304 1.2346 1.2380 1.2381 1.2344 1.1987 1.1620
Table 16 Normalized SIFs of cracks emanating from a square hole in rectangular plate in tension (a/W = 0.5) a/c
1.02 1.04 1.06 1.08 1.1 1.15 1.2 1.5 2.0
H/W 0.5
0.6
0.7
0.8
0.9
1.0
1.5
2.0
2.7294 2.7968 2.8320 2.8434 2.8551 2.8480 2.8232 2.6678 2.4205
2.5352 2.5861 2.6088 2.6116 2.6138 2.5925 2.5577 2.3766 2.1478
2.3478 2.3885 2.4035 2.3999 2.3978 2.3680 2.3296 2.1397 1.9322
2.1511 2.1850 2.1949 2.1892 2.1842 2.1517 2.1134 1.9332 1.7531
1.9554 1.9865 1.9935 1.9885 1.9827 1.9523 1.9189 1.7597 1.6099
1.7829 1.8119 1.8188 1.8155 1.8102 1.7856 1.7575 1.6242 1.5012
1.3813 1.4094 1.4199 1.4223 1.4228 1.4158 1.4054 1.3438 1.2854
1.3195 1.3480 1.3600 1.3630 1.3652 1.3616 1.3545 1.3049 1.2569
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Table 17 Normalized SIFs of cracks emanating from a square hole in rectangular plate in tension (a/W = 0.6) a/c
1.02 1.04 1.06 1.08 1.1 1.15 1.2 1.5 2.0
H/W 0.6
0.7
0.8
0.9
1.0
1.5
2.0
3.6928 3.7733 3.8080 3.8120 3.8145 3.7729 3.7059 3.3500 2.8682
3.4578 3.5130 3.5228 3.5101 3.4926 3.4172 3.3261 2.9111 2.4761
3.0973 3.1345 3.1323 3.1088 3.0845 2.9993 2.9064 2.5130 2.1559
2.7024 2.7291 2.7244 2.7002 2.6774 2.5994 2.5196 2.1911 1.9142
2.3526 2.3755 2.3728 2.3528 2.3351 2.2730 2.2107 1.9532 1.7421
1.6108 1.6390 1.6471 1.6463 1.6430 1.6270 1.6087 1.5149 1.4333
1.5127 1.5421 1.5524 1.5535 1.5530 1.5431 1.5314 1.4597 1.3962
Table 18 Normalized SIFs of cracks emanating from a square hole in rectangular plate in tension (a/W = 0.7) a/c
1.02 1.04 1.06 1.08 1.1 1.15 1.2 1.5 2.0
H/W 0.7
0.8
0.9
1.0
1.5
2.0
5.6584 5.7593 5.7835 5.7478 5.7166 5.5369 5.3084 4.2996 3.2969
5.0653 5.0905 5.0561 4.9786 4.9034 4.6606 4.4069 3.4613 2.7165
4.1905 4.1912 4.1409 4.0607 3.9870 3.7750 3.5687 2.8434 2.3218
3.3975 3.4000 3.3620 3.3017 3.2465 3.0914 2.9448 2.4285 2.0644
1.9630 1.9894 1.9929 1.9843 1.9760 1.9439 1.9108 1.7678 1.6534
1.8099 1.8389 1.8465 1.8433 1.8393 1.8200 1.7990 1.6945 1.6091
Fig. 6 shows the normalized SIFs of cracks emanating from a square hole in infinite plate in tension, from which it can be seen that acm is approximately equal to 1.15. The normalized SIFs as aw = 0.5 and Hw = 0.5 are shown in Fig. 7, from which it is found that acm is approximately equal to 1.1. From Tables 13–18, acm can be determined approximately for given aw and Hw. After introducing the parameter acm, it can be concluded from Tables 13–18 that the square hole has an amplifying effect on the SIFs of the center crack and the amplifying effect is the most obvious at acm.
1.07
Normalized SIFs
1.06 1.05 1.04 1.03 1.02 1.01 1.00 0.99 1
2
3
4
5
6
7
a/c
Fig. 6. Normalized SIFs of cracks emanating from a square hole in infinite plate in tension.
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X. Yan / Engineering Fracture Mechanics 73 (2006) 1743–1754 3.0
Normalized SIFs
2.8
2.6
2.4
2.2
2.0
1
2
3
4
5
6
7
a/c
Fig. 7. Normalized SIFs of cracks emanating from a square hole in rectangular plate in tension (a/W = 0.5, H/W = 0.5).
Acknowledgement Special thanks are due to the National Natural Science Foundation of China (No: 10272037) for supporting the present work. References [1] Bowie OL. Analysis of an infinite plate containing radial cracks originating at the boundary of an internal circular hole. J Math Phys 1956;35:60–71. [2] Newman Jr JC. An improved method of collocation for the stress analysis of cracked plates with various shaped boundaries. NASA TN 1971;D-6376:1–45. [3] Murakami Y. A method of stress intensity factor calculation for the crack emanating from an arbitrarily shaped hole or the crack in the vicinity of an arbitrarily shape hole. Trans Japan Soc Mech Engrs 1978;44(378):423–32. [4] Tweed J, Rooke DP. The distribution of stress near the tip of a radial crack at the edge of a circular hole. Int J Eng Sci 1973;11:1185–95. [5] Isida M, Nakamura Y. Edge cracks originating from an elliptical hole in a wide plate subjected to tension and in-plane shear. Trans Japan Soc Mech Engrs 1980;46:947–56. [6] Yan X. An efficient and accurate numerical method of SIFs calculation of a branched crack, ASME. J Appl Mech 2005;72(3):330–40. [7] Murakami Y. Stress intensity factors handbook. New York: Pergamon Press; 1987. [8] Murakami Y. A simple procedure for the accurate determination of stress intensity factors by finite element method. Engng Fract Mech 1976;8:643–55.
Note
Cracks emanating from circular hole or square hole in rectangular plate in tension Xiangqiao Yan Research Laboratory on Composite Materials, Harbin Institute of Technology, Harbin 150001, PR China Received 10 January 2006; received in revised form 1 February 2006; accepted 1 February 2006 Available online 15 March 2006
Abstract A numerical analysis of cracks emanating from a circular hole (Fig. 1) or a square hole (Fig. 2) in rectangular plate in tension is performed by means of the displacement discontinuity method with crack-tip elements (a boundary element method) presented recently by the author. Detail solutions of the stress intensity factors (SIFs) of the two plane elastic crack problems are given, which can reveal the effect of geometric parameters of the cracked bodies on the SIFs. By comparing the SIFs of the two crack problems with those of the center crack in rectangular plate in tension (Fig. 3), in addition, an effect of the circular hole or the square hole on the SIFs of the center crack is discussed in detail. The numerical results reported here also illustrate that the boundary element method is simple, yet accurate for calculating the SIFs of complex crack problems in finite plate. Ó 2006 Published by Elsevier Ltd. Keywords: Stress intensity factor; Boundary element; Crack-tip element; Displacement discontinuity
1. Introduction Due to the stress concentration effect around the hole, cracks are likely to initiate at the hole under the action of fatigue loading. Consequently, a number of papers dealing with hole edge crack problems are available. Bowie [1] gave solutions of a circular hole with a single edge crack and a pair of symmetrical edge cracks in a plate under tension. Newman [2] by means of the boundary collocation method, and Murakami [3] by using the body force method performed analysis of the tension problem for an elliptical hole with symmetrical edge cracks. Tweed and Rooke [4] used the Mellin transform technique to make analysis of biaxial tensions for a branching crack emanating a circle hole. Isida et al. [5] made an analysis of a slant crack emanating from an elliptical hole under uniaxial tension and shear at infinity by using the body force method. In this paper, a numerical analysis of cracks emanating from a circular hole (Fig. 1) or a square hole (Fig. 2) in rectangular plate in tension is performed by means of the displacement discontinuity method with crack-tip
E-mail address: [email protected] 0013-7944/$ - see front matter Ó 2006 Published by Elsevier Ltd. doi:10.1016/j.engfracmech.2006.02.003
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σ
R 2H a
a 2W σ
Fig. 1. A fair of cracks emanating from a circular hole in rectangular plate in tension.
σ
c
c 2c
2H
a
a 2W σ
Fig. 2. A fair of cracks emanating from a square hole in rectangular plate in tension.
σ
H
2a
H
2W σ Fig. 3. Schematic of a center cracked rectangular plate in tension.
X. Yan / Engineering Fracture Mechanics 73 (2006) 1743–1754
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elements (a boundary element method) presented recently by the author [6]. For the crack problem shown in Fig. 1, the documented Ref. [7] reported the detailed solutions to the SIFs when H/W is very large (larger than 2)(at this time, the effect of H/W on the SIFs is negligible). For cracks emanating from a square hole in infinite plate in tension, the documented Ref. [7] reported also the solutions to the SIFs. But to the author’s knowledge, detail solutions to the SIFs of the two plane elastic problems shown in Figs. 1 and 2 have not been obtained when H/W is smaller than 2. In this paper, specifically, detail solutions of the SIFs of the two plane elastic crack problems are given, which can reveal the effect of geometric parameters of the cracked bodies on the SIFs. By comparing the SIFs of the two crack problems with those of the center crack in rectangular plate in tension (Fig. 3), in addition, an effect of the circular hole or the square hole on the SIFs of the center crack is discussed in detail. The numerical results reported here illustrate also that the boundary element method is simple, yet accurate for calculating the SIFs of complex crack problems in finite plate. 2. Brief description of the numerical approach In this section, the displacement discontinuity method with crack-tip elements presented by the author [6] is described briefly. It consists of the constant displacement discontinuity element presented by Crouch and Starfield [8] and the crack-tip displacement discontinuity elements due to the author. 2.1. Brief introduction of constant displacement discontinuity method [8] The displacement discontinuity Di is defined as the difference in displacement between the two sides of the segment [8]: Dx ¼ ux ðx; 0 Þ ux ðx; 0þ Þ; Dy ¼ uy ðx; 0 Þ uy ðx; 0þ Þ.
ð1Þ
The solution to the subject problem is given by Crouch and Starfield [8]. The displacements and stresses can be written as ux ¼ Dx ½2ð1 mÞF 3 ðx; yÞ yF 5 ðx; yÞ þ Dy ½ð1 2mÞF 2 ðx; yÞ yF 4 ðx; yÞ; uy ¼ Dx ½ð1 2mÞF 2 ðx; yÞ yF 4 ðx; yÞ þ Dy ½2ð1 mÞF 3 ðx; yÞ yF 5 ðx; yÞ;
ð2Þ
and rxx ¼ 2GDx ½2F 4 ðx; yÞ þ yF 6 ðx; yÞ þ 2GDy ½F 5 ðx; yÞ þ yF 7 ðx; yÞ; ryy ¼ 2GDx ½yF 6 ðx; yÞ þ 2GDy ½F 5 ðx; yÞ yF 7 ðx; yÞ; rxy ¼ 2GDx ½F 5 ðx; yÞ þ yF 7 ðx; yÞ þ 2GDy ½yF 6 ðx; yÞ.
ð3Þ
G and v in these equations are shear modulus and Poisson’s ratio, respectively. Functions F2 through F7 are described in Ref. [8]. Eqs. (2) and (3) are used by Crouch and Starfield [8] to set up a constant displacement discontinuity boundary element method. 2.2. Crack-tip displacement discontinuity elements By using Eqs. (2) and (3), recently, the author [6] presented crack-tip displacement discontinuity elements, which can be classified as the left and the right crack-tip displacement discontinuity elements to deal with crack problems in general plane elasticity. The following gives basic formulas of the left crack-tip displacement discontinuity element. For the left crack-tip displacement discontinuity element, its displacement discontinuity functions are chosen as 1 1 aþn 2 aþn 2 Dx ¼ H s ; Dy ¼ H n ; ð4Þ a a
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X. Yan / Engineering Fracture Mechanics 73 (2006) 1743–1754
where Hs and Hn are the tangential and normal displacement discontinuity quantities at the center of the element, respectively. Here, it is noted that the element has the same unknowns as the two-dimensional constant displacement discontinuity element. But it can be seen that the displacement discontinuity functions defined in (4) can model the displacement fields around the crack tip. The stress field determined by the displacement discontinuity functions (4) possesses r1/2 singularity around the crack tip. Based on Eqs. (2) and (3), the displacements and stresses at a point (x, y)) due to the left crack-tip displacement discontinuity element can be obtained, ux ¼ H s ½2ð1 mÞB3 ðx; yÞ yB5 ðx; yÞ þ H n ½ð1 2mÞB2 ðx; yÞ yB4 ðx; yÞ; uy ¼ H s ½ð1 2mÞB2 ðx; yÞ yB4 ðx; yÞ þ H n ½2ð1 mÞB3 ðx; yÞ yB5 ðx; yÞ;
ð5Þ
and rxx ¼ 2GH s ½2B4 ðx; yÞ þ yB6 ðx; yÞ þ 2GH n ½B5 ðx; yÞ þ yB7 ðx; yÞ; ryy ¼ 2GH s ½yB6 ðx; yÞ þ 2GH n ½B5 ðx; yÞ yB7 ðx; yÞ; rxy ¼ 2GH s ½B5 ðx; yÞ þ yB7 ðx; yÞ þ 2GH n ½yB6 ðx; yÞ;
ð6Þ
where functionsB2 through B7 are described in Ref. [6]. 2.3. Implementation of the present numerical approach Crouch and Starfield [8] used Eqs. (2) and (3) to set up constant displacement discontinuity boundary element method. Similarly, we can use Eqs. (5) and (6) to set up boundary element equations associated with the crack-tip elements. The constant displacement discontinuity element together with the crack-tip elements is combined easily to form a very effective numerical approach for calculating the SIFs of general plane cracks. In the boundary element implementation, the left or the right crack-tip element is placed locally at the corresponding left or right crack tip on top of the constant displacement discontinuity elements that cover the entire crack surface and the other boundaries. The method is called a displacement discontinuity method with cracktip elements. 3. Computational formulas of the stress intensity factors and a test example The objective of many analyses of linear elastic crack problems is to obtain the SIFs KI and KII. Based on the displacement field around the crack tip, the following formulas exist pffiffiffiffiffiffi pffiffiffiffiffiffi G 2p G 2p 0:5 limfDy ðrÞ=r g; K II ¼ limfDx ðrÞ=r0:5 g; KI ¼ ð7Þ 4ð1 mÞ r!0 4ð1 mÞ r!0 where Dy(r) and Dx(r) are the normal and shear components of displacement discontinuity at a distance r from the crack tip(s). For the practical purpose, the limits in Eq. (7) can be approximated by simply evaluating the expression for a fixed value of r, small in relation to the size of the crack. By means of the crack-tip displacement discontinuity functions defined in Eq. (4), thus, the approximate formulas of the stress intensity factors KI and KII can be obtained by letting r in Eq. (7) be a, a half length of crack-tip element. pffiffiffiffiffiffi pffiffiffiffiffiffi 2pGH n 2pGH s pffiffiffi ; K II ¼ pffiffiffi . KI ¼ ð8Þ 4ð1 mÞ a 4ð1 mÞ a To illustrate the accuracy of the numerical approach for calculating the SIFs of a complex plane crack problem in finite body, a test example is given here. A center cracked rectangular plate in tension (see Fig. 3) is analyzed by means of the boundary element method. In this analysis, the symmetric conditions of the crack body are used. The following cases are considered a=W ¼ 0:1; 0:2; 0:3; 0:4; 0:5; 0:6; 0:7; H =W ¼ 0:4; 0:5; 0:6; 0:7; 0:8; 0:9; 1:0; 1:5; 2:0;
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Table 1 Normalized SIFs of center cracked rectangular plate in tension a/W
0.1 0.2 0.3 0.4 0.5 0.6 0.7
H/W 0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.5
2.0
1.0613 1.2454 1.5078 1.8302 2.2320 2.7838 3.6246
1.0379 1.1650 1.3594 1.6156 1.9510 2.4046 3.0179
1.0252 1.1219 1.2753 1.4856 1.7605 2.1121 2.5410
1.0186 1.0958 1.2201 1.3918 1.6117 1.8797 2.1935
1.0145 1.0775 1.1789 1.3180 1.4935 1.7040 1.9547
1.0116 1.0638 1.1468 1.2606 1.4034 1.5770 1.7953
1.0101 1.0531 1.1223 1.2167 1.3371 1.4885 1.6906
1.0053 1.0302 1.0699 1.1280 1.2110 1.3199 1.5176
1.0039 1.0269 1.0629 1.1170 1.1964 1.3150 1.5004
Table 2 Normalized SIFs of center cracked rectangular plate in tension reported in Ref. [7] a/W
0.1 0.2 0.3 0.4 0.5 0.6 0.7
H/W 0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.5
1
1.069 1.256 1.520 1.843 2.247 2.806 3.67
1.046 1.175 1.371 1.629 1.967 2.424 3.04
1.033 1.130 1.285 1.497 1.773 2.123 2.55
1.026 1.103 1.228 1.400 1.619 1.883 2.19
1.021 1.083 1.184 1.323 1.496 1.702 1.94
1.017 1.067 1.150 1.262 1.403 1.572 1.78
1.014 1.055 1.123 1.216 1.334 1.481 1.68
1.007 1.029 1.066 1.122 1.203 1.32 –
1.006 1.026 1.058 1.109 1.187 1.303 1.488
Regarding discretization, the number of elements on a half of the crack is 20 and the other boundaries are discretized according to the limitationpthat ffiffiffiffiffiffi all boundary elements have approximately the same length [6]. The calculated SIFs normalized by r pa are given in Table 1. For the comparison purpose, Table 2 lists the numerical results reported in Ref. [7]. It is found from Tables 1 and 2 that the present numerical results are in excellent agreement with those reported in Ref. [7]. 4. Numerical results and discussion In this section, the plane elastic crack problems shown in Figs. 1 and 2 are analyzed by using the displacement discontinuity method with crack-tip elements. 4.1. Cracks emanating from a circular hole in rectangular plate in tension For the purpose of illustrating the accuracy of the numerical results obtained in this subsection, first the following two cases are analyzed for the plane elastic crack problem shown in Fig. 1, W =R ¼ H =R ¼ 10;
a=R ¼ 1:2
W =R ¼ H =R ¼ 10;
a=R ¼ 7.
and Obviously, the former can be regarded approximately as cracks emanating from a circular hole in an infinite plate in tension. While the latter can be regarded approximately as a center crack in a finite square plate in tension, i.e. the effect of the circular hole on the SIFs can be neglected. Regarding discretization, the number of boundary elements on a quarter of the circular hole is 50 and the other boundaries are discretized according to the limitation pffiffiffiffiffiffithat all boundary elements have approximately the same length [6]. The calculated SIFs normalized by r pa for the former and the latter are given, respectively, in Tables 3 and 4. For the comparison purpose, Tables 3 and 4 also list the numerical results reported in Ref. [7]. From Tables 3 and 4, it is found that the numerical results are in very good agreement with those reported in Ref. [7].
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Table 3 Comparison of normalized SIF
a/R = 1.2
Finite plate
Infinite plate [7]
1.0140
0.9851
Table 4 Comparison of normalized SIF
a/R = 7(a/W = 0.7)
Present
Ref [7]
1.7148
1.68
Secondly, we study in detail the plane elastic crack problem shown in Fig. 1. The following cases are considered: a=R ¼ 1:02; 1:04; 1:06; 1:08; 1:1; 1:15; 1:2; 1:5; 2:0; H =W ¼ 0:5; 0:6; 0:7; 0:8; 0:9; 1:0; 1:5; 2:0; a=W ¼ 0:2; 0:3; 0:4; 0:5; 0:6; 0:7. Regarding discretization, the number of boundary elements on a quarter of the circular hole shown in Fig. 1 is varied with a/Rand given in Table 5 and the other boundaries are discretized according to the limitation that pffiffiffiffiffiffi all boundary elements have approximately the same length. The calculated SIFs normalized by r pa are given in Tables 6–11. Now we compare the SIFs listed in Tables 1 and 6–11 to study the effect of the circular hole on the SIFs of the center crack. For this, we denote a/R, a/W and H/W by ar, aw and Hw, respectively, and denote the SIFs listed in Tables 1 and 6–11 by KIcc and K 0Icc , respectively, which can be expressed as K Icc ¼ K Icc ðH w Þ and K 0Icc ¼ K 0Icc ðar ; H w Þ for a given aw. From Tables 1 and 6–11, we find: (1) With increase of ar, K 0Icc fast increases monotonously regardless of the size of aw, and when ar reaches a certain value arc, K 0Icc equals KIcc. Generally, arc depends on Hw for a given aw. For aw = 0.3, for example, arc when Hw = 0.5 and 2.0 is about 1.12 and 1.19, respectively;
Table 5 Variation of number of elements on a quarter of the circular hole shown in Fig. 2 with a/R a/R
Number of elements
1.02
1.04
1.06
1.08
1.1
1.15
1.2
1.5
2.0
250
200
167
175
160
133
100
100
50
Table 6 Normalized SIFs of cracks emanating from a circular hole in rectangular plate in tension (a/W = 0.2) a/R
1.02 1.04 1.06 1.08 1.1 1.15 1.2 1.5 2.0
H/W 0.5
0.6
0.7
0.8
0.9
1.0
0.5638 0.7631 0.8913 0.9822 1.0503 1.1596 1.2212 1.2963 1.2627
0.5362 0.7263 0.8485 0.9357 1.0009 1.1058 1.1650 1.2379 1.2074
0.5201 0.7048 0.8238 0.9088 0.9723 1.0749 1.1326 1.2037 1.1738
0.5083 0.6890 0.8058 0.8893 0.9517 1.0527 1.1093 1.1795 1.1506
0.4982 0.6758 0.7907 0.8730 0.9344 1.0342 1.0903 1.1598 1.1321
0.4894 0.6642 0.7776 0.8586 0.9195 1.0182 1.0740 1.1434 1.1167
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Table 7 Normalized SIFs of cracks emanating from a circular hole in rectangular plate in tension (a/W = 0.3) a/R
1.02 1.04 1.06 1.08 1.1 1.15 1.2 1.5 2.0
H/W 0.5
0.6
0.7
0.8
0.9
1.0
1.5
2.0
0.7197 0.9692 1.1272 1.2376 1.3197 1.4493 1.5206 1.5977 1.5380
0.6680 0.8991 1.0454 1.1477 1.2236 1.3434 1.4088 1.4786 1.4239
0.6353 0.8554 0.9943 1.0921 1.1639 1.2778 1.3398 1.4034 1.3512
0.6084 0.8194 0.9531 1.0472 1.1166 1.2262 1.2857 1.3458 1.2966
0.5836 0.7870 0.9161 1.0073 1.0745 1.1810 1.2393 1.2982 1.2520
0.5614 0.7578 0.8832 0.9719 1.0374 1.1417 1.1990 1.2580 1.2155
0.5022 0.6807 0.7963 0.8781 0.9399 1.0391 1.0952 1.1603 1.1315
0.4902 0.6653 0.7795 0.8595 0.9212 1.0196 1.0761 1.1424 1.1177
Table 8 Normalized SIFs of cracks emanating from a circular hole in rectangular plate in tension (a/W = 0.4) a/R
1.02 1.04 1.06 1.08 1.1 1.15 1.2 1.5 2.0
H/W 0.5
0.6
0.7
0.8
0.9
1.0
1.5
2.0
0.9642 1.2922 1.4950 1.6325 1.7348 1.8924 1.9765 2.0493 1.9444
0.8913 1.1897 1.3735 1.4982 1.5897 1.7298 1.8030 1.8574 1.7565
0.8376 1.1168 1.2883 1.4045 1.4895 1.6183 1.6846 1.7242 1.6253
0.7849 1.0469 1.2083 1.3180 1.3977 1.5188 1.5806 1.6129 1.5193
0.7321 0.9784 1.1305 1.2349 1.3102 1.4255 1.4848 1.5161 1.4320
0.6839 0.9161 1.0605 1.1599 1.2323 1.3433 1.4013 1.4359 1.3622
0.5635 0.7599 0.8850 0.9726 1.0378 1.1410 1.1975 1.2515 1.2095
0.5430 0.7336 0.8557 0.9408 1.0052 1.1072 1.1640 1.2223 1.1870
Table 9 Normalized SIFs of cracks emanating from a circular hole in rectangular plate in tension (a/W = 0.5) a/R
1.02 1.04 1.06 1.08 1.1 1.15 1.2 1.5 2.0
H/W 0.5
0.6
0.7
0.8
0.9
1.0
1.5
2.0
1.3305 1.8697 2.1461 2.3034 2.4334 2.6202 2.7130 2.7502 2.5642
1.2969 1.7119 1.9574 2.1105 2.2249 2.3883 2.4673 2.4727 2.2739
1.2134 1.5953 1.8202 1.9612 2.0645 2.2085 2.2741 2.2448 2.0434
1.1108 1.4604 1.6666 1.7973 1.8910 2.0215 2.0786 2.0345 1.8483
0.9981 1.3168 1.5054 1.6270 1.7135 1.8349 1.8893 1.8509 1.6894
0.8959 1.1851 1.3593 1.4727 1.5542 1.6705 1.7244 1.7019 1.5674
0.6601 0.8836 1.0230 1.1185 1.1887 1.2962 1.3525 1.3859 1.3225
0.6258 0.8395 0.9739 1.0661 1.1347 1.2412 1.2980 1.3411 1.2895
(2) With continuous increase of ar, K 0Icc increases slowly and reaches its maximum K 0Iccm at some value arm; (3) Continuous increase of ar leads to K 0Icc to decrease slowly and K 0Icc approaches KIcc when ar is large enough. Fig. 4 shows the normalized SIFs of cracks emanating from a circular hole in infinite plate in tension (Data in Fig. 4 are from Ref. [7, p. 240]), from which it can be seen that arc and arm are approximately equal to 1.25 and 1.50, respectively. Fig. 5 gives the normalized SIFs of cracks emanating from a circular hole in rectangular plate in tension as aw = 0.4, Hw = 0.5, from which it is found that arc and arm are approximately equal to 1.078
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Table 10 Normalized SIFs of cracks emanating from a circular hole in rectangular plate in tension (a/W = 0.6) a/R
1.02 1.04 1.06 1.08 1.1 1.15 1.2 1.5 2.0
H/W 0.6
0.7
0.8
0.9
1.0
1.5
2.0
2.0121 2.7902 3.1451 3.3150 3.4585 3.6377 3.7076 3.5663 3.1396
1.9728 2.5592 2.8781 3.0503 3.1772 3.3267 3.3752 3.1503 2.7054
1.7724 2.2861 2.5688 2.7237 2.8339 2.9577 2.9887 2.7356 2.3377
1.5221 1.9659 2.2158 2.3533 2.4541 2.5665 2.5951 2.3769 2.0546
1.2886 1.6742 1.8941 2.0227 2.1151 2.2259 2.2610 2.1019 1.8514
0.8130 1.0775 1.2380 1.3449 1.4219 1.5350 1.5898 1.5841 1.4856
0.7565 1.0055 1.1586 1.2610 1.3361 1.4480 1.5050 1.5186 1.4412
Table 11 Normalized SIFs of cracks emanating from a circular hole in rectangular plate in tension (a/W = 0.7) a/R
1.02 1.04 1.06 1.08 1.1 1.15 1.2 1.5 2.0
H/W 0.7
0.8
0.9
1.0
1.5
2.0
3.4981 4.8000 5.3240 5.4848 5.6563 5.7862 5.7663 4.9424 3.8223
3.3016 4.2044 4.6400 4.7955 4.9248 4.9790 4.9060 4.0108 3.0863
2.6998 3.4156 3.7702 3.9096 4.0106 4.0495 3.9809 3.2555 2.5789
2.0987 2.6592 2.9594 3.0934 3.1941 3.2602 3.2300 2.7286 2.2495
1.0654 1.3932 1.5885 1.7120 1.8008 1.9204 1.9692 1.8872 1.7306
0.9741 1.2786 1.4618 1.5803 1.6659 1.7856 1.8391 1.7956 1.6745
1.1 1.0
Normalized SIFs
0.9 0.8 0.7 0.6 0.5 0.4 0.3 1.0
1.5
2.0
2.5
3.0
3.5
4.0
a/R
Fig. 4. Normalized SIFs of cracks emanating from a circular hole in infinite plate in tension.
and 1.50, respectively. From Tables 6–11 and Fig. 4, arc and arm can be determined approximately for given aw and Hw and it is found that, as aw and Hw vary, arc changes from 1.0 to 1.25 but arm almost is a constant, 1.5. After introducing the parameters arc and arm, it can be concluded from Tables 6–11 that: (1) when ar is less than arc, the circular hole has a shielding effect on the crack, and the closer ar is to 1.0 the stronger the effect is; (2) when ar is larger than arc, the circular hole has an amplifying effect on the SIFs of the crack and the amplifying effect is the most obvious at arm.
X. Yan / Engineering Fracture Mechanics 73 (2006) 1743–1754
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2.2
Normalized SIFs
2.0
1.8
1.6
1.4
1.2
1.0
0.8 1
2
3
4
5
6
7
a/R
Fig. 5. Normalized SIFs of cracks emanating from a circular hole in rectangular plate in tension (a/W = 0.4, H/W = 0.5).
4.2. Cracks emanating from a square hole in rectangular plate in tension In this subsection, we study the plane elastic crack problem shown in Fig. 2. The following cases are considered: a=c ¼ 1:02; 1:04; 1:06; 1:08; 1:1; 1:15; 1:2; 1:5; 2:0; 4:0; 7:0; H =W ¼ 0:5; 0:6; 0:7; 0:8; 0:9; 1:0; 1:5; 2:0; a=W ¼ 0:2; 0:3; 0:4; 0:5; 0:6; 0:7. Regarding discretization, the number of boundary elements on a quarter of the square hole shown in Fig. 2 is varied with a/c and given in Table 12 and the other boundaries are discretized according to the limitation that pffiffiffiffiffiffi all boundary elements have approximately the same length. The calculated SIFs normalized by r pa are given in Tables 13–18. Table 12 Variation of number of elements on a quarter of the square hole shown in Fig. 2 with a/c a/c
Number of elements
1.02
1.04
1.06
1.08
1.1
1.15
1.2
1.5
2.0
4.0
7.0
250
200
167
175
160
133
100
100
50
50
50
Table 13 Normalized SIFs of cracks emanating from a square hole in rectangular plate in tension (a/W = 0.2) a/c
1.02 1.04 1.06 1.08 1.1 1.15 1.2 1.5 2.0
H/W 0.5
0.6
0.7
0.8
0.9
1.0
1.2576 1.2882 1.3020 1.3086 1.3124 1.3140 1.3108 1.2779 1.2361
1.1995 1.2285 1.2413 1.2477 1.2512 1.2529 1.2499 1.2204 1.1832
1.1660 1.1944 1.2068 1.2130 1.2164 1.2177 1.2149 1.1863 1.1509
1.1422 1.1698 1.1823 1.1883 1.1917 1.1931 1.1902 1.1626 1.1287
1.1224 1.1498 1.1620 1.1680 1.1713 1.1729 1.1701 1.1434 1.1109
1.1054 1.1324 1.1448 1.1505 1.1541 1.1558 1.1533 1.1276 1.0963
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As done in the subsection above, we compare the SIFs listed in Tables 1 and 13–18 to study the effect of the square hole on the SIFs of the center crack. For this, we denote a/c, a/W and H/W by ac, aw and Hw, respectively, and denote the SIFs listed in Tables 1 13–18 by KIcc and KIsc, respectively, which can be expressed as KIcc = KIcc (Hw) and KIsc = KIsc (ac, Hw) for a given aw. From Tables 1 and 13–18, we find: (1) For any aw, KIsc(ac, Hw) > KIcc(Hw). Only when ac is large enough KIsc (ac, Hw) approaches KIcc(Hw). For aw = 0.5 and Hw = 0.5, for example, KIsc at ac = 7 is 1.9985 (see Fig. 7) which almost equals KIcc(Hw) (= 1.9510). (2) For any given aw there is acm at which KIsc(ac, Hw) has a maximum value. Generally, acm depends on aw and Hw. For aw = 0.5, for example, acm when Hw = 0.5 and 1.0 is 1.1 and 1.06.
Table 14 Normalized SIFs of cracks emanating from a square hole in rectangular plate in tension (a/W = 0.3) a/c
1.02 1.04 1.06 1.08 1.1 1.15 1.2 1.5 2.0
H/W 0.5
0.6
0.7
0.8
0.9
1.0
1.5
2.0
1.5711 1.6084 1.6253 1.6319 1.6367 1.6347 1.6267 1.5690 1.4940
1.4527 1.4863 1.5006 1.5069 1.5103 1.5086 1.5010 1.4499 1.3857
1.5711 1.6084 1.6253 1.6319 1.6367 1.6347 1.4237 1.3748 1.3161
1.3245 1.3543 1.3665 1.3721 1.3743 1.3720 1.3651 1.3177 1.2640
1.2763 1.3052 1.3168 1.3222 1.3242 1.3222 1.3157 1.2714 1.2217
1.2345 1.2625 1.2742 1.2795 1.2817 1.2799 1.2743 1.2329 1.1875
1.1243 1.1517 1.1643 1.1693 1.1730 1.1740 1.1714 1.1421 1.1100
1.1016 1.1293 1.1431 1.1475 1.1523 1.1541 1.1529 1.1259 1.0975
Table 15 Normalized SIFs of cracks emanating from a square hole in rectangular plate in tension (a/W = 0.4) a/c
1.02 1.04 1.06 1.08 1.1 1.15 1.2 1.5 2.0
H/W 0.5
0.6
0.7
0.8
0.9
1.0
1.5
2.0
2.0390 2.0880 2.1096 2.1192 2.1236 2.1187 2.1026 2.0037 1.8676
1.8617 1.9028 1.9197 1.9251 1.9279 1.9192 1.9030 1.8078 1.6903
1.7381 1.7743 1.7877 1.7914 1.7925 1.7818 1.7652 1.6717 1.5656
1.6303 1.6634 1.6747 1.6779 1.6775 1.6667 1.6503 1.5614 1.4657
1.5311 1.5622 1.5725 1.5756 1.5750 1.5647 1.5498 1.4688 1.3846
1.4440 1.4738 1.4840 1.4873 1.4868 1.4782 1.4655 1.3934 1.3205
1.2272 1.2554 1.2670 1.2715 1.2737 1.2718 1.2664 1.2245 1.1821
1.1893 1.2177 1.2304 1.2346 1.2380 1.2381 1.2344 1.1987 1.1620
Table 16 Normalized SIFs of cracks emanating from a square hole in rectangular plate in tension (a/W = 0.5) a/c
1.02 1.04 1.06 1.08 1.1 1.15 1.2 1.5 2.0
H/W 0.5
0.6
0.7
0.8
0.9
1.0
1.5
2.0
2.7294 2.7968 2.8320 2.8434 2.8551 2.8480 2.8232 2.6678 2.4205
2.5352 2.5861 2.6088 2.6116 2.6138 2.5925 2.5577 2.3766 2.1478
2.3478 2.3885 2.4035 2.3999 2.3978 2.3680 2.3296 2.1397 1.9322
2.1511 2.1850 2.1949 2.1892 2.1842 2.1517 2.1134 1.9332 1.7531
1.9554 1.9865 1.9935 1.9885 1.9827 1.9523 1.9189 1.7597 1.6099
1.7829 1.8119 1.8188 1.8155 1.8102 1.7856 1.7575 1.6242 1.5012
1.3813 1.4094 1.4199 1.4223 1.4228 1.4158 1.4054 1.3438 1.2854
1.3195 1.3480 1.3600 1.3630 1.3652 1.3616 1.3545 1.3049 1.2569
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Table 17 Normalized SIFs of cracks emanating from a square hole in rectangular plate in tension (a/W = 0.6) a/c
1.02 1.04 1.06 1.08 1.1 1.15 1.2 1.5 2.0
H/W 0.6
0.7
0.8
0.9
1.0
1.5
2.0
3.6928 3.7733 3.8080 3.8120 3.8145 3.7729 3.7059 3.3500 2.8682
3.4578 3.5130 3.5228 3.5101 3.4926 3.4172 3.3261 2.9111 2.4761
3.0973 3.1345 3.1323 3.1088 3.0845 2.9993 2.9064 2.5130 2.1559
2.7024 2.7291 2.7244 2.7002 2.6774 2.5994 2.5196 2.1911 1.9142
2.3526 2.3755 2.3728 2.3528 2.3351 2.2730 2.2107 1.9532 1.7421
1.6108 1.6390 1.6471 1.6463 1.6430 1.6270 1.6087 1.5149 1.4333
1.5127 1.5421 1.5524 1.5535 1.5530 1.5431 1.5314 1.4597 1.3962
Table 18 Normalized SIFs of cracks emanating from a square hole in rectangular plate in tension (a/W = 0.7) a/c
1.02 1.04 1.06 1.08 1.1 1.15 1.2 1.5 2.0
H/W 0.7
0.8
0.9
1.0
1.5
2.0
5.6584 5.7593 5.7835 5.7478 5.7166 5.5369 5.3084 4.2996 3.2969
5.0653 5.0905 5.0561 4.9786 4.9034 4.6606 4.4069 3.4613 2.7165
4.1905 4.1912 4.1409 4.0607 3.9870 3.7750 3.5687 2.8434 2.3218
3.3975 3.4000 3.3620 3.3017 3.2465 3.0914 2.9448 2.4285 2.0644
1.9630 1.9894 1.9929 1.9843 1.9760 1.9439 1.9108 1.7678 1.6534
1.8099 1.8389 1.8465 1.8433 1.8393 1.8200 1.7990 1.6945 1.6091
Fig. 6 shows the normalized SIFs of cracks emanating from a square hole in infinite plate in tension, from which it can be seen that acm is approximately equal to 1.15. The normalized SIFs as aw = 0.5 and Hw = 0.5 are shown in Fig. 7, from which it is found that acm is approximately equal to 1.1. From Tables 13–18, acm can be determined approximately for given aw and Hw. After introducing the parameter acm, it can be concluded from Tables 13–18 that the square hole has an amplifying effect on the SIFs of the center crack and the amplifying effect is the most obvious at acm.
1.07
Normalized SIFs
1.06 1.05 1.04 1.03 1.02 1.01 1.00 0.99 1
2
3
4
5
6
7
a/c
Fig. 6. Normalized SIFs of cracks emanating from a square hole in infinite plate in tension.
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X. Yan / Engineering Fracture Mechanics 73 (2006) 1743–1754 3.0
Normalized SIFs
2.8
2.6
2.4
2.2
2.0
1
2
3
4
5
6
7
a/c
Fig. 7. Normalized SIFs of cracks emanating from a square hole in rectangular plate in tension (a/W = 0.5, H/W = 0.5).
Acknowledgement Special thanks are due to the National Natural Science Foundation of China (No: 10272037) for supporting the present work. References [1] Bowie OL. Analysis of an infinite plate containing radial cracks originating at the boundary of an internal circular hole. J Math Phys 1956;35:60–71. [2] Newman Jr JC. An improved method of collocation for the stress analysis of cracked plates with various shaped boundaries. NASA TN 1971;D-6376:1–45. [3] Murakami Y. A method of stress intensity factor calculation for the crack emanating from an arbitrarily shaped hole or the crack in the vicinity of an arbitrarily shape hole. Trans Japan Soc Mech Engrs 1978;44(378):423–32. [4] Tweed J, Rooke DP. The distribution of stress near the tip of a radial crack at the edge of a circular hole. Int J Eng Sci 1973;11:1185–95. [5] Isida M, Nakamura Y. Edge cracks originating from an elliptical hole in a wide plate subjected to tension and in-plane shear. Trans Japan Soc Mech Engrs 1980;46:947–56. [6] Yan X. An efficient and accurate numerical method of SIFs calculation of a branched crack, ASME. J Appl Mech 2005;72(3):330–40. [7] Murakami Y. Stress intensity factors handbook. New York: Pergamon Press; 1987. [8] Murakami Y. A simple procedure for the accurate determination of stress intensity factors by finite element method. Engng Fract Mech 1976;8:643–55.