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A new method of plastic zone size determined based on maximum crack opening displacement
Engineering Fracture Mechanics 77 (2010) 2912–2918
Contents lists available at ScienceDirect
Engineering Fracture Mechanics journal homepage: www.elsevier.com/locate/engfracmech
Technical Note
A new method of plastic zone size determined based on maximum crack opening displacement Huang Yi, Chen Jingjie *, Liu Gang School of Naval Architecture, Dalian University of Technology, Dalian 116024, Liaoning, China
a r t i c l e
i n f o
Article history: Received 24 February 2010 Received in revised form 29 June 2010 Accepted 29 June 2010
Keywords: Plastic zone size Maximum crack opening displacement Center cracked plate Finite element analysis
a b s t r a c t This paper presents a new method to determine the crack-tip plastic zone size (Ry) in the center cracked plate in tension. The maximum crack opening displacement (MCOD) is used to estimate Ry in this method. Based on the series of calculation results by finite element analysis, the explicit expression for the crack-tip plastic zone size versus MCOD is fitted by least square method. The expression enables to eliminate the influences of the yield stress of material, crack geometry, plate sizes and transverse stress. Therefore, the presented method of Ry determined by MCOD is suitable to any center crack finite plate of any material under uniaxial or biaxial tension. Ó 2010 Elsevier Ltd. All rights reserved.
1. Introduction Methodologies of failure assessment of cracked component, based on elastic–plastic fracture mechanics (EPFM), are relatively well established [1–3]. In the EPFM, crack-tip opening displacement, crack-tip plastic zone size (Ry) and the range of stress intensity factor have been suggested for characterizing fatigue crack propagation behaviors. Ry has advantages over other parameters [4], which not only facilitates correlation of fast growing cracks but also has an obvious physical interpretation. Many researchers [4–6] consider that Ry is the most suitable parameter that can describe the propagation of fatigue cracks. So, it is important to determine the size of the crack-tip plastic zones when assessing the safety state of cracked component in practice. Several researchers have made attempts for the determination of the plastic zone. Irwin [7] and Dugdale [8] respectively proposed different models to estimate plastic zone size ahead of the crack. However, those models are usually restricted to plates of infinite extent with simple geometrical configurations of crack and do not consider the influence of plate width effect on Ry. In the present paper, the maximum crack opening displacement (MCOD) is introduced to replace the applied load to estimate the Ry for a center cracked plate subjected to biaxial tension loading, as can be seen in Fig. 1a. Then finite element method is employed to calculate series of Ry and corresponding MCOD for the specimens with different widths, different crack lengths and various yield stresses of material under several applied stress levels. A simple relationship between the normalized plastic zone size, Ry/a, and the normalized crack opening displacement, DuE=ars is achieved by the calculated results. Also, the effect of transverse stress on the relationship is considered through changing its value. It is noticed that the relationship is not affected by crack length, plate width, yield stress, applied load and transverse stress. * Corresponding author. Tel.: +86 15040629349; fax: +86 041184706350. E-mail address: [email protected] (C. Jingjie). 0013-7944/$ - see front matter Ó 2010 Elsevier Ltd. All rights reserved. doi:10.1016/j.engfracmech.2010.06.026
H. Yi et al. / Engineering Fracture Mechanics 77 (2010) 2912–2918
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Nomenclature Ry KI
crack-tip plastic zone size mode I stress intensity factor yield stress of material applied uniform tension load stress in x axis direction, and it is also called transverse stress stress in y axis direction ratio of biaxial stress, k = rx/ry half crack length effective half crack length, ae = a + Ry/2 maximum crack opening displacement (MCOD) Young’s modulus half-length of plate half-width of plate
rs r rx ry k a ae Du E L W
Fig. 1. A center cracked plate in tension. (a) The case to biaxial stress; (b) the case to uniaxial stress (transverse stress is zero); (c) the case to uniaxial stress (transverse is compressive stress, which is equal to tensile stress).
2. Theoretical consideration An expression of Ry for perfectly plastic materials under small scale yielding, when the cracked infinite plate is subject to the biaxial uniform tension load, has been derived [7] for plane stress conditions as follows:
Ry ¼
2 1 KI
p rs
ð1Þ
where rs is the yield stress and KI is the mode I stress intensity factor. For a center cracked infinite plate, the stress intensity factor can be estimated by
KI ¼ r
pffiffiffiffiffiffi pa
ð2Þ
where a is the half crack length and r is the applied uniform tension load. Substitute Eq. (2) into Eq. (1), Ry also can be written as
2 r Ry ¼ a
rs
ð3Þ
This relationship shows that Ry is determined by applied load, crack length, and yield stress. When crack length and yield stress are assumed to be constant, the following formula (Eq. (4)) can be obtained.
Ry r
ð4Þ
However, the present investigation attempts to find a new parameter to replace the applied load to estimate the plastic zone size.
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Based on the expression of crack opening displacement far away the crack-tip under small scale yielding condition as shown in Eq. (5), which is proposed by Paris, it can be found that the MCOD is proportional to the applied load r.
Du ¼ 4rae =E
ð5Þ
where Du is the MCOD, E is the elastic modulus, ae is the effective half crack length [9], and it can be expressed as ae ¼ a þ Ry =2. Based on equations above, the relationship can be simply and distinctly expressed as
Du r
ð6Þ
According to Eq. (4) and Eq. (6), we can know that the crack-tip plastic zone size Ry has a certain relationship with MCOD
Ry Du
ð7Þ
Eq. (7) shows that the plastic zone size is related to the maximum crack opening displacement. If the expression of the relationship is given, the plastic zone size can be easily obtained when the maximum crack opening displacement is known. 3. Numerical calculation 3.1. FE model establishment In order to determine the relationship between the Ry and MCOD, finite element simulation is employed to obtain series values of Ry and MCOD. In the numerical calculation, a 2W wide and 2L length plate having a 2a center crack is used to investigate crack-tip plastic zone size under biaxial tension load r, as shown in Fig. 1. The plane stress condition is assumed. Since the plate is symmetric, only one-quarter of the plate is modeled. The eight-node isoparametric quadrilateral element has been utilized. The plastic zone size is dependent on the minimum element size around the crack-tip, and the minimum element size has been widely discussed by the present authors [10–12]. Some accepted recommendations to establish the minimum element size in the model are made. It is commonly admitted that the plastic zone must be divided into no less than 10 linear elements. The minimum finite element mesh size ahead of the crack-tip is 0.05 mm when r ¼ 50 MPa. For low loading levels, finer meshes are used in order to capture the small plastic zone. The material is assumed to have an elastic perfectly plastic stress– strain relationship with a yield stress rs = 235 MPa. The kinematic hardening rule and the Von Mises criteria are employed in the calculations. The Young’s modulus E is 210 GPa and the Poisson’s ratio v is 0.3. 3.2. FE model confirmation According to the finite element model described above, an examination is performed on the calculated plastic zone sizes Ry from finite element models, and a series of r ranging from 10 MPa to 200 MPa with an increasement of 10 MPa are selected to calculate the crack-tip plastic zone size, and other parameters are kept unchanged. It can be seen from Fig. 3 that the calculated results are sufficiently close to those derived from Eq. (1) (Irwin’s formula) under small scale yielding condition. This indicates that the plastic zone sizes calculated by the finite element model are correct and reliable in small scale yielding, and thus the values in large scale yielding can be considered to be correct. To establish the relationship between Ry and MCOD, the values of MCOD should be got from the same models presented above. The results of Ry and MCOD are normalized to expand the application condition and are shown in Fig. 4. The normalized plastic zone size Ry/a, correlates well with the normalized maximum crack opening displacement DuE=ars under the given parameters cases. However, it is difficult to apply the relationship in practice due to the application condition of the infinite plate.
Fig. 2. Finite element model. Fig. 2a gives the whole finite element mesh model (a = 10 mm, W = 150 mm, L = 150 mm), and the half-length and half-width of plate, L and W, are chosen to L = W = 15a, to avoid the end effect. Fig. 2c shows the typical fine mesh configuration near crack-tip surface modeled by ANSYS. In this study, the sizes of element in the fine mesh zone increase linearly along the crack line (see Fig. 2b).
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Fig. 3. Comparison of two results by FEM and Irwin.
Fig. 4. Relationship between Ry/a versus DuE=ars .
It is known that plastic zone size is determined by the applied stress, yield stress of material, crack geometry and specimen sizes. Hence, this paper considers the effects of these parameter on the relationship between Ry/a and DuE=ars .
4. Results and discussion 4.1. Plate width effect and the applied load influence Fig. 2 presents the model employed for finite element simulations, in which, a = 10 mm, L = 150 mm, a/W varies in the range of 0.05–0.5 and the applied load r = 40 MPa, 60 MPa, 80 MPa, 100 MPa. Material parameters are kept unchanged (rs ¼ 235; E = 210 GPa, v = 0.3). When the parameters rs , r, a, E and L are given for specimens with different widths, the MCOD and the Ry have been calculated and the corresponding normalized values are shown in Fig. 5. The results indicate that the relationship between Ry/a versus DuE=ars is virtually independent of the change of plate widths and the applied
Fig. 5. The relationship between Ry/a versus DuE=ars in the case of different W and
r.
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Fig. 6. The relationship between Ry/a versus DuE=ars in the case of different a and
rs .
load. It is expected that the method of Ry estimated by MCOD can eliminate the width effect and is no related to the applied load. 4.2. Crack geometry and yield stress effects To investigate the influence of crack length a and the yield stress rs on the relationship, different half crack length (a = 10–20 mm) and several yield stresses (rs = 235 MPa, 255 MPa, 275 MPa, 300 MPa, 325 MPa, 350 MPa, 375 MPa and 400 MPa), are used in the finite element simulations. Other parameters are taken as L = 150 mm, W = 50 mm and the applied load r is range of 20–100 MPa. The calculating models are similar to the ones mentioned above (refer to Fig. 1). Fig. 6 represents the normalized calculated results. It can be seen that the normalized plastic zone size is a function of the normalized MCOD under different yield stresses and half crack length at the given models. This indicates that the crack length and yield stress have no effect on the relationship between Ry/a and DuE=ars . Meanwhile, results Ry/a are plotted against DuE=ars in Fig. 7, and it can be seen from Fig. 7 that the results can be represented by the same curve. Based on least square method, a simple expression of normalized plastic zone size versus normalized MCOD can be fitted as Eq. (8).
Ry =a ¼ 0:0058ðDuE=ars Þ3 þ 0:0463ðDuE=ars Þ2 þ 0:0099ðDuE=ars Þ 0:0017
ð8Þ
4.3. Transverse stress effect Transverse stress has been found [13,14] to have a significant influence on the plastic zone size. This effect has been evaluated using center cracked plate in tension [15,16]. However, since the transverse stress can change the plastic zone of cracktip. So, the present work also considers the influence the transverse stress on the relationship proposed above. In this section, the transverse stress is the stress in x axis direction, rx , and ry is the stress in y axis direction. The biaxiality can be defined by the ratio
k ¼ rx =ry
ð9Þ
Fig. 7. The relationship between Ry/a versus DuE=ars in all given cases.
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H. Yi et al. / Engineering Fracture Mechanics 77 (2010) 2912–2918 Table 1 The results of Ry calculated by two methods in case of k = 0.a
a
r/MPa
rs/MPa
r0s /MPa
a/W
a/mm
Du/mm
Ry by Eq. (8)/mm
Ry by Irwin/mm
60 60 60 60 60 40 40 40 80 80 80
235 255 275 300 235 325 350 400 375 400 350
346.683 377.150 407.546 445.466 346.683 485.649 523.282 590.512 556.044 593.954 518.074
0.067 0.067 0.067 0.067 0.125 0.1 0.1 0.1 0.1 0.1 0.143
10 10 10 10 17 10 10 16 10 10 17
1.174 102 1.170 102 1.167 102 1.164 102 2.002 102 7.698 103 7.692 103 3.130 102 1.556 102 1.553 102 3.130 102
0.3082 0.2599 0.2225 0.1863 0.5277 0.0694 0.0594 0.4727 0.2129 0.1866 0.4215
0.2995 0.2531 0.2167 0.1814 0.5092 0.0678 0.0584 0.4588 0.2070 0.1814 0.4054
Note that half-length of plate is taken as L = 15a in Table 1.
Table 2 The results of Ry calculated by two methods in case of k = 1.b
b
r/MPa
rs/MPa
r0s /MPa
a/W
a/mm
Du/mm
Ry by Eq. (8)/mm
Ry by Irwin/mm
30 40 50 60 60 80 90 100 80 60 70
305 355 355 275 275 325 400 400 325 300 270
330.541 388.174 394.275 314.608 314.608 373.981 458.375 460.555 373.981 341.425 311.247
0.100 0.100 0.100 0.100 0.083 0.083 0.083 0.083 0.125 0.111 0.091
12 12 15 16 14 15 20 20 17 18 11
6.920 103 9.237 103 1.447 102 1.869 102 1.634 102 2.345 102 3.505 102 3.910 102 2.667 102 2.099 102 1.508 102
0.1011 0.1305 0.2458 0.5904 0.5158 0.6987 0.7817 0.9607 0.7973 0.5643 0.5686
0.0988 0.1274 0.2412 0.5819 0.5092 0.6864 0.7710 0.9429 0.7779 0.5559 0.5564
Note that half-length of plate is taken as L = 15a in Table 2.
Eq. (8) are achieved on the basis of k = 1(rx = ry ; see Fig. 1a), further studies are made to investigate the applicability of the equation when k – 1. For a cracked geometry subjected to unequal biaxial stress (k – 1), the modified yield stress r0s can be expressed by the application of the von Mises yield criterion and Williams’ expression of the elastic stress field [17] for plane stress condition as
r0s ¼
ð1 kÞry þ 2
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4r2s 3ð1 kÞ2 r2y 2
ð10Þ
The case to unequal biaxial stress can be turn into the case to equal biaxial stress by modified yield stress (Eq. 10). That is to say, we can get the plastic zone size under unequal biaxial stress when the Eq. (10) is substituted into Eq. (1). For simplicity, k is set to equal to 0 (see Fig. 1b) and 1 (see Fig. 1c) to assess the applicability of Eq. (8) and the correctness of Eq. (10). Several models and different materials are respectively given in Tables 1 and 2, and the results are also shown in the corresponding tables. Besides, the results of Ry estimated by Eq. (8) are compared with the results by Irwin formula (Eq. (1)). In Tables 1 and 2, it can be seen that the results of Ry calculated by the method proposed in the present paper are very close to those by Irwin. This shows that the Eq. (8) is suitable for the case to uniaxial stress (k = 0 and k = 1) by modifying yield stress of material (Eq. (10)). The relationship between Ry/a versus DuE=ars has no effect on the transverse stress when k – 1.
5. Conclusion The paper focuses on the estimation method of plastic zone size determined by maximum crack opening displace. The 2D elastic–plastic finite element analysis is employed to quantify the plastic zone size and maximum crack opening displacement for a center cracked plate subjected to biaxial tension stress. Based on series of calculation results, a simple relationship between the normalized plastic zone size and the normalized MCOD is proposed by least square method. It is found that the relationship has little effect on many variables such as yield stress, stress applied levels, plate width and crack length. Further, the transverse stress effect on the relationship is taken into account. An expression of modified yield stress is deduced, which enable to turn the case to unequal biaxial stress into the case to equal biaxial stress, and eliminate the influence of
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transverse effect on the relationship. The method proposed in the paper to estimate plastic zone size on the basis of MCOD is suitable for any center cracked plate of various materials subject to different biaxiality stress. References [1] Kumar V, German MD, Shih CF. An engineering approach for elastic-plastic fracture analysis. EPRI final report NP-1931; 1981. [2] Anderson TL. Fracture mechanics: fundamentals and applications, CRC press; 1995. [3] Ainsworth RA, Milne I, Dowling AR, Stewart AT. R6. Assessment of the integrity of structures containing defects. UK: British Energy Generation Ltd. 2001; R6-Revision 4-Amendment 10. [4] Brown MW, de los Rios ER, Miller KJ. A critical comparison of proposed parameters for high-strain fatigue crack growth. ASTM STP 924. 1988. p. 233– 59. [5] Wang GS, Blom AF. A strip model for fatigue crack growth predictions under general load conditions. Engng Fract Mech 1991;40:507–33. [6] Wang GS. The plasticity aspect of fatigue crack growth. Engng Fract Mech 1993;46:909–30. [7] Irwin GR. Plastic zone near a crack and fracture toughness. In: Proc of the 7th sagamore research conference on mechanics & metals behavior of sheet material, vol. 4, New York; 1960. p. 463–78. [8] Dugdale DS. Yielding of steel sheets containing slits. J Mech Phys Solids 1960;8:100–4. [9] Rice JR. An examination of the fracture mechanics energy balance from the point of view of continuum mechanics. Proc first int conf fracture, Sendai 1966;1:309–40. [10] McClung RC, Sehitoglu H. On the finite-element analysis of fatigue crack closure, part-1: basic modeling issues. In: Engng Fract Mech 1989;33:237–52. [11] González-Herrera A, Zapatero J. Influence of minimum element size to determine crack closure stress by the finite element method. Engng Frac Mech 2005;72:337–55. [12] Zapatero J, Moreno B, González-Herrera A. Fatigue crack closure determination by means of finite element analysis. Engng Frac Mech 2008;75:41–57. [13] Harmain GA, Provan JW. Fatigue crack-tip plasticity revisited–the issue of shape addressed. Theor Appl Fract Mech 1997;26:63–79. [14] McClung RC. The influence of applied stress, crack length, and stress intensity factor on crack closure. Metall Trans A 1991;22:1559–71. [15] Fleck NA. Finite element analysis of plasticity-induced crack closure under plane strain conditions. Engng Fract Mech 1986;25:441–9. [16] Kim JH, Lee SB. Fatigue crack opening stress base on the strip-yield model. Theor Appl Frac Mech 2000;34:73–84. [17] Williams ML. On the stress distribution at the base of a stationary crack. J Appl Mech 1957;24:109–14.
Contents lists available at ScienceDirect
Engineering Fracture Mechanics journal homepage: www.elsevier.com/locate/engfracmech
Technical Note
A new method of plastic zone size determined based on maximum crack opening displacement Huang Yi, Chen Jingjie *, Liu Gang School of Naval Architecture, Dalian University of Technology, Dalian 116024, Liaoning, China
a r t i c l e
i n f o
Article history: Received 24 February 2010 Received in revised form 29 June 2010 Accepted 29 June 2010
Keywords: Plastic zone size Maximum crack opening displacement Center cracked plate Finite element analysis
a b s t r a c t This paper presents a new method to determine the crack-tip plastic zone size (Ry) in the center cracked plate in tension. The maximum crack opening displacement (MCOD) is used to estimate Ry in this method. Based on the series of calculation results by finite element analysis, the explicit expression for the crack-tip plastic zone size versus MCOD is fitted by least square method. The expression enables to eliminate the influences of the yield stress of material, crack geometry, plate sizes and transverse stress. Therefore, the presented method of Ry determined by MCOD is suitable to any center crack finite plate of any material under uniaxial or biaxial tension. Ó 2010 Elsevier Ltd. All rights reserved.
1. Introduction Methodologies of failure assessment of cracked component, based on elastic–plastic fracture mechanics (EPFM), are relatively well established [1–3]. In the EPFM, crack-tip opening displacement, crack-tip plastic zone size (Ry) and the range of stress intensity factor have been suggested for characterizing fatigue crack propagation behaviors. Ry has advantages over other parameters [4], which not only facilitates correlation of fast growing cracks but also has an obvious physical interpretation. Many researchers [4–6] consider that Ry is the most suitable parameter that can describe the propagation of fatigue cracks. So, it is important to determine the size of the crack-tip plastic zones when assessing the safety state of cracked component in practice. Several researchers have made attempts for the determination of the plastic zone. Irwin [7] and Dugdale [8] respectively proposed different models to estimate plastic zone size ahead of the crack. However, those models are usually restricted to plates of infinite extent with simple geometrical configurations of crack and do not consider the influence of plate width effect on Ry. In the present paper, the maximum crack opening displacement (MCOD) is introduced to replace the applied load to estimate the Ry for a center cracked plate subjected to biaxial tension loading, as can be seen in Fig. 1a. Then finite element method is employed to calculate series of Ry and corresponding MCOD for the specimens with different widths, different crack lengths and various yield stresses of material under several applied stress levels. A simple relationship between the normalized plastic zone size, Ry/a, and the normalized crack opening displacement, DuE=ars is achieved by the calculated results. Also, the effect of transverse stress on the relationship is considered through changing its value. It is noticed that the relationship is not affected by crack length, plate width, yield stress, applied load and transverse stress. * Corresponding author. Tel.: +86 15040629349; fax: +86 041184706350. E-mail address: [email protected] (C. Jingjie). 0013-7944/$ - see front matter Ó 2010 Elsevier Ltd. All rights reserved. doi:10.1016/j.engfracmech.2010.06.026
H. Yi et al. / Engineering Fracture Mechanics 77 (2010) 2912–2918
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Nomenclature Ry KI
crack-tip plastic zone size mode I stress intensity factor yield stress of material applied uniform tension load stress in x axis direction, and it is also called transverse stress stress in y axis direction ratio of biaxial stress, k = rx/ry half crack length effective half crack length, ae = a + Ry/2 maximum crack opening displacement (MCOD) Young’s modulus half-length of plate half-width of plate
rs r rx ry k a ae Du E L W
Fig. 1. A center cracked plate in tension. (a) The case to biaxial stress; (b) the case to uniaxial stress (transverse stress is zero); (c) the case to uniaxial stress (transverse is compressive stress, which is equal to tensile stress).
2. Theoretical consideration An expression of Ry for perfectly plastic materials under small scale yielding, when the cracked infinite plate is subject to the biaxial uniform tension load, has been derived [7] for plane stress conditions as follows:
Ry ¼
2 1 KI
p rs
ð1Þ
where rs is the yield stress and KI is the mode I stress intensity factor. For a center cracked infinite plate, the stress intensity factor can be estimated by
KI ¼ r
pffiffiffiffiffiffi pa
ð2Þ
where a is the half crack length and r is the applied uniform tension load. Substitute Eq. (2) into Eq. (1), Ry also can be written as
2 r Ry ¼ a
rs
ð3Þ
This relationship shows that Ry is determined by applied load, crack length, and yield stress. When crack length and yield stress are assumed to be constant, the following formula (Eq. (4)) can be obtained.
Ry r
ð4Þ
However, the present investigation attempts to find a new parameter to replace the applied load to estimate the plastic zone size.
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Based on the expression of crack opening displacement far away the crack-tip under small scale yielding condition as shown in Eq. (5), which is proposed by Paris, it can be found that the MCOD is proportional to the applied load r.
Du ¼ 4rae =E
ð5Þ
where Du is the MCOD, E is the elastic modulus, ae is the effective half crack length [9], and it can be expressed as ae ¼ a þ Ry =2. Based on equations above, the relationship can be simply and distinctly expressed as
Du r
ð6Þ
According to Eq. (4) and Eq. (6), we can know that the crack-tip plastic zone size Ry has a certain relationship with MCOD
Ry Du
ð7Þ
Eq. (7) shows that the plastic zone size is related to the maximum crack opening displacement. If the expression of the relationship is given, the plastic zone size can be easily obtained when the maximum crack opening displacement is known. 3. Numerical calculation 3.1. FE model establishment In order to determine the relationship between the Ry and MCOD, finite element simulation is employed to obtain series values of Ry and MCOD. In the numerical calculation, a 2W wide and 2L length plate having a 2a center crack is used to investigate crack-tip plastic zone size under biaxial tension load r, as shown in Fig. 1. The plane stress condition is assumed. Since the plate is symmetric, only one-quarter of the plate is modeled. The eight-node isoparametric quadrilateral element has been utilized. The plastic zone size is dependent on the minimum element size around the crack-tip, and the minimum element size has been widely discussed by the present authors [10–12]. Some accepted recommendations to establish the minimum element size in the model are made. It is commonly admitted that the plastic zone must be divided into no less than 10 linear elements. The minimum finite element mesh size ahead of the crack-tip is 0.05 mm when r ¼ 50 MPa. For low loading levels, finer meshes are used in order to capture the small plastic zone. The material is assumed to have an elastic perfectly plastic stress– strain relationship with a yield stress rs = 235 MPa. The kinematic hardening rule and the Von Mises criteria are employed in the calculations. The Young’s modulus E is 210 GPa and the Poisson’s ratio v is 0.3. 3.2. FE model confirmation According to the finite element model described above, an examination is performed on the calculated plastic zone sizes Ry from finite element models, and a series of r ranging from 10 MPa to 200 MPa with an increasement of 10 MPa are selected to calculate the crack-tip plastic zone size, and other parameters are kept unchanged. It can be seen from Fig. 3 that the calculated results are sufficiently close to those derived from Eq. (1) (Irwin’s formula) under small scale yielding condition. This indicates that the plastic zone sizes calculated by the finite element model are correct and reliable in small scale yielding, and thus the values in large scale yielding can be considered to be correct. To establish the relationship between Ry and MCOD, the values of MCOD should be got from the same models presented above. The results of Ry and MCOD are normalized to expand the application condition and are shown in Fig. 4. The normalized plastic zone size Ry/a, correlates well with the normalized maximum crack opening displacement DuE=ars under the given parameters cases. However, it is difficult to apply the relationship in practice due to the application condition of the infinite plate.
Fig. 2. Finite element model. Fig. 2a gives the whole finite element mesh model (a = 10 mm, W = 150 mm, L = 150 mm), and the half-length and half-width of plate, L and W, are chosen to L = W = 15a, to avoid the end effect. Fig. 2c shows the typical fine mesh configuration near crack-tip surface modeled by ANSYS. In this study, the sizes of element in the fine mesh zone increase linearly along the crack line (see Fig. 2b).
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Fig. 3. Comparison of two results by FEM and Irwin.
Fig. 4. Relationship between Ry/a versus DuE=ars .
It is known that plastic zone size is determined by the applied stress, yield stress of material, crack geometry and specimen sizes. Hence, this paper considers the effects of these parameter on the relationship between Ry/a and DuE=ars .
4. Results and discussion 4.1. Plate width effect and the applied load influence Fig. 2 presents the model employed for finite element simulations, in which, a = 10 mm, L = 150 mm, a/W varies in the range of 0.05–0.5 and the applied load r = 40 MPa, 60 MPa, 80 MPa, 100 MPa. Material parameters are kept unchanged (rs ¼ 235; E = 210 GPa, v = 0.3). When the parameters rs , r, a, E and L are given for specimens with different widths, the MCOD and the Ry have been calculated and the corresponding normalized values are shown in Fig. 5. The results indicate that the relationship between Ry/a versus DuE=ars is virtually independent of the change of plate widths and the applied
Fig. 5. The relationship between Ry/a versus DuE=ars in the case of different W and
r.
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Fig. 6. The relationship between Ry/a versus DuE=ars in the case of different a and
rs .
load. It is expected that the method of Ry estimated by MCOD can eliminate the width effect and is no related to the applied load. 4.2. Crack geometry and yield stress effects To investigate the influence of crack length a and the yield stress rs on the relationship, different half crack length (a = 10–20 mm) and several yield stresses (rs = 235 MPa, 255 MPa, 275 MPa, 300 MPa, 325 MPa, 350 MPa, 375 MPa and 400 MPa), are used in the finite element simulations. Other parameters are taken as L = 150 mm, W = 50 mm and the applied load r is range of 20–100 MPa. The calculating models are similar to the ones mentioned above (refer to Fig. 1). Fig. 6 represents the normalized calculated results. It can be seen that the normalized plastic zone size is a function of the normalized MCOD under different yield stresses and half crack length at the given models. This indicates that the crack length and yield stress have no effect on the relationship between Ry/a and DuE=ars . Meanwhile, results Ry/a are plotted against DuE=ars in Fig. 7, and it can be seen from Fig. 7 that the results can be represented by the same curve. Based on least square method, a simple expression of normalized plastic zone size versus normalized MCOD can be fitted as Eq. (8).
Ry =a ¼ 0:0058ðDuE=ars Þ3 þ 0:0463ðDuE=ars Þ2 þ 0:0099ðDuE=ars Þ 0:0017
ð8Þ
4.3. Transverse stress effect Transverse stress has been found [13,14] to have a significant influence on the plastic zone size. This effect has been evaluated using center cracked plate in tension [15,16]. However, since the transverse stress can change the plastic zone of cracktip. So, the present work also considers the influence the transverse stress on the relationship proposed above. In this section, the transverse stress is the stress in x axis direction, rx , and ry is the stress in y axis direction. The biaxiality can be defined by the ratio
k ¼ rx =ry
ð9Þ
Fig. 7. The relationship between Ry/a versus DuE=ars in all given cases.
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H. Yi et al. / Engineering Fracture Mechanics 77 (2010) 2912–2918 Table 1 The results of Ry calculated by two methods in case of k = 0.a
a
r/MPa
rs/MPa
r0s /MPa
a/W
a/mm
Du/mm
Ry by Eq. (8)/mm
Ry by Irwin/mm
60 60 60 60 60 40 40 40 80 80 80
235 255 275 300 235 325 350 400 375 400 350
346.683 377.150 407.546 445.466 346.683 485.649 523.282 590.512 556.044 593.954 518.074
0.067 0.067 0.067 0.067 0.125 0.1 0.1 0.1 0.1 0.1 0.143
10 10 10 10 17 10 10 16 10 10 17
1.174 102 1.170 102 1.167 102 1.164 102 2.002 102 7.698 103 7.692 103 3.130 102 1.556 102 1.553 102 3.130 102
0.3082 0.2599 0.2225 0.1863 0.5277 0.0694 0.0594 0.4727 0.2129 0.1866 0.4215
0.2995 0.2531 0.2167 0.1814 0.5092 0.0678 0.0584 0.4588 0.2070 0.1814 0.4054
Note that half-length of plate is taken as L = 15a in Table 1.
Table 2 The results of Ry calculated by two methods in case of k = 1.b
b
r/MPa
rs/MPa
r0s /MPa
a/W
a/mm
Du/mm
Ry by Eq. (8)/mm
Ry by Irwin/mm
30 40 50 60 60 80 90 100 80 60 70
305 355 355 275 275 325 400 400 325 300 270
330.541 388.174 394.275 314.608 314.608 373.981 458.375 460.555 373.981 341.425 311.247
0.100 0.100 0.100 0.100 0.083 0.083 0.083 0.083 0.125 0.111 0.091
12 12 15 16 14 15 20 20 17 18 11
6.920 103 9.237 103 1.447 102 1.869 102 1.634 102 2.345 102 3.505 102 3.910 102 2.667 102 2.099 102 1.508 102
0.1011 0.1305 0.2458 0.5904 0.5158 0.6987 0.7817 0.9607 0.7973 0.5643 0.5686
0.0988 0.1274 0.2412 0.5819 0.5092 0.6864 0.7710 0.9429 0.7779 0.5559 0.5564
Note that half-length of plate is taken as L = 15a in Table 2.
Eq. (8) are achieved on the basis of k = 1(rx = ry ; see Fig. 1a), further studies are made to investigate the applicability of the equation when k – 1. For a cracked geometry subjected to unequal biaxial stress (k – 1), the modified yield stress r0s can be expressed by the application of the von Mises yield criterion and Williams’ expression of the elastic stress field [17] for plane stress condition as
r0s ¼
ð1 kÞry þ 2
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4r2s 3ð1 kÞ2 r2y 2
ð10Þ
The case to unequal biaxial stress can be turn into the case to equal biaxial stress by modified yield stress (Eq. 10). That is to say, we can get the plastic zone size under unequal biaxial stress when the Eq. (10) is substituted into Eq. (1). For simplicity, k is set to equal to 0 (see Fig. 1b) and 1 (see Fig. 1c) to assess the applicability of Eq. (8) and the correctness of Eq. (10). Several models and different materials are respectively given in Tables 1 and 2, and the results are also shown in the corresponding tables. Besides, the results of Ry estimated by Eq. (8) are compared with the results by Irwin formula (Eq. (1)). In Tables 1 and 2, it can be seen that the results of Ry calculated by the method proposed in the present paper are very close to those by Irwin. This shows that the Eq. (8) is suitable for the case to uniaxial stress (k = 0 and k = 1) by modifying yield stress of material (Eq. (10)). The relationship between Ry/a versus DuE=ars has no effect on the transverse stress when k – 1.
5. Conclusion The paper focuses on the estimation method of plastic zone size determined by maximum crack opening displace. The 2D elastic–plastic finite element analysis is employed to quantify the plastic zone size and maximum crack opening displacement for a center cracked plate subjected to biaxial tension stress. Based on series of calculation results, a simple relationship between the normalized plastic zone size and the normalized MCOD is proposed by least square method. It is found that the relationship has little effect on many variables such as yield stress, stress applied levels, plate width and crack length. Further, the transverse stress effect on the relationship is taken into account. An expression of modified yield stress is deduced, which enable to turn the case to unequal biaxial stress into the case to equal biaxial stress, and eliminate the influence of
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transverse effect on the relationship. The method proposed in the paper to estimate plastic zone size on the basis of MCOD is suitable for any center cracked plate of various materials subject to different biaxiality stress. References [1] Kumar V, German MD, Shih CF. An engineering approach for elastic-plastic fracture analysis. EPRI final report NP-1931; 1981. [2] Anderson TL. Fracture mechanics: fundamentals and applications, CRC press; 1995. [3] Ainsworth RA, Milne I, Dowling AR, Stewart AT. R6. Assessment of the integrity of structures containing defects. UK: British Energy Generation Ltd. 2001; R6-Revision 4-Amendment 10. [4] Brown MW, de los Rios ER, Miller KJ. A critical comparison of proposed parameters for high-strain fatigue crack growth. ASTM STP 924. 1988. p. 233– 59. [5] Wang GS, Blom AF. A strip model for fatigue crack growth predictions under general load conditions. Engng Fract Mech 1991;40:507–33. [6] Wang GS. The plasticity aspect of fatigue crack growth. Engng Fract Mech 1993;46:909–30. [7] Irwin GR. Plastic zone near a crack and fracture toughness. In: Proc of the 7th sagamore research conference on mechanics & metals behavior of sheet material, vol. 4, New York; 1960. p. 463–78. [8] Dugdale DS. Yielding of steel sheets containing slits. J Mech Phys Solids 1960;8:100–4. [9] Rice JR. An examination of the fracture mechanics energy balance from the point of view of continuum mechanics. Proc first int conf fracture, Sendai 1966;1:309–40. [10] McClung RC, Sehitoglu H. On the finite-element analysis of fatigue crack closure, part-1: basic modeling issues. In: Engng Fract Mech 1989;33:237–52. [11] González-Herrera A, Zapatero J. Influence of minimum element size to determine crack closure stress by the finite element method. Engng Frac Mech 2005;72:337–55. [12] Zapatero J, Moreno B, González-Herrera A. Fatigue crack closure determination by means of finite element analysis. Engng Frac Mech 2008;75:41–57. [13] Harmain GA, Provan JW. Fatigue crack-tip plasticity revisited–the issue of shape addressed. Theor Appl Fract Mech 1997;26:63–79. [14] McClung RC. The influence of applied stress, crack length, and stress intensity factor on crack closure. Metall Trans A 1991;22:1559–71. [15] Fleck NA. Finite element analysis of plasticity-induced crack closure under plane strain conditions. Engng Fract Mech 1986;25:441–9. [16] Kim JH, Lee SB. Fatigue crack opening stress base on the strip-yield model. Theor Appl Frac Mech 2000;34:73–84. [17] Williams ML. On the stress distribution at the base of a stationary crack. J Appl Mech 1957;24:109–14.