Engineering Fracture Mechanics 150 (2015) 19–27
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Technical Note
Effect of initial T-stress on stress intensity factor for a crack in a thin pre-stressed layer Zi-Cheng Jiang a, Guo-Jin Tang a, Xian-Fang Li b,⇑ a b
College of Aerospace Science and Engineering, National University of Defense Technology, Changsha 410073, PR China School of Civil Engineering, Central South University, Changsha 410075, PR China
a r t i c l e
i n f o
Article history: Received 9 May 2015 Received in revised form 11 October 2015 Accepted 12 October 2015 Available online 17 October 2015 Keywords: Initial stress Pre-stressed structure Crack Stress intensity factor Energy release rate
a b s t r a c t Fracture of a pre-stressed thin elastic layer is studied and modelled as bending of double clamped beams with axial force. Using the beam theory, the strain energy in bending state of the beam with an axial force is derived. Explicit expressions for energy release rate and stress intensity factor (SIF) for a crack with initial transverse stress (T-stress) are obtained. Results show that tensile (compressive) initial T-stress decreases (increases) the SIF and impedes (promotes) crack growth. The influence of compressive initial stress is greater than that of tensile initial stress in the same magnitude. Ó 2015 Elsevier Ltd. All rights reserved.
1. Introduction Crack problems have attracted much attention in recent decades. It is recognized in linear elastic fracture mechanics that stress intensity factor (SIF) is a significant fracture parameter to describe singular elastic field near a crack tip. As well known, a transverse stress, called T-stress, parallel to the crack plane does not affect the SIF, but influence the elastic field near the crack tip by changing the nonvanishing term of the elastic field [1]. Therefore, a two-parameter fracture criterion based on a K–T dominating elastic field is more suitable to predict crack growth. The effect of elastic T-stresses on crack initiation direction and yield zone has been analyzed [2–9]. On the other hand, in civil engineering, many practical structures are applied in an environment of action of an initial stress. This class of structures are sometime called pre-stressed structures. In addition, for composite structures such as fiber-reinforced or particle-reinforced structures, the inhomogeneity of materials may give rise to initial stresses or residual stresses due to change in environment temperature [10]. In particular, more applications of initial stresses can be found in micromechanical devices [11] and biological science [12]. For such pre-stressed structures, the study on crack problems is also of importance. However, relevant work is very limited [13]. Soos [14] analyzed stress concentration in a prestressed elastic plane with a crack and obtained the asymptotic behavior of the incremental elastic fields in the close vicinity of the crack tips. For a penny-shaped crack under radial shear, the influence of initial stress on fracture of a cracked halfspace was dealt with using the integral equation method by Nazarenko et al. [15]. Radi et al. further studied the effects of pre-stress on crack-tip fields in an elastic and incompressible solid [16]. For a mode III interface crack in a pre-stressed solid, asymptotic incremental displacement and stress fields in the vicinity of the crack tip were numerical calculated using ⇑ Corresponding author. Tel.: +86 731 8816 7070; fax: +86 731 8557 1736. E-mail address:
[email protected] (X.-F. Li). http://dx.doi.org/10.1016/j.engfracmech.2015.10.034 0013-7944/Ó 2015 Elsevier Ltd. All rights reserved.
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Z.-C. Jiang et al. / Engineering Fracture Mechanics 150 (2015) 19–27
the complex potential method [17]. Akbarov and Turan used a finite element method to study the influence of initial stresses on the SIF for an orthotropic strip [18]. In addition, Corso et al. also investigated stress concentration near a rigid line inclusion in a prestressed elastic plane [19]. Although some studies on the effect of initial stresses on elastic fields near a crack tip have been conducted, there is no information on an explicit dependence of the SIF on initial stress reported. To fill this gap, in this paper we make an attempt to adopt a simple beam theory to achieve this purpose. This paper develops the bending model of double clamped beams subjected to axial loading to simulate a cracked elastic layer with initial stress. An emphasis is placed on the determination of the SIF, which sheds light on the influence of initial stress on the SIF. 2. Theoretical model Consider a cracked elastic layer of thickness 2h subjected to an initial transverse stress (T-stress) along the crack plane. The crack is located at a < x < a and 2a denotes the crack length. Under the action of incremental forces or stresses, as shown in Fig. 1, it is much desired to obtain SIFs near the crack tips. We assume that the initial elastic fields induced by the initial stress are small. In this case, in order to determine the incremental elastic fields of a cracked elastic layer, equilibrium equations for the class of problems can be linearized as
@ r11 @ r13 @2u þ þ r011 2 ¼ 0; @x @z @x 2 @ r13 @ r33 @ w þ þ r011 2 ¼ 0; @x @x @z
ð1Þ ð2Þ
where u and w are the components of incremental displacement along the x and z directions, respectively, r011 is an initial stress along the x direction, and rij is stress tensor. The outline of derivation is given in Appendix A. Although a detailed analysis of a pre-stressed elastic layer with a crack based on a two-dimensional theory of elasticity is possible, but cumbersome, here we propose a simple theoretical model to obtain a dependence of the SIF on the initial Tstress. Consider the case of a thin elastic layer. Since the thickness of the elastic layer is small enough as compared to the crack length 2a, we model this crack as double clamped beams. An advantage of this approach is that an explicit expression for the SIF can be derived in terms of initial stress. To do so, let us simplify the stated problem and use the simplest Euler– Bernoulli theory of beams. For Euler–Bernoulli beams, owing to the hypothesis that the cross-section of the beam remains perpendicular to the neutral surface prior to and posterior to bending deformation, the displacement component u can be represented by the unique independent displacement component w in terms of the following relation
u ¼ z
@w : @x
ð3Þ
Thus multiplying (1) by z and then integrating over cross-section, one gets
M0 Q r011 Iw000 ¼ 0;
ð4Þ
where the prime denotes differentiation with respect to x; I is the second moment of cross-sectional area, and M and Q stand for bending moment and shear force at cross-section, viz.
Z
M¼ A
Z
r11 zdA; Q ¼
A
r13 dA;
ð5Þ
where A is the cross-sectional area. Additionally, integrating (2) over cross-section leads to
Q 0 þ q þ r011 Aw00 ¼ 0;
ð6Þ
where q is distributed loading. From (4) and (6), we eliminate shear force Q and get
M00 þ q þ Pw00 r011 IwIV ¼ 0;
ð7Þ
Fig. 1. A cracked elastic layer with an initial stress r011 along the crack plane; (a) case A: opening of crack under a pair of concentrated forces, (b) case B: opening of crack under uniformly distributed loading.
Z.-C. Jiang et al. / Engineering Fracture Mechanics 150 (2015) 19–27
21
where P ¼ r011 A represents the resultant force owing to the initial stress, P > 0 corresponds to a tensile force and P < 0 to a compressive force, respectively. It is worth noting that for a compressive force, we require that the compressive force is always less than the Euler buckling load to prevent against buckling. The positive direction of some variables are designated in Fig. 2. Now we substitute the well-known relation between bending moment and deflection for Euler–Bernoulli beams
M ¼ EIw00 ;
ð8Þ
into the above-derived Eq. (7), yielding the final governing equation
E þ r011 IwIV Pw00 ¼ q;
ð9Þ
where E is Young’s modulus of the beam. Generally speaking, in practice the initial stress r011 is much less than Young’s modulus. For this reason, we remove the contribution of r011 I and simplify the above equation as
EIwIV Pw00 ¼ q;
ð10Þ
where EI is the bending stiffness of the beam. To acquire the SIF, let us first solve bending of a clamped beam under the action of an axial force P, a concentrated force F at an arbitrary position (for simplicity, in the present paper, the midspan position is assumed), and/or distributed loading q over the whole beam, as shown in Fig. 3. Due to symmetry of the problem, it is sufficient to analyze the half of the beam, 0 < x < L=2; L being the beam span. Thus appropriate boundary conditions are stated as
wjx¼0 ¼ 0; w0 jx¼0 ¼ 0; F w0 jx¼L=2 ¼ 0; Q jx¼L=2 ¼ : 2
ð11Þ ð12Þ
For convenience, we introduce some dimensionless variables as follows
x n¼ ; L
w W¼ ; L
sffiffiffiffiffiffiffiffiffiffi jPjL2 ; k¼ EI
a¼
FL2 ; EI
b¼
qL3 : EI
ð13Þ
Then the original Eq. (10) is rewritten as
W IV k2 W 00 ¼ b; ðP > 0Þ;
W IV þ k2 W 00 ¼ b; ðP < 0Þ
ð14Þ
subjected to
W ð0Þ ¼ 0; W 0 ð0Þ ¼ 0; 1 1 a ¼ 0; W 000 ¼ : W0 2 2 2
ð15Þ ð16Þ
By solving the problem and omitting detailed process, we can obtain the deflection as follow
W ðnÞ ¼
ða þ bÞ cosh 2k a 3
2k
sinh 2k
ðcosh kn 1Þ
ða þ bÞ 2k
3
ðsinh kn knÞ
bn2 2k2
;
ð17Þ
for an axial tensile force P > 0, and
W ðnÞ ¼
ða þ bÞ cos 2k a 3
2k
sin 2k
ðcos kn 1Þ þ
ða þ b Þ 2k
3
ðsin kn knÞ þ
bn2 2k2
;
for an axial compressive force P < 0.
Fig. 2. Positive bending moment and shear force along with the coordinate system.
ð18Þ
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Fig. 3. Schematic of bending of a clamped beam with an axial force and (a) concentrated loading F at midspan and (b) distributed loading q.
3. Energy release rates and stress intensity factors Based on the above bending solution, here we consider a thin elastic layer with a crack lying at the midplane. In addition to an initial T-stress r011 ¼ 2P=A; A being the cross-sectional area of the elastic layer, a pair of concentrated forces F are exerted at the crack center (case A) (see Fig. 1a) and uniformly distributed stress r33 is exerted at the upper and lower surfaces (case B) (see Fig. 1b). For case A, meaning q ¼ 0, with the above solution (17) or (18), we need to compute the total strain energy stored in a deformed beam, which can be achieved by performing the following integral
Z
L
U¼ 0
M2 EI dx ¼ L 2EI
Z
1=2
2 W 00 dn;
ð19Þ
0
where the strain energy due to extensional or shrinking of the beam arising from the axial force has been neglected, since it is small enough as compared to that due to bending of the beam. Upon substitution of (17) or (18) into (19) one gets the total strain energy U:
U¼
k k 1 sinh ; 2 2 cosh2 k 8k3 L 4
a2 EI
for P > 0, or
U¼
a2 EI k 3
8k L 2
sin
k 1 ; 2 cos2 4k
ð20Þ
ð21Þ
for P < 0. Note that the above-derived energy does not coincide with that calculated by FwðL=2Þ=2; wðL=2Þ being the deflection of the loading position, although they are identical in the absence of axial force. For case B, meaning F ¼ 0, with the above solution (17) or (18), we can similarly compute the total strain energy stored in a deformed beam through (19). Upon substitution of (17) or (18) into (19) one gets the total strain energy U:
U¼
b2 EI
k2
4k4 L 4sinh2 2k
þ
! k k coth 2 ; 2 2
ð22Þ
for P > 0, or
! k k U¼ 4 þ cot 2 ; 2 4k L 4 sin2 2k 2 b2 EI
k2
ð23Þ
for P < 0. Note that for case B (Fig. 1b), the strain energy calculated here only corresponds to that of the cracked elastic layer when the crack surface is loaded by r33 , and the energy independent of the crack length has been disregarded. Next we apply the above-derived strain energy to determine energy release rate (ERR) and stress intensity factor (SIF) of a thin elastic layer with a crack of length 2a lying at the midplane. Using the symmetry of the upper and lower half-layers, the strain energy stored in the cracked layer is twice as much as that for a clamped beam. Then the cracked elastic layer has energy 2U. In accordance with the energy balance point of view in the classical linear elastic fracture mechanics, the driving force of crack growth results from the released energy when crack extends per unit length. Knowledge of the strain energy permits us to evaluate ERR G for crack propagation through the ratio of the stored elastic energy to the area of the advance region. Thus for a crack in a two-dimensional elastic layer of breadth b, the ERR is therefore calculated by
G¼
dð2U Þ : 2bda
ð24Þ
Notice that the factor 2 in the denominator is due to two crack tips. 3
Keeping L ¼ 2a; P ¼ hbr011 I ¼ bh =12 for a rectangular beam in mind and inserting (20) into (24), after some algebra we get the ERR
Z.-C. Jiang et al. / Engineering Fracture Mechanics 150 (2015) 19–27
G¼ for
ð25Þ
r011 > 0 and G¼
for
2 sinh k20 a F ; 3 b 3 2k0 Eh cosh k20 a 3a
23
2 sin k20 a F ; 3 b cos3 k20 a 2k0 Eh 3a
ð26Þ
r011 < 0, where k0 ¼
rffiffiffiffiffiffi jP j : EI
ð27Þ
If only distributed loading is prescribed, we similarly obtain an expression for the ERR G
G¼ for
3
Eh k40
ðk0 a coth k0 a 1Þ 1
12r233 3
Eh k40
ðk0 a cot k0 a 1Þ 1
!
k20 a2 2
sinh k0 a
r011 > 0 and G¼
for
12r233
k20 a2
;
ð28Þ
!
2
sin k0 a
;
ð29Þ
r011 < 0, where we have made a substitution of q ¼ r33 b. Making use of the well-known formula between the stress intensity factor (SIF) K and ERR G for plane stress state [20]
K¼
pffiffiffiffiffiffi EG;
ð30Þ
one immediately obtains the SIF as
8 rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k a > sinh 02 > aF 6 > < 2hb h k acosh3 k0 a; 0 2 K¼ rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k0 a > sin > aF 6 2 > ; : 2hb h 3 k0 a k0 a cos
r011 > 0
;
ð31Þ
r011 < 0;
2
for case A where a pair of concentrated forces are applied at the crack center and
8 rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi > k20 a2 2a2 r33 > 3 k0 a coth k0 a1 > 1 ; 4 4 2 < h h k0 a sinh k0 a K¼ ffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi > 2 k2 a2 > k0 a1 > 1 sin02 k a ; : 2a hr33 3h k0 a cot k4 a4 0
0
r011 > 0
;
ð32Þ
r < 0; 0 11
for case B where uniformly distributed loading at the crack surfaces. The obtained results (31) and (32) describe the effect of the initial stress on the SIF when concentrated and uniformly distributed loading are applied, respectively. As a check, if setting k0 ! 0, the above-derived results reduce to
rffiffiffi 3 ; h
K¼
aF 2hb
K¼
2a2 r33 h
and
rffiffiffiffiffiffi 1 ; 3h
ð33Þ
ð34Þ
which are in exact agreement with the classical results in the absence of initial stress [21]. 4. Results and discussion In this section, the influence of the initial stress on the SIF is examined. The variation of the normalized SIF, K=K c , against the ratio of the initial stress to Young’s modulus is displayed for a=h ¼ 5 when concentrated forces and uniformly distributed loading are applied in Fig. 4, where K c is the corresponding classical SIF, defined by (33) and (34) for cases A and B, respectively. From Fig. 4, it is seen that the SIF monotonically decreases with the initial stress rising, regardless of given concentrated or distributed loadings. In other words, when an initial stress is compressive, the SIF is larger than that without initial stress, implying that a compressive initial stress promotes crack growth. However, when an initial stress is tensile, the SIF is lower than that without initial stress, implying that a tensile initial stress impedes crack growth. Although previous studies [15,18] based on different approaches have indicated similar conclusion, only numerical results have bee presented to show
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Fig. 4. The variation of the normalized SIF against the ratio of the initial stress to Young’s modulus
r011 =E for a=h ¼ 5.
the dependence of the SIFs on initial stress. However, here we provide some explicit exact expressions for calculating SIFs. Of course, the present model is based on the simplest Euler–Bernoulli beam theory, which essentially is also an approximation of a two-dimensional theory for a thin elastic layer. Furthermore, the influence of a compressive initial stress is greater than that of a tensile initial stress in the same magnitude. For instance, if choosing an initial stress as r011 =E ¼ 0:001, we find that for case A with a=h ¼ 10, the normalized SIF becomes 0.827 for a tensile initial stress and 1.236 for a compressive initial stress, decreasing 17.3% and increasing 23.6%, respectively. It indicates that a crack lying in the area of compressive initial stresses is more sensitive than it in the area of tensile initial stresses. For a prescribed tensile initial stress r011 =E ¼ 0:001, Fig. 5 shows the variation of the normalized SIF as a function of a=h. For comparison, Fig. 6 displays the variation of the normalized SIF as a function of a=h for a prescribed compressive initial stress r011 =E ¼ 0:001. From Figs. 5 and 6, when a=h rises, the effect of initial stress becomes greater. Although we have obtained exact expressions for the SIF, these expressions contain trigonometric and hyperbolic functions. Considering a special case of sufficiently small values of (32) for the SIF as
r011 a2 =h2 E, we can approximate the exact expressions (31) and
vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi u " 0 2 4 # 0 2 4 r r Fa u 3 a 51 a 11 t 1 11 ; K¼ þ 2hb h h 5 h E E
Fig. 5. The variation of the normalized SIF as a function of a=h for a prescribed tensile initial stress
ð35Þ
r011 =E ¼ 0:001.
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Fig. 6. The variation of the normalized SIF as a function of a=h for a prescribed compressive initial stress
r011 =E ¼ 0:001.
Table 1 Exact and approximate results of the normalized SIF. a=h
r33 is given
F is given Exact (31)
Appr. (35)
Exact (32)
Appr. (36)
0.992 0.952 0.827 0.668
0.992 0.952 0.838 0.785
0.994 0.961 0.861 0.733
0.994 0.962 0.868 0.809
1.008 1.052 1.236 1.680
1.008 1.052 1.226 1.554
1.006 1.042 1.188 1.539
1.006 1.041 1.181 1.447
r011 =E ¼ 103 2 5 10 15
r011 =E ¼ 103 2 5 10 15
for case A, and
vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u " 0 2 4 # 0 2 2a2 r33 u a t 1 1 16 r11 a þ 1296 r11 ; K¼ 3h 5 E h 175 h h E
ð36Þ
for case B. Table 1 shows a comparison of the exact SIFs with their approximations, where in the calculations we have taken r011 =E ¼ 103 . From Table 1, we find that the above approximations give accurate enough results as compared to the exact ones. The relative errors of the approximate SIFs do not exceed 1.3%, even for a=h ¼ 10. This shows that for small
r011 a2 =h2 E, these approximate expressions for the SIF can be employed and satisfactory accuracy can be achieved. 2 Of course, when a=h is further raised, the accuracy of approximations deteriorates since r011 a2 =h E is close to or even exceed 2 0 2 1. For large values of r11 a =h E, it is natural to choose the exact expressions (31) and (32) to determine the SIF. values of
5. Conclusions This paper dealt with a pre-stressed elastic layer with a crack and mainly analyzed the effect of initial T-stress on the SIF near the crack tips. The approach used here is based on the simplest Euler–Bernoulli beam theory. The original crack problem was modelled as bending of double-clamped beams with an axial force. The strain energy stored a deformed beam was calculated and the ERR as well as the SIF of a crack in a pre-stressed elastic layer was obtained explicitly. Numerical calculations of the SIF were performed and obtained results show that the initial stress strongly affects the SIF. A tensile initial T-stress decreases the SIF and a compressive initial stress increases the SIF. It infers that the apparent fracture toughness of a pre-
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Z.-C. Jiang et al. / Engineering Fracture Mechanics 150 (2015) 19–27
stressed solid is enhanced for a tensile initial T-stress area and weakened for a compressive initial T-stress area. If considering cohesive force at the crack tip (see e.g. [22]), the present method can be further extended to treat a generalized Dugdale–Barenblat model. Acknowledgement This work was supported by the Faculty Research Foundation of Central South University, PR China (No. 2013JSJJ020). Appendix A For an elastic medium with initial stresses, denoted as r0ij , the initial strains vanish. Under the action of additional loading, there are incremental stresses and strains, denoted as rij and eij , in the elastic medium. Applying the principle of virtual work according to total stresses rij þ r0ij , we have
Z V
rij þ r0ij deij dV
Z Z 0 f j þ f j duj dV
@V
V
g j þ g 0j duj dð@V Þ ¼ 0;
ðA1Þ
where d denotes the variation of a function, V is the volume occupied by an elastic medium and @V is the boundary surface of V; f j and g j are the density of body forces and surface forces, respectively, and a quantity with the superscript 0 stands for the corresponding one associated with the initial state. For the time being, we use the following geometric nonlinearity relation
eij ¼
1 ui;j þ uj;i þ uk;i uk;j ; 2
ðA2Þ
where a comma denotes differentiation with respect to the space variables following the comma, and the Einstein convention of summation over repeated lower-case indices has been used. Substituting (A2) into (A1) leads to
Z V
rij deij þ r0ij uk;j duk;i dV
Z
Z
V
f j duj dV
@V
g j duj dð@V Þ ¼ 0:
ðA3Þ
0
where the equilibrium equations of initial state, i.e. r0ij;i þ f j ¼ 0 and r0ij ni ¼ g 0j , have been used. Now we only consider the case of small deformation. After employing the divergence theorem, we have
Z V
rij deij þ r0ij uk;j duk;i dV ¼
Z
@V
Z
rij þ r0il uj;l ni duj dð@V Þ
V
rij;i þ r0il uj;il duj dV;
ðA4Þ
where ni ði ¼ 1; 2; 3Þ are the unit outer normal vector to the surface @V. Thus (A3) becomes
Z
@V
rij þ r0il uj;l ni g j duj dð@V Þ
Z
V
rij;i þ r0il uj;il þ f j duj dV ¼ 0:
ðA5Þ
Utilizing the variational principle, we get the equilibrium equations of an elastic medium with initial stresses
rij;i þ r0il uj;il þ f j ¼ 0;
ðA6Þ
subjected to boundary conditions
rij þ r0il uj;l ni ¼ g j ; or uj ¼ uj ;
ðA7Þ
where uj denotes prescribed displacement components. Taking r0ij ¼ 0 except for r011 – 0 and neglecting body forces, if plane elastic problems in the xoz-plane are restricted, from (A6) we get the equilibrium Eqs. (1) and (2). References [1] Williams ML. On the stress distribution at the base of stationary crack. J Appl Mech 1957;24:109–14. [2] Larsson SG, Carlsson AJ. Influence of non-singular stress terms and specimen geometry on small scale yielding at crack tips in elastic–plastic materials. J Mech Phys Solids 1973;21:263–77. [3] Ayatollahi MR, Pavier MJ, Smith DJ. Determination of T-stress from finite element analysis for mode I and mixed mode I/II loading. Int J Fract 1998;91:283–98. [4] Ayatollahi MR, Pavier MJ, Smith DJ. Mode I cracks subjected to large T-stresses. Int J Fract 2002;117:159–74. [5] Fett T, Rizzi G, Bahr H-A. Green’s functions for the T-stress of small kink and fork cracks. Engng Fract Mech 2006;73:1426–35. [6] Li X-F, Xu LR. T-stresses across static crack kinking. ASME J Appl Mech 2007;74:181–90. [7] Tvergaard V. Effect of T-stress on crack growth under mixed mode I–III loading. Int J Solids Struct 2008;45:5181–8. [8] Li X-F. Dynamic T-stress for a mode-I crack in an infinite elastic plane. ASME J Appl Mech 2007;74:378–81. [9] Li X-F, Liu G-L, Lee KY. Effects of T-stresses on fracture initiation for a closed crack in compression with frictional crack faces. Int J Fract 2009;160:19–30.
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