Fatigue growth simulation of part-through flaws in thick-walled pipes under rotary bending

International Journal of Fatigue 22 (2000) 1–9 www.elsevier.com/locate/ijfatigue

Fatigue growth simulation of part-through flaws in thick-walled pipes under rotary bending Andrea Carpinteri *, Roberto Brighenti, Andrea Spagnoli Department of Civil Engineering, University of Parma, Parco Area delle Scienze 181/A, 43100 Parma, Italy Received 19 May 1999; received in revised form 20 September 1999; accepted 20 September 1999

Abstract The fatigue growth of a circumferential external surface flaw in a thick-walled round pipe subjected to rotary bending is simulated numerically. A three-dimensional finite element analysis is carried out to obtain the stress–strain field of the cracked structural component for any position of the flaw. The crack front is assumed to have an elliptical-arc shape during the whole propagation, as has been deduced from several experimental investigations. The results for rotary bending are compared to those for cyclic bending.  1999 Elsevier Science Ltd. All rights reserved. Keywords: Round pipe; Circumferential external surface flaw; Rotary bending; SIF (Stress–Intensity Factors); Fatigue crack growth; Fatigue fracture mechanics

1. Introduction The fatigue behaviour of surface flaws in round bars has been examined by several authors (for example, see [1–7]), and a few studies have been carried out on circumferential edge flaws in pipes [1,2,8–14]. Structural hollow cylinders are subjected to cyclic loading in many practical applications. For example, tubular components are frequently used as columns supporting offshore structures, and fatigue loads resulting from wave action are significant on them [15]. In order to analyse the fatigue strength of hollow cylinders, rotary bending tests are usually performed, that is to say, the bending moment M is applied to a pipe while it rotates around its axis. Consequently, the position of a surface flaw continuously changes due to the rotational movement of the cylinder. The same problem but for round bars has recently been examined in [16,17]. In the present paper, a circumferential external surface flaw in a metallic round pipe under rotary bending is considered (Fig. 1). The crack front is assumed to be an elliptical arc during the whole propagation, as several

experimental investigations have shown [18,19], and therefore the flaw configuration can be described by the relative crack depth x=a/t in correspondence to the deepest point A on the crack front, and the flaw aspect ratio a=a/b (Fig. 2). The position of the flaw is identified by the parameter J which measures the angle between the loading axis l–l (perpendicular to the vector M) and the symmetry axis c–c; such angle is assumed to be positive if it is clockwise from l to c. The ratio R*=R/t between the internal radius of the pipe and its wall thickness is assumed to be equal to 1 (thick-walled pipe), but the results presented in the following change only slightly for 0.5ⱕR*ⱕ2. The finite element method and the superposition principle are applied to compute the stress–intensity factor (SIF) distribution along the front of the defect for any value of the angle J. The fatigue crack growth is numerically simulated by employing a two-parameter theoretical model [3] based on the Paris–Erdogan law. It is shown that these flaws tend to follow preferred propagation paths in the diagram of a against x, with negligible differences when comparing the results obtained for rotary bending to those for cyclic bending.

* Corresponding author. Tel.: +39-0521-905922; fax: +39-0521905924. E-mail address: [email protected] (A. Carpinteri) 0142-1123/00/$ - see front matter  1999 Elsevier Science Ltd. All rights reserved. PII: S 0 1 4 2 - 1 1 2 3 ( 9 9 ) 0 0 1 1 5 - 2

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Nomenclature a a,b c–c D f1, f2 h

crack depth for the deepest point A on the flaw front semi-axes of the ellipse geometrical axis of symmetry external diameter of the pipe influence functions for case nos. 1 and 2, respectively distance of the intersection point B (between crack front and external surface) from the axis of symmetry KI,M, K*I,M stress–intensity factor (SIF) and dimensionless SIF under bending, respectively l–l axis of loading M total bending moment Mx, My components of M along the x- and y-axis, respectively R internal radius of the pipe R*=R/t dimensionless radius of the pipe Rf =smin/smax=loading ratio t wall thickness of the pipe x,y principal axes a=a/b aspect ratio of the elliptical-arc crack front at crack aspect ratio of transition b intersection angle between the crack front and external surface ⌬sM,⌬KI,M stress range and stress–intensity factor range sM maximum bending stress for an uncracked round pipe J angular coordinate of the generic position of the symmetry axis c–c with respect to the loading axis l–l JKmax value of the rotation angle J for which the stress–intensity factor at the generic point P on the crack front attains its maximum value x=a/t relative crack depth for point A on the flaw front z,z*=z/h coordinate and normalized coordinate of the generic point P along the crack front, respectively

Fig. 1.

Part-through cracked round pipe under rotary bending.

冑 (J⫽90°,x,a,z∗)⫽f (x,a,z∗) s 冑pa,

2. Mechanical behaviour of cracked pipes

KI(1)(J⫽0°,x,a,z∗)⫽f1(x,a,z∗) sM pa,

(1)

2.1. Bending moments Mx and My

KI(2)

(2)

Consider a thick-walled pipe with an elliptical-arc surface flaw in the centre-line cross section. The generic point P along the crack front is identified by the normalized coordinate z*=z/h, that ranges from 0 to 2 (Fig. 2). The stress–intensity factors under bending moment M for two particular cases (J=0°, i.e. M⬅Mx, case no. 1; J=90°, i.e. M⬅My, case no. 2) can be calculated at point P through the following expressions:

where sM=M/[p((D/2)4⫺R4)/2D] is the maximum bending stress due to M for an uncracked round pipe, while f1 and f2 are the dimensionless stress–intensity factors. The values of f1 (case no. 1) have been reported in [13], whereas the values of f2 (case no. 2) are computed here from the displacements near the flaw. Such displacements are obtained from a three-dimensional finite element analysis, by employing 1840 20-node solid

2

M

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Fig. 2.

3

Components of the bending moment M along the principal axes x and y, and geometrical parameters.

elements and 8881 nodes to model a quarter of the pipe (Fig. 3). Collapsed quarter-point finite elements (wedge elements) are used near the crack front to induce a square-root singularity of the stress–strain field; such elements do not generally produce reliable results in two small zones near points B and B⬘ (see Fig. 2) [4,20] and, therefore, f2 is only determined for 0.10ⱕz*ⱕ1.90. All

the results deduced are shown in Table 1, where x ranges from 0.1 to 0.8, while a is made to vary from 0.0 to 1.2. 2.2. Rotary bending as a superposition of Mx and My In general, the vector M under rotary bending is inclined with respect to the principal axes x and y (Fig. 2), and can be resolved into two components, Mx and My: Mx⫽M cosJ

(3)

My⫽M sinJ Consequently, the stress–intensity factor for this case can be obtained from a simple linear combination of the above SIFs [Eqs. (1) and (2)] related to the two particular situations previously described (case nos 1 and 2, respectively): KI,M(J,x,a,z∗)⫽KI(Mx)⫹KI(My)

(4)

⫽cosJ KI(1)(J⫽0°,x,a,z∗)⫹sinJ KI(2)(J⫽90°,x,a,z∗) If all the terms of such an equation are divided by sM√pa, the following dimensionless expression is obtained: K ∗I,M(J,x,a,z∗)⫽f1(x,a,z∗) cosJ⫹f2(x,a,z∗) sinJ

Fig. 3.

Finite element mesh for a quarter of the cracked pipe.

(5)

The above dimensionless SIF along the crack front is shown in Fig. 4 for x=0.5 and different values of J, in the case of a equal to 0.2 (a) or 1.0 (b). Note that the maximum (or minimum) value of a given curve is attained for z∗=z¯ ∗, where z¯ ∗ strongly depends on the ∗ parameter J. The negative values of K I,M in Fig. 4, determined by considering the problem as a linear-elastic one, represent the compressive stress condition (the same remark can be made for the negative SIFs in Fig. 6), and will be neglected in the following when the fatigue crack growth is examined. ∗ Then the stress–intensity factor K I,M against z* is plotted in Fig. 5 for x=0.2 or 0.8 and J=30° or 60°, with the flaw aspect ratio a ranging from 0.0 to 1.2. It can

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Table 1 Influence function f2 (J=90°) along the crack front. Points from A through to C⬘ have a coordinate x*=1.00, 1.25, 1.50, 1.75 and 1.90, respectively R*=R/t=1 a=a/b 0.0

0.2

0.4

0.6

0.8

1.0

1.2

A L⬘ J⬘ N⬘ C⬘ A L⬘ J⬘ N⬘ C⬘ A L⬘ J⬘ N⬘ C⬘ A L⬘ J⬘ N⬘ C⬘ A L⬘ J⬘ N⬘ C⬘ A L⬘ J⬘ N⬘ C⬘ A L⬘ J⬘ N⬘ C⬘

0.1

0.2

0.3

0.4

0.5

0.6

0.7

x=a/t 0.8

0.000 0.034 0.062 0.078 0.068 0.000 0.022 0.041 0.051 0.046 0.000 0.011 0.021 0.028 0.030 0.000 0.006 0.012 0.017 0.019 0.000 0.004 0.008 0.012 0.013 0.000 0.003 0.005 0.008 0.010 0.000 0.002 0.004 0.006 0.008

0.000 0.048 0.088 0.112 0.100 0.000 0.038 0.069 0.088 0.080 0.000 0.022 0.041 0.055 0.058 0.000 0.013 0.024 0.035 0.039 0.000 0.008 0.016 0.023 0.027 0.000 0.005 0.011 0.017 0.020 0.000 0.004 0.008 0.013 0.016

0.000 0.057 0.107 0.137 0.127 0.000 0.048 0.090 0.115 0.108 0.000 0.031 0.059 0.079 0.083 0.000 0.019 0.036 0.051 0.058 0.000 0.012 0.024 0.035 0.041 0.000 0.008 0.017 0.025 0.030 0.000 0.006 0.012 0.019 0.023

0.000 0.066 0.123 0.160 0.151 0.000 0.057 0.107 0.139 0.133 0.000 0.039 0.074 0.101 0.105 0.000 0.025 0.048 0.068 0.076 0.000 0.016 0.032 0.046 0.054 0.000 0.011 0.022 0.034 0.040 0.000 0.008 0.016 0.025 0.031

0.000 0.070 0.134 0.177 0.171 0.000 0.064 0.120 0.160 0.156 0.000 0.047 0.089 0.122 0.128 0.000 0.031 0.059 0.085 0.096 0.000 0.020 0.040 0.059 0.069 0.000 0.014 0.028 0.042 0.051 0.000 0.010 0.021 0.032 0.040

0.000 0.075 0.143 0.192 0.190 0.000 0.069 0.131 0.176 0.175 0.000 0.053 0.101 0.139 0.147 0.000 0.036 0.070 0.100 0.114 0.000 0.024 0.047 0.070 0.083 0.000 0.017 0.033 0.051 0.061 0.000 0.012 0.025 0.039 0.047

0.000 0.079 0.152 0.205 0.207 0.000 0.073 0.141 0.191 0.194 0.000 0.058 0.112 0.155 0.165 0.000 0.041 0.080 0.114 0.131 0.000 0.028 0.055 0.081 0.096 0.000 0.019 0.039 0.059 0.072 0.000 0.014 0.029 0.045 0.055

0.000 0.082 0.159 0.218 0.224 0.000 0.076 0.149 0.204 0.211 0.000 0.062 0.122 0.170 0.183 0.000 0.045 0.089 0.128 0.148 0.000 0.031 0.062 0.092 0.110 0.000 0.022 0.044 0.067 0.082 0.000 0.016 0.033 0.051 0.063

be observed that the curves present a downward concavity for low values of a, while the direction of the concavity changes for a greater than a transition value, at. Such a crack aspect ratio of transition decreases by increasing x: for instance, at is equal to about 0.8 for x=0.2, and about 0.6 for x=0.8 (Fig. 5). 2.3. Maximum stress–intensity factor at the generic point P along the crack front ∗ The dimensionless SIF K I,M against the rotation angle J is displayed in Fig. 6 for x=0.5 and different values of z*, in the case of a equal to 0.2 (a) or 1.0 (b). Note that the curves for high values of a [for example, a=1.0 in Fig. 6(b)] are very close to one another, that is to say, K ∗I,M is almost independent of z* in this case. On the other hand, the curves for low values of a [see a=0.2 in Fig. 6(a)] are more distant from one another than in the previous case.

The maximum value of a generic curve in Fig. 6 is attained for a characteristic rotation angle, called JKmax in the following, which depends on the parameter z*. In general, for a flaw configuration described by a given couple of values (x,a) and subjected to rotary bending, the maximum stress–intensity factor at point P, defined by the coordinate z*, is attained in correspondence to the angle JKmax, which can be computed by equating the following derivative to zero: ∂KI,M(J,x,a,z∗) ⫽⫺sinJK maxKI(1)(J⫽0°) ∂J

(6)

⫹cosJK maxKI(2)(J⫽90°)⫽0 that is

冉

JK max⫽arctg

KI(2)(J=90°,x,a,z∗) KI(1)(J=0°,x,a,z∗)

冊

(7)

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and in a dimensionless form (dividing all the terms by sM√pa): K ∗I,max(x,a,z∗)⫽f1(x,a,z∗)cosJK max

(9)

⫹f2(x,a,z∗)sinJK max

3. Fatigue behaviour under rotary bending 3.1. Theoretical background As is well-known, for surface flaws in round bars under cyclic bending loading, the flaw aspect ratio a is a function of the relative crack depth x [3–7], and an analogous conclusion has also been drawn for partthrough cracks in round pipes under the same type of loading [13]. Such analyses have considered a bending moment acting about a principal axis of the cylinder cross-section. Only a few studies have theoretically examined the fatigue behaviour of cracked bars subjected to rotary bending [16,17]. Now the propagation paths of circumferential surface flaws in thick-walled pipes under rotary bending are deduced in the diagram of a against x by applying a two-parameter theoretical model [3]. According to such a model, the crack front lying on an ellipse with semiaxes a and b grows after one cyclic loading step to the new configuration described by the following equation: x2 y2 ⫹ ⫽1 (b∗)2 (a∗)2

Fig. 4. Dimensionless stress–intensity factor along the crack front for x=0.5 and different values of J, in the case of (a) a=0.2 and (b) a=1.0.

⫽arctg

冉

f2(x,a,z∗) f1(x,a,z∗)

冊

By analysing this equation, it can be concluded that such a characteristic rotation angle depends on the point considered (i.e. z*), and this dependency is stronger for low values of a [Fig. 7(a)] than for high values of a [Fig. 7(b)]. Furthermore, it can be remarked that, for any couple of values (a, z*), JKmax increases by increasing x. Finally, the maximum SIF at the generic point P being examined can be computed from Eqs. (4) and (7): KI,max(x,a,z∗)⫽cosJK maxKI(1)(J⫽0°,x,a,z∗) ⫹senJK maxKI(2)(J⫽90°,x,a,z∗)

(8)

(10)

where the two unknowns a* and b* (semi-axes of the new crack front) can be determined through the condition that the coordinates of the new points A* and C*, obtained from those of the old points A and C by employing the Paris–Erdogan law, must satisfy Eq. (10). Details of this procedure have been discussed in [3]. In the following examples, the material constants Q and m of the fatigue law (da/dN=Q(⌬KI)m, with da/dN in mm cycle⫺1 and the SIF range ⌬KI in N mm⫺3/2) are assumed to be equal to 1.64×10⫺10 and 2, respectively. Consider a flaw configuration described by a fixed couple of values (x,a), subjected to rotary bending. The stress–intensity factor at a given point (z*) on the crack front ranges from a maximum value, KI,max(z*), which has previously been determined [see Eq. (8)], to a minimum value which is equal to ⫺KI,max(z*) (e.g. see Fig. 6). If the negative stresses are neglected, the loading ratio Rf is equal to 0, and the SIF range to be inserted into the Paris–Erdogan law is equal to: ⌬KI,rotary(z∗)⫽KI,max(z∗)

(11)

By assuming that the whole stress range affected the fatigue process, the SIF range would be equal to

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Fig. 5. Dimensionless stress–intensity factor along the crack front for different values of a, in the case of x=0.2 or 0.8 and J=30° or 60°.

2KI,max(z*). Consequently, for a given initial flaw, the number of loading cycles to reach a certain crack front configuration would obviously differ from that of the previous case, but the flaw propagation paths in the diagram of a against x would not change, as has been discussed in [3]. Under cyclic bending with a given value, ␽¯ , of the angle between l- and c-axis, the SIF range at the same point considered above (z*) is equal to: ¯ ,z∗) ⌬KI,cyclic(␽¯ ,z∗)⫽(1⫺Rf)KI,M(␽ (12) where KI,M(␽¯ ,z∗) is computed from Eq. (4). In the following, the crack growth under rotary bending is compared to that under cyclic bending with Rf=0 and J=␽¯ =0°. 3.2. Surface flaw propagation and intersection angle b The flaw propagation paths for rotary bending with stress range ⌬sM=100 N mm⫺2 (continuous curves), deduced through the theoretical model proposed in [3], are displayed in Fig. 8. The initial crack configurations considered have a relative crack depth x0 equal to 0.1 and a crack aspect ratio a0 equal to 0.001 (nearly straight

front, case no. 1), 1.2 (case no. 7) or an intermediate value (cases nos 2–6). The crack growth paths under cyclic bending with Rf=0 and J=0° (dashed curves in Fig. 8) are almost coincident with those under rotary bending. That occurs because, for a given point (z*) on a certain flaw (x,a), the maximum value of SIF under rotary bending is attained for a small value of J and is only slightly greater than the maximum SIF computed for J=0° under cyclic bending. This remark is valid for any point and flaw configuration (e.g. see Fig. 6), except point A (z*=1.00) where the maximum SIFs for the two loading cases above coincide to each other. Note that, for round bars, the flaw propagation paths under cyclic bending remarkably differ from those under rotary bending, as has been concluded in [17]. All the crack growth paths in Fig. 8 tend to converge to a horizontal asymptote with a between 0.7 and 0.8, that is, the fronts of the seven flaws examined become nearly circular-arc-shaped. Obviously the theoretical propagation paths are real only for x0ⱕxⱕxf, where xf is the relative crack depth for which KI,M is equal to the fatigue fracture toughness KfC of the metallic material of the pipe. The equations of two representative continuous

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Fig. 8. Crack growth paths for seven initial flaws under rotary bending or cyclic bending with Rf=0 and J=0°.

for point 6 (0.1; 1.0), i.e. an initial circular-arc-fronted flaw, a⫽1.093⫺1.317x⫹3.712x2⫺5.260x3⫹2.597x4. Fig. 6. Dimensionless stress–intensity factor against rotation angle for x=0.5 and different values of z*, in the case of (a) a=0.2 and (b) a=1.0.

(14)

The intersection angle b (Fig. 2) is displayed in Fig. 9 for rotary bending (continuous curves), in the case of the seven crack growth paths plotted in Fig. 8. It can be observed that b tends quite rapidly to about 90°,

Fig. 7. Rotation angle JKmax (for which KI,max is attained at a given point, z*) against normalized coordinate z* for different values of x, in the case of (a) a=0.2 and (b) a=1.0.

curves in Fig. 8 (in the range 0.1ⱕξⱕ0.8) are reported here: for point 1 with coordinates (x0, a0) = (0.1; 0.001), i.e. an initial straight-fronted flaw, a⫽⫺0.221⫹3.112x⫺3.800x2⫹1.627x3;

(13)

especially for high values of a0. The intersection angle b for cyclic bending (dashed curves in Fig. 9) is almost coincident with that for rotary bending, and that occurs for the same reasons discussed above, when the crack growth paths have been analysed. Finally, the fatigue propagation under rotary bending for two initial defects considered in Figs. 8 and 9 is also

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A. Carpinteri et al. / International Journal of Fatigue 22 (2000) 1–9

subjected to rotary bending has been simulated numerically.

Fig. 9. Crack front intersection angle for the crack growth paths in Fig. 8, in the case of rotary bending or cyclic bending with Rf=0 and J=0°.

analysed in Fig. 10. For both cases examined, the parameters x0 and a0 are enclosed within parentheses, and the flaw propagation stages are plotted for x from 0.1 to 0.8, with relative crack depth increments equal to 0.1. Furthermore, the loading cycle numbers (divided by 103) to reach the stages x=0.4 and 0.8, respectively, are shown. Note that, for a given value of x0 (for example, x0=0.1 in Fig. 10), each propagation stage is reached after a cycle number which increases by increasing a0. 4. Conclusions The behaviour of an elliptical-arc circumferential external flaw in a thick-walled round pipe (0.5ⱕR*ⱕ2)

1. The stress–intensity factor along the crack front can be determined for any rotation angle J between the loading axis l and the symmetry axis c, through a linear combination of the influence functions f1 and f2. These dimensionless SIFs, which are related to two particular cases (J=0°, case no. 1; J=90°, case no. 2), have been computed by applying the finite element method. 2. For any value of the relative crack depth x=a/t and the rotation angle J, a transition is observed in the diagram of the stress–intensity factor against the normalized coordinate z*=z/h: the different curves present a downward concavity for low values of a=a/b, while the direction of the concavity changes for a greater than a transition value, at, which decreases by increasing x. 3. For a given couple of values (x,a), the maximum SIF is attained in correspondence to a value of z* which strongly depends on the rotation angle J. On the other hand, the maximum stress–intensity factor at a given point (z*) on the crack front and the rotation angle for which such a maximum value of SIF is attained can be computed if the influence functions f1 and f2 are known. 4. The crack growth paths under rotary bending have been deduced by employing a two-parameter theoretical model discussed in [3]. Such paths are almost coincident with those under cyclic bending with Rf=0 and J=0°. During crack propagation, the elliptical-arc front of the flaw becomes nearly circular-arc-shaped, and the intersection angle b between the crack front and the external surface tends to values very close to 90°.

Fig. 10. Fatigue propagation stages under rotary bending for the initial defects nos 1 and 6 in Fig. 8. The loading cycle numbers (divided by 103) to reach the stages x=0.4 and 0.8, respectively, are also displayed.

A. Carpinteri et al. / International Journal of Fatigue 22 (2000) 1–9

Acknowledgements [9]

The authors gratefully acknowledge the research support for this work provided by the Italian Ministry for University and Technological and Scientific Research (MURST) and the Italian National Research Council (CNR).

[10] [11]

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