Stress Intensity Factors for Cracked Rectangular Cross-Section Thin-Walled Tubes Y.J

Engineering Fracture Mechanics 71 (2004) 1501–1513 www.elsevier.com/locate/engfracmech

Stress intensity factors for cracked rectangular cross-section thin-walled tubes Y.J. Xie *, X.H. Wang, Y.C. Lin Department of Mechanical Engineering, Liaoning University of Petroleum & Chemical Technology, 1 West Dandong Road, Fushun 113001, LN, PR China Received 14 January 2003; received in revised form 24 June 2003; accepted 3 July 2003

Abstract For cracked structural rectangular thin-walled tubes, an exact and very simple method to determine the stress intensity factors has been proposed based on a new concept of crack surface widening energy release rate. Unlike the classical crack extension energy release rate, the crack surface widening energy release rate can be expressed by the G- integral and elementary strength theory of materials for slender cracked structures. From present discussions, a series of new and exact solutions of stress intensity factors are derived for cracked rectangular and square tubes. The present method can also be applied to cracked polygon thin-walled tubes. Ó 2003 Elsevier Ltd. All rights reserved.

Keywords: Stress intensity factor; Fracture; Mechanics; Tubes

1. Introduction

As a typical and important engineering structural components, the square, rectangular and polygon thin-walled tubes are widely used. The cracks in these thin-walled tubes have naturally received consid- erable attentions. However, because the thin-walled tubes belong to three-dimensional finite boundary problems or beam-like structures, it is very difficult to get the exact solutions of stress intensity factor from the existing classical methods. But, the endeavor to find simple and effective methodology for slender members has never stopped [1,2]. In recent years, G-integral was proposed [3], which came from conservation law and the concept of crack mouth widening energy release rate. The basic characters of this method are simple and within the framework of elementary strength theory of materials. For the two-dimensional elasto-static boundary value problems, the conservation law Jk have two components [4–7]. If crack surface is parallel to axis x1, J1 can be used as J-integral theory to describe crack extension energy release rate and J2 as G -integral theory to describe the crack mouth widening energy release rate [3]. The natures of two forms of energy release rate are the same, i.e., the energy release rate per

* Corresponding author. E-mail address: [email protected] (Y.J. Xie).

unit translation of crack boundary along different direction x1 or x2. J- and G -integral theories yield KI using different components of stress and strains for the same crack problems. To determine the KI of cracked beams, for example, stress and strain fields needed in J-integral method should be usually deter- mined by numerical analysis, but the stress and strain fields needed in G-integral method can be given by the bending theory in mechanics of materials. For finite boundary crack problems, the G-integral method is usually simpler than J-integral in estimates of stress intensity factors. For three-dimensional problems, conservation law Jk has three components. When axis x2 is perpendicular to the crack surface, the component J2 can be also used as G -integral to express the crack surface widening energy release rate and then to determine stress intensity factors. The purpose of present work is to propose a method and technique to determine a series of stress intensity factors for cracked rectangular and polygon thin-walled tubes.

2. Three-dimensional G -integral

Consider a three-dimensional deformation ﬁeld for which the displacement ui depends on x1, x2 and x3. From the conservation law, the three-dimensional G-integral or crack surface widening energy release rate, similar to the two-dimensional case [3], can be deﬁned as ZZ G ¼ ðwn2 Tiui;2ÞdX: ð1Þ X

The X in Eq. (1) is one of the crack surfaces. It is perpendicular to x2-axis. It can be also a curved surface and its boundary is the rim of the crack. This curved surface and crack surface may form a closed one [10,11]. w is the strain energy density; Ti the stress vector acting on the outer side of X; n the unit outward normal to X and n2 the projection of n on x2; ui;2 ¼ oui=ox2. Referring Fig. 2, the meaning of G -integral is the energy release rate per unit translation of crack surface Xdi in x2-direction or crack surface widening energy release rate when it is used to solve crack problems. For a two-dimensional deformation ﬁeld, the counterpart of Eq. (1), the crack mouth widening energy release rate [3], is Z G ¼ ðwn2 Tiui;2Þds; ð2Þ s where s in Eq. (2) is a curve in the x1x2 plane. For mode I crack with a unit thickness, shown in Fig. 1, let s in Eq. (2) represent to the path sdef next to the upper crack tip region. Note that sde is a straight line and sef is a quarter of a circle. Along path s and s , Eq. (2) yields [3] Z de ef

ðwn2 Tiui;2Þds ¼ 0 ð3Þ sde

n e

x2 n f d x1

Crack tip

Fig. 1. Integration path next to crack tip region. Y.J. Xie et al. / Engineering Fracture Mechanics 71 (2004) 1501–1513 1503 and Z 2 2 ð1 l ÞKI ðwn2 Tiui;2Þds ¼ ðplane strainÞ; ð4Þ sef 2pE where l is PoissonÕs ratio, E theH elastic modulus and KI the stress intensity factor for mode I loads. For a closed path s ¼ sde þ sef sdf , ðwn2 Tiui;2Þds ¼ 0 from the conservation law. It follows that I Z s Z Z 2 2 ð1 l ÞKI ðwn2 Tiui;2Þds ¼ ðwn2 Tiui;2Þds ðwn2 Tiui;2Þds ¼ ðwn2 Tiui;2Þds ¼ 0: sdef sdf 2pE sdf s ð5Þ Then Z 2 2 ð1 l ÞKI G ¼ ðwn2 Tiui;2Þds ¼ : ð6Þ sdf 2pE

Eq. (6) can be interpreted as the energy release per unit moving of boundary sdf in x2-direction or the crack mouth widening energy release rate.

3. Single-edge crack in rectangular thin-walled tube under lateral loads and tension

A cracked rectangular thin-walled tube is shown in Fig. 2. The x1x2 plane in Fig. 2 is a plane of symmetry for cracked tube, and all loads act in this plane; hence the bending deﬂections will take place in this plane also. The q is transverse load per unit length; Q the transverse shear force (Qþ ¼ 0 for mode I crack); M the bending moment; N the axial force; yc the position of neutral axis in cracked cross-sectional area and yc ¼ 0:5hða=bÞ=ð1 þ h=b a=bÞ. The symbol Ô+Õ denotes the cracked cross-section; Ô)Õ the remote uncracked cross-section. In this section, the energy release rate of crack surface widening will be derived by the G- integral method and elementary method in mechanics of materials respectively.

3.1. G-integral method for single-edge crack

In cracked cross-sectional area there are a crack and two two-dimensional singular stress ﬁelds discussed in above section next to the crack tip zones. If Xdfgi denotes the right surface of crack, from Eqs. (1) and (6) the crack surface widening energy release rate can be given by (see Figs. 1–3)

b Q- M- N t M+ Q+ y X N c 2 h X3

X3 i d 2a i d X1 Ωdi, crack surface X1

Fig. 2. Cracked tube with one-half symmetry subjected to bending and tension for mode I crack. 1504 Y.J. Xie et al. / Engineering Fracture Mechanics 71 (2004) 1501–1513 ZZ Z ZZ ZZ 2 2 tð1 l ÞKI G ¼ ðwn2 Tiui;2ÞdX ¼ 2t ðwn2 Tiui;2Þds þ wdX ¼ þ wdX: Xdfgi sdf Xfg pE Xfg ð7Þ

This expression means the energy release rate per unit translation of crack surface Xdfgi in x2-direction or the crack surface widening energy release rate of half cracked tube in Fig. 2.

3.2. Elementary method for single-edge crack

Considering the crack in tube as an elliptical hole shown in Fig. 4 under the condition of c ! 0 [3,8–10], from bending theory, the strain energy is Z Z N 2 c N 2 dx c ðM þ Qþx 0:5qx2Þ2 U ¼ ðl cÞþ 2 þ 2 2 dx 2EA 2EA c ða=b; x =cÞ 2EIc ða=b; x =cÞ 2 Z 0 1 2 0 2 2 lc 2 2 ðM þ Q x2 0:5qx2Þ þ dx2; ð8Þ 0 2EI

1 3 3 2 where A ¼ 2ðb þ hÞt, I ¼ 6 ½th þ bt þ 3bth for remote uncracked cross-section and

i h Ω g Crack surface dfgi motion n x2 f Crack e d

Fig. 3. Local wall of tube around crack.

Section A-A M+ Q+ Q- M- N yc X2 N 2a X2

A A

c 0 c 0 l X1 X3

Fig. 4. An elliptical hole model for cracked tubes (c ! 0) [8]. Y.J. Xie et al. / Engineering Fracture Mechanics 71 (2004) 1501–1513 1505 rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a x a x 2 h c ; 2 ¼ 1 1 2 1 þ ð9Þ 1 b c b c b rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a x 3 þðt=bÞ2 a x 2 3 ða=bÞ2ð1 x2=c2Þ c ; 2 ¼ 1 1 2 p2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi : 2 b c 2 b c 2 2 2 3 þ h=b þðt=bÞ 3 þ h=b þðt=bÞ 1 þ h=b ða=bÞ 1 x2=c ð10Þ Making use of ClapeyronÕs theorem, the work of the external loads, V ¼ 2U. Potential energy P ¼ U V . Then the crack surface widening energy release rate can be given by the following equation from elementary theory in mechanics of materials: h i o 2 þ 2 2 2 P N a ðM Þ a ½M þ Q l 0:5ql G ¼ lim ¼ c1 1 þ c2 ; ð11Þ c!0 oc 2EA b 2EI b 2EI where by introducing the integral variable n ¼ x2=c and b ¼ 1 þ h=b, sffiffiffiffiffiffiffiffiffiffiffiffi1 Z a 1 2 b þ a dn qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2b1 1 b bÀÁ1p c1 ¼ ¼ ÀÁ ÀÁarctan ; ð12Þ 2 a a b 0 c1ða=b; nÞ a 2 a b1 b 2 b b b1 b Z 1 a dn bt 2 þ b1 c2 ¼ ¼ b 0 c2ða=b; nÞ 2ð1 þ 2b1Þ 2 vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 u 2 6 u1 þ 2ð1þ2b1Þ a 7 3b1ð2 þ b1Þ 6 1 t b1ð2þb1Þ b p 7 þ ÀÁ4 rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi arctan 5: ð13Þ 2 a 2ÀÁ 2ð1þ2b1Þ a 2ð1 þ 2b Þ 1 2b 2 1 4 1 b 4 þ 1 a b1ð2þb1Þ b 1 2 2þb b b1 1 Additionally, from equilibrium condition, Eq. (11) can be rearranged as h i N 2 a ðM þÞ2 a ½M þ Ny 2 G ¼ c 1 þ c c : ð14Þ 2EA 1 b 2EI 2 b 2EI

3.3. Stress intensity factors

Eqs. (7) and (14) are the crack surface widening energy release rate based on diﬀerent deﬁnitions, but the result should be the same. Let Eq. (7) be equal to Eq. (14), it follows that ZZ 2 2 2 h i þ 2 þ 2 tð1 l ÞKI N a ðM Þ a ½M Nyc þ wdX ¼ c1 1 þ c2 : ð15Þ pE Xfg 2EA b 2EI b 2EI Because of free action of crack surface, the integral in left-hand side of Eq. (15) is a small quantity, which can be neglected. Hence, stress intensity factor is ( " #) 1=2 p N 2 a ðM þÞ2 a ðM þ Ny Þ2 K ¼ c 1 þ c c : ð16Þ I 2tð1 l2Þ A 1 b I 2 b I

Eq. (16) indicates that stress intensity factor for cracked tubes depend only on axial force N, bending þ moment M in cracked cross-sectional area, geometrical factors c1ða=bÞ and c2ða=bÞ. It represents actually a series of closed form solutions of stress intensity factors for cracked rectangular thin-walled tubes under diﬀerent loads such as bending, tension, three-point-bending, distributed load and combined loads. 1506 Y.J. Xie et al. / Engineering Fracture Mechanics 71 (2004) 1501–1513

1.2

Present method 1.0 FEM

0.8 1/2 b) π (

o 0.6 σ / I K 0.4

0.2 Normalized stress intensity factor

0.0 0.0 0.1 0.2 0.3 0.4 0.5 Normalized crack length a/b

Fig. 5. Stress intensity factor for single-edge cracked tube under bending, h ¼ b and t=b ¼ 0:1.

In the case of pure bending, from Eq. (16), following stress intensity factor can be given by pffiffiffiffiffiffi a K ¼ r pbf ; ð17Þ I 0 b þ where r0 ¼ M ðh=2Þ=I and the normalized stress intensity factor " # 1=2 a 3 þ h=b þðt=bÞ2 a f ¼ c 1 : ð18Þ b 3ð1 l2Þ 2 b

Fig. 5 compares the results calculated by Eq. (18) against the results by finite element method (FEM). In FEM analysis, ANSYS6.1 Finite Element Package had been used and the eight-node structural shell element had been selected. Around the crack tip, we used a highly refined mesh where the tip was surrounded by six-node triangular quarter-point elements. More than 10,000 nodes and 7000 elements had been used for meshing a quart of symmetrical structures. Stress intensity factors were computed using a displacement extrapolation method with the crack-flank displacements.

4. Double-edge cracks in rectangular thin-walled tube

4.1. Double-edge cracks in rectangular thin-walled tube under bending

A rectangular thin-walled tube with double-edge cracks is illustrated in Fig. 6. There are four singular stress ﬁelds next to the upper and lower tips of two cracks. Though tensile stress exists below the neutral axis and compressive stress above, but ðKIÞlow ¼ðKIÞupper ¼ KI. Then the crack surface widening energy release rate due to the two cracks can be given by (see Figs. 1, 3 and 6) ZZ Z ZZ G ¼ 2 ðwn2 Tiui;2ÞdX ¼ 4t ðwn2 Tiui;2Þds þ 2 wdX X s X dfgi ZZ df fg 2tð1 l2ÞK2 ¼ I þ 2 wdX: ð19Þ pE Xfg Y.J. Xie et al. / Engineering Fracture Mechanics 71 (2004) 1501–1513 1507

- - t M+ Q+ Q M

X2 h 2a X 3

c 0 X1 l X1

Fig. 6. Double-edge cracked tube with one-half symmetry subject to lateral loads.

Because the tubes can be regarded as a slender member shown in Fig. 6, the strain energy is Z Z lc 2 2 c þ þ 2 2 ðM þ Q x2 0:5qx2Þ ðM Q x2 0:5qx2Þ U ¼ dx2 þ dx2; ð20Þ 0 2EI 0 2EIc3ða=h; x2=cÞ where a x 8 a 3 x 2 3=2 c ; 2 ¼ 1 ÀÁ 1 2 : ð21Þ 3 h c b t2 h c 1 þ h 3 þ h2 From the bending theory of beams, the crack surface widening energy release rate is h i o 2 2 þ 2 þ 2 P ðM þ Q l 0:5ql Þ ðM Þ a ðM Þ a G ¼ lim ¼ þ c3 ¼ c3 1 ; ð22Þ c!0 oc 2EI 2EI h 2EI h where ÀÁ "#ÀÁ Z 3 3 2 a 1 dn 3p a 16 8 a c ¼ 1 þ ÀÁh þ ÀÁh : ð23Þ 3 h c ða=h; nÞ 2 b t2 35 b t2 0 3 1 þ h 3 þ h2 1 þ h 3 þ h2 Let Eq. (19) be equal to Eq. (22), it yields pffiffiffiffiffiffi a K ¼ r phf ; ð24Þ I 0 h

þ where r0 ¼ M ðh=2Þ=I and normalized stress intensity factor " ÀÁ !# 3 1=2 a 1 p a 3 2048 a f ¼ 1 þ ÀÁh : ð25Þ h 2 ð1 l2Þ h 105p b t2 1 þ h 3 þ h2 Fig. 7 gives a comparison between Eq. (25) and FEM for pure bending. Another case of double-edge cracked rectangular thin-walled tube is shown in Fig. 8. There are also four singular plane strain fields in regions next to the crack tips. By similar procedure of Eq. (19) to Eq. (25), following result can be found: 0 1 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiÀÁ 2 2 þ2 a 2tð1 l ÞK M B 2 1 þ b2 p C I ¼ @ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiÀÁarctan ÀÁb ÀÁ 1A; ð26Þ 2 a a pE 2EI 2 a 1 b2 b 2b2 b b2 1 b2 b 1508 Y.J. Xie et al. / Engineering Fracture Mechanics 71 (2004) 1501–1513

0.5

0.4 Present method FEM

1/2 0.3 h) π ( o σ / I

K 0.2

0.1 Normalized stress intensity factor

0.0 0.0 0.1 0.2 0.3 0.4 0.5 Normalized crack length a/b

Fig. 7. Stress intensity factor for double-side cracked tube under bending, h ¼ b and t=b ¼ 0:1.

q b

- + + Q- M t M Q Ω c2

X2 h X3 Ωin Ω+ + - i d Ω Ω Ω out Ωc1 c 0 X1 l X1

Fig. 8. Another case of double-edge cracked square thin-walled tube. where t2 23þ h2 b : 27 2 ¼ h t2 ð Þ 3 þ b þ h2 Then pffiffiffiffiffiffi a K ¼ r pbf ð28Þ I 0 b and the normalized stress intensity factor 2 0 13 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiÀÁ 1=2 h t2 a a 6 3 þ b þ h2 B 2 1 þ b2 b p C7 f ¼ 4 @ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiÀÁarctan ÀÁ ÀÁ 1A5 : ð29Þ 2 2 a a b 6ð1 l Þ 2 a 1 b2 b 2b2 b b2 1 b2 b Fig. 9 compares the results calculated by Eq. (29) against that by FEM for bending. Y.J. Xie et al. / Engineering Fracture Mechanics 71 (2004) 1501–1513 1509

1.5

Present method 1.2 FEM

1/2 0.9 b) π ( o σ /

I 0.6 K

0.3 Normalized stress intensity factor

0.0 0.0 0.1 0.2 0.3 0.4 0.5 Normalized crack length a/b

Fig. 9. Stress intensity factor for double-side (top and bottom) cracked tube under bending, h ¼ b and t=b ¼ 0:1.

4.2. Double-edge cracks in rectangular thin-walled tube under tension

If a double-edge cracked rectangular thin-walled tube as shown in Fig. 8 is subjected to tension, the following crack mouth widening energy release rates can be given by G-integral and elementary method respectively: ZZ 2tð1 l2ÞK2 G ¼ I þ 2 wdX ð30Þ pE Xfg and 0 1 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiÀÁ 2 2 a N B b1 b1 þ 2 b b1p C G ¼ @ ÀÁqffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiÀÁarctan ÀÁ ÀÁ 1A: ð31Þ 2 a a 2EA a 2 a b1 2 b 4 b b b1 4 b

Then, the stress intensity factor pffiffiffiffiffiffi a K ¼ r pbf ; ð32Þ I 0 b where r0 ¼ N=A and normalized stress intensity factor 2 0 13 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiÀÁ 1=2 2 a a 6 b1 B b1 b1 þ 2 b b1p C7 f ¼ 4 @ ÀÁqffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiÀÁarctan ÀÁ ÀÁ 1A5 : ð33Þ 2 2 a a b 2ð1 l Þ a 2 a b1 2 b 4 b b b1 4 b

Comparisons between Eq. (33) and FEM are shown in Fig. 10. 1510 Y.J. Xie et al. / Engineering Fracture Mechanics 71 (2004) 1501–1513

1.0

Present method 0.8 FEM

1/2 0.6 b) π ( o σ / I 0.4 K

0.2 Normalized stress intensity factor

0.0 0.0 0.1 0.2 0.3 0.4 0.5 Normalized crack length a/b

Fig. 10. Stress intensity factor for double-side cracked tube under tension, h ¼ b and t=b ¼ 0:1.

5. Multi-cracks in square thin-walled tube under tension

This section will discuss two cases of square tubes with multi-cracks under tension as shown in Fig. 11. At first, for the case of four-edge cracks in Fig. 11(a), there are eight and same singular plane strain fields next to crack tips. The two forms of crack surface widening energy release rate are respectively ZZ 4tð1 l2ÞK2 G ¼ I þ 4 wdX ð34Þ pE Xfg and 0 1 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiÀÁ N 2 B 1 1 þ 2 a p C G ¼ @ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiÀÁarctan ÀÁb ÀÁ 1A: ð35Þ 2 a a 2EA a a 1 2 b 4 b b 1 4 b Then, stress intensity factor is pffiffiffiffiffiffi a K ¼ r pbf ; ð36Þ I 0 b

b b 2a a

t a t

b 2a X3 b X3

(a)X1 (b) X1

Fig. 11. Two cases of multi-crack conﬁgurations for square tube under tension: (a) four-edge cracked square tubes; (b) four-corner cracked square tubes. Y.J. Xie et al. / Engineering Fracture Mechanics 71 (2004) 1501–1513 1511

4.0

3.2 Present method FEM for edge cracks FEM for corner cracks 1/2 2.4 b) π ( o σ / I

K 1.6

0.8 Normalized stress intensity factor

0.0 0.0 0.1 0.2 0.3 0.4 0.5 Normalized crack length a/b

Fig. 12. Stress intensity factor for four-edge and four-corner cracked tube under tension, t=b ¼ 0:1. where r ¼ N=A and normalized stress intensity factor 0 2 0 13 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiÀÁ 1=2 a a 6 1 B 1 1 þ 2 b p C7 f ¼ 4 @ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiÀÁarctan ÀÁ ÀÁ 1A5 : ð37Þ 2 2 a a b 2ð1 l Þ a a 1 2 b 4 b b 1 4 b For four-corner cracked square tube under tension as shown in Fig. 11(b), it is not difficult to get the stress intensity factor by using the method in above sections. It is interested to note that the expressions of stress intensity factors in this case are the same with Eqs. (36) and (37). Eq. (37) is plotted and compared with FEM as shown in Fig. 12 for two cases of crack configurations.

6. A further theoretical illustration on present method

In fact, the present method in above sections can be evolved from the conservation law by means of basic assumptions and expressions of stress and strain in elementary strength theory of materials. Following is the details. For three-dimensional problems, if there is not any cave or crack in a closed surface Xclosed, the conservation law can be expressed as [4–7] ZZ

ðwn2 Tiui;2ÞdX ¼ 0; ð38Þ Xclosed which will be used to analyze crack problems in this section. Take a simple case of double-edge cracked tube under tension in Fig. 8 for example. Consider a closed þ surface Xclosed ¼ X þ X þ Xin þ Xout þ Xc1 þ Xc2 in Fig. 8. X denotes the remote uncracked cross- þ sectional area; X the cross-sectional area of crack ligament. Xin and Xout are inner and outer surfaces respectively; Xc1 and Xc2 the crack surfaces. Along these surfaces, following integral results can be found by using the elementary strength theory: ZZ

ðwn2 Tiui;2ÞdX ¼ 0; ð39Þ Xin 1512 Y.J. Xie et al. / Engineering Fracture Mechanics 71 (2004) 1501–1513 ZZ

ðwn2 Tiui;2ÞdX ¼ 0; ð40Þ Xout ZZ ðwn2 Tiui;2ÞdA ¼ w Nu~2;2; ð41Þ X ZZ þ þ ðwn2 Tiui;2ÞdA ¼ðw Nu~2;2Þ; ð42Þ Xþ where the u~i shows the displacement of neutral axis. This displacement can be given by elementary strength theory of materials. w is the strain energy density per unit length of tube. Note that n in above Eqs. (39)– (42) is outward and normal to the integral surfaces. If crack surfaces Xc1 and Xc2 are divided into three parts in Fig. 3, i.e., Xc1ðor Xc2Þ¼Xdf þ Xfg þ Xgi in which Xdf and Xgi belong to K-dominant regions of crack tip, from Eqs. (6) and (7), the following expressions can be given: ZZ ZZ 2 2 2tð1 l ÞKI ðwn2 Tiui;2ÞdA ¼ 2 wdX; ð43Þ Xc1þXc2 pE Xfg where the outward normal n of Xc1 and Xc2 is opposite to x2. Note that Xfg is free action and just outside K- dominant region. Also, in elementary strength theory of materials, one of the basic assumptions states that the transverse normal stresses r11 and r33 are never considered. This assumption does not cause significant error in relevant results. Hence, outside the K-dominant area r11 and r33 can be neglected for cracked box beams. Then the integral in right-hand side of Eq. (43) is negligible from the point of view of elementary theory. þ For the closed surface Xclosed ¼ X þ X þ Xin þ Xout þ Xc1 þ Xc2 in Eq. (38), the integral over such Xclosed must vanish. Substituting Eqs. (39)–(43) into Eq. (38), Eq. (38) can be rearranged as ZZ 2 2 2tð1 l ÞKI þ þ N þ þ 2 wdX ¼ðw Nu~2;2Þðw Nu~2;2Þ¼ ðu~2;2 u~2;2Þ: ð44Þ pE Xfg 2 At remote uncracked cross-section, it is not difficult to get the axial strain from elementary theory N u~ ¼ : ð45Þ 2;2 EA þ At the cracked cross-section, axial strain u~2;2 can be found [3,8] from the limit of average strain of beam with elliptical boundary and varying cross-section shown in Fig. 8: Z c þ 1 N dx2 u~2;2 ¼ lim c!0 c EA½1 2ða=bÞð1 x2=c2Þ=b 0 0 2 1 1 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiÀÁ 2 a N B b b1 þ 2 b1p C ¼ @ ÀÁqffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1 ÀÁarctan ÀÁb ÀÁA: ð46Þ 2 a a EA a 2 a b1 2 b 4 b b b1 4 b Substituting Eqs. (45) and (46) into Eq. (44), it will yield Eq. (33). The above discussion indicates that the basic idea of present method is to treat the conservation integral or conservation law expressed by Eq. (38) using elementary mechanics except the K-dominant regions for cracked slender structural members. Hence, this method has strong character of elementary method. Some errors are inevitable, but it is good enough for practical engineering applications. The method in this section facilitates to track where the errors come from. The physical meaning of left-hand side of Eq. (44) is Y.J. Xie et al. / Engineering Fracture Mechanics 71 (2004) 1501–1513 1513 energy release rate per unit translation of crack surfaces Xc1 and Xc2 in x2-direction. This is the physical and mathematical base of present method in previous sections.

7. Conclusions

In engineering structures a large number of thin-walled structural tubing are used. The existence of defects and cracks is usually inevitable and should cause unexpected failure. The relevant parameters are the stress intensity factors, which represent intensification of the stress fields in vicinity the crack tip. Be- cause the rectangular thin-walled tube belong to three-dimensional structures, any crack configuration would be difficult to treat using classical method. In present paper, an analytical method has been devel- oped to estimate the stress intensity factors for cracked rectangular tubes. A series of new solutions of stress intensity factors have been derived. Unlike classical method, the G-integral theory can be applied in an extremely simple manner not only to the cracked rectangular thin-walled tubes, but also to the cracked polygon thin-walled tubes.

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