Analysis of composite finite wedges under anti-plane shear

ARTICLE IN PRESS International Journal of Mechanical Sciences 51 (2009) 583–597

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Analysis of composite finite wedges under anti-plane shear Chih-Hao Chen a,, Chein-Lee Wang a, Chien-Chung Ke b a b

Department of Resources Engineering, National Cheng Kung University, Tainan 701, Taiwan, ROC Geotechnical Engineering Research Center, Sinotech Engineering Consultants, Inc., Taipei 110, Taiwan, ROC

a r t i c l e in fo

abstract

Article history: Received 22 December 2008 Received in revised form 2 June 2009 Accepted 5 June 2009 Available online 12 June 2009

Problems of composite finite wedges under anti-plane shear applied on a circular arc are analyzed in this study. The considered conditions of radial edges are free–free, free–fixed, and fixed–fixed. A procedure that uses the finite Mellin transform and the Laplace transform is developed to solve these problems. Explicit solutions for displacement and stress fields are derived. Stress intensity factors (SIFs) of composite circular shafts with an interfacial edge crack are extracted from the derived stress fields, and the distributions for various loading angles are presented and discussed. It was found that if the loading angles are the same, free–free and fixed–fixed edge problems can be degenerated into single material problems. Uniform stresses were found along the interface in free–free and fixed–fixed edge problems. Solutions of a general loading case deduced from the derived results compare well with those obtained from finite element (FE) analyses. & 2009 Elsevier Ltd. All rights reserved.

Keywords: Composite Finite wedge Anti-plane shear Stress intensity factor

1. Introduction Wedge problems have been investigated by many researchers. In-plane stress analysis of an isotropic wedge was first conducted by Tranter [1] using the Mellin transform of the Airy stress function in cylindrical coordinates. Thereafter, a number of associated in-plane problems were extended to bi-materials and anisotropic materials, with emphasis placed on the order of the stress singularity and the stress distribution near the wedge apex [2–9]. For these problems, out-of-plane normal stresses/strains are relative to in-plane stresses/ strains, and out-of-plane (anti-plane) shear stresses/strains are zero, which is due to the plane stress/strain condition being used. Studies which treated out-of-plane shear stresses/strains with zero in-plane stresses/strains used the generalized plane strain condition [10]. The wedges were usually regarded as sectors with infinite/finite radii using cylindrical coordinates. For cases with infinite radii [11–16], the order of the stress singularity as well as the stress distributions for a range near the wedge apex was determined. The corresponding SIFs were calculated using the assumption that the problems were in infinite domains. Nonetheless, the obtained SIF from the analysis of an infinite domain should be calibrated by a suitable geometric function to make it fit for an application. The true full-field stress distributions can be affected by all the boundaries in addition to the locations near the wedge apex. Accordingly, wedges with finite radii have been considered [17–21]. Stress analysis of a finite domain is helpful for the comprehensive understanding of a problem. Although there are some numerical methods [22,23] that can be used to determine the full-field stress distributions for a wedge problem, analytical expressions for a finite wedge under anti-plane shear deformation can be obtained using the methods of the integral transform [18]. SIF can then be represented by an equation with geometric factors for convenience. All previous studies, except for particular cases in Shahani [20] and Lin and Ma [21], considered a group of problems with materials subjected to anti-plane shear loads in radial directions. The conditions of the circular arc of the wedge were fixed or traction free. As a result, the solutions are accessible after the integral transforms. However, wedges with finite radii may be subjected to loads on the circular boundaries, especially in cases of crack problems. To this end, the present study considers another group of finite wedge problems. The tractions are subjected to a circular arc, and traction free or fixed conditions are imposed on the radial edges of the wedge. In the present work, a procedure that uses the finite Mellin transform in conjunction with the Laplace transform is proposed for solving the stated problems. Based on the proposed procedure with the prescribed point loads, fundamental solutions for this type of anti-plane shear problem of a composite finite wedge as well as the corresponding SIFs are obtained. Solutions for related general loading

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E-mail addresses: [email protected] (C.-H. Chen), [email protected] (C.-L. Wang), [email protected] (C.-C. Ke). 0020-7403/$ - see front matter & 2009 Elsevier Ltd. All rights reserved. doi:10.1016/j.ijmecsci.2009.06.002

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Nomenclature a AI, AII, AIII BI, BII, BIII c

radius of the finite wedge coefficients of solution in the Mellin domain coefficients of solution in the Mellin domain strip of regularity in the complex plane of the Mellin domain DI, DII, DIII determinants of two-by-two matrices (characteristic functions) differential of DII D0 II f(j) function for the finite Mellin transformation nðjÞ function after applying the finite Mellin transformaf2 tion F concentrated anti-plane shear force FE finite element ðIIÞ ðIIIÞ , H , H normalized mode III stress intensity factors HðIÞ III III III j material number (j ¼ 1, 2) ðIIÞ ðIIIÞ mode III stress intensity factors K ðIÞ III , K III , K III L[] applying the Laplace transform applying the inverse Laplace transform L1[] applying the finite Mellin transform of the second M2[] kind applying the inverse finite Mellin transform of the M1 2 [] second kind ODE ordinary differential equation p parameter in the kernel of the finite Mellin transform negative and positive roots of DL ¼ 0 (L ¼ I, II, III) p, p+ smallest positive root of DL ¼ 0 (L ¼ I, II, III) p1, L positive roots of DII ¼ 0 (n ¼ 1, 2, 3, y) pn, II P anti-plane shear pressure r cylindrical coordinate axis R ratio of material constants (m1/m2) Re[] real part s parameter in the kernel of the Laplace transform SIF(s) stress intensity factor(s) (III) normalized uniform stresses T(I) III , TIII v strip of regularity in the complex plane of the Laplace domain

(j) (j) W(j), W(j) I , WII , WIII anti-plane displacements of jth material nðjÞ nðjÞ W 2 , W 2;I anti-plane displacements of jth material after applying the Mellin transform of the second kind

¯ n2;IðjÞ anti-plane displacements of jth material after ¯ n2 ðjÞ , W W applying the Mellin transform of the second kind and the Laplace transform anti-plane displacement of jth material for the W ðjÞ arc;L studied case with Problem L (L ¼ I, II, III) z cylindrical polar coordinate axis r2 Laplacian operator a apex angle of each material of the finite wedge b loading angle of 1st material of the finite wedge g loading angle of 2nd material of the finite wedge d Dirac-delta function m1, m2, mj shear moduli y cylindrical polar coordinate axis L problem L (L ¼ I, II, III) j2, j1 upper and lower bounds of b loading range for the studied case x2, x1 upper and lower bounds of g loading range for the studied case trz, tyz shear stresses in the rz-direction and in the yz-direction ðjÞ ðjÞ ðjÞ tðjÞ rz , trz;I , trz;II , trz;III shear stresses in the rz-direction of jth material ðjÞ ðjÞ ðjÞ tðjÞ yz , tyz;I , tyz;II , tyz;III shear stresses in the yz-direction of jth material tðjÞ shear stress in the rz-direction of jth material for the arc rz;L studied case with Problem L (L ¼ I, II, III) tðjÞ shear stress in the yz-direction of jth material for the arc yz;L studied case with Problem L (L ¼ I, II, III) OðjÞ normalized displacement of jth material for the studied L case with Problem L (L ¼ I, II, III) GðjÞ normalized shear stress in the rz-direction of jth L material for the studied case with Problem L (L ¼ I, II, III) YðjÞ normalized shear stress in the yz-direction of jth L material for the studied case with Problem L (L ¼ I, II, III)

problems can be obtained by integrating the derived fundamental solutions over the range of interest, as shown in Section 3. The results are compared with those obtained from FE analyses.

2. Problem formulation and solutions Fig. 1 shows the composite wedge, with a finite radius a, considered in this study. The apex angle of each material is a, and the shear moduli are m1 and m2, respectively. An infinite length along the z-axis perpendicular to the plane is assumed. Accordingly, the problem belongs to the plane deformation type. The only displacement component in the z-direction is a function relative to the in-plane coordinates (r, y). In this study, the circular arc of the wedge (r ¼ a) is subjected to a pair of anti-plane concentrated forces, F, at angles y ¼ b and y ¼ g. The shear stresses on the z-axis, trz and tyz, are the remaining components in the constitutive equations, which can be expressed as

tðjÞ rz ðr; yÞ ¼ mj tðjÞ yz ðr; yÞ ¼

@W ðjÞ ðr; yÞ @r

mj @W ðjÞ ðr; yÞ r

@y

(1)

(2)

where W is the displacement in the z-axis, and j ¼ 1, 2 for the jth material. The static equilibrium equation in the absence of body forces is @tðjÞ ðr; yÞ @½rtðjÞ rz ðr; yÞ ¼0 þ yz @y @r

(3)

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Fig. 1. Schematic view of the considered composite wedge subjected to a pair of concentrated anti-plane shear loads on a circular arc.

By substituting Eqs. (1) and (2) into Eq. (3), the equilibrium equation can be reduced to

r2 W ðjÞ ðr; yÞ ¼ 0

(4)

where

r2 ¼

@2 1@ 1 @2 þ þ 2 2 2 r @r r @y @r

denotes the Laplacian operator. The Mellin transform can be employed for solving Eq. (4). To deal with the traction boundary on the circular arc, the finite Mellin transform of the second kind is adopted [24]: Z a ðjÞ nðjÞ ðjÞ ða2p r p1 þ r p1 Þf ðr; yÞdr (5) M 2 ½f ðr; yÞ; p ¼ f 2 ðp; yÞ ¼ 0

where p is a complex transform parameter. The inversion formula is in the form: nðjÞ

ðjÞ

M 1 2 ½f 2 ðp; yÞ; r ¼ f ðr; yÞ ¼

1 2pi

Z

cþi1

ðjÞ

r p f ðp; yÞdp

(6)

ci1

pffiffiffiffiffiffiffi where i ¼ 1 and the constant Re[p] ¼ c defines the path of integration in the complex plane. Applying Eq. (5) to Eq. (4) yields ! 2 d @W ðjÞ ða; yÞ 2 W 2nðjÞ ðp; yÞ þ 2apþ1 þ p ¼0 2 @r dy

(7)

provided " lim ða2p r pþ1 þ r pþ1 Þ r!0

# @W ðjÞ ðr; yÞ þ pða2p r p  rp ÞW ðjÞ ðr; yÞ ¼ 0 @r

(8)

Eq. (8) can be used to define the path of the line integral Re[p] ¼ c in Eq. (6); i.e., the strip of regularity should be determined from Eq. (8) such that the integral in Eq. (5) exists. The prescribed boundary conditions are:

tðjÞ rz ða; yÞ ¼ F½ð2  jÞdðy  bÞ þ ð1  jÞdðy þ gÞ

(9a)

W ðjÞ ð0; yÞ ¼ 0

(9b)

where bra, gra, arp, and d denotes the Dirac-delta function. Applying Eq. (9) to Eq. (7) with the aid of Eq. (1) gives ! 2 d 2Fapþ1 2 W 2nðjÞ ðp; yÞ þ þ p ½ð2  jÞdðy  bÞ þ ð1  jÞdðy þ gÞ ¼ 0 2 mj dy

(10)

Eq. (10) is an ODE with a non-homogeneous term. Mathematically, this equation has a solution which is related to a trigonometric function that repeats periodically. Thus, this study regards the boundary-value problem of this kind as an initial-value problem.

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The Laplace transform is [24] nðjÞ

¯ 2 ðp; sÞ ¼ L½W 2nðjÞ ðp; yÞ; s ¼ W

Z

1 0

esy W n2 ðjÞ ðp; yÞdy

where s is the transform parameter. The inverse of the Laplace transform in terms of Eq. (11) is given by Z vþi1 1 ¯ n2 ðjÞ ðp; sÞds ¯ 2nðjÞ ðp; sÞ; y ¼ W n2 ðjÞ ðp; yÞ ¼ L1 ½W esy W 2pi vi1

(11)

(12)

where the constant Re[s] ¼ v defines the path of integration in the complex plane. Taking the Laplace transform on both sides of Eq. (10) gives nðjÞ

¯ 2 ðp; sÞ  sW n2 ðjÞ ðp; 0Þ þ ð1Þj ðs2 þ p2 ÞW

@W 2nðjÞ ðp; 0Þ 2Fapþ1 ¼ ½ðj  2Þesb þ ðj  1Þesg  @y mj

(13)

where W n2 ðjÞ ðp; 0Þ and @W n2 ðjÞ ðp; 0Þ=@y are the initial boundary conditions in the Laplace domain to be determined, or the undetermined coefficients in the Mellin domain; they can be deduced from the boundary conditions prescribed on the radial edges (i.e., boundary conditions at y ¼ 7a). Detailed solutions are presented below. 2.1. Problem I: Free–free edge Traction-free radial edges are considered in Problem I. The corresponding boundary conditions are: ð2Þ tð1Þ yz;I ðr; aÞ ¼ tyz;I ðr; aÞ ¼ 0

(14a)

ð2Þ W ð1Þ I ðr; 0Þ ¼ W I ðr; 0Þ

(14b)

ð2Þ tð1Þ yz;I ðr; 0Þ ¼ tyz;I ðr; 0Þ

(14c)

where the subscript I denotes Problem I in this study. After applying the finite Mellin transform (Eq. (5)) to Eq. (14), the radial boundary conditions are: ð2Þ M2 ½tð1Þ yz;I ðr; aÞ ¼ M 2 ½tyz;I ðr; aÞ ¼ 0

(15a)

ð2Þ M2 ½W ð1Þ I ðr; 0Þ ¼ M 2 ½W I ðr; 0Þ

(15b)

ð2Þ M2 ½r tð1Þ yz;I ðr; 0Þ ¼ M 2 ½r tyz;I ðr; 0Þ

(15c)

To solve Eq. (13), the following coefficients are assumed based on Eqs. (15b) and (15c): ð2Þ M2 ½W ð1Þ I ðr; 0Þ ¼ M 2 ½W I ðr; 0Þ ¼ AI

(16)

ð2Þ M2 ½r tð1Þ yz;I ðr; 0Þ ¼ M 2 ½r tyz;I ðr; 0Þ ¼ BI

(17)

After applying Eq. (17) with the aid of Eq. (2), the following relationship is obtained: nðjÞ @W 2;I ðp; 0Þ

@y

¼

BI

mj

Substituting Eqs. (16) and (18) into Eq. (13) gives       s 1 1 1 2Fapþ1 ¯ n2;IðjÞ ðp; sÞ ¼ W AI þ ð1Þjþ1 BI þ 2 ½ðj  2Þesb þ ðj  1Þesg  2 2 2 2 2 mj s þ p mj s þp s þp Applying the inverse Laplace transform (Eq. (12)) to Eq. (19) yields ( ) ð2  jÞHðy  bÞ sin½pðy  bÞ sinðpyÞ 2Fapþ1 nðjÞ W 2;I ðp; yÞ ¼ AI cosðpyÞ þ BI  þð1  jÞHðy þ gÞ sin½pðy þ gÞ mj p mj p

(18)

(19)

(20)

where H is the Heaviside function. After applying Eq. (15a) with the aid of Eq. (2), the coefficients AI and BI can be solved: AI ¼

Fapþ1 cosðpaÞfcos½pða  gÞ  cos½pða  bÞg DI ðpÞ

(21)

BI ¼

Fapþ1 p sinðpaÞfm1 cos½pða  gÞ þ m2 cos½pða  bÞg DI ðpÞ

(22)

where DI ðpÞ ¼ pðm1 þ m2 Þ sinð2paÞ

(23)

The zeros of DI(p) determine the order of the stress singularity at the wedge apex. The orders of the stress singularity for bi-material wedge problems under anti-plane shear deformation were discussed in previous studies [12,16]. The behavior of the stress singularity

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depends on the conditions of the radial edges. The characteristic functions, Eq. (23), as well as those in the following sections, obtained in this study are identical to those in previous studies since the same conditions of radial edges are used. By applying the inverse finite Mellin transform (Eq. (6)) in conjunction with the residue theorem, the displacement field is derived as W ðjÞ I ðr; yÞ ¼

2Fa ðm1 þ m2 Þp 8 9        2 ð2n  1Þp ð2n  1Þp ð2n  1Þp r ð2n1Þp=2a > > j1 j2 > > > > þ R sin g sin sin b y R > 1 > < 2n  1 = 2a 2a 2a a X       np=a

> 1 > npg npb npy r > n¼1 > > > cos > > : n cos a  cos a ; a a

(24)

where R¼

m1 m2

(25)

It should be noted that the path of integration for inversion integrals is within the strip of regularity Re[p]o0, but we replace p (negative p) by p+ (positive p) here. The corresponding stress fields are:

tðjÞ rz;I ðr; yÞ ¼

tðjÞ yz;I ðr; yÞ ¼

2F ðR þ 1Þa 8 9       ð2n  1Þp ð2n  1Þp ð2n  1Þp r ðð2n1Þp=2aÞ1 > > > > > > þ sin sin g R sin b y > 1 > < = 2a 2a 2a a X       ðnp=aÞ1

> > npg npb npy r > 2j n¼1 > > > cos cos  cos > > : R ; a a a a 2F ðR þ 1Þa 8 9       ð2n  1Þp ð2n  1Þp ð2n  1Þp r ðð2n1Þp=2aÞ1 > > > > > > þ sin cos g R sin b y > 1 > < = 2a 2a 2a a X       ðnp=aÞ1

> > npg npb npy r > 2j n¼1 > > > sin cos  cos > > : þR ; a a a a

(26)

(27)

For R ¼ 1 and b ¼ g, the solution is identical to that for the anti-symmetrical problem of the single material case in Shahani [20], which was obtained by the separation of variables. When a ¼ p, a uniform tðjÞ rz;I is found at y ¼ 0: lim tðjÞ rz;I ðr; 0Þ ¼ r!0

2FR2j ½cosðbÞ  cosðgÞ ðR þ 1Þp

(28)

This shows that the uniform tðjÞ rz;I results from the discrepancy of the two loading angles. The following normalized expression is used for this uniform stress: T ðIÞ III ¼

ðR þ 1Þp 2FR2j

lim tðjÞ rz;III ðr; 0Þ

(29)

r!0

(I) where the superscript I denotes Problem I here. The T(I) III distribution for g versus b is plotted in Fig. 2. If b ¼ g and y ¼ 0, TIII is reduced to ðjÞ and W ðjÞ . This implies that if b ¼ g , the composite material problem can be regarded as two single material problems zero, and so are trz;I I for which the interface is fixed. The following SIF definition is used for an interfacial crack problem in this study:

ðLÞ ¼ lim K III r!0

pffiffiffiffiffiffi 2pr 1p1;L tðjÞ yz;I ðr; 0Þ

(30)

where L ¼ I, II and III for Problems I, II, and III, respectively. P1,L is the smallest positive root for DL(p) ¼ 0. The SIF of Problem I (a ¼ p) is K ðIÞ III ¼

2F ðR þ 1Þ

rffiffiffiffiffiffi 2a

p

R sin

hgi 2

þ sin

 

b 2

(31)

The normalized SIF for Problem I can be defined by HðIÞ III ¼

K ðIÞ III ðR þ 1Þ 2F

rffiffiffiffiffiffi

p

2a

(32)

(I) The H(I) III distributions for g versus b with R ¼ 0.1, 1, and 10 are plotted in Fig. 3. It can be seen that b dominates the magnitude of HIII for R ¼ 0.1 and that g dominates the magnitude of H(I) III for R ¼ 10. This indicates that the loading angle of a stiffer material has a lesser effect on SIF. In addition, b and g contribute equally to H(I) III for R ¼ 1, as expected.

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Fig. 2. T(I) III distribution for g versus b in Problem I.

Fig. 3. H(I) III distributions for g versus b in Problem I. (a) R ¼ 0.1, (b) R ¼ 1, (c) R ¼ 10.

2.2. Problem II: Free–fixed edge The considered boundary conditions in the Mellin domain for Problem II are: ð2Þ M2 ½tð1Þ yz;II ðr; aÞ ¼ M 2 ½W II ðr; aÞ ¼ 0

(33a)

ð2Þ M2 ½W ð1Þ II ðr; 0Þ ¼ M 2 ½W II ðr; 0Þ ¼ AII

(33b)

ARTICLE IN PRESS C.-H. Chen et al. / International Journal of Mechanical Sciences 51 (2009) 583–597 ð2Þ M 2 ½r tð1Þ yz;II ðr; 0Þ ¼ M 2 ½r tyz;II ðr; 0Þ ¼ BII

589

(33c)

where the subscript II denotes Problem II in this study. Similar to Problem I, the coefficients in the Mellin domain for Problem II are determined: AII ¼

2Fapþ1 fcosðpaÞ sin½pða  gÞ  sinðpaÞ cos½pða  bÞg DII ðpÞ

(34)

BII ¼

2Fapþ1 pfm1 sinðpaÞ sin½pða  gÞ  m2 cosðpaÞ cos½pða  bÞg DII ðpÞ

(35)

where DII ðpÞ ¼ p½m1 sin2 ðpaÞ  m2 cos2 ðpaÞ

(36)

The displacement field is derived as ) 8( 9 3 2 sinðpaÞ cos½pða  bÞ > > > > > > cosðp y Þ > > 7 6 > > 1 6 <  cosðpaÞ sin½pða  gÞ = r p 7 X 7 6 2Fa ( ) W ðjÞ ðr; y Þ ¼ 7 6 0 j1 II > > 7 6 a ðpÞ D sinðp a Þ sin½pð a  g Þ R > > II n¼1 4 > > 5 > > sinðp y Þ  > > : ; Rj2 cosðpaÞ cos½pða  bÞ

(37) p¼pn;II

where D0II ðpÞ ¼ paðm1 þ m2 Þ sinð2paÞ pn;II ¼

pn;II ¼

1

"

1

a

# rffiffiffi! 1 þ ðn  1Þp R

tan1

a

(38)

" 1

np  tan

rffiffiffi!# 1 R

for n ¼ 1; 3; 5; . . .

(39a)

for n ¼ 2; 4; 6; . . .

(39b)

Note that D0 II(pn,II)a0 can be obtained if D0 II(pn,II) ¼ 0. The corresponding stress fields are: ) 8( 9 3 2 sinðpaÞ cos½pða  bÞ > > > > > > cosðp y Þ > > 7 6 > 1 62Fpm > <  cosðpaÞ sin½pða  gÞ = r p1 7 X 7 6 j ðjÞ ( j1 ) trz;II ðr; yÞ ¼ 7 6 0 > 7 6 D ðpÞ > a R sinðpaÞ sin½pða  gÞ > > n¼1 4 II > > 5 > > sinðp y Þ  > > j2 : ; R cosðpaÞ cos½pða  bÞ

(40)

p¼pn;II

) 8( 9 3 sinðpaÞ cos½pða  bÞ > > > > > > sinðp y Þ > > 7 6 > 1 62Fpm > <  cosðpaÞ sin½pða  gÞ = r p1 7 X j 7 6 ðjÞ ( ) tyz;II ðr; yÞ ¼ 7 6 0 > 7 6 DII ðpÞ > a Rj1 sinðpaÞ sin½pða  gÞ > > n¼1 4 > > 5 > > cos p ð y Þ þ > > : ; Rj2 cosðpaÞ cos½pða  bÞ 2

1

(41) p¼pn;II

pffiffiffiffiffiffiffiffiffi ð 1=RÞ  1. For R ¼ 1 and a ¼ 1/2p (the case of a single material with

The smallest admissible order of stress for Problem II is 1=a tan a crack), the square-root singularity is recovered. Using the definition in Eq. (30), the SIF of Problem II (a ¼ p) is rffiffiffiffi Fa1p1;II 2 pffiffiffi ðIIÞ K III ¼ f R½cosðp1;II bÞ  cosðp1;II gÞ þ sinðp1;II bÞ þ R sinðp1;II gÞg ðR þ 1Þ p p ffiffiffiffiffiffiffiffiffi  1=R . The normalized SIF for Problem II can be defined by where p1;II ¼ 1=ptan1 rffiffiffiffi ðIIÞ K ðR þ 1Þ p ðIIÞ ¼ III 1p HIII 2 Fa 1;II

(42)

(43)

(II) The H(II) III distributions for g versus b with R ¼ 0.1, 1, and 10 are plotted in Fig. 4. The figure shows that increasing b has little effect on HIII (II) for constant g when R ¼ 10. If m1bm2, HIII is reduced to a function with respect to g.

2.3. Problem III: Fixed–fixed edge The considered boundary conditions for Problem III are: ð2Þ M 2 ½W ð1Þ III ðr; aÞ ¼ M 2 ½W III ðr; aÞ ¼ 0

(44a)

ð2Þ M 2 ½W ð1Þ III ðr; 0Þ ¼ M 2 ½W III ðr; 0Þ ¼ AIII

(44b)

ð2Þ M 2 ½r tð1Þ yz;III ðr; 0Þ ¼ M 2 ½r tyz;III ðr; 0Þ ¼ BIII

(44c)

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Fig. 4. T(II) III distributions for g versus b in Problem II. (a) R ¼ 0.1, (b) R ¼ 1, (c) R ¼ 10.

where the subscript III denotes Problem III in this study. Similar to Problem I, the coefficients in the Mellin domain for Problem III are determined: AIII ¼

Fapþ1 sinðpaÞfsin½pða  bÞ  sin½pða  gÞg DIII ðpÞ

(45)

BIII ¼

Fapþ1 p cosðpaÞfm1 sin½pða  gÞ þ m2 sin½pða  bÞg DIII ðpÞ

(46)

where DIII ðpÞ ¼ pðm1 þ m2 Þ sinð2paÞ

(47)

Eq. (47) is equal to Eq. (23), so the order of the stress singularity is the same as that in Problem I. The displacement and stress fields are derived as W ðjÞ III ðr; yÞ ¼

tðjÞ rz;III ðr; yÞ ¼

2Fa ðm1 þ m2 Þp 8 9        2 ð2n  1Þp ð2n  1Þp ð2n  1Þp r ð2n1Þp=2a > > > > > >  cos cos g b y cos > 1 > < 2n  1 = 2a 2a 2a a X       



n p = a > 1 j1 > npg npb npy r > n¼1 > > > þ Rj2 sin sin sin > > : þn R ; a a a a 2F ðR þ 1Þa 8 9        ð2n  1Þp ð2n  1Þp ð2n  1Þp r ðð2n1Þp=2aÞ1 > > 2j > > > >  cos cos g R cos b y > 1 > < = 2a 2a 2a a X       



ðn p = a Þ1 > > > n¼1 > > þ R sin npg þ sin npb sin npy r > > > : ; a a a a

(48)

(49)

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Fig. 5. T(III) III distributions for g versus b in Problem III. (a) R ¼ 0.1, (b) R ¼ 1, (c) R ¼ 10.

Fig. 6. Studied case with a pair of shear pressures.

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tðjÞ yz;III ðr; yÞ ¼

2F ðR þ 1Þa 8 9        ð2n  1Þp ð2n  1Þp ð2n  1Þp r ðð2n1Þp=2aÞ1 > > > > > >  cos sin g R2j cos b y > > 1 < = 2a 2a 2a a X        npg

r ðnp=aÞ1 > > n p b n p y > n¼1 > > > cos > > :  R sin a þ sin a ; a a

(50)

(j) If b ¼ g and y ¼ 0, tðjÞ rz;III and WIII are reduced to zero, which is similar to the results obtained in Problem I. In this case, the problem can be regarded as two single material problems for which both radial edges are fixed. When a ¼ p, a uniform tðjÞ yz;III is found at y ¼ 0 as

lim tðjÞ yz;III ðr; 0Þ ¼ r!0

2F ½R sinðgÞ þ sinðbÞ ðR þ 1Þp

(51)

The following normalized expression is used for this uniform stress: T ðIIIÞ III ¼

ðR þ 1Þp lim tðjÞ ðr; 0Þ r!0 yz;III 2F

(52)

The T(III) distributions for g versus b with R ¼ 0.1, 1, and 10 are plotted in Fig. 5. Symmetric behavior of T(III) can be observed for III III 0rbr901 and 901rbr1801 with R ¼ 0.1; and for 0rgr901 and 901rgr1801 with R ¼ 10. The loading angle of the softer material dominates the magnitude of T(III) III . Using the definition in Eq. (30), the SIF for this problem is K ðIIIÞ III ¼ 0

(53)

K(III) III ¼ 0 because the traditional definition of SIF, Eq. (30), is according to the radial direction parallel to crack surfaces. Nonetheless, it should be noted that a singular tðjÞ yz;III occurs in directions other than y ¼ 0.

Fig. 7. Finite element meshes used for the case shown in Fig. 6.

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3. A case study In this section, a case with a pair of anti-plane shear pressures, P, shown in Fig. 6, is studied. The solutions for this case can be obtained by integrating the fundamental solutions (derived in Section 2) with respect to the loading range as W ðjÞ arc;L ðr; yÞ ¼

Z g¼x2 Z

tðjÞ arc rz;L ðr; yÞ ¼

tðjÞ ðr; yÞ ¼ arc yz;L

g¼x1

b¼j2

b¼j1

Z g¼x2 Z g¼x1

b¼j2

b¼j1

Z g¼x2 Z g¼x1

W ðjÞ L ðr; yÞdb dg

b¼j2 b¼j1

(54a)

tðjÞ rz;L ðr; yÞdb dg

(54b)

tðjÞ yz;L ðr; yÞdb dg

(54c)

where x2x1 ¼ j2j1, and L ¼ I, II, and III for Problems I, II, and III, respectively. The explicit results are given in Appendix. Finite element results are used to compare the derived solutions. The FE program ANSYS was employed to compute displacement and stress solutions of a special case. The following geometric parameters were assumed: a ¼ 1, a ¼ 601, j1 ¼ 451, j2 ¼ 601, x1 ¼ 301, and x2 ¼ 451; material constants: m1 ¼ 4e9 and m2 ¼ 8e9; and arc line loads: P ¼ 1e6. A 3-D model with 8616 twenty-node isoparametric brick elements was used. Only one degree of freedom in the z-direction (the in-plane degrees of freedom were set to zero) was considered for modeling. The FE mesh is shown in Fig. 7. Surface loads were applied parallel to the z-direction on the area of the arc plane from y ¼ j1 to y ¼ j2 and from y ¼ x1 to y ¼ x2. Since the in-plane degrees of freedom were fixed, a uniform outcome in the z-direction was expected (each depth has the same outcome). To compare the results in a more

Fig. 8. Normalized solutions of the studied case compared with FE results for Problem I.

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general form, we defined normalized expressions for the displacement and stresses as

OðjÞ L ¼

ðm1 þ m2 Þ ðjÞ W arc;L ðr; yÞ Paa

(55a)

GðjÞ L ¼

ðR þ 1Þ ðjÞ tarc rz;L ðr; yÞ P

(55b)

YðjÞ L ¼

ðR þ 1Þ ðjÞ tarc yz;L ðr; yÞ P

(55c)

It should be noted that the multiplier for the normalization is the constant obtained after integration in Eq. (54). The FE results for Problems I, II, and III are compared with the analytical solutions (expanding Eq. (55) to 10,000 terms) in Figs. 8–10. Good agreement is achieved. Except for some critical locations for which FE fails to predict the correct stress fields, the discrepancy is well below 1%. These results verify that the fundamental solution derived in this study is appropriate and reliable.

4. Conclusions Problems of composite finite wedges under anti-plane shear applied on a circular arc were solved using the finite Mellin transform in conjunction with the Laplace transform. The full-field solutions of displacements and stresses for various boundary edges, namely, free–free, free–fixed, and fixed–fixed, were derived explicitly. The SIF distributions of three considered problems for composite circular shafts with an interfacial crack are presented and discussed. For loads applied at equal angles, free–free and fixed–fixed edge problems can be degenerated into single material problems. Uniform stresses were found along the interface in free–free and fixed–fixed edge problems. A case with general loads was calculated and the results compared well with those obtained from FE analyses. The derived fundamental solutions can be used for further investigations.

Fig. 9. Normalized solutions of the studied case compared with FE results for Problem II.

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Fig. 10. Normalized solutions of the studied case compared with FE results for Problem III.

Appendix A. Explicit solutions for the studied case For Problem I (free–free edge): 8 9   9 8 ð2n  1Þp > > > > j1 > > > > R cos x > > > >     ‘ > > 2< = > 2a 2 ð2n  1Þp r ð2n1Þp=2a > > > > >   sin > > y > > 1 X 2 = ð2n  1Þp > 2a a j2 > > Paa X 2ð1Þ‘þ1 < 2n  1 > ðjÞ > > j cos þR ‘ ; : W arc;I ðr; yÞ ¼ 2 a 2 > > ðm1 þ m2 Þ n¼1 ‘¼1 p > > >     >     np=a > > 2

> > > > 1 n p x n pj n p y r > > ‘ ‘ >þ > cos  sin sin > > : ; a a a n a

(A.1)

  9 8 8 9 ð2n  1Þp > > > > > > > > x Rcos > >     ‘ > > < = > 2a > > 2 ð2n  1Þp r ðð2n1Þp=2aÞ1 > > >   sin > > y > > 1 X 2 = ‘þ1 < 2n  1 > ð2n  1Þ p X > 2 a a P 2ð1Þ > þcos > ðjÞ > > j : ‘ ; tarc rz;I ðr; yÞ ¼ 2a > > p ðR þ 1Þ n¼1 ‘¼1 > > > >    ðnp=aÞ1    > > npj  > > > > 1 n p x n p y r ‘ 2j > > ‘ > > cos  sin sin : þR ; a a a n a

(A.2)

  9 8 8 9 ð2n  1Þp > > > > > > > > x R cos > >     ‘ > > < =

> 2a > > ð2n  1Þp 2 r ðð2n1Þp=2aÞ1 > > >   > > cos y > > 1 2 < = ‘þ1 ð2n  1Þp X X > > 2 a 2n  1 a P 2ð1Þ > > ðjÞ > > þ cos j : ‘ ; tarc yz;I ðr; yÞ ¼ 2 a > > p ðR þ 1Þ n¼1 ‘¼1 > > > >    ðnp=aÞ1    > > npj  > > > > 1 npx‘ n p y r 2j > > ‘ > > sin  sin R sin : ; a a a n a

(A.3)

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For Problem II (free–fixed edge): ) 8( 9 sinðpm aÞ sin½pm ða  j‘ Þ > > > > > > cosðpm yÞ > > > > a Þ cos½p ð a  x Þ þ cosðp 1 2 ‘ < = r pm ‘þ1 m m XX Pa 2ð1Þ ðjÞ ( j1 ) W arc;II ðr; yÞ ¼ 2 > R sinðpm aÞ cos½pm ða  x‘ Þ ðm1 þ m2 Þ m¼1 ‘¼1 pm a sinð2pm aÞ > > > a >þ > > sinðpm yÞ > > > : ; þRj2 cosðpm aÞ sin½pm ða  j‘ Þ ) 8( 9 R sinðpm aÞ cos½pm ða  x‘ Þ > > > > > > sinðp y Þ > > m > > 1 X 2 < þ cosðpm aÞ sin½pm ða  j‘ Þ = r pm 1 ‘þ1 X Pa 2ð1Þ ðjÞ ( )  tarc rz;II ðr; yÞ ¼ sinðp a Þ sin½p ð a  j Þ > ðR þ 1Þ m¼1 ‘¼1 pm a sin 2pm a > m ‘ > > a >  m > þR2j > cosðpm yÞ > > > : ; þ cos pm a cos½pm ða  x‘ Þ ) 8( 9 R sinðpm aÞ cos½pm ða  x‘ Þ > > > > > > cosðp y Þ > > m > > þ cosðpm aÞ sin½pm ða  j‘ Þ 1 X 2 < = r pm 1 ‘þ1 X Pa 2ð1Þ ðjÞ ( ) tarc yz;II ðr; yÞ ¼ > sinðp a Þ sin½p ð a  j Þ ðR þ 1Þ m¼1 ‘¼1 pm a sinð2pm aÞ > m m ‘ > > a > > > sinðpm yÞ > R2j > > : ; þ cosðpm aÞ cos½pm ða  x‘ Þ

(A.4)

(A.5)

(A.6)

where " # 8 rffiffiffi! > 1 1 > 1 > ; ¼ tan p þ ðn  1Þ p > > R < m¼2n1 a " ! # rffiffiffi pm ¼ > 1 1 > > > pm¼2n ¼ np  tan1 ; > : a R

n2N

(A.7)

For Problem III (fixed–fixed edge): 8 9   9 8 ð2n  1Þp > > > > > > > sin x > > > >   ð2n1Þp=2a > 2 < ‘ > > = > > 2 a 2 ð2n  1Þ p r > > > >   > > y cos > > 1 2 < = ‘þ1 ð2n  1Þ p > > 2 a 2n  1 a X X > > Pa a 2ð1Þ ðjÞ > > j  sin ‘ : ; W arc;III ðr; yÞ ¼ 2 a 2 > > ðm1 þ m2 Þ n¼1 ‘¼1 p > > >     np=a > >  2  > npj  > > > > 1 n p x n p y r > > ‘ j1 j2 ‘ > > sin þ R þ R cos cos > > : ; a a a n a

(A.8)

  9 8 8 9 ð2n  1Þp > > > > > > > > x sin > >     ‘ > > < = > 2a > > 2 ð2n  1Þp r ðð2n1Þp=2aÞ1 > > > 2j   > > R y cos > > 1 X 2 = ‘þ1 < ð2n  1Þ p X > > 2 a 2n  1 a P 2ð1Þ >  sin > ðjÞ > > j ‘ ; : tarc rz;III ðr; yÞ ¼ 2a > > p ðR þ 1Þ n¼1 ‘¼1 > > > >     ðnp=aÞ1   > > npj  > > > > 1 n p x n p y r ‘ > > ‘ > > sin þ cos þ R cos : ; a a a n a

(A.9)

  9 8 8 9 ð2n  1Þp > > > > > > > > x sin > >     > ‘ > < = > 2a > > 2 ð2n  1Þp r ðð2n1Þp=2aÞ1 > 2j > >   > > R y sin > > 1 X 2 = ‘þ1 < ð2n  1Þ p X > > 2 a 2n  1 a P 2ð1Þ >  sin > ðjÞ > > j ‘ ; : tarc yz;III ðr; yÞ ¼ 2a > > p ðR þ 1Þ n¼1 ‘¼1 > > > >     ðnp=aÞ1   > > npj  > > > > 1 n p x n p y r ‘ > > ‘ > > cos þ cos þ R cos : ; a a a n a

(A.10)

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