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Stress analysis of a cylindrical bar with a spherical cavity or rigid inclusion by the eigenfunction expansion variational method
International Journal of Engineering Science 42 (2004) 325–338 www.elsevier.com/locate/ijengsci
Stress analysis of a cylindrical bar with a spherical cavity or rigid inclusion by the eigenfunction expansion variational method Y.Z. Chen Division of Engineering Mechanics, Jiangsu University, Zhenjiang, Jiangsu 212013, PR China Received 30 December 2002; received in revised form 27 June 2003; accepted 28 July 2003
Abstract Axisymmetric tension problem of a round bar containing a spherical cavity or a rigid inclusion is considered in this paper. The bar has a finite length. In order to solve the problem, an eigenfunction expansion form is suggested. The eigenfunction always satisfies the governing equations of elasticity and the traction free condition on the surface of sphere, or the fixed displacement condition on the surface of sphere. The undetermined coefficients in the eigenfunction expansion form are determined by the use of the variational method in elasticity. The whole process of solution is called the eigenfunction expansion variational method (abbreviated as EEVM) in this paper. The solutions for two cases, one for the bar containing a spherical cavity, other for the bar containing a spherical inclusion, are obtained. Finally, some numerical examples are given and some stress concentration factors for the problem are presented. Ó 2003 Elsevier Ltd. All rights reserved. Keywords: Eigenfunction expansion variational method; Notch problem; Rigid inclusion problem
1. Introduction Since a reliable design of the machine part depends on knowledge of stress concentration analysis in the theory of elasticity, the stress concentration problems have received much attention and have been widely studied. Some earlier works in this field were initiated by Neuber [1]. Also, the three dimensional stress concentration problem was summarized by Sternberg [2].
E-mail address: [email protected] (Y.Z. Chen). 0020-7225/$ - see front matter Ó 2003 Elsevier Ltd. All rights reserved. doi:10.1016/j.ijengsci.2003.07.001
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Y.Z. Chen / International Journal of Engineering Science 42 (2004) 325–338
It is easy to see that the three dimensional stress concentration problem possesses an inherent difficulty and very few problems in this field have been solved. An earlier investigation of the torsion problems of the round bar weakened by cavities can be found from [3,4]. Stress concentration in the neighborhood of nearly spherical cavities was analyzed in [5]. More recently, a three dimensional stress concentration problem of infinite elastic medium containing cavities was solved in [6]. However, in the above mentioned works, only the problem for an infinite medium containing the spherical cavities was considered. Therefore, the problem for a finite body containing a cavity is becoming more important to solve. In this paper, the tension problem of the round bar having a finite length and containing a spherical cavity or rigid inclusion will be discussed. The following analysis mainly depends on two types of particular elastic solutions for bodies bounded by an inner sphere surface. The first type of particular solutions is composed of a series with positive power of R. Therefore, it can be used in the region inside a sphere surface with radius R1 . In addition, the second type of particular solutions is composed of a series with negative power of R. Similarly, it can be used to the region outside of a sphere surface with the radius R0 . The finite body containing a cavity is always embedded in the region R0 6 R 6 R1 . Therefore, both types of the particular solutions can be used to construct the solution of a finite body containing the cavity. These two types of the particular solutions have been derived in [7], and they take the series form with some undetermined coefficients. After appropriately choosing the undetermined coefficients in both types of the particular solutions, a linear combination of the two types will give one more particular solution which is called the eigenfunction expansion form in this paper. The obtained form has the following properties: (a) each term in the form always satisfies the governing equations of elasticity, (b) each term in the form satisfies the traction free condition on the inner surface of the sphere, i.e. on the surface of the cavity. In order to satisfy the boundary value condition on the tops and the cylindrical surface of the round bar, the variational method in elasticity is used, and the undetermined coefficients in the eigenfunction expansion form can be obtained immediately. The whole process of solving the boundary value problem is thus called the eigenfunction expansion variational method (abbreviated as EEFM) in this paper. The EEVM was used to solve: (a) the crack problem in a finite plate [8], (b) the crack problem in a bonded materials [9], (c) the notch problem in a finite plate [10], (d) the periodic crack problem in an infinite plate [11] and (e) the 3-D crack problem in an infinite body [12]. Meantime, 3-D elastic analysis singularity at polyhedral corner points was studied [13]. If the finite body contains a spherical rigid inclusion, similar analysis and derivation can be carried out without any difficulty. In literature, if one assume that the investigated function satisfies the governing equation exactly and does not satisfy the boundary condition, the relevant solution method is generally called the Trefftz method. An example based on the method can be found in [14, Section 118]. However, in the above-mentioned EEVM method, in addition to satisfying the governing equation of elasticity the assumed function also satisfies a portion of boundary condition. In this sense, the EEVM has a little modification to the Trefftz method. It is not easy to determine the investigator who first proposed this idea. In the authorÕs knowledge, Williams proposed an expansion form of Airy stress function to solve the edge crack problem and his contribution became an earlier version of EEVM [15]. Note that, the mentioned expansion form satisfies the biharmonic equation as well as the traction free boundary condition on the crack face.
Y.Z. Chen / International Journal of Engineering Science 42 (2004) 325–338
327
2. Formulation of the eigenfunction expansion variational method in elasticity Obviously, the idea of EEVM is a general one and can be used in many cases. Thus, formulation of EEVM is written out in a general case. It is assumed that there is a three dimensional body containing a spherical cavity with the traction free condition (Fig. 1). In addition, we suppose the traction free condition on the surface of the cavity is satisfied by the following system of the displacement and stress fields ðmÞ
ui ;
ðmÞ
m ¼ 1; 2; 3; . . .
rij ;
ð1Þ
which are some particular solution of elasticity. This system of the displacement and stress fields is called the eigenfunction expansion form in this paper. Below, the problem of utilizing the generalized variational principle is discussed [16,17]. In present case, the generalized variational principle can be expressed as follows. If the stress components rij satisfy the traction free condition on the cavity surface sc (Fig. 1), the actual solution can be obtained by the stationary condition of a functional defined as Z Z Z Z Z Z Z Z pi ui ds P¼ Aðeij Þ dv rij ½eij ðui;j þ uj;i Þ=2 dv R R st Z Z rij nj ðui ui Þ ds ð2Þ su
In the functional (2) the independent quantities subject to variation are the displacement components ui , the strain components eij , and the stress components rij . In Eq. (2) the tensor notation is used, R is the region occupied by the elastic medium containing a cavity, st is the portion of the outer boundary in which the external tractions pi are given, su is the portion of the outer boundary in which the displacement components ui are given, nj is the direction cosine of the outer normal to the boundary, and Aðeij Þ is the strain energy function.
z
θ
R
R0 sc
y
su φ
st
x(r)
Fig. 1. A three-dimensional body containing a spherical cavity.
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Y.Z. Chen / International Journal of Engineering Science 42 (2004) 325–338
Since the displacement–strain relations are satisfied by the eigenfunction expansion form, the functional P in Eq. (2) can be simplified as Z Z Z Z Z Z Z Aðeij Þ dv rij nj ðui ui Þ ds ð3Þ P¼ pi ui ds R
st
su
Because the stress components rij derived from the eigenfunction expansion form satisfy the traction free condition, i.e. rij nj ¼ 0 on the cavity surface sc in Fig. 1. ClapeyronÕs theorem yields Z Z Z Z Z 1 Aðeij Þ dv ¼ rij nj ui ds ð4Þ 2 R st þsu Substituting Eq. (4) into Eq. (3), we find that the functional P finally becomes Z Z Z Z Z Z Z Z 1 1 pi ui ds þ P¼ rij nj ui ds rij nj ui ds rij nj ui ds 2 2 st su st su
ð5Þ
Naturally, we can use the following eigenfunction expansion form ui ¼
M X
ðmÞ
M X
rij ¼
Xm ui ;
m¼1
ðmÞ
ð6Þ
Xm rij
m¼1
as an approximate solution for the boundary value problem. By substituting Eq. (6) in Eq. (5) and requiring oP=oXk ¼ 0;
k ¼ 1; 2; . . . ; M
ð7Þ
a linear system is obtained as follows M X
Akm Xm ¼ Bk ;
k ¼ 1; 2; . . . ; M
ð8Þ
m¼1
where Akm ¼ Amk ¼
Z Z
ðkÞ
ðmÞ
rij nj ui ds
Z Z
st
Bk ¼
Z Z st
ðkÞ pi ui ds
ðmÞ
ðkÞ
rij nj ui ds
ðk; m ¼ 1; 2; . . . ; MÞ
ð9Þ
su
Z Z
ðkÞ
rij nj ui ds
ðk ¼ 1; 2; . . . ; MÞ
ð10Þ
su
Thus, the undetermined coefficients Xm can be evaluated from the above equation. From the above mentioned analysis we see that the main work in the following investigation is to find the eigenfunction expansion form for the proposed problem. A particular case is worth discussing. If there is no portion of su , thus s ¼ st and the boundary value problem is a pure stress boundary value problem. In this case, Eqs. (9) and (10) become
Y.Z. Chen / International Journal of Engineering Science 42 (2004) 325–338
Akm ¼ Amk ¼
Z Z
ðmÞ
ðkÞ
rij nj ui ds
ðk; m ¼ 1; 2; . . . ; MÞ
329
ð11Þ
s
Bk ¼
Z Z
ðkÞ
pi ui ds ðk ¼ 1; 2; . . . ; MÞ
ð12Þ
s
Substituting Eqs. (11) and (12) into Eq. (8) gives Z Z
M X s
! ðmÞ rij nj Xm
ðkÞ
pi ui ds ¼ 0
ðk ¼ 1; 2; . . . ; MÞ
ð13Þ
m¼1
Clearly, this equation can be considered as the projection of the contour traction residual on the kth displacement (k ¼ 1; 2; . . . ; M) being equal to zero. Finally, if the eigenfunction expansion form satisfies the fixed displacement condition on sc , all the above equations can be used for the problem of a round bar containing a spherical rigid inclusion. 3. Round bar containing a spherical cavity Consider a round bar with finite length 2H and outer radius B in Fig. 2. The bar contains a spherical cavity with radius R0 and is stretched by a tension p in z-direction. Therefore, the boundary value conditions can be expressed by rRR jR¼R0 ¼ 0;
rRh jR¼R0 ¼ 0
ðon the surface of cavity; expressed in spherical coordinatesÞ rzz jz¼H ¼ p;
rrz jz¼H ¼ 0
ðon the tops of bar; expressed in cylindrical coordinatesÞ rrr jr¼B ¼ 0;
ð14Þ
ð15Þ
rrz jr¼B ¼ 0
ðon the cylindrical surface of bar; expressed in cylindrical coordinatesÞ
ð16Þ
In the present study, the appropriate series solution has been given in [7], and can be expressed in the spherical coordinates as 2GuR ¼ 2GðuRð1Þ þ uRð2Þ Þ 2GuRð1Þ ¼ 2A0 ð1 2mÞR þ
1 X
An ðn þ 1Þðn 2 þ 4mÞRnþ1 þ Bn nRn1 Pn ðlÞ
n¼2;4
2GuRð2Þ ¼ D0 =R2 þ
1 X n¼2;4
Cn nðn þ 3 4mÞ=Rn Dn ðn þ 1Þ=Rnþ2 Pn ðlÞ
ð17Þ
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Y.Z. Chen / International Journal of Engineering Science 42 (2004) 325–338 p
B z
θ 2H
R
R0 y φ x (r)
p
Fig. 2. Round bar containing a spherical cavity or rigid inclusion.
2Guh ¼ 2Gðuhð1Þ þ uhð2Þ Þ 1 X An ðn þ 5 4mÞRnþ1 Bn Rn1 Pn0 ðlÞ sin h 2Guhð1Þ ¼ 2Guhð2Þ ¼
n¼2;4 1 X
ð18Þ
Cn ðn 4 þ 4mÞ=Rn Dn =Rnþ2 Pn0 ðlÞ sin h
n¼2;4
rRR ¼ rRRð1Þ þ rRRð2Þ rRRð1Þ ¼ 2A0 ð1 þ mÞ þ
1 X An ðn þ 1Þðn2 n 2 2mÞRn þ Bn nðn 1ÞRn2 Pn ðlÞ n¼2;4
rRRð2Þ ¼ 2D0 =R3 þ
1 X
ð19Þ
Cn nðn2 þ 3n 2mÞ=Rnþ1 þ Dn ðn þ 1Þðn þ 2Þ=Rnþ3 Pn ðlÞ
n¼2;4
rRh ¼ rRhð1Þ þ rRhð2Þ 1 X rRhð1Þ ¼ An ðn2 þ 2n 1 þ 2mÞRn Bn ðn 1ÞRn2 Pn0 ðlÞ sin h rRhð2Þ ¼
n¼2;4 1 X n¼2;4
Cn ðn2 2 þ 2mÞ=Rnþ1 þ Dn ðn þ 2Þ=Rnþ3 Pn0 ðlÞ sin h
ð20Þ
Y.Z. Chen / International Journal of Engineering Science 42 (2004) 325–338
331
rhh ¼ rhhð1Þ þ rhhð2Þ 1 X
rhhð1Þ ¼ 2A0 ð1 þ mÞ þ
An ðn þ 1Þðn2 þ 4n þ 2 þ 2mÞRn Bn n2 Rn2 Pn ðlÞ
n¼2;4
þ
1 X
An ðn þ 5 4mÞRn þ Bn Rn2 Pn0 ðlÞ cos h
n¼2;4
rhhð2Þ ¼ D0 =R3 þ
1 h X
2
i
ð21Þ
Cn nðn2 2n 1 þ 2mÞ=Rnþ1 Dn ðn þ 1Þ =Rnþ3 Pn ðlÞ
n¼2;4
þ
1 X
Cn ð n þ 4 4mÞ=Rnþ1 þ Dn =Rnþ3 Pn0 ðlÞ cos h
n¼2;4
r// ¼ r//ð1Þ þ r//ð2Þ r//ð1Þ ¼ 2A0 ð1 þ mÞ þ
1 X
An ðn þ 1Þðn 2 ð4n þ 2ÞmÞRn þ Bn nRn2 Pn ðlÞ
n¼2;4
þ
1 X
An ðn þ 5 4mÞRn Bn Rn2 Pn0 ðlÞ cos h
n¼2;4
r//ð2Þ
ð22Þ
1 X ¼ D0 =R3 þ Cn nðn þ 3 ð4n þ 2ÞmÞ=Rnþ1 Dn ðn þ 1Þ=Rnþ3 Pn ðlÞ n¼2;4
þ
1 X
Cn ðn 4 þ 4mÞ=Rnþ1 Dn =Rnþ3 Pn0 ðlÞ cos h
n¼2;4
In above mentioned equations, G denotes the shear modulus of elasticity, m the PoissonÕs ratio, l ¼ cos h, and Pn ðlÞ the Legendre polynomials. Concept for obtaining the series solution shown by Eqs. (17)–(22) is introduced below. In 3-D elasticity, the displacements can be expressed by PapkovichÕs presentation where a harmonic vector and a harmonic scalar are involved [7]. In fact, the harmonic vector (three harmonic functions) as well as the harmonic scalar are some harmonic functions. In the spherical coordinates, method of separation of variables will lead to a set of harmonic functions. Substituting those harmonic functions into PapkovichÕs presentation, one will get some particular solutions for 3-D elasticity in spherical coordinates. The obtained solutions are expressed in the form of Eqs. (17)–(22). To satisfy Eq. (14), i.e. the traction free condition on the surface of cavity, substituting Eqs. (19) and (20) into Eq. (14) yields D0 ¼ A0 ð1 þ mÞR30
ð23Þ
and An =Dn þ ðn 1Þðn þ 2Þð2n þ 1ÞR2n1 Bn =Dn Cn ¼ ðn2 1Þðn þ 2Þð2n þ 3ÞR2nþ1 0 0 Dn ¼ ð2n5 þ 5n4 5n2 þ 6n þ 4 4nðn þ 1Þm2 ÞR2nþ3 An =Dn 0 þ nðn 1Þð2n 1Þðn þ 2ÞR2nþ3 Bn =Dn 0
ð24Þ
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Y.Z. Chen / International Journal of Engineering Science 42 (2004) 325–338
where Dn ¼ ðn þ 2Þ 2ðn2 þ n þ 1Þ ð4n þ 2Þm
ð25Þ
Finally, after substituting Eqs. (23) and (24) into Eqs. (17)–(22), the eigenfunction expansion form is obtainable. Below, the boundary value problem shown in Fig. 2 is considered. In the process of using Eqs. (8), (11) and (12), the following points need to be mentioned: (a) In this case, the right hand terms of Eqs. (15) and (16) (rzz jz¼H ¼ p, rrz jz¼H ¼ 0, rrr jr¼B ¼ 0, rrz jr¼B ¼ 0) serve as the applied traction pi in Eq. (12). (b) After substituting Eqs. (23) and (24) into Eqs. (17)–(22), only the undetermined coefficients A0 ; A2 ; A4 ; . . . B2 ; B4 ; . . . are involved in Eqs. (17)–(22). These coefficients A0 ; A2 ; A4 ; . . . B2 ; B4 ; . . . serve as the coefficients Xm in Eq. (8). ðkÞ (c) In Eqs. (17)–(22), the functions after the coefficients A0 ; A2 ; A4 ; . . . B2 ; B4 ; . . . serve as the ui , ðkÞ rij in (11) and (12). After using Eqs. (8), (11), (12) and (17)–(25), the calculation is straightforward with no difficulty. The detailed computations will not be cited here. In this computation, we take m ¼ 1=3 and M ¼ 15 terms (M––the number of truncated terms) are truncated in the expansion form Eqs. (17)– (22). The calculated results for the stresses on the surface of cavity are expressed as follows rhh jR¼R0 ¼ f1 ðR0 =B; H =B; hÞp r// jR¼R0 ¼ f2 ðR0 =B; H =B; hÞp 1=2 2 2 req ¼ rhh rhh r// þ r//
ð26Þ ¼ f3 ðR0 =B; H=B; hÞp
R¼R0
where req denotes the equivalent stress used the Mises yield criterion of plasticity. The calculated f1 , f2 and f3 values are plotted in Figs. 3 and 4 and listed in Table 1, respectively. For comparison, the exact solution for an infinite body containing the spherical cavity under the remote tension rz ¼ p is also introduced [7] 27 15m 15 2 rhh jR¼R0 ¼ F1 ðhÞp ¼ cos h p 2ð7 5mÞ 7 5m 3 15m 15m 2 cos h p r// jR¼R0 ¼ F2 ðhÞp ¼ 2ð7 5mÞ 7 5m 1=2 ¼ F3 ðhÞp req ¼ r2hh rhh r// þ r2//
ð27Þ
R¼R0
This case must be equivalent to the R0 =B ! 0 case in the present study. The results from Eq. (27) (R0 =B ¼ 0) are listed in Table 1, and several calculated results for f1 , f2 and f3 values (R0 =B > 0) are also listed in Table 1.
Y.Z. Chen / International Journal of Engineering Science 42 (2004) 325–338
333
3.5 3.0
H/B=2.0
2.5
f 1 (R0/B,H/B,θ)
2.0
R0/B=0.7
f 2 (R0/B,H/B,θ)
1.5
0.1
1.0
0.4
0.5 0.0 -0.5 0.7
-1.0 -1.5 0
10
20
30
40 50 θ (degree)
60
70
80
90
Fig. 3. Calculated values for f1 ðR0 =B; H =B; hÞ and f2 ðR0 =B; H =B; hÞ in Eq. (26) in the case of H =B ¼ 2:0.
3.5 R0/B=0.7
3.0
H/B=2.0
2.5
f (R0/B,H/B,θ) 3
2.0 1.5 0.1
0.4
1.0 0.5 0.0 0
10
20
30
40
50
60
70
80
90
θ (degree)
Fig. 4. Calculated values for f3 ðR0 =B; H =B; hÞ in Eq. (26) in the case of H =B ¼ 2:0.
Table 1 Calculated results of f1 ðR0 =B; H=B; hÞ, f2 ðR0 =B; H =B; hÞ and f3 ðR0 =B; H =B; hÞ in Eq. (26) in the case of H =B ¼ 1:0 R0 =B
0
f1 jh¼0° f1 jh¼90° f2 jh¼0° f2 jh¼90° f3 jh¼0° f3 jh¼90°
)0.750 2.063a )0.750a 0.188a 0.750a 1.975a
a
Exact solution.
a
0.1
0.2
0.3
0.4
0.5
0.6
0.7
)0.753 2.067 )0.753 0.187 0.753 1.981
)0.777 2.100 )0.777 0.181 0.777 2.015
)0.837 2.182 )0.837 0.166 0.837 2.104
)0.949 2.334 )0.949 0.129 0.949 2.272
)1.134 2.577 )1.134 0.053 1.134 2.551
)1.409 2.941 )1.409 )0.095 1.409 2.989
)1.769 3.478 )1.769 )0.379 1.769 3.682
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Y.Z. Chen / International Journal of Engineering Science 42 (2004) 325–338
Table 2 Calculated results of f1 ðR0 =B; H =B; hÞ, f2 ðR0 =B; H =B; hÞ and f3 ðR0 =B; H=B; hÞ in Eq. (26) in the case of R0 =B ¼ 0:5 h 0°
15°
30°
45°
60°
75°
90°
H =B ¼ 1:0 case f1 ðM ¼ 15Þ f1 ðM ¼ 11Þ f1 ðM ¼ 7Þ f2 ðM ¼ 15Þ f2 ðM ¼ 11Þ f2 ðM ¼ 7Þ f3 ðM ¼ 15Þ f3 ðM ¼ 11Þ f3 ðM ¼ 7Þ
)1.134 )1.120 )1.081 )1.134 )1.120 )1.081 1.134 1.120 1.081
)0.832 )0.828 )0.815 )1.032 )1.024 )0.996 0.948 0.941 0.919
)0.083 )0.091 )0.117 )0.783 )0.782 )0.774 0.745 0.740 0.723
0.820 0.815 0.785 )0.490 )0.490 )0.490 1.146 1.141 1.114
1.675 1.677 1.662 )0.221 )0.220 )0.218 1.796 1.797 1.781
2.325 2.326 2.309 )0.023 )0.022 )0.019 2.336 2.337 2.319
2.577 2.574 2.550 0.053 0.053 0.055 2.551 2.548 2.523
H =B ¼ 2:0 case f1 ðM ¼ 15Þ f1 ðM ¼ 11Þ f1 ðM ¼ 7Þ f2 ðM ¼ 15Þ f2 ðM ¼ 11Þ f2 ðM ¼ 7Þ f3 ðM ¼ 15Þ f3 ðM ¼ 11Þ f3 ðM ¼ 7Þ
)0.854 )0.864 )0.855 )0.854 )0.864 )0.855 0.854 0.864 0.855
)0.666 )0.670 )0.655 )0.796 )0.803 )0.790 0.740 0.746 0.732
)0.140 )0.127 )0.102 )0.634 )0.633 )0.612 0.577 0.580 0.568
0.623 0.649 0.675 )0.402 )0.389 )0.361 0.894 0.908 0.911
1.456 1.471 1.481 )0.150 )0.132 )0.102 1.537 1.541 1.534
2.123 2.107 2.091 0.050 0.066 0.095 2.098 2.074 2.045
2.381 2.347 2.319 0.128 0.141 0.168 2.320 2.280 2.239
M––number of the truncated terms.
As expected, from Table 1 we see that the difference between two cases R0 =B ¼ 0 (exact solution) and R0 =B ¼ 0:1 (numerical solution using EEVM) is small. Also, the obtained results are useful to assess the strength of a body weakened by the spherical cavity. From Fig. 4 or Table 1 we also see that the most dangerous point is always located at the position h ¼ 90°. To investigate the effect of the ratio H =B to the solution, several calculated values of f1 ðR0 =B; H =B; hÞ, f2 ðR0 =B; H=B; hÞ and f3 ðR0 =B; H =B; hÞ, under the conditions of R0 =B ¼ 0:5 and M ¼ 15, are listed in the first two columns of Table 2. From Table 2 we see that, if H =B is changed from H =B ¼ 2:0 to 1.0, the equivalent stress req at the most dangerous point is elevated from 2.320p to 2.551p. The convergence of the numerical solution is also studied. To this end, several particular values of f1 ðR0 =B; H =B; hÞ, f2 ðR0 =B; H =B; hÞ and f3 ðR0 =B; H =B; hÞ under the conditions R0 =B ¼ 0:5 and H =B ¼ 2, are also listed in last three columns of Table 2. From Table 2 we see that, if the number of the truncated terms in the eigenfunction expansion form is changed from M ¼ 7 to 15, convergence can be found from the listed results. 4. Round bar containing a spherical rigid inclusion In this section, we consider the problem for the round bar containing a spherical rigid inclusion, as shown in Fig. 2. The bar is also stretched by a tension p in the z-direction. In this case, the series solution, from Eqs. (17)–(22), can also be used.
Y.Z. Chen / International Journal of Engineering Science 42 (2004) 325–338
335
Nevertheless, the fixed displacement condition on the surface of the rigid inclusion will lead to uR jR¼R0 ¼ 0;
uh jR¼R0 ¼ 0
ð28Þ
Substituting Eqs. (17) and (18) into Eq. (28) yields D0 ¼ 2ð1 2mÞR30 A0
ð29Þ
and An =dn þ ð2n þ 1ÞR2n1 Bn =dn Cn ¼ ðn þ 1Þð2n þ 3ÞR2nþ1 0 0 Dn ¼ ð2n þ 1Þðn2 þ n þ 8 24m þ 16m2 ÞR2nþ3 An =dn þ nð2n 1ÞR2nþ1 Bn =dn 0 0
ð30Þ
where dn ¼ 2ð3n þ 2Þ þ 4ð2n þ 1Þm
ð31Þ
As before, after substituting Eqs. (29) and (30) into Eqs. (17)–(22), the eigenfunction expansion form is obtainable. In the same condition of computation, the calculated results for the stresses on the surface of the rigid sphere can be expressed as rRR jR¼R0 ¼ g1 ðR0 =B; H =B; hÞp rRh jR¼R0 ¼ g2 ðR0 =B; H =B; hÞp rhh jR¼R0 ¼ r// jR¼R0 ¼ g3 ðR0 =B; H =B; hÞp 1=2 req ¼ ðg1 g3 Þ2 þ 3g22 ¼ g4 ðR0 =B; H =B; hÞp
ð32Þ
where req denotes the equivalent stress used in the Mises yield criterion of plasticity. Some calculated g1 , g2 , g3 and g4 values are plotted in Figs. 5 and 6 and listed in Table 3, respectively. For comparison, the exact solution for an infinite body containing the spherical rigid inclusion is also cited as follows [7] 2ð1 2mÞ 7 5m 15ð1 mÞ 2 þ cos h p rRR jR¼R0 ¼ G1 ðhÞp ¼ 3ð1 þ mÞ 24 30m 8 10m 15ð1 mÞ sin h cos h p rRh jR¼R0 ¼ G2 ðhÞp ¼ 8 10m ð33Þ 1 2m 8 25m 15m 2 þ þ cos h p rhh jR¼R0 ¼ r// jR¼R0 ¼ G3 ðhÞp ¼ 3ð1 þ mÞ 24 30m 8 10m 1=2 2 req ¼ ðG1 G3 Þ þ 3G22 ¼ G4 ðR0 =B; H =B; hÞp As expected, from Table 3 we see that the difference between two cases R0 =B ¼ 0 (exact solution) and R0 =B ¼ 0:1 (numerical solution using EEVM) is small. Also, the obtained results are
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Y.Z. Chen / International Journal of Engineering Science 42 (2004) 325–338 2.0 1.5
g (R /B,H/B,θ) 1 0
1.0
g2 (R /B,H/B,θ) - 0
0.5 R0/B=0.1
0.4
0.0
0.7
-0.5 -1.0 H/B=2.0 -1.5 0
10
20
30
40
50
60
70
80
90
θ ( degree ) Fig. 5. Calculated values for g1 ðR0 =B; H =B; hÞ and g2 ðR0 =B; H =B; hÞ in Eq. (32) in the case of H =B ¼ 2:0.
2.0 R0/B=0.1 H/B=2.0
1.5 0.4
1.0
0.7 0.5 g3(R0/B,H/B,θ)
0.0
g4(R0/B,H/B,θ) - -0.5 0
10
20
30
40
50
60
70
80
90
θ ( degree )
Fig. 6. Calculated values for g3 ðR0 =B; H =B; hÞ and g4 ðR0 =B; H =B; hÞ in Eq. (32) in the case of H =B ¼ 2:0.
useful to assess the strength of body containing the spherical rigid inclusion. From Fig. 6 or Table 3 we see that the most dangerous point is always located at the vicinity of h ¼ 45°.
5. Remarks A simple, yet efficient, method is presented for the solution of the axisymmetric tension problem of the round bar containing a spherical cavity or rigid inclusion. It is well known that, if the problem for the infinite body containing the spherical cavity is concerned, sometimes, the
Y.Z. Chen / International Journal of Engineering Science 42 (2004) 325–338
337
Table 3 Calculated results of g1 ðR0 =B; H =B; hÞ, g2 ðR0 =B; H =B; hÞ, g3 ðR0 =B; H =B; hÞ and g4 ðR0 =B; H =B; hÞ in Eq. (32) in the case of H =B ¼ 2:0 R0 =B g1 jh¼0° g1 jh¼45° g1 jh¼90° g2 jh¼0° g2 jh¼45° g2 jh¼90° g3 jh¼0° g3 jh¼45° g3 jh¼90° g4 jh¼0° g4 jh¼45° g4 jh¼90° a
0 a
1.929 0.857a )0.214a 0.0 a )1.071a 0.0a 0.964a 0.429a )0.107a 0.964a 1.905a 0.107a
0.1
0.2
0.3
0.4
0.5
0.6
0.7
1.937 0.857 )0.223 0.0 )1.080 0.0 0.968 0.428 )0.111 0.968 1.919 0.111
1.923 0.855 )0.215 0.0 )1.069 0.0 0.961 0.428 )0.108 0.961 1.900 0.108
1.889 0.848 )0.194 0.0 )1.043 0.0 0.945 0.424 )0.097 0.945 1.855 0.097
1.834 0.831 )0.158 0.0 )1.001 0.0 0.917 0.415 )0.079 0.917 1.783 0.079
1.762 0.798 )0.108 0.0 )0.949 0.0 0.881 0.399 )0.054 0.881 1.692 0.054
1.679 0.748 )0.050 0.0 )0.893 0.0 0.839 0.374 )0.025 0.839 1.591 0.025
1.588 0.680 0.006 0.0 )0.836 0.0 0.794 0.340 0.003 0.794 1.488 0.003
Exact solution.
integral transform method is used. However, it is difficult to use the integral transform method to the problem of the finite body with complex geometry. Therefore, the EEVM can be considered as a complementary approach to some classical approaches mainly used for the case of infinite medium. Essentially, the EEVM belongs to the category of the TrefftzÕs method. That is to say, one seeks the approximate solution within the range of the solutions which satisfy the governing equations of elasticity exactly. However, as shown in Eqs. (8)–(10), the EEVM can be used in all cases, including the displacement boundary value condition and the mixed boundary value condition. Finally, the results obtained from two cases, one for containing spherical cavity and other for containing the spherical rigid inclusion, are quite different. In the first case, if the cavity is larger, then the bar weakened by the cavity is more dangerous, where the req curve corresponding R0 =B ¼ 0:7 is located in a higher position (Fig. 4). On the contrary, in the second case, if the spherical rigid inclusion is larger, then the bar is safer, where the req curve corresponding R0 =B ¼ 0:7 is located in a lower position (Fig. 6). This situation can be seen from Figs. 4 and 6, respectively.
Acknowledgement The research project is supported by the National Natural Science Foundation of China.
References [1] [2] [3] [4]
H. Neuber, Kerbspannungsle, Springer-Verlag, Berlin, 1958. E. Sternberg, Applied Mechanics Review 11 (1958) 1. C.B. Ling, Quarterly Applied Mathematics 10 (1952) 149. S.C. Das, ASME Journal of Applied Mechanics 21 (1954) 83.
338 [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17]
Y.Z. Chen / International Journal of Engineering Science 42 (2004) 325–338 G.P. Grechka, D.I. Chernopickii, Soviet Applied Mechanics 19 (1984) 1068. M. Hamada, Bulletin of the JSME 29 (1986) 8. A.E. LurÕe, Three-dimensional Problems of the Theory of Elasticity, Interscience publishers, New York, 1964. Y.Z. Chen, Engineering Fracture Mechanics 17 (1983) 387. W.C. Wang, J.T. Chen, Computer Method in Applied Mechanics and Engineering 73 (1989) 153. Y.Z. Chen, Computer Method in Applied Mechanics and Engineering 42 (1984) 57. Y.Z. Chen, K.Y. Lee, ASME Journal of Applied Mechanics 69 (2002) 552. Y.Z. Chen, K.Y. Lee, International Journal of Solids and Structures 39 (2002) 5787. E. Glushov, N. Glushkova, O. Lapina, International Journal of Solids and Structures 36 (1999) 1105. I.S. Sokolnikoff, Mathematical Theory of Elasticity, McGraw-Hill, New York, 1956. M.L. Williams, ASME Journal of Applied Mechanics 24 (1957) 109. H.C. Hu, Scintia Sinica 4 (1955) 33 (in Chinese). K. Washisu, Variational Methods in Elasticity and Plasticity, Pergamon, Oxford, 1975.
Stress analysis of a cylindrical bar with a spherical cavity or rigid inclusion by the eigenfunction expansion variational method Y.Z. Chen Division of Engineering Mechanics, Jiangsu University, Zhenjiang, Jiangsu 212013, PR China Received 30 December 2002; received in revised form 27 June 2003; accepted 28 July 2003
Abstract Axisymmetric tension problem of a round bar containing a spherical cavity or a rigid inclusion is considered in this paper. The bar has a finite length. In order to solve the problem, an eigenfunction expansion form is suggested. The eigenfunction always satisfies the governing equations of elasticity and the traction free condition on the surface of sphere, or the fixed displacement condition on the surface of sphere. The undetermined coefficients in the eigenfunction expansion form are determined by the use of the variational method in elasticity. The whole process of solution is called the eigenfunction expansion variational method (abbreviated as EEVM) in this paper. The solutions for two cases, one for the bar containing a spherical cavity, other for the bar containing a spherical inclusion, are obtained. Finally, some numerical examples are given and some stress concentration factors for the problem are presented. Ó 2003 Elsevier Ltd. All rights reserved. Keywords: Eigenfunction expansion variational method; Notch problem; Rigid inclusion problem
1. Introduction Since a reliable design of the machine part depends on knowledge of stress concentration analysis in the theory of elasticity, the stress concentration problems have received much attention and have been widely studied. Some earlier works in this field were initiated by Neuber [1]. Also, the three dimensional stress concentration problem was summarized by Sternberg [2].
E-mail address: [email protected] (Y.Z. Chen). 0020-7225/$ - see front matter Ó 2003 Elsevier Ltd. All rights reserved. doi:10.1016/j.ijengsci.2003.07.001
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Y.Z. Chen / International Journal of Engineering Science 42 (2004) 325–338
It is easy to see that the three dimensional stress concentration problem possesses an inherent difficulty and very few problems in this field have been solved. An earlier investigation of the torsion problems of the round bar weakened by cavities can be found from [3,4]. Stress concentration in the neighborhood of nearly spherical cavities was analyzed in [5]. More recently, a three dimensional stress concentration problem of infinite elastic medium containing cavities was solved in [6]. However, in the above mentioned works, only the problem for an infinite medium containing the spherical cavities was considered. Therefore, the problem for a finite body containing a cavity is becoming more important to solve. In this paper, the tension problem of the round bar having a finite length and containing a spherical cavity or rigid inclusion will be discussed. The following analysis mainly depends on two types of particular elastic solutions for bodies bounded by an inner sphere surface. The first type of particular solutions is composed of a series with positive power of R. Therefore, it can be used in the region inside a sphere surface with radius R1 . In addition, the second type of particular solutions is composed of a series with negative power of R. Similarly, it can be used to the region outside of a sphere surface with the radius R0 . The finite body containing a cavity is always embedded in the region R0 6 R 6 R1 . Therefore, both types of the particular solutions can be used to construct the solution of a finite body containing the cavity. These two types of the particular solutions have been derived in [7], and they take the series form with some undetermined coefficients. After appropriately choosing the undetermined coefficients in both types of the particular solutions, a linear combination of the two types will give one more particular solution which is called the eigenfunction expansion form in this paper. The obtained form has the following properties: (a) each term in the form always satisfies the governing equations of elasticity, (b) each term in the form satisfies the traction free condition on the inner surface of the sphere, i.e. on the surface of the cavity. In order to satisfy the boundary value condition on the tops and the cylindrical surface of the round bar, the variational method in elasticity is used, and the undetermined coefficients in the eigenfunction expansion form can be obtained immediately. The whole process of solving the boundary value problem is thus called the eigenfunction expansion variational method (abbreviated as EEFM) in this paper. The EEVM was used to solve: (a) the crack problem in a finite plate [8], (b) the crack problem in a bonded materials [9], (c) the notch problem in a finite plate [10], (d) the periodic crack problem in an infinite plate [11] and (e) the 3-D crack problem in an infinite body [12]. Meantime, 3-D elastic analysis singularity at polyhedral corner points was studied [13]. If the finite body contains a spherical rigid inclusion, similar analysis and derivation can be carried out without any difficulty. In literature, if one assume that the investigated function satisfies the governing equation exactly and does not satisfy the boundary condition, the relevant solution method is generally called the Trefftz method. An example based on the method can be found in [14, Section 118]. However, in the above-mentioned EEVM method, in addition to satisfying the governing equation of elasticity the assumed function also satisfies a portion of boundary condition. In this sense, the EEVM has a little modification to the Trefftz method. It is not easy to determine the investigator who first proposed this idea. In the authorÕs knowledge, Williams proposed an expansion form of Airy stress function to solve the edge crack problem and his contribution became an earlier version of EEVM [15]. Note that, the mentioned expansion form satisfies the biharmonic equation as well as the traction free boundary condition on the crack face.
Y.Z. Chen / International Journal of Engineering Science 42 (2004) 325–338
327
2. Formulation of the eigenfunction expansion variational method in elasticity Obviously, the idea of EEVM is a general one and can be used in many cases. Thus, formulation of EEVM is written out in a general case. It is assumed that there is a three dimensional body containing a spherical cavity with the traction free condition (Fig. 1). In addition, we suppose the traction free condition on the surface of the cavity is satisfied by the following system of the displacement and stress fields ðmÞ
ui ;
ðmÞ
m ¼ 1; 2; 3; . . .
rij ;
ð1Þ
which are some particular solution of elasticity. This system of the displacement and stress fields is called the eigenfunction expansion form in this paper. Below, the problem of utilizing the generalized variational principle is discussed [16,17]. In present case, the generalized variational principle can be expressed as follows. If the stress components rij satisfy the traction free condition on the cavity surface sc (Fig. 1), the actual solution can be obtained by the stationary condition of a functional defined as Z Z Z Z Z Z Z Z pi ui ds P¼ Aðeij Þ dv rij ½eij ðui;j þ uj;i Þ=2 dv R R st Z Z rij nj ðui ui Þ ds ð2Þ su
In the functional (2) the independent quantities subject to variation are the displacement components ui , the strain components eij , and the stress components rij . In Eq. (2) the tensor notation is used, R is the region occupied by the elastic medium containing a cavity, st is the portion of the outer boundary in which the external tractions pi are given, su is the portion of the outer boundary in which the displacement components ui are given, nj is the direction cosine of the outer normal to the boundary, and Aðeij Þ is the strain energy function.
z
θ
R
R0 sc
y
su φ
st
x(r)
Fig. 1. A three-dimensional body containing a spherical cavity.
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Y.Z. Chen / International Journal of Engineering Science 42 (2004) 325–338
Since the displacement–strain relations are satisfied by the eigenfunction expansion form, the functional P in Eq. (2) can be simplified as Z Z Z Z Z Z Z Aðeij Þ dv rij nj ðui ui Þ ds ð3Þ P¼ pi ui ds R
st
su
Because the stress components rij derived from the eigenfunction expansion form satisfy the traction free condition, i.e. rij nj ¼ 0 on the cavity surface sc in Fig. 1. ClapeyronÕs theorem yields Z Z Z Z Z 1 Aðeij Þ dv ¼ rij nj ui ds ð4Þ 2 R st þsu Substituting Eq. (4) into Eq. (3), we find that the functional P finally becomes Z Z Z Z Z Z Z Z 1 1 pi ui ds þ P¼ rij nj ui ds rij nj ui ds rij nj ui ds 2 2 st su st su
ð5Þ
Naturally, we can use the following eigenfunction expansion form ui ¼
M X
ðmÞ
M X
rij ¼
Xm ui ;
m¼1
ðmÞ
ð6Þ
Xm rij
m¼1
as an approximate solution for the boundary value problem. By substituting Eq. (6) in Eq. (5) and requiring oP=oXk ¼ 0;
k ¼ 1; 2; . . . ; M
ð7Þ
a linear system is obtained as follows M X
Akm Xm ¼ Bk ;
k ¼ 1; 2; . . . ; M
ð8Þ
m¼1
where Akm ¼ Amk ¼
Z Z
ðkÞ
ðmÞ
rij nj ui ds
Z Z
st
Bk ¼
Z Z st
ðkÞ pi ui ds
ðmÞ
ðkÞ
rij nj ui ds
ðk; m ¼ 1; 2; . . . ; MÞ
ð9Þ
su
Z Z
ðkÞ
rij nj ui ds
ðk ¼ 1; 2; . . . ; MÞ
ð10Þ
su
Thus, the undetermined coefficients Xm can be evaluated from the above equation. From the above mentioned analysis we see that the main work in the following investigation is to find the eigenfunction expansion form for the proposed problem. A particular case is worth discussing. If there is no portion of su , thus s ¼ st and the boundary value problem is a pure stress boundary value problem. In this case, Eqs. (9) and (10) become
Y.Z. Chen / International Journal of Engineering Science 42 (2004) 325–338
Akm ¼ Amk ¼
Z Z
ðmÞ
ðkÞ
rij nj ui ds
ðk; m ¼ 1; 2; . . . ; MÞ
329
ð11Þ
s
Bk ¼
Z Z
ðkÞ
pi ui ds ðk ¼ 1; 2; . . . ; MÞ
ð12Þ
s
Substituting Eqs. (11) and (12) into Eq. (8) gives Z Z
M X s
! ðmÞ rij nj Xm
ðkÞ
pi ui ds ¼ 0
ðk ¼ 1; 2; . . . ; MÞ
ð13Þ
m¼1
Clearly, this equation can be considered as the projection of the contour traction residual on the kth displacement (k ¼ 1; 2; . . . ; M) being equal to zero. Finally, if the eigenfunction expansion form satisfies the fixed displacement condition on sc , all the above equations can be used for the problem of a round bar containing a spherical rigid inclusion. 3. Round bar containing a spherical cavity Consider a round bar with finite length 2H and outer radius B in Fig. 2. The bar contains a spherical cavity with radius R0 and is stretched by a tension p in z-direction. Therefore, the boundary value conditions can be expressed by rRR jR¼R0 ¼ 0;
rRh jR¼R0 ¼ 0
ðon the surface of cavity; expressed in spherical coordinatesÞ rzz jz¼H ¼ p;
rrz jz¼H ¼ 0
ðon the tops of bar; expressed in cylindrical coordinatesÞ rrr jr¼B ¼ 0;
ð14Þ
ð15Þ
rrz jr¼B ¼ 0
ðon the cylindrical surface of bar; expressed in cylindrical coordinatesÞ
ð16Þ
In the present study, the appropriate series solution has been given in [7], and can be expressed in the spherical coordinates as 2GuR ¼ 2GðuRð1Þ þ uRð2Þ Þ 2GuRð1Þ ¼ 2A0 ð1 2mÞR þ
1 X
An ðn þ 1Þðn 2 þ 4mÞRnþ1 þ Bn nRn1 Pn ðlÞ
n¼2;4
2GuRð2Þ ¼ D0 =R2 þ
1 X n¼2;4
Cn nðn þ 3 4mÞ=Rn Dn ðn þ 1Þ=Rnþ2 Pn ðlÞ
ð17Þ
330
Y.Z. Chen / International Journal of Engineering Science 42 (2004) 325–338 p
B z
θ 2H
R
R0 y φ x (r)
p
Fig. 2. Round bar containing a spherical cavity or rigid inclusion.
2Guh ¼ 2Gðuhð1Þ þ uhð2Þ Þ 1 X An ðn þ 5 4mÞRnþ1 Bn Rn1 Pn0 ðlÞ sin h 2Guhð1Þ ¼ 2Guhð2Þ ¼
n¼2;4 1 X
ð18Þ
Cn ðn 4 þ 4mÞ=Rn Dn =Rnþ2 Pn0 ðlÞ sin h
n¼2;4
rRR ¼ rRRð1Þ þ rRRð2Þ rRRð1Þ ¼ 2A0 ð1 þ mÞ þ
1 X An ðn þ 1Þðn2 n 2 2mÞRn þ Bn nðn 1ÞRn2 Pn ðlÞ n¼2;4
rRRð2Þ ¼ 2D0 =R3 þ
1 X
ð19Þ
Cn nðn2 þ 3n 2mÞ=Rnþ1 þ Dn ðn þ 1Þðn þ 2Þ=Rnþ3 Pn ðlÞ
n¼2;4
rRh ¼ rRhð1Þ þ rRhð2Þ 1 X rRhð1Þ ¼ An ðn2 þ 2n 1 þ 2mÞRn Bn ðn 1ÞRn2 Pn0 ðlÞ sin h rRhð2Þ ¼
n¼2;4 1 X n¼2;4
Cn ðn2 2 þ 2mÞ=Rnþ1 þ Dn ðn þ 2Þ=Rnþ3 Pn0 ðlÞ sin h
ð20Þ
Y.Z. Chen / International Journal of Engineering Science 42 (2004) 325–338
331
rhh ¼ rhhð1Þ þ rhhð2Þ 1 X
rhhð1Þ ¼ 2A0 ð1 þ mÞ þ
An ðn þ 1Þðn2 þ 4n þ 2 þ 2mÞRn Bn n2 Rn2 Pn ðlÞ
n¼2;4
þ
1 X
An ðn þ 5 4mÞRn þ Bn Rn2 Pn0 ðlÞ cos h
n¼2;4
rhhð2Þ ¼ D0 =R3 þ
1 h X
2
i
ð21Þ
Cn nðn2 2n 1 þ 2mÞ=Rnþ1 Dn ðn þ 1Þ =Rnþ3 Pn ðlÞ
n¼2;4
þ
1 X
Cn ð n þ 4 4mÞ=Rnþ1 þ Dn =Rnþ3 Pn0 ðlÞ cos h
n¼2;4
r// ¼ r//ð1Þ þ r//ð2Þ r//ð1Þ ¼ 2A0 ð1 þ mÞ þ
1 X
An ðn þ 1Þðn 2 ð4n þ 2ÞmÞRn þ Bn nRn2 Pn ðlÞ
n¼2;4
þ
1 X
An ðn þ 5 4mÞRn Bn Rn2 Pn0 ðlÞ cos h
n¼2;4
r//ð2Þ
ð22Þ
1 X ¼ D0 =R3 þ Cn nðn þ 3 ð4n þ 2ÞmÞ=Rnþ1 Dn ðn þ 1Þ=Rnþ3 Pn ðlÞ n¼2;4
þ
1 X
Cn ðn 4 þ 4mÞ=Rnþ1 Dn =Rnþ3 Pn0 ðlÞ cos h
n¼2;4
In above mentioned equations, G denotes the shear modulus of elasticity, m the PoissonÕs ratio, l ¼ cos h, and Pn ðlÞ the Legendre polynomials. Concept for obtaining the series solution shown by Eqs. (17)–(22) is introduced below. In 3-D elasticity, the displacements can be expressed by PapkovichÕs presentation where a harmonic vector and a harmonic scalar are involved [7]. In fact, the harmonic vector (three harmonic functions) as well as the harmonic scalar are some harmonic functions. In the spherical coordinates, method of separation of variables will lead to a set of harmonic functions. Substituting those harmonic functions into PapkovichÕs presentation, one will get some particular solutions for 3-D elasticity in spherical coordinates. The obtained solutions are expressed in the form of Eqs. (17)–(22). To satisfy Eq. (14), i.e. the traction free condition on the surface of cavity, substituting Eqs. (19) and (20) into Eq. (14) yields D0 ¼ A0 ð1 þ mÞR30
ð23Þ
and An =Dn þ ðn 1Þðn þ 2Þð2n þ 1ÞR2n1 Bn =Dn Cn ¼ ðn2 1Þðn þ 2Þð2n þ 3ÞR2nþ1 0 0 Dn ¼ ð2n5 þ 5n4 5n2 þ 6n þ 4 4nðn þ 1Þm2 ÞR2nþ3 An =Dn 0 þ nðn 1Þð2n 1Þðn þ 2ÞR2nþ3 Bn =Dn 0
ð24Þ
332
Y.Z. Chen / International Journal of Engineering Science 42 (2004) 325–338
where Dn ¼ ðn þ 2Þ 2ðn2 þ n þ 1Þ ð4n þ 2Þm
ð25Þ
Finally, after substituting Eqs. (23) and (24) into Eqs. (17)–(22), the eigenfunction expansion form is obtainable. Below, the boundary value problem shown in Fig. 2 is considered. In the process of using Eqs. (8), (11) and (12), the following points need to be mentioned: (a) In this case, the right hand terms of Eqs. (15) and (16) (rzz jz¼H ¼ p, rrz jz¼H ¼ 0, rrr jr¼B ¼ 0, rrz jr¼B ¼ 0) serve as the applied traction pi in Eq. (12). (b) After substituting Eqs. (23) and (24) into Eqs. (17)–(22), only the undetermined coefficients A0 ; A2 ; A4 ; . . . B2 ; B4 ; . . . are involved in Eqs. (17)–(22). These coefficients A0 ; A2 ; A4 ; . . . B2 ; B4 ; . . . serve as the coefficients Xm in Eq. (8). ðkÞ (c) In Eqs. (17)–(22), the functions after the coefficients A0 ; A2 ; A4 ; . . . B2 ; B4 ; . . . serve as the ui , ðkÞ rij in (11) and (12). After using Eqs. (8), (11), (12) and (17)–(25), the calculation is straightforward with no difficulty. The detailed computations will not be cited here. In this computation, we take m ¼ 1=3 and M ¼ 15 terms (M––the number of truncated terms) are truncated in the expansion form Eqs. (17)– (22). The calculated results for the stresses on the surface of cavity are expressed as follows rhh jR¼R0 ¼ f1 ðR0 =B; H =B; hÞp r// jR¼R0 ¼ f2 ðR0 =B; H =B; hÞp 1=2 2 2 req ¼ rhh rhh r// þ r//
ð26Þ ¼ f3 ðR0 =B; H=B; hÞp
R¼R0
where req denotes the equivalent stress used the Mises yield criterion of plasticity. The calculated f1 , f2 and f3 values are plotted in Figs. 3 and 4 and listed in Table 1, respectively. For comparison, the exact solution for an infinite body containing the spherical cavity under the remote tension rz ¼ p is also introduced [7] 27 15m 15 2 rhh jR¼R0 ¼ F1 ðhÞp ¼ cos h p 2ð7 5mÞ 7 5m 3 15m 15m 2 cos h p r// jR¼R0 ¼ F2 ðhÞp ¼ 2ð7 5mÞ 7 5m 1=2 ¼ F3 ðhÞp req ¼ r2hh rhh r// þ r2//
ð27Þ
R¼R0
This case must be equivalent to the R0 =B ! 0 case in the present study. The results from Eq. (27) (R0 =B ¼ 0) are listed in Table 1, and several calculated results for f1 , f2 and f3 values (R0 =B > 0) are also listed in Table 1.
Y.Z. Chen / International Journal of Engineering Science 42 (2004) 325–338
333
3.5 3.0
H/B=2.0
2.5
f 1 (R0/B,H/B,θ)
2.0
R0/B=0.7
f 2 (R0/B,H/B,θ)
1.5
0.1
1.0
0.4
0.5 0.0 -0.5 0.7
-1.0 -1.5 0
10
20
30
40 50 θ (degree)
60
70
80
90
Fig. 3. Calculated values for f1 ðR0 =B; H =B; hÞ and f2 ðR0 =B; H =B; hÞ in Eq. (26) in the case of H =B ¼ 2:0.
3.5 R0/B=0.7
3.0
H/B=2.0
2.5
f (R0/B,H/B,θ) 3
2.0 1.5 0.1
0.4
1.0 0.5 0.0 0
10
20
30
40
50
60
70
80
90
θ (degree)
Fig. 4. Calculated values for f3 ðR0 =B; H =B; hÞ in Eq. (26) in the case of H =B ¼ 2:0.
Table 1 Calculated results of f1 ðR0 =B; H=B; hÞ, f2 ðR0 =B; H =B; hÞ and f3 ðR0 =B; H =B; hÞ in Eq. (26) in the case of H =B ¼ 1:0 R0 =B
0
f1 jh¼0° f1 jh¼90° f2 jh¼0° f2 jh¼90° f3 jh¼0° f3 jh¼90°
)0.750 2.063a )0.750a 0.188a 0.750a 1.975a
a
Exact solution.
a
0.1
0.2
0.3
0.4
0.5
0.6
0.7
)0.753 2.067 )0.753 0.187 0.753 1.981
)0.777 2.100 )0.777 0.181 0.777 2.015
)0.837 2.182 )0.837 0.166 0.837 2.104
)0.949 2.334 )0.949 0.129 0.949 2.272
)1.134 2.577 )1.134 0.053 1.134 2.551
)1.409 2.941 )1.409 )0.095 1.409 2.989
)1.769 3.478 )1.769 )0.379 1.769 3.682
334
Y.Z. Chen / International Journal of Engineering Science 42 (2004) 325–338
Table 2 Calculated results of f1 ðR0 =B; H =B; hÞ, f2 ðR0 =B; H =B; hÞ and f3 ðR0 =B; H=B; hÞ in Eq. (26) in the case of R0 =B ¼ 0:5 h 0°
15°
30°
45°
60°
75°
90°
H =B ¼ 1:0 case f1 ðM ¼ 15Þ f1 ðM ¼ 11Þ f1 ðM ¼ 7Þ f2 ðM ¼ 15Þ f2 ðM ¼ 11Þ f2 ðM ¼ 7Þ f3 ðM ¼ 15Þ f3 ðM ¼ 11Þ f3 ðM ¼ 7Þ
)1.134 )1.120 )1.081 )1.134 )1.120 )1.081 1.134 1.120 1.081
)0.832 )0.828 )0.815 )1.032 )1.024 )0.996 0.948 0.941 0.919
)0.083 )0.091 )0.117 )0.783 )0.782 )0.774 0.745 0.740 0.723
0.820 0.815 0.785 )0.490 )0.490 )0.490 1.146 1.141 1.114
1.675 1.677 1.662 )0.221 )0.220 )0.218 1.796 1.797 1.781
2.325 2.326 2.309 )0.023 )0.022 )0.019 2.336 2.337 2.319
2.577 2.574 2.550 0.053 0.053 0.055 2.551 2.548 2.523
H =B ¼ 2:0 case f1 ðM ¼ 15Þ f1 ðM ¼ 11Þ f1 ðM ¼ 7Þ f2 ðM ¼ 15Þ f2 ðM ¼ 11Þ f2 ðM ¼ 7Þ f3 ðM ¼ 15Þ f3 ðM ¼ 11Þ f3 ðM ¼ 7Þ
)0.854 )0.864 )0.855 )0.854 )0.864 )0.855 0.854 0.864 0.855
)0.666 )0.670 )0.655 )0.796 )0.803 )0.790 0.740 0.746 0.732
)0.140 )0.127 )0.102 )0.634 )0.633 )0.612 0.577 0.580 0.568
0.623 0.649 0.675 )0.402 )0.389 )0.361 0.894 0.908 0.911
1.456 1.471 1.481 )0.150 )0.132 )0.102 1.537 1.541 1.534
2.123 2.107 2.091 0.050 0.066 0.095 2.098 2.074 2.045
2.381 2.347 2.319 0.128 0.141 0.168 2.320 2.280 2.239
M––number of the truncated terms.
As expected, from Table 1 we see that the difference between two cases R0 =B ¼ 0 (exact solution) and R0 =B ¼ 0:1 (numerical solution using EEVM) is small. Also, the obtained results are useful to assess the strength of a body weakened by the spherical cavity. From Fig. 4 or Table 1 we also see that the most dangerous point is always located at the position h ¼ 90°. To investigate the effect of the ratio H =B to the solution, several calculated values of f1 ðR0 =B; H =B; hÞ, f2 ðR0 =B; H=B; hÞ and f3 ðR0 =B; H =B; hÞ, under the conditions of R0 =B ¼ 0:5 and M ¼ 15, are listed in the first two columns of Table 2. From Table 2 we see that, if H =B is changed from H =B ¼ 2:0 to 1.0, the equivalent stress req at the most dangerous point is elevated from 2.320p to 2.551p. The convergence of the numerical solution is also studied. To this end, several particular values of f1 ðR0 =B; H =B; hÞ, f2 ðR0 =B; H =B; hÞ and f3 ðR0 =B; H =B; hÞ under the conditions R0 =B ¼ 0:5 and H =B ¼ 2, are also listed in last three columns of Table 2. From Table 2 we see that, if the number of the truncated terms in the eigenfunction expansion form is changed from M ¼ 7 to 15, convergence can be found from the listed results. 4. Round bar containing a spherical rigid inclusion In this section, we consider the problem for the round bar containing a spherical rigid inclusion, as shown in Fig. 2. The bar is also stretched by a tension p in the z-direction. In this case, the series solution, from Eqs. (17)–(22), can also be used.
Y.Z. Chen / International Journal of Engineering Science 42 (2004) 325–338
335
Nevertheless, the fixed displacement condition on the surface of the rigid inclusion will lead to uR jR¼R0 ¼ 0;
uh jR¼R0 ¼ 0
ð28Þ
Substituting Eqs. (17) and (18) into Eq. (28) yields D0 ¼ 2ð1 2mÞR30 A0
ð29Þ
and An =dn þ ð2n þ 1ÞR2n1 Bn =dn Cn ¼ ðn þ 1Þð2n þ 3ÞR2nþ1 0 0 Dn ¼ ð2n þ 1Þðn2 þ n þ 8 24m þ 16m2 ÞR2nþ3 An =dn þ nð2n 1ÞR2nþ1 Bn =dn 0 0
ð30Þ
where dn ¼ 2ð3n þ 2Þ þ 4ð2n þ 1Þm
ð31Þ
As before, after substituting Eqs. (29) and (30) into Eqs. (17)–(22), the eigenfunction expansion form is obtainable. In the same condition of computation, the calculated results for the stresses on the surface of the rigid sphere can be expressed as rRR jR¼R0 ¼ g1 ðR0 =B; H =B; hÞp rRh jR¼R0 ¼ g2 ðR0 =B; H =B; hÞp rhh jR¼R0 ¼ r// jR¼R0 ¼ g3 ðR0 =B; H =B; hÞp 1=2 req ¼ ðg1 g3 Þ2 þ 3g22 ¼ g4 ðR0 =B; H =B; hÞp
ð32Þ
where req denotes the equivalent stress used in the Mises yield criterion of plasticity. Some calculated g1 , g2 , g3 and g4 values are plotted in Figs. 5 and 6 and listed in Table 3, respectively. For comparison, the exact solution for an infinite body containing the spherical rigid inclusion is also cited as follows [7] 2ð1 2mÞ 7 5m 15ð1 mÞ 2 þ cos h p rRR jR¼R0 ¼ G1 ðhÞp ¼ 3ð1 þ mÞ 24 30m 8 10m 15ð1 mÞ sin h cos h p rRh jR¼R0 ¼ G2 ðhÞp ¼ 8 10m ð33Þ 1 2m 8 25m 15m 2 þ þ cos h p rhh jR¼R0 ¼ r// jR¼R0 ¼ G3 ðhÞp ¼ 3ð1 þ mÞ 24 30m 8 10m 1=2 2 req ¼ ðG1 G3 Þ þ 3G22 ¼ G4 ðR0 =B; H =B; hÞp As expected, from Table 3 we see that the difference between two cases R0 =B ¼ 0 (exact solution) and R0 =B ¼ 0:1 (numerical solution using EEVM) is small. Also, the obtained results are
336
Y.Z. Chen / International Journal of Engineering Science 42 (2004) 325–338 2.0 1.5
g (R /B,H/B,θ) 1 0
1.0
g2 (R /B,H/B,θ) - 0
0.5 R0/B=0.1
0.4
0.0
0.7
-0.5 -1.0 H/B=2.0 -1.5 0
10
20
30
40
50
60
70
80
90
θ ( degree ) Fig. 5. Calculated values for g1 ðR0 =B; H =B; hÞ and g2 ðR0 =B; H =B; hÞ in Eq. (32) in the case of H =B ¼ 2:0.
2.0 R0/B=0.1 H/B=2.0
1.5 0.4
1.0
0.7 0.5 g3(R0/B,H/B,θ)
0.0
g4(R0/B,H/B,θ) - -0.5 0
10
20
30
40
50
60
70
80
90
θ ( degree )
Fig. 6. Calculated values for g3 ðR0 =B; H =B; hÞ and g4 ðR0 =B; H =B; hÞ in Eq. (32) in the case of H =B ¼ 2:0.
useful to assess the strength of body containing the spherical rigid inclusion. From Fig. 6 or Table 3 we see that the most dangerous point is always located at the vicinity of h ¼ 45°.
5. Remarks A simple, yet efficient, method is presented for the solution of the axisymmetric tension problem of the round bar containing a spherical cavity or rigid inclusion. It is well known that, if the problem for the infinite body containing the spherical cavity is concerned, sometimes, the
Y.Z. Chen / International Journal of Engineering Science 42 (2004) 325–338
337
Table 3 Calculated results of g1 ðR0 =B; H =B; hÞ, g2 ðR0 =B; H =B; hÞ, g3 ðR0 =B; H =B; hÞ and g4 ðR0 =B; H =B; hÞ in Eq. (32) in the case of H =B ¼ 2:0 R0 =B g1 jh¼0° g1 jh¼45° g1 jh¼90° g2 jh¼0° g2 jh¼45° g2 jh¼90° g3 jh¼0° g3 jh¼45° g3 jh¼90° g4 jh¼0° g4 jh¼45° g4 jh¼90° a
0 a
1.929 0.857a )0.214a 0.0 a )1.071a 0.0a 0.964a 0.429a )0.107a 0.964a 1.905a 0.107a
0.1
0.2
0.3
0.4
0.5
0.6
0.7
1.937 0.857 )0.223 0.0 )1.080 0.0 0.968 0.428 )0.111 0.968 1.919 0.111
1.923 0.855 )0.215 0.0 )1.069 0.0 0.961 0.428 )0.108 0.961 1.900 0.108
1.889 0.848 )0.194 0.0 )1.043 0.0 0.945 0.424 )0.097 0.945 1.855 0.097
1.834 0.831 )0.158 0.0 )1.001 0.0 0.917 0.415 )0.079 0.917 1.783 0.079
1.762 0.798 )0.108 0.0 )0.949 0.0 0.881 0.399 )0.054 0.881 1.692 0.054
1.679 0.748 )0.050 0.0 )0.893 0.0 0.839 0.374 )0.025 0.839 1.591 0.025
1.588 0.680 0.006 0.0 )0.836 0.0 0.794 0.340 0.003 0.794 1.488 0.003
Exact solution.
integral transform method is used. However, it is difficult to use the integral transform method to the problem of the finite body with complex geometry. Therefore, the EEVM can be considered as a complementary approach to some classical approaches mainly used for the case of infinite medium. Essentially, the EEVM belongs to the category of the TrefftzÕs method. That is to say, one seeks the approximate solution within the range of the solutions which satisfy the governing equations of elasticity exactly. However, as shown in Eqs. (8)–(10), the EEVM can be used in all cases, including the displacement boundary value condition and the mixed boundary value condition. Finally, the results obtained from two cases, one for containing spherical cavity and other for containing the spherical rigid inclusion, are quite different. In the first case, if the cavity is larger, then the bar weakened by the cavity is more dangerous, where the req curve corresponding R0 =B ¼ 0:7 is located in a higher position (Fig. 4). On the contrary, in the second case, if the spherical rigid inclusion is larger, then the bar is safer, where the req curve corresponding R0 =B ¼ 0:7 is located in a lower position (Fig. 6). This situation can be seen from Figs. 4 and 6, respectively.
Acknowledgement The research project is supported by the National Natural Science Foundation of China.
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