composite structures tailoring for reducing...analytically ﬁnd the stress concentration factor, k,...

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Composite Structures

journal homepage: www.elsevier.com/locate/compstruct

Material tailoring for reducing stress concentration factor at a circular holein a functionally graded material (FGM) panel

G.J. Niea,b,⁎, Z. Zhonga, R.C. Batrab

a School of Aerospace Engineering and Applied Mechanics, Tongji University, Shanghai 200092, ChinabDepartment of Biomedical Engineering and Mechanics, Virginia Polytechnic Institute and State University, Blacksburg, VA 24061, USA

A R T I C L E I N F O

Keywords:Functionally graded materialsCircular holeStress concentration factorMaterial tailoring

A B S T R A C T

By assuming that Young’s modulus and Poisson’s ratio of a linearly elastic and isotropic material vary along theradial direction in a panel with a circular hole and deformed by a far field uniaxial tensile traction, we firstanalytically find the stress concentration factor, K, at the hole. The problem is solved by superposing solutions oftwo problems – one of uniform biaxial tension and the other of pure shear. The solutions of the first and thesecond problem are, respectively, in terms of hypergeometric functions and Frobenius series. Subsequently, weanalytically study the material tailoring problem for uniform biaxial tension, and give explicit variation ofYoung’s modulus to achieve a prespecified K. For the panel loaded by a far field uniaxial tensile traction, weshow that the K can be reduced by a factor of about 8 by appropriately grading Young’s modulus and Poisson’sratio in the radial direction. By plotting K versus the two inhomogeneity parameters, we solve the materialtailoring problem for a panel loaded with a far field uniaxial traction. The analytical results should serve asbenchmarks for verifying the accuracy of approximate/numerical solutions for an inhomogeneous panel.

1. Introduction

Even though the mechanical behavior of an inhomogeneous mate-rial has been studied since 1950’s, there has been tremendous activityin this field during the last three decades [1–6]. A heterogeneous ma-terial with continuous spatial variation of material parameters is oftencalled a functionally graded material (FGM). With the availability of 3-D printing for manufacturing materials with complex microstructures,it is now feasible to fabricate structures to have the optimum stress andstrain distributions for enhancing their mechanical properties underprescribed loads [7,8]. One such problem is controlling the stressconcentration factor, K, around a circular hole in a panel.

It is well known that K at a circular hole in an infinite panel com-posed of a homogeneous, isotropic and linearly elastic material de-formed in uniaxial tension equals 3 [9]. For orthotropic materialsLekhnitskii et al. [10] have deduced K for an infinite plate containing acircular hole and deformed by remote uniaxial tensile tractions. Basedon Lekhnitskii’s solution of the plane elastostatics problem using com-plex variables, Britt [11], Tenchev et al. [12] and Xu et al. [13,14],respectively, found K for anisotropic rectangular panels with centrallylocated circular and elliptical cutouts, laminated composites with cir-cular holes, and composite laminates with either an elliptical hole ormultiple holes.

Authors of Refs. [15–18] employed the finite element method (FEM)to evaluate stresses in composite laminates with circular holes. Kubairand Bhanu-Chandar [19] and Enab [20] using the FEM found that K issignificantly influenced by the spatial variation of the material in-homogeneity. One needs a very fine mesh near the hole and conductconvergence studies to deduce reasonably accurate values of K that canbe an arduous task.

Huang and Haftka [21], and Cho and Rowlands [22,23] optimizedfiber orientations near a hole to minimize K and increase the load-carrying capacity of composite laminates. Lopes et al. [24] and Gomeset al. [25] found fiber orientation angles and their volume fractionseither to minimize the peak stress around cutouts or to maximize thebuckling and the first-ply failure load of composite panels. They pointedout that the optimum variable-stiffness designs with a central hole canhave nearly the same initial buckling loads as panels with the samevolume fractions of fibers but no hole.

By dividing an inhomogeneous material panel into a series of piece-wise homogeneous layers and using the method of complex variables,Yang et al. [26,27], Yang and Gao [28] and Kushwaha and Saini [29]have shown that when Young’s modulus decreases with the distancefrom the hole boundary, K > 3. Mohammadi et al. [30] analyticallyfound K around a circular hole in an infinite FGM plate subjected touniform biaxial tension and pure shear by assuming that both Young’s

https://doi.org/10.1016/j.compstruct.2018.08.078Received 28 June 2018; Received in revised form 1 August 2018; Accepted 27 August 2018

⁎ Corresponding author at: School of Aerospace Engineering and Applied Mechanics, Tongji University, Shanghai 200092, China.E-mail address: [email protected] (G.J. Nie).

Composite Structures 205 (2018) 49–57

Available online 03 September 20180263-8223/ © 2018 Elsevier Ltd. All rights reserved.

T

modulus and Poisson’s ratio vary exponentially in the radial direction.By assuming that Young’s modulus has a power law variation in theradial direction and Poisson’s ratio is a constant, Sburlati [31] studiedthe elastic response of an FGM annular ring inserted in a hole of ahomogeneous plate. Kubair [32] used the method of separation ofvariables to find closed-form expressions for stresses and displacementsin FGM plates with and without holes under anti-plane shear loadingand used a non-traditional definition of K.

We note that there are a limited number of analytical studies on thestress concentration around a circular hole in isotropic FGM panels.Furthermore, there are no results on material tailoring for reducing K.We analytical (i) find K at a circular hole in an isotropic FGM panelunder a far field uniaxial tensile traction, (ii) analyze the material tai-loring problem for uniform biaxial tension loading, and (iii) investigatethe effect of material inhomogeneity parameters on K. For far fielduniform tensile loading, we provide a response function to estimateinhomogeneity parameters for a desired value of K.

The rest of the paper is organized as follows. Sections 2 and 3, re-spectively, give the formulation and the solution of the direct problemin which we analyze deformations of the panel under prescribed farfield surface tractions. Section 3 is divided into three subsections thatprovide details of deformations under uniform biaxial tension, pureshear and uniaxial tension, respectively. In Section 4 we analyticallysolve the material tailoring problem for uniform biaxial tensionloading. Section 5 provides numerical results that establish the accu-racy and the convergence of the series solution for the pure shearloading, and delineate effects of the variation of the material propertieson K and stress distributions. Conclusions of the work are summarizedin Section 6.

2. Formulation of the direct problem

We consider an isotropic and linearly elastic FGM panel with acircular hole of radius a subjected to a far-field uniaxial traction σ0, asshown in Fig. 1(a). We analyze how the radial variation in Young’smodulus and Poisson’s ratio affects the stress concentration at the holeperiphery for plane stress deformations of the panel. We solve theproblem by superposing solutions of two problems – biaxial tension andpure shear, as shown in Fig. 1(b) and (c).

We use cylindrical coordinates (r, θ) with origin at the hole center todescribe the panel deformations. In the absence of body forces, equili-brium equations are

σ

0,

0,

σr r θ

σ σr

σr r

σθ r rθ

1

1 2

rr rθ rr θθ

rθ θθ

+ + =

+ + =

∂

∂

∂

∂

−

∂

∂

∂

∂ (1a,b)

where σrr , σθθ and σrθ are stress components. Hooke’s law relatingstresses to infinitesimal strains, ε ε ε, ,rr θθ rθ, is

εE r

σ v r σ εE r

σ v r σ εv r

E rσ1

( )[ ( ) ], 1

( )[ ( ) ], 2[1 ( )]

( ).rr rr θθ θθ θθ rr rθ rθ= − = − =

+

(2a-c)

We assume that Young’s modulus, E r( ), and Poisson’s ratio, v r( ),are given by either

(i) general power-law variations

E r E β ra

v r v β ra

n( ) [1 ( ) ], ( ) [1 ( ) ], with 0n n1 2= + = + <∞ ∞ (3a)

or

(ii) exponential and power-law variations

E r E exp γ ra

v r v γ ra

( ) [ ( ) ], ( ) [1 ( ) ],11

21= = +∞

−∞

−

(3b)

where E E rlim ( )r

=∞→∞

and v v rlim ( )r

=∞→∞

, β β,1 2 and γ γ,1 2

( β β γ γ1 { , , , } 11 2 1 2− < < ) are real numbers, and n is a negative integerwhich helps find an analytical solution of the problem. For a homo-geneous material, β β γ γ 01 2 1 2= = = = . The variations of E with r a/ forn 3, 5= − − and β 0.9, 0.91 = − depicted in Fig. 2(a) reveal thatE r E( )/ →∞ 1 as r a/ →5. Similarly, the variation of E with r a/ forγ 0.9, 0.91 = − depicted in Fig. 2(b) implies that E r E( )/ →∞ 1 as r a/ →50.

The far-field boundary conditions for the biaxial tension and thepure shear problems are:

σ r σlim ( )2r

rr0=

→∞ (4a)

σ r θ σ θ σ r θ σ θlim ( , )2

cos 2 , lim ( , )2

sin 2r

rrr

rθ0 0= = −

→∞ →∞ (4b)

and boundary conditions at the hole periphery are

σ a θ σ a θ( , ) 0, ( , ) 0.rr rθ= = (4c)

When solving the problem for stresses, we employ the followingcompatibility equation:

εr r

εθ r

εr r

εr r

εr θ r

εθ

1 2 1 1 1θθ rr θθ rr rθ rθ2

2 2

2

2

2

2∂

∂+

∂

∂+

∂

∂−

∂

∂=

∂

∂ ∂+

∂

∂ (5)

3. Solution of the direct problem

We analytically solve the problem for the uniform biaxial tension insubsection 3.1, and use the Frobenius series to analyze the problem forpure shear loading in subsection 3.2. By superposing solutions of thesetwo problems, we obtain the solution for the uniaxial tension problemin subsection 3.3.

3.1. Uniform biaxial tension

We note that the problem geometry, the material properties and the

Fig. 1. Schematic sketch of a panel with a circular hole subjected to a uniform far-field tensile traction, and its split into two problems.

G.J. Nie et al. Composite Structures 205 (2018) 49–57

50

far field loading are axisymmetric. Accordingly, we assume panel’sdeformations to be axisymmetric for which equilibrium Eq. (1) reducesto

σ σ r dσdrθθ rr

rr= +(6)

Thus knowing σrr , one can find σθθ from Eq. (6).We introduce a non-dimensional parameter, ρ r

a= , and substitutefor strains from Eq. (2) into the compatibility condition (5) to get

d σdρ ρ

EE

dσdρ

vEρE

EρE

vρ

σ( 3 ) ( ) 0rr rrrr

2

2 + −′

+′−

′−

′=

(7)

where ( ) ( )ddρ

′ = .

3.1.1. General power-law variations of E r( ) and v r( )given by Eq. (3a)Substituting for E r( ) and v r( ) from Eq. (3a) into Eq. (7) and solving

the resulting equation, we get

σ Cρ

C β β, for 0, 0rr12 2 1 2= + = =

(8a)

σ Cρ

F t t t β ρ C F t t t β ρ n( , ; ; ) ( , ; ; ), for 3rrn n1

2 2 1 1 2 3 1 2 2 1 4 5 6 1= − + − ≤ −

(8b)

σ Cρβ

F t t tρβ

Cρβ

F t t tρβ

n

( ) ( , ; 2 ; ) ( ) ( , ; 2 ; ),

for 1,

rrt t

11

2 1 7 8 81

21

2 1 9 10 101

7 9= − + −

= − (8c)

σ Cρβ

F t t tρβ

Cρβ

F t t tρβ

n

( ) ( , ; ; ) ( ) ( , ; ; ),

for 2

rrt t

1

2

12 1 12 11 13

2

12

2

12 1 12 14 15

2

1

11 14= − + − −

= − (8d)

Here C1 and C2 are constants to be determined from boundary condi-tions (4a) and (4c), F a b c z a b c z k( , ; ; ) ( ) ( ) /( ) / !k k k k

k2 1 0= ∑ =

∞ is a hy-pergeometric function, and

t H H t H H tn

t H H t H H t

nt H t H t H t

H t H t H t H t H t

H Hn

vn

ββ

Hn

H

nH v

ββ

H v ββ

, , 1 2 , , ,

1 2 , 32

, 12

, 32

,

12

, 1 , , 1 2 , 1 ,

1 2 , 1 ( 1) 14

, 12

1 ,

12

1 , 54

( 1) , 12 2

( 1) .

1 2 1 2 2 1 3 4 3 1 5 3 1 6

7 4 8 4 9 4 10

4 11 5 12 5 13 5 14 5 15

5 1 22

12 3

42

15

2

1

= − = + = − = − = +

= + = − − = − = − +

= + = − − = − = − = − +

= + = + − + = − −

= − + = − − = − −

∞

∞∞

(9)

For β 01 = , the solution can be expressed in terms of Bessel functionsJ z( )n and Y z( )n , respectively, of the first and the second kind, and themodified Bessel functions I z( )n and K z( )n .

We note that the solution (8a) for the homogeneous material agreeswith that given in Timoshenko’s book [9]. Constants Ci, i=1, 2,…

appearing in different equations do not, in general, have the same va-lues.

3.1.2. Exponential and power-law variations of E r( ) and v r( ) given by Eq.(3b)

Substituting for E r( ) and v r( ) from Eq. (3b) into Eq. (7) and solvingthe resulting equation, we get

σρ

tρ

F t tρ

C U t tρ

C1 exp( )[ ( ; 3; ) ( , 3, ) ]rr 218

1 1 1716

1 1716

2= +(10)

Here C1 and C2 are constants to be determined from boundary condi-tions (4a) and (4c), F a b z( ; ; )1 1 and U a b z( , , ) are the confluent hy-pergeometric functions defined as F a b z a b z k( ; ; ) ( ) /( ) / !k k k

k1 1 0= ∑ =

∞ ,U a b z a e t t dt( , , ) 1/Γ( ) (1 )zt a b a

01 1∫= +

∞ − − − − , where Γ a( ) is the Eulergamma function, and

t γ γ v γ tv γ v γ

tt

γ t( 4 ) , 3

2(2 1) 2

2,

216 1 1 2 171 2

118

1 16= + = +

− −=

−∞

∞ ∞

(11)

For γ 01 = , the solution can be expressed in terms of Bessel functionsJ z( )n and Y z( )n , respectively, of the first and the second kind.

3.2. Pure shear deformations

We introduce an Airy stress function, φ r θ( , ), and note that

σr

φr r

φθ

σφ

rσ

r rφθ

1 1 , , ( 1 ),rr θθ rθ2

2

2

2

2=∂

∂+

∂

∂=

∂

∂= −

∂

∂

∂

∂ (12)

identically satisfy equilibrium Eq. (1). In view of far-field condition(4b), we assume that

φ r θ φ r cos θ( , ) ( ) 22= (13)

From Eqs. (2), (5), (12) and (13), we find that the compatibilitycondition reduces to

d φdρ

z ρd φdρ

z ρd φdρ

z ρdφdρ

z ρ φ( ) ( ) ( ) ( ) 04

24 3

32

3 2

22

2 12

0 2+ + + + =(14)

where

z ρ vEρ E

v Eρ E

v Eρ E

Eρ E

vρ

( ) 4 8 ( ) 8 12 4 ,0 2

2

2 2 2 3 2= −″

+′

−′ ′

−′

+″

z ρ vEρE

v EρE

v EρE

Eρ E

vρ ρ

z ρ

EE

EE

vEρE

EρE

vρ ρ

( ) 2 ( ) 2 9 9 , ( )

2( ) 2 9 ,

12

2 2 3 2

2

2 2

=″

−′

+′ ′

+′

−″

+

= −″

+′

+′−

′−

′−

z ρρ

EE

( ) 2 2 .3 = −′

(15)

We solve the 4th order ordinary differential Eq. (14) with variablecoefficients by the method of Frobenius series.

Fig. 2. Variation of Young’s modulus with r/a for (a) two values of n and two values of β1 in Eqs. (3a), and (b) two values of γ1 in Eq. (3b).


51

3.2.1. General power-law variations of E r( ) and v r( )given by Eq. (3a)We assume that

φ ρ ρ b s ρ b s( ) ( )( ) , with ( ) 1s

k

M

kn k

20

0∑= == (16)

where b k M, 0, 1, 2, ,k = ⋯ , are given by the recurrence formulae, thevariable s is to be determined, and M equals the total number of termsin the series.

Substituting from Eqs. (3a) and (16) into Eq. (14) and equating tozero coefficients of different powers of ρ, we get recurrence formulae((17a) and (17b)) for bk and the indicial Eq. (17c) for s:

b s B s b sn s n s n s n s

( ) ( ) ( )( 4)( 2)( )( 2)1

1 0=+ − + − + + + (17a)

b sB s b s B s b s

nk n s nk n s nk n s nk n s

k M

( )( ) ( ) ( ) ( )

( 2 4)( 2 2)( 2 )( 2 2)

, 0, 1, 2, , 2

k

k k

2

2 3 1=+

+ + − + + − + + + + +

= ⋯ −

+

+

(17b)

s s s s( 4)( 2)( 2) 0− − + = (17c)

The lengthy expressions for B s i( ) ( 1, 2, 3)i = are listed in AppendixA.

Roots, s 4, 2, 0, 2= − , of the indicial Eq. (17c) are independent ofthe inhomogeneity parameters β β,1 2 and n. According to the Frobeniusmethod [33], solutions of the stress function for different values of n are

φ ρ C ρ b s ρ C b s ρ

Cρ

b s ρ C ρ b s ρ n

n n n

( ) ( )| ( ) ( )| ( )

( )| ( ) ( )| ( ) for

0 and 1, 2, 4

k

M

k sn k

k

M

k sn k

k

M

k sn k

k

M

k sn k

2 12

02 2

00

32

02 4

4

04

∑ ∑

∑ ∑

= +

+ +

< ≠ − ≠ − ≠ −

=

=

=

=

=

=−

=

=

(18a)

φ ρ C ρ d b sds

ρ db sds

ρ b s ρ

C d sb sds

ρ sb s ρ

Cρ

b s ρ C ρ b s ρ n

n

( ) { { ( ) 2 ln ( ) (ln ) ( )} ( ) }

{ { [ ( )] ln [ ( )]} ( ) }

( )| ( ) ( )| ( ) for

1, and 2

k

Mk k

ks

n k

k

Mk

ks

n k

k

M

k sn k

k

M

k sn k

2 12

0

2

22

2

20 0

32

02 4

4

04

∑

∑

∑ ∑

=∼

+∼

+∼

+ +

+ +

= − = −

= =

= =

=

=−

=

=

(18b)

φ ρ C ρ d s b sds

ρ s b s ρ

C b s ρ Cρ

b s ρ

C ρ b s ρ n

( ) { { [( 2) ( )] ln [( 2) ( )]} ( ) }

( )| ( ) ( )| ( )

( )| ( ) for 4

k

Mk

ks

n k

k

M

k sn k

k

M

k sn k

k

M

k sn k

2 12

0 2

20

032

02

44

04

∑

∑ ∑

∑

=−

+ −

+ +

+ = −

= =

=

=

=

=−

=

=(18c)

where b s s b s( ) ( 2) ( )k k2∼

= − and C i, 1, 2, 3, 4,i = are constants to be de-termined from boundary conditions (4b) and (4c).

3.2.2. Exponential and power-law variations of E r( ) and v r( )given by Eq.(3b)

For E r( ) and v r( ) given by Eq. (3b) we also use the method ofFrobenius series to solve Eq. (14). Letting n 1= − in Eq. (16), we have

φ ρ ρ b s ρ b s( ) ( )( ) , with ( ) 1s

k

M

kk

20

10∑= =

=

−

(19)

Substituting from Eqs. (3b) and (19) into Eq. (14), and equating tozero coefficients of different powers of ρ, we get following recurrence

formulae ((20a), (20b), (20c)) for b s( )k and the indicial equation (20d)for s:

b s D s b ss s s s

( ) ( ) ( )( 5)( 3)( 1)( 1)1

1 0=− − − + (20a)

b s D s b s D s b ss s s s

( ) ( ) ( ) ( ) ( )( 6)( 4)( 2)2

2 0 3 1=+

− − − (20b)

b s D s b s D s b s D s b sk s k s k s k s

k

M

( ) ( ) ( ) ( ) ( ) ( ) ( )( 1 )( 3 )( 5 )( 7 )

,

0, 1, 2, , 3

kk k k

34 5 1 6 2=

+ +

+ − + − + − + −

= ⋯ −

++ +

(20c)

s s s s( 4)( 2)( 2) 0− − + = (20d)

The lengthy expressions for D s i( ) ( 1, 2, ,6)i = ⋯ are listed inAppendix B.

The indicial Eq. (20d) is the same as the indicial Eq. (17c) and isindependent of the inhomogeneity parameters γ1 and γ2. Thus

φ ρ C ρ d b sds

ρ db sds

ρ b s ρ

C d sb sds

ρ sb s ρ

Cρ

b s ρ C ρ b s ρ

( ) { { ( ) 2 ln ( ) (ln ) ( )} ( ) }

{ { [ ( )] ln [ ( )]} ( ) }

( )| ( ) ( )| ( )

k

Mk k

ks

k

k

Mk

ks

k

k

M

k sk

k

M

k sk

2 12

0

2

22

2

1

20 0

1

32

02

14

4

04

1

∑

∑

∑ ∑

=∼

+∼

+∼

+ +

+ +

= =

−

= =

−

=

=−−

=

=−

(21)

where b s s b s( ) ( 2) ( )k k2∼

= − and C i, 1, 2, 3, 4,i = are constants to be de-termined from boundary conditions (4b) and (4c).

From the expression (Eq. (18) or Eq. (21)) of the stress functionφ ρ( )2 , we find stresses by using Eqs. (12) and (13).

3.3. Stress concentration factor, K, for the uniaxial tension problem

We note that for the uniform biaxial tension and the pure shearproblems, σθθ has the maximum value at θ π

2= . Denoting by K0, K2 andK, respectively, the stress concentration factor corresponding to thebiaxial tension, the pure shear and the uniaxial tension problems (seeFig. 1), we have [34]

Kσ ρ

σK

σ ρ θ

σK K K2 ( )|

,2 ( , )|

,2

θθ ρ θθ ρ θ0

1

02

1,

0

0 2π2= = =

+= = =

(22a, b, c)

4. Analytical solution of material tailoring for uniform biaxialtension

We can analytically solve the material tailoring problem for uniformbiaxial tension loading but not for the simple shear problem since itssolution is in terms of Frobenius series. Assume that the stress state is

σ σ ρ σ σ m ρ m2

(1 ),2

[1 ( 1) ] with 0rrm

θθm0 0= − = − + < (23)

For m 1= − , σθθσ20= is a constant in the panel and equals the far

field axial stress. Also, the boundary condition at the hole periphery isexactly satisfied. For the stress state shown in Eq. (23), Eq. (22a) gives

K m 00 = − > (24)

We now find E ρ( ) and v ρ( ) to achieve K m0 = − . Substituting for thestress σrr from Eq. (23) into Eq. (7), we obtain

ρ mE ρE ρ

ρE ρE ρ

v ρ v ρ m m ρ[ ( 1) 1]( )( )

( 1)[( )( )

( ) ( )] ( 2) 0m m m 1+ −′

− −′

− ′ − + =−

(25)

We have one ordinary differential equation for two unknown


52

functions, E ρ( ) and v ρ( ). One way to solve the problem is to assumepolynomial expressions for E ρ( ) and v ρ( ), and find unknowns in themby minimizing the value of the square of the left hand side of Eq. (25).Here, we assume that v ρ v( ) 0= =a constant, and solve Eq. (25) for E ρ( ).Eq. (25) has the solution

E ρ Em ρ v ρ

m( ) (

( 1) (1 ) 1)

m m

00 m

m v2

0 1=+ + − − +

− +(26)

where E E ρ( )|ρ0 1= = . We note that

E E ρ E vm

lim ( ) ( 1 )ρ

00 m

m v2

0 1= =−

∞→∞

+− +

(27)

Thus, for m 2= − , Eqs. (24) and (26) gives E ρ E( ) 0= and K 20 =

which equals that for a homogeneous material.For K0= 1, 2, 3 and 4, the required variations of E ρ E( )/ 0 taking

v 0.250 = displayed in Fig. 3, respectively, have

E E/ 3.161, 1.0, 0.54, 0.3570 =∞ . Thus dE ρdρ ρ

( )

1=> 0 (<0) for K 20 <

(> 2).

5. Numerical results

We first establish in subsection 5.1 that the Frobenius series solutionconverges to an analytical solution of the problem, and then in sub-section 5.2 combine it with the analytical solution of the biaxial tensionproblem to get a solution of the far field simple tensile loading. We useit in subsection 5.2.2 to ascertain the effect of material inhomogeneityparameters on the stress concentration factor, K.

5.1. Accuracy and convergence of the solution for the pure shear problem

We first find the number of terms in the Frobenius series solutionthat give a converged solution for the pure shear problem. We note thatwhen only v varies, i.e., β 01 = in Eq. (3a) or γ 01 = in Eq. (3b), theanalytical solution (14) for the general power-law variations of E r( )and v r( ) given by Eq. (3a) is

φ ρ Cv β ρ

nF d d

v β ρn

Cv β ρ

nF d d

v β ρn

Cv β ρ

nF d d

v β ρn

C F d dv β ρ

n

( ) ( ) ( ; ; )

( ) ( ; ; )

( ) ( ; ; ) ( ; ; )

n n

n n

n n n

2 12

2 3 1 22

22

2 3 3 42

32

2 3 5 62

4 2 3 7 82

n

n

n

4

2

2

=

+

+ +

∞ ∞

∞ ∞

∞ ∞ ∞−

(28a)

The solution for the exponential and the power-law variations ofE r( ) and v r( ) given by Eq. (3b) is

φ ρ Cv γ

ρF d d

v γρ

C Gv γ

ρd d

C Gv γ

ρd d

C Gv γ

ρd d

( ) ( ) ( ; ; ) ( ,0, 2, 4, 2

)

( ,2, 0, 2, 4

)

( ,4, 2, 0, 2

)

2 12 2

2 3 9 102

2 2422 2 11 12

3 2432 2 11 12

4 2442 2 11 12

= − + −− −

+− −

+ −− −

∞ ∞ ∞

∞

∞

(28b)

where constants C i, 1, 2, 3, 4,i = are determined from boundary

Fig. 3. The required variation of Young’s modulus with the radial coordinate toachieve a given stress concentration factor K0(v0= 0.25).

Table 1Convergence of the stress concentration factor for different terms in the series solution (v 0.25=∞ ).

K2 M 5= M 10= M 15= M 20= M 25=

n 5= − β β0.5, 0.51 2= = 5.36106 5.36248 5.36247 5.36247 5.36247

β β0.5, 0.51 2= − = − 2.29613 2.29783 2.29784 2.29784 2.29784

n 7= − β β0.5, 0.51 2= = − 5.47874 5.47945 5.47944 5.47944 5.47944

β β0.5, 0.51 2= − = 2.23846 2.23930 2.23930 2.23930 2.23930

Table 2For n β5, 01= − = comparison of the stress concentration factor obtained fromthe series solution and the exact solution (v 0.25=∞ ).

K2 β 0.72 = − β 0.52 = − β 0.32 = − β 0.32 = β 0.52 = β 0.72 =

Exact 4.02733 4.01951 4.0117 3.98832 3.98055 3.97278Series 4.02733 4.01951 4.0117 3.98832 3.98055 3.97278

Table 3For γ 01 = comparison of the stress concentration factor obtained from theseries solution and the exact solution (v 0.3=∞ ).

K2 γ 0.82 = − γ 0.32 = − γ 0.32 = γ 0.82 =

Exact 4.17719 4.06563 3.93534 3.82967Series 4.17719 4.06563 3.93534 3.82967

Table 4Variation of the stress concentration factor, K , with β β,1 2 and n for far fielduniaxial tension (v 0.25=∞ ).

β1 n 5= − n 3= −

β 0.52 = β 0.52 = − β 0.02 = β 0.52 = β 0.52 = − β 0.02 =

-0.9 0.38716 0.38308 0.38508 0.42322 0.41982 0.42143-0.7 1.08318 1.07383 1.07841 1.14830 1.14194 1.14488-0.5 1.70091 1.68839 1.69450 1.76402 1.75669 1.76000-0.3 2.25883 2.24431 2.25138 2.30254 2.29504 2.29835-0.1 2.76874 2.75294 2.76061 2.78232 2.77500 2.778160.0 3.00836 2.99212 3.00000 3.00410 2.99695 3.000000.1 3.23889 3.22231 3.23035 3.21539 3.20842 3.211360.3 3.67540 3.65837 3.66661 3.61017 3.60363 3.606330.5 4.08298 4.06572 4.07405 3.97288 3.96679 3.969250.7 4.46532 4.44803 4.45636 4.30825 4.30262 4.304830.9 4.82544 4.80824 4.81652 4.62002 4.61482 4.61681


53

conditions (4b) and (4c), G xa ab b(

, ,, , )pq

mn p

q

1

1

…

…is the Meijer G function,

and d N N{ , }n n112

3 12

3= + − + + , d {1 , 1 , 1 }n n n2

2 4 6= + + + , d3 =

N N{ , }n n12

1 12

1+ − + + , d {1 , 1 , 1 }n n n4

2 2 4= − + + , d5 = N{ n

12

3− −

N, }n12

3− + , d {1 , 1 , 1 }n n n6

6 4 2= − − − , d7 = N N{ , }n n

12

1 12

1− − − + ,

d {1 , 1n n84 2

= − − , 1 }n2

+ , d i9

7 232=

− , d i10

7 232=

+ , d i i11

( 23 )2= −

− ,

d i i12

( 23 )2=

+ , with N n nn

12 122

2=

+ − .For a homogeneous material, i.e., β β 01 2= = in Eq. (3a) or

γ γ 01 2= = in Eq. (3b), the solution (14) is

φ ρ C ρ C ρ C Cρ

( )2 14

22

342= + + +

(29)

One gets the stress concentration factor, K 42 = , for the pure shearproblem.

Values of K2 listed in Table 1 for different values of M suggest thatM=5 in the series solution gives a reasonably accurate value of K2, andthe series solution rapidly converges for the considered values of n, β1and β2.

For general power-law variations of E r( ) and v r( ) given by Eq. (3a),values of K2 listed in Table 2 for different values of β2 computed usingM=20 in the series solution and those from the analytical solution (Eq.(28a)) ensure accuracy of the series solution in Eq. (18). Similarly, forexponential and power-law variations of E r( ) and v r( ) given by Eq.(3b), the closeness of values of K2 listed in Table 3 for different values ofγ2 computed using M=10 in the series solution and those from theanalytical solution (Eq. (28b)) verify the accuracy of the series solutionin Eq. (21). Thus values of M to get a converged solution depend uponvalues of n, β1, β2, γ1 and γ2.

5.2. Far field uniaxial tensile loading

5.2.1. Stress concentration factorsFor the uniaxial loading, values of K for different values of β β,1 2

and n in Eq. (3a) listed below in Table 4 reveal that

(i) for β 01 > (β 01 < ) the K for the FGM panel is larger (smaller) thanthat for the homogeneous material panel,

(ii) when the material parameter monotonically increases (β 01 < )with the distance from the hole center, the presently computedreduction in K for the FGM panel agrees with that reported in [19],and

(iii) the maximum (minimum) value of K , equals 4.83 (0.39) that is1.61 (0.13) times the value of K for a homogeneous material.

Thus, the stress concentration factor can be reduced by a factor ofabout 8 (= 3/0.39) by properly grading Young’s modulus and Poisson’sratio in the radial direction.

For different values of β1 and β2, the variation of K with n is shownin Fig. 4. Here we have tacitly assumed that K continuously dependsupon n since the analytical solutions has been found only for integervalues of n. It is seen from Fig. 4 that K increases (decreases) with anincrease of the absolute value of n for β 01 > (β 01 < ). From values of Klisted in Table 4 and plotted in Fig. 4, we conclude that the value ofβ2(β1) that affects the radial variation of Poisson’s ratio (Young’smodulus) has very little (significant) influence on K.

Fig. 4. Variation of the stress concentration factor, K, with n for v∞=0.25.

Fig. 5. Variation of the stress concentration factor, K, on the β1 n-plane forv∞=0.25.

Table 5Variation of the stress concentration factor, K , with γ1 and γ2 (v 0.3=∞ ).

γ1 γ2

-0.9 -0.5 0.0 0.5 0.9

-0.9 1.76807 1.76009 1.75397 1.75240 1.75461-0.7 1.99928 1.99106 1.98474 1.98304 1.98520-0.5 2.25523 2.24687 2.24041 2.23860 2.24068-0.3 2.53766 2.52926 2.52273 2.52082 2.522790.3 3.56100 3.55312 3.54688 3.54481 3.546280.5 3.96624 3.95876 3.95280 3.95073 3.951990.7 4.40609 4.39912 4.39352 4.39148 4.392510.9 4.88193 4.87556 4.87039 4.86841 4.86919

Table 6aVariation of the stress concentration factor, K , with dE ρ

dρ ρ

( )

1=(E ρ( )| 1.9ρ 1 == ).

β 0.91 = n 1= − γ 0.641 = β 0.91 = n 2= − β 0.91 = n 3= − β 0.91 = n 4= − β 0.91 = n 5= −

dE ρdρ ρ

( )

1=

-0.90 -1.22 -1.80 -2.70 -3.60 -4.50

K 4.2260 4.2617 4.4516 4.6160 4.7286 4.8135


54

In order to better show the influence of the variation of E on K, weset β 02 = and find a least-squares fit to obtain the following relation-ship among K, n, and β1.

K β n β n β n β2.944 1.907 0.023 0.581 0.001 0.104 while 01 12 2

1 2= + − − − − =

(30)

The variation of K, on the β1n-plane, is displayed in Fig. 5. One canfind values of these two variables for a desired value of K, and hencedesign the FGM panel.

For E and v varying according to Eq. (3b), values of K for differentvalues of γ1 and γ2 are listed in Table 5. We note that for γ 01 > (γ 01 < ) Kfor a FGM panel is larger (smaller) than that for a homogeneous ma-terial panel. The maximum (minimum) value of K , equals 4.88 (1.75)that is 1.63 (0.58) times the value of K for a homogeneous materialpanel. Values of γ2 in the range [−0.9, 0.9] have a negligible effect onK.

Based on the data in Table 5, we also find a least-squares fit toobtain the relationship between K and γ1 as

K γ γ γ γ3.000 1.704 0.385 0.034 while 01 12

13

2= + + + = (31)

For a given value of K, the nonlinear algebraic Eq. (31) can beiteratively solved forγ1.

5.2.2. Dependence of K upon E ρ( )|ρ 1= and dE ρdρ ρ

( )

1=We now investigate the dependence of K upon E ρ( )|ρ 1= anddE ρ

dρ ρ

( )

1=. We set v =0.3, and list values of K in Tables 6a and 6b and 7a

and 7b for two arbitrarily chosen values of E ρ( )|ρ 1= and dE ρdρ ρ

( )

1=.

We conclude from values of K listed in Tables 6a and 6b that K

slowly decreases with an increase in dE ρdρ ρ

( )

1=for both values of E ρ( )|ρ 1=

and it is less than that of a homogeneous material when 0dE ρdρ ρ

( )

1>

=.

That is, the signs of dE ρdρ ρ

( )

1=are opposite for simultaneously increasing

values of K andK0. Values of K listed in Tables 7a and 7b suggest that Kincreases gradually with an increase in the value of E ρ( )|ρ 1= for the two

values of dE ρdρ ρ

( )

1=. Based on these observations, we recommend that

0dE ρdρ ρ

( )

1>

=for FGM panels.

5.2.3. Stress distributionsFor the FGM material defined by Eq. (3a), we have plotted in

Fig. 6(a)–(c) the variation of the non-dimensional stresses along a radialdirection for specific values of the angular coordinate, θ, and the hoopstress vs. θ for ρ 1= .

It is clear that on the line θ π/2= , for positive (negative) values of β1 andβ2 the hoop stress is maximum (minimum) on the hole surface. However,when β1 and β2 have opposite signs, then the sign of β1 in the expression forYoung’s modulus determines the value of the hoop stress. For β 0.91 = − , thehoop stress is nearly uniform on the hole surface. The decay of the hoop stresswith ρ near the hole in the FGM panel is steeper than that in the homo-geneous material panel. Stresses are essentially constant for ρ > 5.

Even though the radial variation of Poisson’s ratio has very littleeffect on the stress distributions and hence the stress concentrationfactor, as demonstrated in [35,36], it noticeably affects the displace-ment field.

6. Conclusions

We have analytically found the stress concentration factor and thestress distributions in a panel with a circular hole and made of an

Table 6bVariation of the stress concentration factor, K , with dE ρ

dρ ρ

( )

1=(E ρ( )| 0.4ρ 1 == ).

γ 0.921 = − β 0.61 = − n 1= − β 0.61 = − n 2= − β 0.61 = − n 3= − β 0.61 = − n 4= − β 0.61 = − n 5= −

dE ρdρ ρ

( )

1=

0.37 0.60 1.20 1.80 2.40 3.00

K 1.7362 1.7641 1.5304 1.4644 1.4241 1.3961

Fig. 6. For n=−5, v∞=0.25, variation of the (a) radial stress on the lineθ=0, (b) the hoop stress on the line θ= π/2, and (c) the hoop stress on thehole circumference ρ=1.


55

isotropic functionally graded material. Young’s modulus, E, andPoisson’s ratio are assumed to vary only in the radial direction either asgeneral power-law functions or the former exponentially and the lattera power law. Variation of E governs stresses near the hole periphery andtheir rate of decay along a radial line. The stress concentration factor, K,can be decreased by a factor of 8 by suitably grading the materialproperties. Values of K slowly decrease with an increase of the value ofdE ρ

dρ ρ

( )

1=, and it is less than that for a homogeneous material when

0dE ρdρ ρ

( )

1>

=. Furthermore, K increases gradually with an increase in

the value of E ρ( )|ρ 1= .These results suggest that one can tailor the radial variation of E to

achieve a pre-specified value of K, and essentially a uniform hoop stressin the entire panel. For the material problem an explicit expression for

E ρ( ) is given for remote biaxial loading of a panel.For remote uniaxial tensile loading, we have synthesized the com-

puted results to express K in terms of the material inhomogeneityparameters. One can use these relations to find values of the in-homogeneity parameters for a desired value of K.

Acknowledgements

GJN's and ZZ's work was supported by the National Natural ScienceFoundation of China (Nos. 11772232, 11372225 and 11072177). RCB’swork was partially supported by the US Office of Naval Research GrantN00014-18-1-2548 with Dr. Y. D. S. Rajapakse as the ProgramManager. GJN is visiting RCB's group during 2018.

Appendix A

Expressions for B s i( ) ( 1, 2, 3)i = appearing in Eqs. (17a) and (17b) are given below.

B s n s ns n s s ns n s s ns s n v nv β nsv n sv ns v β nv n v n sv nsv ns v β( ) (12 32 6 8 5 8 2 2 4 4 ) (2 ) (4 4 2 )12 2 2 2 2 3 3 4 2

12 2

12 2 2

2= − − − + − + + + − + − + − − + − + − +∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞

B s n kn kn k n kn k n k n k n k n k n s β

kns ns n s kn s k n s kn s k n s k n s s kns β

kn s ns n s k n s s ns kns nv n v kn v β nsv s kn v k n v n sv kn sv ns v β

nv n v β β k n v kn v kn v nsv n sv kn sv ns v β β

( ) (12 16 6 4 5 4 2 16 )

(8 6 10 12 2 6 4 4 12 )

(6 5 6 4 2 4 4 4 2 ) (2 2 )

(4 4 ) ( 2 2 2 )

22 2 2 3 2 3 3 3 2 4 3 4 4 4

12

2 2 2 2 3 2 3 3 3 2 212

2 2 2 2 2 2 2 2 3 3 3 2 212 4 3 2 3 2 2 2

12

21 2

2 3 2 3 2 2 21 2

= − − + + − + − + − −

+ − + − + − + − + +

+ − − − + + − − − + + − + − + − −

+ + + − − − − + +

∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞

∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞

B s n n kn kn k n n kn k n k n n s β

ns kns k n k n k n n s kn s k n s kn s β

s k n s k n s ns kns n s kn s k n s s β ns kns s β

n n kn n kn k n ns n s v β kn s nsv β n kn n kn k n s ns s kns nv β

( ) (2 20 32 10 8 2 13 19 8 32 )

(10 16 5 6 2 13 38 24 10 )

(8 18 8 19 24 5 18 12 8 ) (6 8 2 )

(6 4 2 2 3 2 3 ) (2 ) (4 6 2 2 3 2 3 2 )

32 2 2 2 3 3 2 3 3 3 4

1

2 4 3 4 4 4 2 2 2 2 31

2 2 3 3 3 2 2 2 2 2 2 2 2 2 31

3 3 41

2 2 3 3 2 3 21 1

2 2 2 2 22

= − − + + + + + + + −

+ + − − − + + + −

+ − − + + − − − + − + +

+ − + − − − + − − + + − − + + + − + + +∞ ∞ ∞

Appendix B

Expressions for D s i( ) ( 1, 2, ,6)i = ⋯ appearing in Eqs. (20a)–(20c) are given below.

D s s v γ s v γ s v γ s s v γ( ) ( 2 8 12) (5 3 ) ( 6) ( 3 8)13

1 12

12

2= − + − + − + + − − +∞ ∞ ∞ ∞

D s sv v s s γ s s v γ γ( ) ( 4 ) ( 5 16)22

12 2

1 2= − − + + − +∞ ∞ ∞

D s s v γ s v γ s v γ s s v γ( ) ( 2 12 9) (12 ) (5 13) ( 5 12)33

12

1 12

2= − + − + + − + − − +∞ ∞ ∞ ∞

D s s k v γ γ( ) ( 4)4 12

2= − − ∞

D s s k v γ s v γ k v s γ k s v γ γ k s s v γ γ( ) ( 5 2) ( 3) ( 2 3) (7 2 ) ( 7 22)52 2

12

12

12

1 22 2

1 2= − + − + + + − − + + − + + − +∞ ∞ ∞ ∞ ∞

D s k s γ v γ s v γ k v s γ sk v γ s v γ k s v γ k s s v γ

k s v γ

( ) (2 2 ) 18( 1) (18 ) (18 6 ) 2 (18 ) (7 43) (6 7 43) ( 7 18)

(7 2 )6

3 31 1

21

21 1 1

21

2 22

2

= − + + + + + + − − + − + + + + − + − +

− −

∞ ∞ ∞ ∞ ∞ ∞ ∞

∞

Table 7bVariation of the stress concentration factor, K , with E ρ( )|ρ 1= ( 0.36dE ρ

dρ ρ

( )

1=

=).

γ 0.811 = − β 0.361 = − n 1= − β 0.181 = − n 2= − β 0.121 = − n 3= − β 0.091 = − n 4= − β 0.071 = − n 5= −

E ρ( )|ρ 1= 0.45 0.64 0.82 0.88 0.91 0.93

K 1.8593 2.4469 2.6201 2.7326 2.7918 2.8288

Table 7aVariation of the stress concentration factor, K , with E ρ( )|ρ 1= ( 0.9dE ρ

dρ ρ

( )

1= −

=).

β 0.91 = n 1= − γ 0.531 = β 0.451 = n 2= − β 0.31 = n 3= − β 0.231 = n 4= − β 0.181 = n 5= −

E ρ( )|ρ 1= 1.90 1.70 1.45 1.30 1.23 1.18

K 4.2260 4.0163 3.8004 3.6059 3.4874 3.4080


56

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composite structures tailoring for reducing...analytically ﬁnd the stress concentration factor, k,...

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