(7), 939–954 (2015) Applied Mathematics and Mechanics ...

Appl. Math. Mech. -Engl. Ed., 35(7), 939–954 (2015)DOI 10.1007/s10483-015-1959-9c©Shanghai University and Springer-Verlag

Berlin Heidelberg 2015

Applied Mathematicsand Mechanics(English Edition)

Thermo-electromechanical behavior of functionally gradedpiezoelectric hollow cylinder under non-axisymmetric loads∗

A. ATRIAN1,†, J. JAFARI FESHARAKI1, S. H. NOURBAKHSH2

1. Department of Mechanical Engineering, Najafabad Branch,

Islamic Azad University, Isfahan 85141-43131, Iran;

2. Faculty of Engineering, Shahrekord University, Shahrekord 88179-53884, Iran

Abstract This paper presents an analytical solution of a thick walled cylinder com-posed of a functionally graded piezoelectric material (FGPM) and subjected to a uniformelectric field and non-axisymmetric thermo-mechanical loads. All material properties, ex-cept Poisson’s ratio that is assumed to be constant, obey the same power law. An exactsolution for the resulting Navier equations is developed by the separation of variables andcomplex Fourier series. Stress and strain distributions and a displacement field throughthe cylinder are obtained by this technique. To examine the analytical approach, differentexamples are solved by this method, and the results are discussed.

Key words functionally graded piezoelectric material (FGPM), thermoelasticity, Fourierseries, non-axisymmetric

Chinese Library Classification O343.8, O3452010 Mathematics Subject Classification 74B05, 74F05, 74E05

1 Introduction

Functionally graded materials (FGMs) are composed of two various materials, for which thecontinuous transition form one to the other is in a specific gradient. Continuous variations ofthe material parameters are usually assumed to be in power law or exponential functions alongthe radial direction. FGMs are so beneficial in high-technology industries such as aircraft andaerospace because of possessing properties of both their constitutive materials and possibilityto define the final desired specifications. The combination of the FGMs and the piezoelectricmaterials produces functionally graded piezoelectric materials (FGPMs) with more sophisti-cated applications. Piezoelectric materials due to their direct and inverse effects have differentapplications as sensors and actuators. The primary goal in designing such materials is to takeadvantage of the desirable properties[1–3]. Although a large number of studies have been de-veloped in the case of symmetric behavior[4–5] of FGMs, there are not enough researches innon-axisymmetric fields. Displacement and stress distributions under different loads are typicalresults presented in these research works.

Electro-magneto-elastic behaviors of functionally graded piezoelectric solid cylinder andsphere were studied by Dai et al.[4] using the infinitesimal theory of electro-magneto-elasticity.Alibeigloo[1], Alibeigloo and Chen[5], and Alibeigloo and Nouri[6] used the state space method

∗ Received Jul. 24, 2014 / Revised Nov. 26, 2014† Corresponding author, E-mail: [email protected]

940 A. ATRIAN, J. JAFARI FESHARAKI, and S. H. NOURBAKHSH

to obtain thermo-elastic behavior of functionally graded cylindrical shell bonded to thin piezo-electric layers. Tutuncu and Temel[7] analyzed some FGM structures like cylinders, disks, andspheres using the plane elasticity theory and the complementary functions method. Their novelapproach reduces the boundary value problem to an initial-value problem which can be solvedaccurately by one of the many efficient methods such as the Runge-Kutta approach. Arefi andRahimi[8] also developed a three-dimensional multi-field formulation of a functionally gradedpiezoelectric thick shell of revolution by the tensor analysis. Chen and Shi[9] applied the theoryof elasticity using the Airy stress function to obtain the exact solution of cylinder subjectedto thermal and electric loadings. For comparison purposes, numerical results were also carriedout. With this regard, Akbarzadeh et al.[10] used the hybrid Fourier-Laplace transform methodto evaluate dynamic response of a simply supported functionally graded rectangular plate sub-jected to a lateral thermo-mechanical loading. Using the theory of electro-thermo-elasticity, Daiet al.[11] obtained the analytical solution of FGPM hollow cylinder rotating about its axis ata constant angular velocity. Peng and Li[12] also presented the distribution of thermal stressesand radial displacement numerically by converting the resulting boundary value problem to aFredholm integral equation. Dumir et al.[13] considered separation of variables to obtain anexact piezoelastic solution for an infinitely long, simply-supported, orthotropic, piezoelectric,radially polarized, and circular cylindrical shell panel in a cylindrical bending under pressureand electrostatic excitation. They applied the trigonometric Fourier series for displacementsand electric potential to satisfy the boundary conditions. The electro-thermo-mechanical so-lution of a radially polarized cylindrical shell was also presented by Dube et al.[14] using theseparation of variables technique.

In the case of non-axisymmetric loadings, some studies can be named. Tokovyy and Ma[15]

carried out a solution based upon the direct integration method for a radially inhomogeneoushollow cylinder under non-axisymmetric conditions. Poultangari et al.[16] investigated ana-lytically functionally graded hollow spheres under non-axisymmetric thermo-mechanical loads.They used the Legendre polynomials and the system of Euler differential equations to solvethe Navier equations. Shao et al.[17] obtained analytical solutions of time-dependent temper-ature and thermo-mechanical stresses through the functionally graded hollow cylinder undernon-axisymmetric mechanical and transient thermal loads by the complex Fourier series andLaplace transform techniques. Jabbari et al.[18] also presented the general solution using theseparation of variables and complex Fourier series for functionally graded hollow cylinders undernon-axisymmetric and steady-state thermo-mechanical loads. For functionally graded eccentricand non-axisymmetrically loaded circular cylinders, Batra and Nie[19] investigated plane straindeformations analytically. They used Frobenius series to solve the resulting fourth-order ordi-nary differential equations.

Yang and Gao[20] developed two-dimensional analysis of thermal stresses in a functionallygraded plate with a circular hole based on the complex variable method. Jafari-Fesharaki etal.[21–22] also applied the separation of variables and complex Fourier series to get stress andstrain fields through an FGPM cylinder under non-axisymmetric electro-mechanical loads.

In this research work, two-dimensional (r, θ) distributions of displacement and stress in athick walled FGPM cylinder placed in a uniform electric field and under non-axisymmetricthermo-mechanical loads are evaluated analytically for different boundary conditions. Eventu-ally, a number of examples are considered to check the solution.

2 Statement of problem

Consider a thick walled functionally graded piezoelectric cylinder with a and b as the innerand outer radii, respectively. All the material properties, except Poisson’s ratio (ν) which isassumed to be constant, are approximated to obey the same power law through the thickness.These parameters are Young’s modulus and the thermal expansion, dielectric, pyroelectric, and

Thermo-electromechanical behavior of functionally graded piezoelectric hollow cylinder 941

piezoelectric coefficients, respectively,⎧⎨

⎩

E = E0rβ, α = α0r

β, η = η0rβ,

prr = prr0r2β, pθθ = pθθ0r

2β,err = err0r

β, erθ = erθ0rβ, eθθ = eθθ0r

β.(1)

As mentioned before, to check the solution procedure, some examples will be carried out. Thematerial chosen for this purpose is PZT-4, and its constants are[23]

⎧⎪⎨

⎪⎩

υ = 0.36, E0 = 82.9 GPa, α0 = 1.97 × 10−6 1/K, η0 = 6.5 × 10−9 C2/(N · m2),prr0 = 5.4 × 10−5 C/(m2 · K), pθθ0 = 5.4 × 10−5 C/(m2 · K),err0 = −5.2 C/m2, erθ0 = −5.2 C/m2, eθθ0 = 12.7 C/m2.

(2)

3 Derivation of basic formulations

To govern the basic equations, in the absence of body forces, consider the equilibrium andMaxwell equations in the cylindrical coordinates (r, θ), respectively, as[5]

⎧⎪⎨

⎪⎩

∂σrr

∂r+

1r

∂σrθ

∂θ+

σrr − σθθ

r= 0,

∂σrθ

∂r+

1r

∂σθθ

∂θ+

2σrθ

r= 0,

(3a)

∂Dr

∂r+

Dr

r+

1r

∂Dθ

∂θ= 0. (3b)

Here, σij are the stress tensor components, and Di are the electrical displacement components.The strain-displacement and constitutive equations are expressed as[24]

⎧⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎩

εrr =∂U

∂r,

εθθ =U

r+

1r

∂V

∂θ,

εrθ =12

(1r

∂U

∂θ+

∂V

∂r− V

r

),

(4)

⎧⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎩

σrr = (λ + 2μ)εrr + λεθθ − (3λ + 2μ)α T (r, θ) + err∂ϕ

∂r,

σθθ = (λ + 2μ)εθθ + λεrr − (3λ + 2μ)α T (r, θ) + erθ∂ϕ

∂r,

σrθ = 2μεrθ,

(5)

where U and V are the displacement components in the radial and circumferential directions,respectively. ϕ is the electric potential, and T is the thermal distribution in the thickness ofcylinder. λ and μ are Lame’ coefficients as follows:

λ =Eν

(1 + ν)(1 − 2ν), μ =

E

2(1 + ν). (6)

Moreover, the electrical displacements Dr and Dθ are represented as[5]

Dr = errεrr + erθεθθ − η∂ϕ

∂r+ prr T (r, θ),

Dθ = 2eθθεrθ + pθθ T (r, θ).(7)


Equation (3) after replacement from Eqs. (4)−(7) and (1) becomes as below. All of the constantsin these three partial differential equations are explained in Appendix A.

∂2U

∂r2+ m1

1r

∂U

∂r+ m2

1r2

U + m31r2

∂V

∂θ+ m4

1r

∂2V

∂r∂θ

+ m51r2

∂2U

∂θ2+ m6

1r

∂ϕ

∂r+ m7

∂2ϕ

∂r2= m8r

β−1T + m9rβ ∂T

∂r, (8a)

∂2V

∂r2+ m10

1r

∂V

∂r+ m11

1r2

V + m121r2

∂U

∂θ+ m13

1r

∂2U

∂r∂θ

+ m141r2

∂2V

∂θ2+ m15

1r

∂2ϕ

∂r∂θ= m16r

β−1 ∂T

∂θ, (8b)

∂2ϕ

∂r2+ m17

1r

∂ϕ

∂r+ m18

1r2

∂2U

∂θ2+ m19

∂2U

∂r2+ m20

1r

∂U

∂r+ m21

1r2

U

+ m221r

∂2V

∂r∂θ+ m23

1r2

∂V

∂θ= m24r

β−1T + m25rβ ∂T

∂r+ m26r

β−1 ∂T

∂θ. (8c)

4 Solution procedure

To solve Eq. (8), U , V , and ϕ are considered to be in the complex Fourier series form as

U(r, θ) =+∞∑

n=−∞Un(r) exp(inθ), (9a)

V (r, θ) =+∞∑

n=−∞Vn(r) exp(inθ), (9b)

ϕ(r, θ) =0∑

n=0

ϕn(r) exp(inθ) = ϕ0(r), (9c)

where Un(r) and Vn(r) are the coefficients of the complex Fourier series of U(r, θ) and V (r, θ),respectively, i.e.,

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

Un(r) =12π

∫ +π

−π

U(r, θ) exp(−inθ)dθ,

Vn(r) =12π

∫ +π

−π

V (r, θ) exp(−inθ)dθ.

(10)

Due to the considered symmetric condition, the electric field is independent of the circumfer-ential direction. Therefore, Eq. (9c) is valid only for n = 0, and Eq. (8c) will be excluded fromthe system of Eq. (8) for n �= 0.

The steady state heat conduction problem has already been solved by Jabbari et al.[18] andis in the form of the complex Fourier series as follows:

⎧⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎩

T (r, θ) =+∞∑

n=−∞Tn(r) exp(inθ),

Tn(r) = Dn1rγn1 + Dn2r

γn2 ,

γn1,2 = −β

2±

√β2

4+ n2.

(11)


After substitution of Eqs. (9a), (9b), and (10) into Eqs. (8a) and (8b) and omitting the termexp (inθ), Eq. (8) can be written as

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

∂2Un

∂r2+ m1

1r

∂Un

∂r+ m2

1r2

Un + m31r2

(in)Vn + m41r(in)

∂Vn

∂r

+m51r2

(−n2)Un = m8rβ−1Tn + m9r

β ∂Tn

∂r,

∂2Vn

∂r2+ m10

1r

∂Vn

∂r+ m11

1r2

Vn + m121r2

(in)Un + m131r(in)

∂Un

∂r

+m141r2

(−n2)Vn = m16rβ−1(in)Tn.

(12)

The above two equations yield a system of ordinary differential equations having two generaland particular solutions which may be written in the following forms:

{Un(r) = Ug

n(r) + Upn(r),

Vn(r) = V gn (r) + V p

n (r).(13)

A general solution is conducted with the assumptions of{

Ugn(r) = Arζ ,

V gn (r) = Brζ .

(14)

The replacement of Eq. (14) into Eq. (12) expresses{

A(ζ(ζ − 1) + m1ζ + m2 − m5n

2)

+ B (m3(in) + m4ζ(in)) = 0,

A ((m12 + m13ζ)(in)) + B(ζ(ζ − 1) + m10ζ + m11 − m14n

2)

= 0.(15)

In order to find A and B and to use them for calculating nontrivial solutions of Ugn(r) and

V gn (r), the determinant of coefficients in the system of Eq. (15) must be set to be zero. It is due

to dependency of these two equations which must express eventually two related distributionsfor U and V , i.e.,

∣∣∣∣∣

ζ(ζ − 1) + m1ζ + m2 − m5n2 m3(in) + m4ζ(in)

(m12 + m13ζ)(in) ζ(ζ − 1) + m10ζ + m11 − m14n2

∣∣∣∣∣= 0. (16)

The concluded fourth-order equation has four roots of ζn1 to ζn4 which give the general solutionas follows:

⎧⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎩

Ugn(r) =

4∑

j=1

Anjrζnj,

V gn (r) =

4∑

j=1

Bnjrζnj =

4∑

j=1

NnjAnjrζnj,

(17)

where Nnj is the relation between the constants Anj and Bnj and is obtained from Eq. (15) as

Nnj =− (

ζnj(ζnj − 1) + m1ζnj + m2 − m5n2)

(m3 + m4ζnj)in, j = 1, 2, 3, 4. (18)

The particular solutions of Eq. (12) are assumed in the forms of

Upn(r) = Hn1r

γn1+β+1 + Hn2rγn2+β+1,

V pn (r) = Hn3r

γn1+β+1 + Hn4rγn2+β+1.

(19)


Substituting Eq. (19) and the second equation of Eq. (11) into Eq. (12) yields⎧⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎩

h1Hn1rγn1+β−1 + h2Hn2r

γn2+β−1 + h3Hn3rγn1+β−1 + h4Hn4r

γn2+β−1

= h5rγn1+β−1 + h6r

γn2+β−1,

h7Hn1rγn1+β−1 + h8Hn2r

γn2+β−1 + h9Hn3rγn1+β−1 + h10Hn4r

γn2+β−1

= h11rγn1+β−1 + h12r

γn2+β−1,

(20)

where the constants h1, h2, · · · , h12 are presented in Appendix A. Equating the coefficients ofthe identical powers gives two systems of algebraic equations expressing Hn1 to Hn4. RewritingEq. (13), the complete solutions for Un(r) and Vn(r) then become

⎧⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎩

Un(r) = Ugn(r) + Up

n(r) =4∑

j=1

Anjrζnj + (Hn1r

γn1+β+1 + Hn2rγn2+β+1),

Vn(r) = V gn (r) + V p

n (r) =4∑

j=1

NnjAnjrζnj + (Hn3r

γn1+β+1 + Hn4rγn2+β+1).

(21)

For n = 0, the relations between U − V and ϕ− V are lost, and the system of Eq. (8) becomesjust an uncoupled equation in terms of V and two other coupled equations as follows. Therefore,it is necessary to obtain the separate solution for n = 0. Inserting n = 0 into Eqs. (9) and (11)and substituting them in Eq. (8), one obtains⎧⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎩

∂2U0

∂r2+ m1

1r

∂U0

∂r+ m2

1r2

U0 + m61r

∂φ0

∂r+ m7

∂2φ0

∂r2= m8r

β−1T0 + m9rβ ∂T0

∂r,

∂2V0

∂r2+ m10

1r

∂V0

∂r+ m11

1r2

V0 = 0,

∂2ϕ0

∂r2+ m17

1r

∂ϕ0

∂r+ m19

∂2U0

∂r2+ m20

1r

∂Un

∂r+ m21

1r2

U0 = m24rβ−1T0 + m25r

β ∂T0

∂r.

(22)

It is clear that the previous process should be repeated in order to reach the solutions for n = 0as

⎧⎪⎪⎨

⎪⎪⎩

U0(r) = Ug0 (r) + Up

0 (r),

V0(r, θ) = V g0 (r),

ϕ0(r) = ϕg0(r) + ϕp

0(r).

(23)

The uncoupled equation, the second equation of Eq. (22), is an Euler equation, which can besolved by the assumption of V0 = rζ0 ,

⎧⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎩

ζ20 + ζ0(m10 − 1) + m11 = 0,

ζ01,2 = −(m10 − 1)/2 ± √(m10 − 1)2/4 − m11,

V0(r) =2∑

j=1

B0jrζ0j.

(24)

For the general solution of remained equations of (22), the same procedure as done for obtainingEq. (17) must be performed. By assumption of Ug

0 = A0rζ0 and ϕg

0 = C0rζ0 , the general solution

for n = 0 would be⎧⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎩

Ug0 (r) =

4∑

j=1

A0jrζ0j,

ϕg0(r) =

4∑

j=1

C0jrζ0j =

4∑

j=1

M0jA0jrζ0j,

(25)


where

M0j = −ζ0j(ζ0j − 1) + m1ζ0j + m2

m6ζ0j + m7ζ0j(ζ0j − 1), j = 1, 2, 3, 4. (26)

Similarly, the particular solutions of the first and third equations of Eq. (22) are supposed inthe forms of

{Up

0 (r) = H01rγ01+β+1 + H02r

γ02+β+1,

ϕp0(r) = H05r

γ01+β+1 + H06rγ02+β+1.

(27)

Substituting the above equations in two pointed equations of Eq. (22) leads to

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

h01H01rγ01+β−1 + h02H02r

γ02+β−1 + h05H05rγ01+β−1 + h06H06r

γ02+β−1

= h07rγ01+β−1 + h08r

γ02+β−1,

h017H01rγ01+β−1 + h018H02r

γ02+β−1 + h021H05rγ01+β−1 + h022H06r

γ02+β−1

= h023rγ01+β−1 + h024r

γ02+β−1.

(28)

The used constants of h0i are also presented in Appendix A. H01, H02, H05, and H06 areobtained by equating the coefficients of the identical powers. Now, the solutions for U(r, θ),V (r, θ), and ϕ0(r) are completed and can be written as

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

U(r, θ) =4∑

j=1

A0jrζ0j + (H01r

γ01+β+1 + H02rγ02+β+1)

++∞∑

n=−∞,n �=0

( 4∑

j=1

Anjrζnj + (Hn1r

γn1+β+1 + Hn2rγn2+β+1 )

)einθ,

V (r, θ) =2∑

j=1

B0jrζ0j

++∞∑

n=−∞,n �=0

( 4∑

j=1

NnjAnjrζnj + (Hn3r

γn1+β+1 + Hn4rγn2+β+1)

)einθ,

ϕ(r) =4∑

j=1

M0jA0jrζ0j + (H05r

γ01+β+1 + H06rγ02+β+1).

(29)

Substituting these relations into Eqs. (4) and (5) gives the strain and stress distributions in thefunctionally graded piezoelectric hollow cylinder. The coefficients Ki are also stated in Ap-pendix A.

εrr =∂U

∂r

=4∑

j=1

A0jζ0jrζ0j−1+H01(γ01+β+1)rγ01+β+H02(γ02+β+1)rγ02+β

++∞∑

n=−∞,n �=0

( 6∑

j=1

Anjζnjrζnj−1 + Hn1(γn1 + β + 1)rγn1+β

+ Hn2(γn2 + β + 1)rγn2+β)

exp(inθ), (30a)


εθθ =U

r+

1r

∂V

∂θ

=4∑

j=1

A0jrζ0j−1 + H01r

γ01+β + H02rγ02+β

++∞∑

n=−∞,n �=0

( 4∑

j=1

Anj(1 + inNnj)rζnj−1 + (Hn1 + inHn3)rγn1+β

+ (Hn2 + inHn4)rγn2+β)

exp(inθ), (30b)

εrθ =12

(1r

∂U

∂θ+

∂V

∂r− V

r

)

=12

2∑

j=1

B0j(ζ0j − 1)rζ0j−1

+12

+∞∑

n=−∞,n �=0

( 4∑

j=1

Anj(in + ζnjNnj − Nnj)rζnj−1

+ (inHn1 + (γn1 + β)Hn3)rγn1+β

+ (inHn2 + (γn2 + β)Hn4)rγn2+β)

exp(inθ), (30c)

σrr = (λ + 2μ)εrr + λεθθ − (3λ + 2μ)αT (r, θ) + err∂φ

∂r

=4∑

j=1

A0jK2rζ0j+β−1 + H01K3r

γ01+2β + H02K4rγ02+β

++∞∑

n=−∞,n �=0

( 4∑

j=1

AnjK5rζnj+β−1 + K6r

γn1+2β + K7rγn2+2β

)exp(inθ)

− K8

+∞∑

n=−∞

(Dn1r

2β+γn1 + Dn2r2β+γn2

)exp(inθ)

+ err0

( 4∑

j=1

M0jA0jζ0jrζ0j+β−1 + K9H05r

γ01+2β + K10H06r2β+γ02

), (31a)

σθθ = (λ + 2μ)εθθ + λεrr − (3λ + 2μ)αT (r, θ) + erθ∂φ

∂r

=4∑

j=1

A0jK11rζ0j+β−1 + H01K12r

γ01+2β + H02K13rγ02+β

++∞∑

n=−∞,n �=0

( 4∑

j=1

AnjK14rζnj+β−1 + K15r

γn1+2β + K16rγn2+2β

)exp(inθ)

− K8

+∞∑

n=−∞

(Dn1r

2β+γn1 + Dn2r2β+γn2

)exp(inθ)

+ erθ0

( 4∑

j=1

M0jA0jζ0jrζ0j+β−1 + K9H05r

γ01+2β + K10H06r2β+γ02

), (31b)


σrθ = 2μεrθ

=E0

2(1 + υ)

2∑

j=1

B0j(ζ0j − 1)rζ0j+β−1

+E0

2(1 + υ)

+∞∑

n=−∞,n �=0

( 4∑

j=1

Anj(in + ζnjNnj − Nnj)rζnj+β−1

+ (inHn1 + (γn1 + β)Hn3)rγn1+2β + (inHn2 + (γn2 + β)Hn4)rγn2+2β)

exp(inθ). (31c)

5 Results and discussion

The analytical solution can be checked for some examples. The first example is adoptedto compare the results of the proposed solution with the data reported by Jafari-Fesharaki etal.[21]. A hollow cylinder with the inner and outer radii equal to 0.5 m and 2 m, respectively,and made of PZT-4 is considered for this example. The assumed boundary conditions are alsolisted in Table 1. In this order, the temperature for current analysis at inner and outer radiimust be equated to zero. Figure 1 depicts the identical variation of the electric potential forvarious values of material inhomogeneity (β) with the graph presented in Fig. 2 of Ref. [21].

Table 1 Assumed boundary conditions

Example U(b, θ) V (b, θ) σrr(a, θ) σrθ(a, θ) ϕ(a, θ) ϕ(b, θ) T (a, θ) T (b, θ)

1 0 0 −115 GPa 0 3.8×109 V 0 0 0

2 0 0 40cos2 θ MPa 0 100 V 0 0 0

3 0 0 40cos2 θ MPa 0 100 V 0 100cos2 θ ◦C 0

Fig. 1 Electric potential variations throughthickness for various material inho-mogeneities (Example 1)

Fig. 2 Electric radial displacement in FGPMhollow cylinder (Example 2)

For the second and third examples, the previously mentioned material parameters of PZT-4 are taken into account. Furthermore, the internal and external radii of functionally gradedpiezoelectric hollow cylinder are also assumed to be 0.5 m and 0.8 m, respectively, and the power-law index of material is chosen as β = 1. At the second example, the effects of two-dimensional


electro-mechanical fields are shown, while for the third one, the thermo-electromechanical be-havior of the cylinder is discussed. The considered boundary conditions for these examples arealso expressed in Table 1. These conditions are selected in a manner to monitor the effects oftemperature better. It is also examined that after 35 numbers of terms the series expansion isconverged.

Figures 2 and 3 show the distributions of the radial component of electric displacement forboth examples, and it is clear that temperature can decrease its amplitude. While for thecircumferential components, temperature does not have such an effect, as shown in Figs. 4 and5. Due to the assumed boundary conditions, radial stresses have a harmonic pattern at theinner surface of cylinder and continue to decrease along the radial direction. These variationscan be witnessed in both Figs. 6 and 7, and clearly no significant discrepancies exist whentemperature enhances. Figures 8 and 9 also represent the shear stress components through thecylinder. The behavior of this stress component has turned from the inner surface, which isconstant because of considered boundary conditions, to harmonic pattern at outer layers.

Fig. 3 Electric radial displacement inFGPM hollow cylinder (Example 3)

Fig. 4 Electric circumferential displacement inFGPM hollow cylinder (Example 2)

Fig. 5 Electric circumferential displacementinFGPMhollowcylinder(Example 3)

Fig. 6 Radial stress in FGPM hollow cylinder(Example 2)

As previously mentioned about the temperature effects on the radial electric displacement,this parameter has the same effect on the circumferential stress and decreases the harmonicmanner of this stress (see Figs. 10 and 11). Figures 12 and 13 show a harmonic pattern forthe radial displacements in Examples 2 and 3, respectively. Although the cylinder has radialand circumferential stresses at its outer surface, there are not any displacement in that sectionwhich is in accord with the boundary conditions. Moreover, the same variations for displacement


Fig. 7 Radial stress in FGPM hollow cylin-der (Example 3)

Fig. 8 Shear stress in FGPM hollow cylinder(Example 2)

Fig. 9 Shear stress in FGPM hollow cylinder (Example 3)

components in the circumferential direction can be observed (see Figs. 14 and 15). Knowing theFGPM cylinder behavior and its responses against different field exposures makes one capableof applying these structures more beneficial in different industries as sensors and actuators.

Fig. 10 Circumferential stress in FGPM hollow cylinder (Example 2)

6 Conclusions

The exact two-dimensional solution of a functionally graded piezoelectric hollow cylinderunder non-axisymmetric loads, based on the direct method using the complex Fourier series,


Fig. 11 Circumferential stress in FGPM hollow cylinder (Example 3)

Fig. 12 Radial displacement in FGPM hollow cylinder (Example 2)

Fig. 13 Radial displacement in FGPM hollow cylinder (Example 3)

Fig. 14 Circumferential displacement in FGPM hollow cylinder (Example 2)


Fig. 15 Circumferential displacement in FGPM hollow cylinder (Example 3)

is conducted in current research work. Generality of this method and its possibility to exertdifferent electric, thermal, and mechanical fields without any mathematical limitation are ofgreat importance. Obtaining the displacement and stress distributions through the cylindermakes it possible to control and optimize applications of such parts better in different industrialcases.

References

[1] Alibeigloo, A. Thermoelastic solution for static deformations of functionally graded cylindricalshell bonded to thin piezoelectric layers. Composite Structures, 93(2), 961–972 (2010)

[2] Loghman, A., Ghorbanpour-Arani, A., Amir, S., and Vajedi, A. Magnetothermoelastic creepanalysis of functionally graded cylinders. International Journal of Pressure Vessels and Piping,87(7), 389–395 (2010)

[3] Rahimi, G. H., Arefi, M., and Khoshgoftar, M. J. Application and analysis of functionally gradedpiezoelectrical rotating cylinder as mechanical sensor subjected to pressure and thermal loads. Ap-plied Mathematics and Mechanics (English Edition), 32(8), 997–1008 (2011) DOI 10.1007/s10483-011-1475-6

[4] Dai, H., Fu, Y., and Yang, J. Electromagnetoelastic behaviors of functionally graded piezoelectricsolid cylinder and sphere. Acta Mechanica Sinica, 23(1), 55–63 (2007)

[5] Alibeigloo, A. and Chen, W. Q. Elasticity solution for an FGM cylindrical panel integrated withpiezoelectric layers. European Journal of Mechanics A/Solids, 29, 714–723 (2010)

[6] Alibeigloo, A. and Nouri, V. Static analysis of functionally graded cylindrical shell with piezo-electric layers using differential quadrature method. Composite Structures, 92, 1775–1785 (2010)

[7] Tutuncu, N. and Temel, B. A novel approach to stress analysis of pressurized FGM cylinders,disks and spheres. Composite Structures, 91(3), 385–390 (2009)

[8] Arefi, M. and Rahimi, G. Three-dimensional multi-field equations of a functionally graded piezo-electric thick shell with variable thickness, curvature and arbitrary nonhomogeneity. Acta Me-chanica, 223(1), 63–79 (2012)

[9] Chen, Y. and Shi, Z. F. Analysis of a functionally graded piezothermoelastic hollow cylinder.Journal of Zhejiang University-Science A, 6(9), 956–961 (2005)

[10] Akbarzadeh, A. H., Abbasi, M., Hosseini-Zad, S. K., and Eslami, M. R. Dynamic analysis offunctionally graded plates using the hybrid Fourier-Laplace transform under thermomechanicalloading. Meccanica, 46(7), 1373–1392 (2011)

[11] Dai, H. L., Dai, T., and Zheng, H. Y. Stresses distributions in a rotating functionally gradedpiezoelectric hollow cylinder. Meccanica, 47(2), 423–436 (2012)

[12] Peng, X. L. and Li, X. F. Thermoelastic analysis of a cylindrical vessel of functionally gradedmaterials. International Journal of Pressure Vessels and Piping, 87(5), 203–210 (2010)


[13] Dumir, P. C., Dube, G. P., and Kapuria, S. Exact piezoelastic solution of simply-supportedorthotropic circular cylindrical panel in cylindrical bending. International Journal of Solids andStructures, 34(6), 685–702 (1997)

[14] Dube, G. P., Kapuria, S., and Dumir, P. C. Exact piezothermoelastic solution of simply-supportedorthotropic flat panel in cylindrical bending. International Journal of Mechanical Sciences, 38(11),1161–1177 (1996)

[15] Tokovyy, Y. V. and Ma, C. C. Analysis of 2D non-axisymmetric elasticity and thermoelasticityproblems for radially inhomogeneous hollow cylinders. Journal of Engineering Mathematics, 61(2),171–184 (2008)

[16] Poultangari, R., Jabbari, M., and Eslami, M. R. Functionally graded hollow spheres under non-axisymmetric thermo-mechanical loads. International Journal of Pressure Vessels and Piping,85(5), 295–305 (2008)

[17] Shao, Z. S., Ang, K. K., Reddy, J. N., and Wang, T. J. Nonaxisymmetric thermomechanicalanalysis of functionally graded hollow cylinders. Journal of Thermal Stresses, 31(6), 515–536(2008)

[18] Jabbari, M., Sohrabpour, S., and Eslami, M. R. General solution for mechanical and thermalstresses in a functionally graded hollow cylinder due to nonaxisymmetric steady-state loads. Jour-nal of Applied Mechanics, 70(1), 111–118 (2003)

[19] Batra, R. C. and Nie, G. J. Analytical solutions for functionally graded incompressible eccen-tric and non-axisymmetrically loaded circular cylinders. Composite Structures, 92(5), 1229–1245(2010)

[20] Yang, Q. and Gao, C. F. Non-axisymmetric thermal stress around a circular hole in a functionallygraded infinite plate. Journal of Thermal Stresses, 33(4), 318–334 (2010)

[21] Jafari-Fesharaki, J., Jafari-Fesharaki, V., Yazdipoor, M., and Razavian, B. Two-dimensional solu-tion for electro-mechanical behavior of functionally graded piezoelectric hollow cylinder. AppliedMathematical Modelling, 36(11), 5521–5533 (2012)

[22] Jafari-Fesharaki, J., Golabi, S., and Atrian, A. Effect of Lorentz force on non-axisymmetricthermo-mechanical behavior of functionally graded hollow cylinder. Advanced Materials Research,328-330, 1094–1097 (2011)

[23] Akbari-Alashti, R. and Khorsand, M. Three-dimensional thermo-elastic analysis of a functionallygraded cylindrical shell with piezoelectric layers by differential quadrature method. InternationalJournal of Pressure Vessels and Piping, 88(5-7), 167–180 (2011)

[24] Dai, H. L., Hong, L., Fu, Y. M., and Xiao, X. Analytical solution for electromagnetothermoelasticbehaviors of a functionally graded piezoelectric hollow cylinder. Applied Mathematical Modelling,34(2), 343–357 (2010)

Appendix A

m1 = m10 = m17 = −m11 = β + 1,

m2 =βν

1 − ν− 1, m3 =

2ν(β + 1) − 1

2(1 − ν)− 1,

m4 =1

2(1 − ν), m5 =

1

m14= (1 − 2ν)m4,

m6 =(err0(β + 1) − erθ0)(1 + ν)(1 − 2ν)

(1 − ν)E0,

m7 =err0(1 + ν)(1 − 2ν)

(1 − ν)E0, m8 =

2βα0(1 + ν)

1 − ν,


m9 =m8

2β, m12 = m1 +

1

m5, m13 =

1

1 − 2ν,

m15 =2erθ0(1 + ν)

E0, m16 =

2α0(1 + ν)

1 − 2ν, m18 =

−eθθ0

η0,

m19 =m21

β=

−erθ0

η0, m20 =

−err0(1 + β) − erθ0

η0, m22 = m19 + m18,

m23 =−erθ0β + eθθ0

η0, m24 =

prr0(1 + 2)β

η0, m25 =

prr0

η0, m26 =

pθθ0

η0,

h1 = (γn1 + β + 1)(γn1 + β) + m1(γn1 + β + 1) + m2 − m5n2,

h2 = (γn2 + β + 1)(γn2 + β) + m1(γn2 + β + 1) + m2 − m5n2,

h3 = m3(in) + m4(in)(γn1 + β + 1),

h4 = m3(in) + m4(in)(γn2 + β + 1),

h5 = Dn1(m8 + m9γn1), h11 = Dn1(m16(in)),

h6 = Dn2(m8 + m9γn2), h12 = Dn2(m16(in)),

h7 = m12(in) + m13(in)(γn1 + β + 1),

h8 = m12(in) + m13(in)(γn2 + β + 1),

h9 = (γn1 + β + 1)(γn1 + β) + m10(γn1 + β + 1) + m11 − m14n2,

h10 = (γn2 + β + 1)(γn2 + β) + m10(γn2 + β + 1) + m11 − m14n2,

h01 = (γ01 + β + 1)(γ01 + β) + m1(γ01 + β + 1) + m2,

h02 = (γ02 + β + 1)(γ02 + β) + m1(γ02 + β + 1) + m2,

h05 = m6(γ01 + β + 1) + m7(γ01 + β + 1)(γ01 + β),

h06 = m6(γ02 + β + 1) + m7(γ02 + β + 1)(γ02 + β),

h07 = D01(m8 + m9γ01), h023 = D01(m24 + m25γ01),

h08 = D02(m8 + m9γ02), h024 = D02(m24 + m25γ02),

h017 = m19(γ01 + β + 1)(γ01 + β) + m20(γ01 + β + 1) + m21,

h018 = m19(γn0 + β + 1)(γ02 + β) + m20(γ02 + β + 1) + m21,

h021 = (γ01 + β + 1)(γ01 + β) + m17(γ01 + β + 1),

h022 = (γ02 + β + 1)(γ02 + β) + m17(γ02 + β + 1),

K1 =E0

(1 + υ)(1− 2υ), K2 = K1(ζ0j(1 − υ) + υ), j = 1, 2, 3, 4,

K3 = K1((γ01 + β + 1)(1 − υ) + υ), K4 = K1((γ02 + β + 1)(1 − υ) + υ),

K5 = K1(ζnj(1 − υ) + (1 + inNnj)υ), j = 1, 2, 3, 4,

K6 = K1(Hn1(γn1 + β + 1)(1 − υ) + (Hn1 + inHn3)υ),

K7 = K1(Hn2(γn2 + β + 1)(1 − υ) + (Hn2 + inHn4)υ),


K8 = K1α0(1 + υ), K9 = γ01 + β + 1, K10 = γ02 + β + 1,

K11 = K1(ζ0jυ + 1 − υ), j = 1, 2, 3, 4, K12 = K1((γ01 + β + 1)υ + 1 − υ),

K13 = K1((γ02 + β + 1)υ + 1 − υ),

K14 = K1(ζnjυ + (1 + inNnj)(1 − υ)), j = 1, 2, 3, 4,

K15 = K1(Hn1(γn1 + β + 1)υ + (Hn1 + inHn3)(1 − υ)),

K16 = K1(Hn2(γn2 + β + 1)υ + (Hn2 + inHn4)(1 − υ)).

(7), 939–954 (2015) Applied Mathematics and Mechanics ...

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