Thermo-electromechanical behavior of functionally graded piezoelectric hollow cylinder under non-axisymmetric loads

(1)

DOI 10.1007/s10483-015-1959-9

cShanghai University and Springer-Verlag Berlin Heidelberg 2015

and Mechanics

(English Edition)

Thermo-electromechanical behavior of functionally graded piezoelectric 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 composed of a functionally graded piezoelectric material (FGPM) and subjected to a uniform electric field and non-axisymmetric thermo-mechanical loads. All material properties, except Poisson’s ratio that is assumed to be constant, obey the same power law. An exact solution for the resulting Navier equations is developed by the separation of variables and complex Fourier series. Stress and strain distributions and a displacement field through the cylinder are obtained by this technique. To examine the analytical approach, different examples are solved by this method, and the results are discussed.

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

Chinese Library Classification O343.8, O345

2010 Mathematics Subject Classification 74B05, 74F05, 74E05

1 Introduction

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

Electro-magneto-elastic behaviors of functionally graded piezoelectric solid cylinder and sphere 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]

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to obtain thermo-elastic behavior of functionally graded cylindrical shell bonded to thin piezoelectric layers. Tutuncu and Temel^[7] analyzed some FGM structures like cylinders, disks, and spheres using the plane elasticity theory and the complementary functions method. Their novel approach reduces the boundary value problem to an initial-value problem which can be solved accurately by one of the many efficient methods such as the Runge-Kutta approach. Arefi and Rahimi^[8] also developed a three-dimensional multi-field formulation of a functionally graded piezoelectric thick shell of revolution by the tensor analysis. Chen and Shi^[9]applied the theory of elasticity using the Airy stress function to obtain the exact solution of cylinder subjected to thermal and electric loadings. For comparison purposes, numerical results were also carried out. With this regard, Akbarzadeh et al.^[10]used the hybrid Fourier-Laplace transform method to evaluate dynamic response of a simply supported functionally graded rectangular plate subjected to a lateral thermo-mechanical loading. Using the theory of electro-thermo-elasticity, Dai et al.^[11] obtained the analytical solution of FGPM hollow cylinder rotating about its axis at a constant angular velocity. Peng and Li^[12] also presented the distribution of thermal stresses and radial displacement numerically by converting the resulting boundary value problem to a Fredholm integral equation. Dumir et al.^[13] considered separation of variables to obtain an exact piezoelastic solution for an infinitely long, simply-supported, orthotropic, piezoelectric, radially polarized, and circular cylindrical shell panel in a cylindrical bending under pressure and electrostatic excitation. They applied the trigonometric Fourier series for displacements and electric potential to satisfy the boundary conditions. The electro-thermo-mechanical solution of a radially polarized cylindrical shell was also presented by Dube et al.^[14] using the separation 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 inhomogeneous hollow cylinder under non-axisymmetric conditions. Poultangari et al.^[16] investigated analytically functionally graded hollow spheres under non-axisymmetric thermo-mechanical loads.

They used the Legendre polynomials and the system of Euler diﬀerential equations to solve the Navier equations. Shao et al.^[17] obtained analytical solutions of time-dependent temperature and thermo-mechanical stresses through the functionally graded hollow cylinder under non-axisymmetric mechanical and transient thermal loads by the complex Fourier series and Laplace transform techniques. Jabbari et al.^[18] also presented the general solution using the separation of variables and complex Fourier series for functionally graded hollow cylinders under non-axisymmetric and steady-state thermo-mechanical loads. For functionally graded eccentric and non-axisymmetrically loaded circular cylinders, Batra and Nie^[19]investigated plane strain deformations analytically. They used Frobenius series to solve the resulting fourth-order ordinary diﬀerential equations.

Yang and Gao^[20] developed two-dimensional analysis of thermal stresses in a functionally graded plate with a circular hole based on the complex variable method. Jafari-Fesharaki et al.^[21–22] also applied the separation of variables and complex Fourier series to get stress and strain 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 a thick walled FGPM cylinder placed in a uniform electric ﬁeld and under non-axisymmetric thermo-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 withaandbas the inner and outer radii, respectively. All the material properties, except Poisson’s ratio (ν) which is assumed 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

(3)

piezoelectric coeﬃcients, respectively,

⎧⎨

⎩

E=E0r^β, α=α0r^β, η=η0r^β, prr=prr0r²^β, pθθ=pθθ0r²^β,

err =err0r^β, erθ=erθ0r^β, eθθ=eθθ0r^β.

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

⎧⎪

⎨

⎪⎩

υ= 0.36, E0= 82.9 GPa, α0= 1.97×10⁻⁶1/K, η0= 6.5×10⁻⁹C²/(N·m²), prr0= 5.4×10⁻⁵C/(m²·K), pθθ0= 5.4×10⁻⁵C/(m²·K),

e_rr0=−5.2 C/m², e_rθ0=−5.2 C/m², e_θθ0= 12.7 C/m².

(2)

3 Derivation of basic formulations

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

⎧⎪

⎨

⎪⎩

∂σ_rr

∂r +1 r

∂σ_rθ

∂θ +σ_rr−σ_θθ r = 0,

∂σrθ

∂r +1 r

∂σθθ

∂θ +2σrθ

r = 0,

(3a)

∂Dr

∂r +Dr

r +1 r

∂Dθ

∂θ = 0. (3b)

Here,σij are the stress tensor components, andDi are the electrical displacement components.

The strain-displacement and constitutive equations are expressed as^[24]

⎧⎪

⎪⎪

⎨

⎪⎪

⎪⎩

ε_rr = ∂U

∂r, εθθ= U

r +1 r

∂V

∂θ, εrθ= 1

2 1

r

∂U

∂θ +∂V

∂r −V r

,

(4)

⎧⎪

⎪⎪

⎪⎨

⎪⎪

⎩

σ_rr = (λ+ 2μ)ε_rr+λε_θθ−(3λ+ 2μ)α T(r, θ) +e_rr∂ϕ

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

∂r, σrθ= 2μεrθ,

(5)

whereU 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 of cylinder. λandμare Lame’ coeﬃcients as follows:

λ= Eν

(1 +ν)(1−2ν), μ= E

2(1 +ν). (6)

Moreover, the electrical displacementsD_r andD_θ are represented as^[5]

Dr=errεrr+erθεθθ−η∂ϕ

∂r +prrT(r, θ), Dθ= 2eθθεrθ+pθθT(r, θ).

(7)

(4)

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

∂²U

∂r² +m11 r

∂U

∂r +m2 1

r²U+m3 1 r²

∂V

∂θ +m41 r

∂²V

∂r∂θ +m51

r²

∂²U

∂θ² +m61 r

∂ϕ

∂r +m7∂²ϕ

∂r² =m8r^β−¹T+m9r^β∂T

∂r, (8a)

∂²V

∂r² +m101 r

∂V

∂r +m11 1

r²V +m12 1 r²

∂U

∂θ +m131 r

∂²U

∂r∂θ +m141

r²

∂²V

∂θ² +m151 r

∂²ϕ

∂r∂θ =m16r^β−¹∂T

∂θ, (8b)

∂²ϕ

∂r² +m171 r

∂ϕ

∂r +m181 r²

∂²U

∂θ² +m19∂²U

∂r² +m201 r

∂U

∂r +m21 1 r²U +m221

r

∂²V

∂r∂θ+m23 1 r²

∂V

∂θ =m24r^β−¹T +m25r^β∂T

∂r +m26r^β−¹∂T

∂θ. (8c)

4 Solution procedure

To solve Eq. (8),U,V, andϕare 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)

whereUn(r) andVn(r) are the coeﬃcients of the complex Fourier series ofU(r, θ) andV(r, θ), respectively, i.e.,

⎧⎪

⎪⎪

⎨

⎪⎪

⎪⎩

U_n(r) = 1 2π

+π

−π U(r, θ) exp(−inθ)dθ, Vn(r) = 1

2π

+π

−π V(r, θ) exp(−inθ)dθ.

(10)

Due to the considered symmetric condition, the electric ﬁeld is independent of the circumferential direction. Therefore, Eq. (9c) is valid only forn= 0, and Eq. (8c) will be excluded from the system of Eq. (8) forn= 0.

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

⎧⎪

⎪⎪

⎪⎨

⎪⎪

⎩

T(r, θ) =

+∞

n=−∞

Tn(r) exp(inθ), Tn(r) =Dn1r^γⁿ¹+Dn2r^γⁿ², γ_n1,2=−β

2 ± β²

4 +n².

(11)

(5)

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

⎧⎪

⎪⎪

⎪⎨

⎪⎪

⎩

∂²Un

∂r² +m11 r

∂Un

∂r +m2 1

r²Un+m31

r²(in)Vn+m41

r(in)∂Vn

∂r +m5 1

r²(−n²)U_n=m8r^β−¹T_n+m9r^β∂T_n

∂r ,

∂²Vn

∂r² +m101 r

∂Vn

∂r +m11 1

r²Vn+m12 1

r²(in)Un+m131

r(in)∂Un

∂r +m14 1

r²(−n²)Vn=m16r^β−¹(in)Tn.

(12)

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

U_n(r) =U_n^g(r) +U_n^p(r),

Vn(r) =V_n^g(r) +V_n^p(r). (13) A general solution is conducted with the assumptions of

U_n^g(r) =Ar^ζ,

V_n^g(r) =Br^ζ. (14)

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

ζ(ζ−1) +m1ζ+m2−m5n²

+B(m3(in) +m4ζ(in)) = 0, A((m12+m13ζ)(in)) +B

ζ(ζ−1) +m10ζ+m11−m14n²

= 0. (15)

In order to ﬁnd A and B and to use them for calculating nontrivial solutions of U_n^g(r) and V_n^g(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 distributions forU 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 ofζn1toζn4which give the general solution as follows:

⎧⎪

⎪⎪

⎪⎨

⎪⎪

⎩

U_n^g(r) = 4 j=1

Anjr^ζ^nj,

V_n^g(r) = 4 j=1

Bnjr^ζ^nj = 4 j=1

NnjAnjr^ζ^nj,

(17)

whereNnj is the relation between the constantsAnj andBnj and is obtained from Eq. (15) as Nnj = −

ζ_nj(ζ_nj−1) +m1ζ_nj+m2−m5n²

(m3+m4ζnj)in , j= 1,2,3,4. (18) The particular solutions of Eq. (12) are assumed in the forms of

U_n^p(r) =Hn1r^γⁿ¹⁺^β⁺¹+Hn2r^γⁿ²⁺^β⁺¹,

V_n^p(r) =Hn3r^γⁿ¹⁺^β⁺¹+Hn4r^γⁿ²⁺^β⁺¹. (19)

(6)

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

⎧⎪

⎪⎪

⎨

⎪⎪

⎪⎩

h1Hn1r^γⁿ¹⁺^β−¹+h2Hn2r^γⁿ²⁺^β−¹+h3Hn3r^γⁿ¹⁺^β−¹+h4Hn4r^γⁿ²⁺^β−¹

=h5r^γⁿ¹⁺^β−¹+h6r^γⁿ²⁺^β−¹,

h7Hn1r^γⁿ¹⁺^β−¹+h8Hn2r^γⁿ²⁺^β−¹+h9Hn3r^γⁿ¹⁺^β−¹+h10Hn4r^γⁿ²⁺^β−¹

=h11r^γⁿ¹⁺^β−¹+h12r^γⁿ²⁺^β−¹,

(20)

where the constantsh1, h2,· · ·,h12 are presented in Appendix A. Equating the coeﬃcients of the identical powers gives two systems of algebraic equations expressingH_n1toH_n4. Rewriting Eq. (13), the complete solutions forUn(r) andVn(r) then become

⎧⎪

⎪⎪

⎪⎨

⎪⎪

⎩

U_n(r) =U_n^g(r) +U_n^p(r) = 4 j=1

A_njr^ζ^nj + (H_n1r^γⁿ¹⁺^β⁺¹+H_n2r^γⁿ²⁺^β⁺¹), V_n(r) =V_n^g(r) +V_n^p(r) =

4 j=1

N_njA_njr^ζ^nj + (H_n3r^γⁿ¹⁺^β⁺¹+H_n4r^γⁿ²⁺^β⁺¹).

(21)

Forn= 0, the relations betweenU−V andϕ−V are lost, and the system of Eq. (8) becomes just an uncoupled equation in terms ofV and two other coupled equations as follows. Therefore, it is necessary to obtain the separate solution forn= 0. Insertingn= 0 into Eqs. (9) and (11) and substituting them in Eq. (8), one obtains

⎧⎪

⎪⎪

⎨

⎪⎪

⎪⎩

∂²U0

∂r² +m11 r

∂U0

∂r +m2 1

r²U0+m61 r

∂φ0

∂r +m7∂²φ0

∂r² =m8r^β−¹T0+m9r^β∂T0

∂r ,

∂²V0

∂r² +m101 r

∂V0

∂r +m11 1

r²V0= 0,

∂²ϕ0

∂r² +m171 r

∂ϕ0

∂r +m19∂²U0

∂r² +m201 r

∂U_n

∂r +m21 1

r²U0=m24r^β−¹T0 +m25r^β∂T0

∂r . (22)

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

as ⎧

⎪⎪

⎨

⎪⎪

⎩

U0(r) =U0^g(r) +U0^p(r), V0(r, θ) =V0^g(r), ϕ0(r) =ϕ^g0(r) +ϕ^p0(r).

(23)

The uncoupled equation, the second equation of Eq. (22), is an Euler equation, which can be solved by the assumption ofV0=r^ζ⁰,

⎧⎪

⎪⎪

⎪⎨

⎪⎪

⎩

ζ0²+ζ0(m10−1) +m11= 0, ζ01,2=−(m10−1)/2±

(m10−1)²/4−m11, V0(r) =

2 j=1

B0jr^ζ^0j.

(24)

For the general solution of remained equations of (22), the same procedure as done for obtaining Eq. (17) must be performed. By assumption ofU0^g=A0r^ζ⁰andϕ^g0=C0r^ζ⁰, the general solution forn= 0 would be

⎧⎪

⎪⎪

⎪⎨

⎪⎪

⎩

U0^g(r) = 4 j=1

A0jr^ζ^0j,

ϕ^g0(r) = 4 j=1

C0jr^ζ^0j = 4 j=1

M0jA0jr^ζ^0j,

(25)

(7)

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 in the forms of

U0^p(r) =H01r^γ⁰¹⁺^β⁺¹+H02r^γ⁰²⁺^β⁺¹,

ϕ^p₀(r) =H05r^γ⁰¹⁺^β⁺¹+H06r^γ⁰²⁺^β⁺¹. (27) Substituting the above equations in two pointed equations of Eq. (22) leads to

⎧⎪

⎪⎪

⎨

⎪⎪

⎪⎩

h01H01r^γ⁰¹⁺^β−¹+h02H02r^γ⁰²⁺^β−¹+h05H05r^γ⁰¹⁺^β−¹+h06H06r^γ⁰²⁺^β−¹

=h07r^γ⁰¹⁺^β−¹+h08r^γ⁰²⁺^β−¹,

h017H01r^γ⁰¹⁺^β−¹+h018H02r^γ⁰²⁺^β−¹+h021H05r^γ⁰¹⁺^β−¹+h022H06r^γ⁰²⁺^β−¹

=h023r^γ⁰¹⁺^β−¹+h024r^γ⁰²⁺^β−¹.

(28)

The used constants of h0i are also presented in Appendix A. H01, H02, H05, and H06 are obtained by equating the coeﬃcients of the identical powers. Now, the solutions for U(r, θ), V(r, θ), andϕ0(r) are completed and can be written as

⎧⎪

⎪⎪

⎨

⎪⎪

⎪⎩

U(r, θ) = 4 j=1

A0jr^ζ^0j + (H01r^γ⁰¹⁺^β⁺¹+H02r^γ⁰²⁺^β⁺¹)

+

+∞

n=−∞,n=0

⁴

j=1

Anjr^ζ^nj+ (Hn1r^γⁿ¹⁺^β⁺¹+Hn2r^γⁿ²⁺^β⁺¹ )

eⁱ^nθ, V(r, θ) =

2 j=1

B0jr^ζ^0j

+

+∞

n=−∞,n=0

⁴

j=1

N_njA_njr^ζ^nj+ (H_n3r^γⁿ¹⁺^β⁺¹+H_n4r^γⁿ²⁺^β⁺¹)

eⁱ^nθ, ϕ(r) =

4 j=1

M0jA0jr^ζ^0j + (H05r^γ⁰¹⁺^β⁺¹+H06r^γ⁰²⁺^β⁺¹).

(29)

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

εrr=∂U

∂r

= 4 j=1

A0jζ0jr^ζ^0j⁻¹+H01(γ01+β+1)r^γ⁰¹⁺^β+H02(γ02+β+1)r^γ⁰²⁺^β

+

+∞

n=−∞,n=0

⁶

j=1

A_njζ_njr^ζ^nj⁻¹+H_n1(γ_n1+β+ 1)r^γⁿ¹⁺^β

+H_n2(γ_n2+β+ 1)r^γⁿ²⁺^β

exp(inθ), (30a)

(8)

ε_θθ= U r +1

r

∂V

∂θ

= 4 j=1

A0jr^ζ^0j−1+H01r^γ⁰¹⁺^β+H02r^γ⁰²⁺^β

+

+∞

n=−∞,n=0

⁴

j=1

A_nj(1 + inN_nj)r^ζ^nj⁻¹+ (H_n1+ inH_n3)r^γⁿ¹⁺^β

+ (Hn2+ inHn4)r^γⁿ²⁺^β

exp(inθ), (30b)

εrθ= 1 2

1 r

∂U

∂θ +∂V

∂r −V r

= 1 2

2 j=1

B0j(ζ0j−1)r^ζ^0j⁻¹

+1 2

+∞

n=−∞,n=0

⁴

j=1

Anj(in+ζnjNnj−Nnj)r^ζ^nj⁻¹

+ (inHn1+ (γn1+β)Hn3)r^γⁿ¹⁺^β + (inHn2+ (γn2+β)Hn4)r^γⁿ²⁺^β

exp(inθ), (30c)

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

∂r

= 4 j=1

A0jK2r^ζ^0j⁺^β−¹+H01K3r^γ⁰¹⁺²^β+H02K4r^γ⁰²⁺^β

+

+∞

n=−∞,n=0

⁴

j=1

AnjK5r^ζ^nj⁺^β−¹+K6r^γⁿ¹⁺²^β+K7r^γⁿ²⁺²^β

exp(inθ)

−K8 +∞

n=−∞

Dn1r²^β⁺^γⁿ¹+Dn2r²^β⁺^γⁿ²

exp(inθ)

+err0

⁴

j=1

M0jA0jζ0jr^ζ^0j⁺^β−¹+K9H05r^γ⁰¹⁺²^β+K10H06r²^β⁺^γ⁰²

, (31a)

σ_θθ= (λ+ 2μ)ε_θθ+λε_rr−(3λ+ 2μ)αT(r, θ) +e_rθ∂φ

∂r

= 4 j=1

A0jK11r^ζ^0j⁺^β−¹+H01K12r^γ⁰¹⁺²^β+H02K13r^γ⁰²⁺^β

+

+∞

n=−∞,n=0

⁴

j=1

AnjK14r^ζ^nj⁺^β−¹+K15r^γⁿ¹⁺²^β+K16r^γⁿ²⁺²^β

exp(inθ)

−K8 +∞

n=−∞

Dn1r²^β⁺^γⁿ¹+Dn2r²^β⁺^γⁿ²

exp(inθ)

+erθ0

⁴

j=1

M0jA0jζ0jr^ζ^0j⁺^β−¹+K9H05r^γ⁰¹⁺²^β+K10H06r²^β⁺^γ⁰²

, (31b)

(9)

σrθ= 2μεrθ

= E0

2(1 +υ) 2 j=1

B0j(ζ0j−1)r^ζ^0j⁺^β−¹

+ E0

2(1 +υ)

+∞

n=−∞,n=0

⁴

j=1

Anj(in+ζnjNnj−Nnj)r^ζ^nj⁺^β−¹

+ (inHn1+ (γn1+β)Hn3)r^γⁿ¹⁺²^β+ (inHn2+ (γn2+β)Hn4)r^γⁿ²⁺²^β

exp(inθ). (31c)

5 Results and discussion

The analytical solution can be checked for some examples. The ﬁrst example is adopted to compare the results of the proposed solution with the data reported by Jafari-Fesharaki et al.^[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 also listed in Table 1. In this order, the temperature for current analysis at inner and outer radii must be equated to zero. Figure 1 depicts the identical variation of the electric potential for various 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 40cos²θMPa 0 100 V 0 0 0

3 0 0 40cos²θMPa 0 100 V 0 100cos²θ^◦C 0

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

Fig. 2 Electric radial displacement in FGPM hollow 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 graded piezoelectric hollow cylinder are also assumed to be 0.5 m and 0.8 m, respectively, and the power- law index of material is chosen asβ= 1. At the second example, the eﬀects of two-dimensional

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

Figures 2 and 3 show the distributions of the radial component of electric displacement for both examples, and it is clear that temperature can decrease its amplitude. While for the circumferential components, temperature does not have such an eﬀect, as shown in Figs. 4 and 5. Due to the assumed boundary conditions, radial stresses have a harmonic pattern at the inner surface of cylinder and continue to decrease along the radial direction. These variations can be witnessed in both Figs. 6 and 7, and clearly no signiﬁcant discrepancies exist when temperature enhances. Figures 8 and 9 also represent the shear stress components through the cylinder. The behavior of this stress component has turned from the inner surface, which is constant because of considered boundary conditions, to harmonic pattern at outer layers.

Fig. 3 Electric radial displacement in FGPM hollow cylinder (Example 3)

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

Fig. 5 Electric circumferential displacement in FGPM hollow cylinder (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 harmonic manner of this stress (see Figs. 10 and 11). Figures 12 and 13 show a harmonic pattern for the radial displacements in Examples 2 and 3, respectively. Although the cylinder has radial and circumferential stresses at its outer surface, there are not any displacement in that section which is in accord with the boundary conditions. Moreover, the same variations for displacement

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Fig. 7 Radial stress in FGPM hollow cylinder (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 the FGPM cylinder behavior and its responses against different field exposures makes one capable of 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 cylinder under non-axisymmetric loads, based on the direct method using the complex Fourier series,

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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)

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

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

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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 ,