Vibration Analysis of the Piezoelectric, Piezomagnetic Materials in Spherical Symmetry by Co. SEP

Communications in Control Science and Engineering (CCSE) Volume 1 Issue 1, January 2013

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Vibration Analysis of the Piezoelectric, Piezomagnetic Materials in Spherical Symmetry Qun Guan Institute of Civil Engineering and Architecture, Hefei University of Technology, Hefei, Anhui, 230009, China grz@hfut.edu.cn Abstract Considering that the piezoelectric, piezomagnetic materials are in the spherical coordinate system and not accounting the body weight, body electric charge and body electric current, from the motion equation, gradient equation and the piezoelectric, piezomagnetic constructive equation, the steady-state solutions of variables such as stress, strain, displacement, electric displacement, electric field intensity, electric potential, magnetic intensity, magnetic potential under additional stimulations are deduced, thereof it can provide good theoretical basement for the dynamic control of the piezoelectric, piezomagnetic materials in the space spherical symmetry. Keywords Piezoelectric; Piezomagnetic; Spherical Symmetry; Steady-state Solution

Introduction Piezoelectric, piezomagnetic materials are receiving increasingly extensive application in the vibration, survey and control fields because of their higher electromechanical conversion efficiency, big stress, high energy density and rapid mechanical response. Piezoelectric, piezomagnetic materials can be used in making instruments of diverse forms and shapes, which generates extension, compression, distortion and such deformations by making use of the variation of additional electric field and magnetic field. These instruments can be utilized for mechanics sensors, which surveys static stress, vibration stress, distortion stress, acceleration and other physics variables, and can also make into slight displacement control machines of simple structure and diverse forms which are extensively used in excessively delicate location, laser tiny process, numeral control machine,

robot, valve control and so on. In recent years, some scholars have studied extensively piezoelectric materials of electromechanical coupling effects. Wang has given the general solutions of space axial symmetry problem in piezoelectric materials, and Yao has given the general solutions of spherical symmetry problem in piezoelectric materials, Wang has also given the state space solutions of the space axial symmetry problem in layered piezoelectric materials, Ding and other scholars have systematically studied the general solutions of transversely isotropic piezoelectric materials. But for the solutions of piezoelectric, piezomagnetic, electromagnetic coupling elastic media, limited study has been done. E.Pan has given the fundamental solutions of static and dynamic analysis about transversely isotropic piezoelectric, piezomagnetic elastic plate with simply supported, and Li has given the steady solutions of the spherical symmetry problem in piezoelectric sphere shell, in addition Zhu has given the three dimensions solutions of the free vibration about the electromagnetic laminated plate, Liu has given the important properties of Green function of two dimensions piezoelectric, piezomagnetic, electromagnetic anisotropic elastic media. This paper considers that piezoelectric, piezomagnetic, electromagnetic coupling elastic medias are in the spherical coordinate system and from the fundamental equations of the linear electromagnetic elastic solid, the steady solutions of stress, strain, displacement, electro-displacement, electric field strength, electric potential, magnetic strength, magnetic potential and other variables under additional stimulations are deduced, therefore they can provide good theoretical basement for the dynamic control of piezoelectric, piezomagnetic materials in the spherical symmetry.

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Communications in Control Science and Engineering (CCSE) Volume 1 Issue 1, January 2013

Fundamental Equations For materials of piezoelectric, piezomagnetic, electromagnetic coupling elastic media, not accounting the body weight, body electric charge and body electric current, their fundamental equations[13]-[15] in the spherical coordinate are as follows: Motion equation： ∂Dr 2 + Dr = 0 ∂r r

∂u ∂σ r 2 + (σ r − σ θ ) = ρ 2r r ∂t ∂r

the additional stimulations are simple harmonic, that is:

∂Br 2 + Br = 0 ∂r r

（1）

Er = −

εφ = εθ =

∂ϕ ∂r

Hr = −

ur r

∂ψ ∂r

（2）

Constructive equations：

σ r = c11

∂ur u + 2c12 r − e31Er − q31H r ∂r r

（3）

（4）

∂ur u + 2e31 r + ε 33 Er + d 33 H r ∂r r

∂ur u + 2q31 r + d 33 Er + µ33 H r ∂r r （6）

Here: ρ is

the

mass

density of piezoelectric, σ piezomagnetic material； r , σ θ are respectively radial

stress and loop stress； ε r , ε θ are respectively radial

strain and loop strain； u r , Dr , Br are respectively radial displacement, radial electric displacement and magnetism ； E r , H r are respectively electric field strength and magnetic field strength ； ϕ ,ψ are respectively electric potential and magnetic potential； c, e, q, d , ε , µ are respectively elastic, piezoelectric, piezo-magnetic and electromagnetic coupling constants. One rank ordinary differential equation set （ 1 ） — （ 6 ） has totally 11 unknown variables ： σ r , σ θ , ε r , ε θ , Dr , E r , Br , H r , u r , ϕ ,ψ ， and they are all functions of r and t .

density, electric potential maximal value and magnetic potential maximal value, ω is the frequency of additional stimulation. Now consider the steady response solutions of piezoelectric, piezomagnetic elastic materials in spherical symmetry and their responding steady solutions can also be expressed to be harmonic as follows:

u r (r , t ) = u~ (r )e iωt

φ ( r , t ) = φ ( r ) e i ωt

Other unknown functions can be responsibly expressed as σ r (r ), σθ (r ), εr (r ), εθ (r ), D r (r ), E r (r ), B r (r ), H r (r ) , then multiply

（5）

Br = q33

~ Ψ (t ) = Ψe iωt

~ Φ (t ) = Φe iωt

ψ (r , t ) = ψ~ (r )e iωt

u ∂u u σ φ = σ θ = c12（ r + r ）+ c11 r − e31 Er − q31 H r r r ∂r

Dr = e33

q(t ) = q~e iωt

~ ~ Here q~ , Φ and Ψ are respectively additional load

Gradient equations： ∂ur ∂r

This paper mainly studies the steady response solutions of piezoelectric, piezomagnetic elastic materials in spherical symmetry under the stimulations of surface force q(t ) , additional electric potential Φ(t ) and magnetic potential Ψ (t ) . Suppose that

εr =

Steady Response Solutions in Spherical Symmetry

e iωt factor. Therefore,

substituting the simple harmony of each unknown function into the above descriptive equation, then the dynamic response of piezoelectric, piezomagnetic elastic materials in spherical symmetry can be transformed to the solutions of ~ ~ ~ ~ ~ σ (r ), σ (r ), ε (r ), ε (r ), D (r ), θ

~ ~ ~ ~ ~ ~ (r ), Φ : E r (r ), B r (r ), H r (r ), u r r ( r ), Ψr ( r )

Integrating the 2nd and the 3rd equ in equation（1），the following are obtained ： Dr =

a1 r2

Br =

a2 r2

（7）

Here： a1 , a 2 are integral constants。Eliminating Dr and

Br in （5）、（6）：

du~ u~ s a + s a ~ Er (r ) = s1 r + 2 s2 r − 3 1 2 4 2 dr r r

u~ s a + s a du~ ~ H r (r ) = s5 r + 2 s6 r − 4 1 2 7 2 dr r r

（8）

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Substituting equations （7）、（8）into equations（3）、（4）：

d 33 s0

e33 µ33 − d 33 q33 s0

J v ( b2 r ) cos(νπ ) − J −v ( b2 r )

s5 =

Substituting（12）into equation（10），comparing the coefficients of each rank power of r on the two bottoms of the equation，we can get the expression of special solutions about equation（10）：

g1 =

− c12ε 33 µ33 + c12 d 332 − e31e33 µ33 + e31d 33 q31 − q31q33ε 33 + q31d 33e33 s0

d 33 s0

− c11ε 33 µ33 − c12 d 33 µ33 + c11d 332 + c12 d 332 − 2e312 µ33 + 2e31d 33q31 − 2q312 ε 33 + 2q31d 33e31 s0

−e d +q ε g 4 = 31 33 31 33 s0

∞

u~2 (r ) = b3 ∑ c 2 n r 2 n −1

Substituting equation（9）into the first equation of equation（1）,we can get： b b d u~ 2 du~ + + (b2 − 12 )u~ = 33 dr 2 r dr r r

（10）

Here： 2 （ 2 g 2−2 g1）； b = ρω ； b = 2( g 3 a1 + g 4 a2 ) b1 = 3 2 g5 g5 g5

Equation （ 10 ） is two rank unhomogeneous Bessel equation，the general solutions of its corresponding homogeneous Bessel equation（10）are [16]： ~ ( r ) = r 1−ν [ a J ( b r ) + a Y ( b r )] u r 3 ν 2 4 ν 2

（11） Here： a 3 、 a 4 are await constants， ν=

1 1 + 4b1 , 、 Jν ( b2 r ) 、 Yν ( b2 r ) are 2

（13）

n =0

Here：

c0 = −1 / b1 n

− c ε µ + c d 2 − e 2 µ + 2e33 d 33 q33 − q332 ε 33 g 5 = 11 33 33 11 33 33 33 s0 q ε −e d e d −q ε g 7 = 33 33 33 33 g 6 = 33 33 33 33 s0 s0

（12）

n =0

ε 33q31 − d 33e31 s0

e µ −q d g 3 = 31 33 31 33 s0

The special solutions of equation （10）corresponding unhomogeneous equation can be solved by Frobenius progression： ∞

ε 33q33 − d 33e33 s0 s7 =

sin(νπ )

u~2 (r ) = b3 ∑ c n r n −1

s6 =

g2 =

b r (−1) k ( 2 )ν + 2 k ! ( 1 ) 2 Γ ν + + k k k =0

（9）

µ e31µ33 − d 33 q31 s3 = 33 s0 s0

s4 = −

rank：

∞

Yν ( b2 r ) =

s 0 = d 332 − ε 33 µ 33 s1 =

Jν ( b2 r ) = ∑

Where：

s2 =

respectively the first sort, the second sort of Bessel function of

du~ u~ g a + g a σ~r (r ) = g1 r + g 2 r + 3 1 2 4 2 dr r r ~ ~ g a +g a du u σ~θ (r ) = g 5 r + 2 g1 r + 6 1 2 7 2 dr r r

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c 2 n = (−b2 ) n ∏ k =1

1 ，n 2k (2k − 1) − b1

= 1、2、3

From the general solutions equation （ 11 ） of the homogeneous equation and the special solutions equation（13）of the unhomogeneous equation we can get the whole solutions of equation（19）as follows：

u~ (r ) = u~1 (r ) + u~2 (r ) = b3 b4 + a3 D1 + a 4 D2 （14） Here： ∞

b4 = ∑ c 2 n r 2 n −1 n =0

D1 = r 1−ν Jν ( b2 r )

D2 = r 1−ν Yν ( b2 r ) Substituting the above equation into equa（2）, equ（8） and equ （ 9 ） we can get the expressions of ε (r ), ε (r ), σ (r ), σ (r ), E (r ), H (r ), φ ( r ), and ψ (r ) r

which are expressed by the general solution of displacement u~ (r ) ,as follows：

ε~r (r ) = b3b5 + a3 [νrD1 +

ν b2 r

D1 − D3 ]

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+ a 4 [νrD2 +

ν b2 r

Communications in Control Science and Engineering (CCSE) Volume 1 Issue 1, January 2013

D2 − D4 ]

（15）

From the second row of the grad equation（2），we can get：

3 1 − 2 s2a3[21−ν Γ( ) − ν 2−ν (1 + b2 )Γ( )] 2 2 φr ( r ) = 1 + 2ν 2 −ν ( b2 ) Γ( ) 2 3 2 2−ν Γ( ) 2 − s1a3ν 2ν − 1 3−ν ) ( b2 ) Γ( 2 ~

ε~θ (r ) = b3b4 + a3 D1 + a 4 D2 r

σ~r (r ) = g5 b3b5 + g5 a3 (νrD1 + + g5a 4 (νrD2 +

ν b2 r

（16）

ν b2 r

D1 − D3 )

D2 − D4 )

g a +g a + 6 1 2 7 2 r

g b b + g1a3 D1 + g1a4 D2 +2 1 3 4 r （17）

σ~θ (r ) = g1b3b5 + g1a3 (νrD1 + + g1a 4 (νrD2 +

ν b2 r

− D1 − D3 )

D2 − D4 )

s2b3b4 + s2a3 D1 + s2a4 D2 r

3 1 − 2 s6a3[21−ν Γ( ) −ν 2 −ν (1 + b2 )Γ( )] 2 2 ψ~r (r ) = 1 + 2ν 2−ν ( b2 ) Γ( ) 2

3 2 2−ν Γ( ) 2 − s5a3ν 2ν − 1 3−ν ) ( b2 ) Γ( 2 2 s a b ( A cosνπ + A6 ) − 6 4 2 3 b2 sinνπ

ν ~ E r (r ) = s1a3 (νrD1 + D1 − D3 ) b2 r

2 s2a4 ( A3 cosνπ − A6 −νA2 cosνπ + νA5 ) sinνπ cn r 2 n −1 s3 a1 + s4 a2 − 2 s2b3 ∑ − + a5 r n = 0 2n − 1

（18）

b2 r

b2 sinνπ

∞

g3a1 + g 4 a 2 r2

s1a 4 ( b2 +ν )( A1 cosνπ − A4 )

（21）

g2b3b4 + g2a3 D1 + g2a4 D2 r

+ s1a 4 (νrD2 +

−

s3a1 + s4 a 2 r2

−

2 s6 a 4 (νA2 cosνπ −νA5 ) b2 b2 sinνπ

（22）

（19）

ν ~ H r(r ) = s5a3 (νrD1 + D1 − D3 ) b2 r ν + s 5 a 4 (νrD2 + D2 − D4 ) b2 r

−

（20） Here：

b5 = ∑ (2n − 1)c 2 n r 2 n − 2 D3 = r 1−ν Jν +1 ( b2 r ) n =0

D4 = r 1−ν Yν +1 ( b2 r )

b2 sin νπ

cn r 2 n −1 s4 a1 + s7 a2 − + a6 r n = 0 2n − 1 ∞

s6b3b4 + s6 a3 D1 + s6 a4 D2 s4 a1 + s7 a2 − r r2

∞

s 5 ( b2 +ν )( A1 cosνπ − A4 )

− 2 s6b3 ∑ Here：

3 1 2 2−ν Γ( ) 2 −ν Γ( ) 2 2 A2 = A1 = ( −ν +1) 2ν − 1 1 + 2ν 3−ν ( b2 ) Γ( ) Γ( b2 ) 2 2 3 21−ν Γ( ) 2 A3 = + 2ν 1 ( b2 ) 2−ν Γ( ) 2

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1 − 2ν ) 2 A4 = 1 ( b2 )1−ν Γ( ) 2 2 −ν Γ(

3 − 2ν ) 2 A5 = 1 ( b2 ) 3−ν Γ(− ) 2 2 2−ν Γ(

1 − 2ν ) 2 A6 = 1 ( b2 ) 2−ν Γ(− ) 2 2 2−ν Γ(

Here：a5 , a 6 are integral constants. So far，11 unknown variables of equations from（1 ）to（6）have been deduced，these solutions contain 6 unknown integral constants a1 , a 2 , a3 , a 4 , a5 , a 6 ，which can de determined by the boundary conditions of stress, displacement, electric field, electromagnetism. Boundary Conditions For piezoelectric, piezomagnetic and electro-magnetic coupling elastic materials ， there exist mechanics boundary condition, electrics boundary condition, electromagnetic field boundary conditions. Because of different applied state conditions, they may be under different boundary conditions.

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+ − dψ~ r dψ~ r ~ + ~ − ,ψ r = ψ r = dn dn

（28）

“＋、－”in equations（27）and（28）respectively indicate that in the subject and out the subject. Applied Examples Since the piezoelectric, piezomagnetic materials can make

into instruments of diverse shapes, select the piezoelectric, piezomagnetic spherical valve as the example. From the above obtained general solutions of the dynamic problem about spherical symmetry, firstly work out the integral calculus constant a 2 through the second equation of equa（7） under magnetic field boundary condition（1）. If the considered is under the stress boundary condition, the displacement boundary condition and the electrics short circuit，firstly work out integral calculus constants a1 , a 3 , a 4 ,from equ（14）, （17）（ , 18），then work out a 5 , according to the electric potential condition（21）, and finally work out a 6 from

1. Mechanics boundary condition

equ（22）in the light of the magnetic field boundary condition（2）. If the considered is under the stress boundary condition, displacement boundary condition and electrics open circuit, firstly work out integral

（1）Stress boundary conditions:

calculus constants a1 , a 2 , a 3 , a 4 from the first equation

σ~r = q~ , σ~θ = 0

（23）

（2）Displacement boundary condition：

= ur 0= or ur ur (0) 2. Electrics boundary condition （1）Electrics short circuit：

（25）

（2）Electrics open circuit：

= D r 0= or D r D r (0)

（26）

a5 in terms of electrics potential（21），finally

（1）There exist in the endless or on the surface of closed magnetism boundary：

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（27）

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