Vibrations of Finite Bodies

Abstract

In this chapter we study vibrations of finite piezoelectric bodies. In some cases, for example, thickness vibrations of unbounded plates, although the in-plane dimensions of the plates are infinite, what matters is the finite plate thickness. Sects. 5.2 and 5.7 are on antiplane problems of polarized ceramics for which the notation in Sect. 2.9 is followed. The solutions in Sects. 5.1, 5.2, 5.3, 5.4, 5.5, 5.6, and 5.7 are exact. Those in Sects. 5.8, 5.9, and 5.10 are approximate.

In this chapter we study vibrations of finite piezoelectric bodies. In some cases, for example, thickness vibrations of unbounded plates, although the in-plane dimensions of the plates are infinite, what matters is the finite plate thickness. Sects. 5.2 and 5.7 are on antiplane problems of polarized ceramics for which the notation in Sect. 2.9 is followed. The solutions in Sects. 5.1, 5.2, 5.3, 5.4, 5.5, 5.6, and 5.7 are exact. Those in Sects. 5.8, 5.9, and 5.10 are approximate.

5.1 Thickness Stretch of a Ceramic Plate Through e₃₃

Thickness vibrations plates are motions spatially varying along the plate thickness only. Solutions to thickness vibrations of piezoelectric plates can be obtained in a general manner [1]. In this section we discuss a special case, that is, the thickness-stretch or thickness-extensional vibration of a ceramic plate. Consider the ceramic plate poled along the x₃ axis in Fig. 5.1. The plate is bounded by two planes at x₃ = ±h which are traction free and electroded. A time-harmonic voltage is applied across the plate thickness.

Fig. 5.1

An electroded ceramic plate with thickness poling

For polarized ceramics the governing equations are those in Sect. 2.9. The boundary conditions are:

$$ {\displaystyle \begin{array}{l}{T}_{3j}=0,\kern1em {x}_3=\pm h,\\ {}\varphi \left({x}_3=h\right)-\varphi \left({x}_3=-h\right)=V\exp \left( i\omega \kern0.1em t\right).\end{array}} $$

(5.1)

Consider thickness-stretch motions described by

$$ {\displaystyle \begin{array}{l}{u}_3={u}_3\left({x}_3\right)\exp \left( i\omega \kern0.2em t\right),\kern1em {u}_1={u}_2=0,\\ {}\varphi =\varphi \left({x}_3\right)\exp \left( i\omega \kern0.2em t\right).\end{array}} $$

(5.2)

The nontrivial components of strain and electric field are

$$ {S}_{33}={u}_{3,3},\kern1em {E}_3=-{\varphi}_{,3}, $$

(5.3)

where the time-harmonic factor has been dropped. The nontrivial stress and electric displacement components are

$$ {\displaystyle \begin{array}{l}{T}_{11}={T}_{22}={c}_{13}{u}_{3,3}+{e}_{31}{\varphi}_{,3}\\ {}{T}_{33}={c}_{33}{u}_{3,3}+{e}_{33}{\varphi}_{,3},\\ {}{D}_3={e}_{33}{u}_{3,3}-{\varepsilon}_{33}{\varphi}_{,3}.\end{array}} $$

(5.4)

In terms of the displacement and potential, the equations of motion and charge are

$$ {\displaystyle \begin{array}{l}{c}_{33}{u}_{3,33}+{e}_{33}{\varphi}_{,33}=-{\rho \omega}^2{u}_3,\\ {}{e}_{33}{u}_{3,33}-{\varepsilon}_{33}{\varphi}_{,33}=0.\end{array}} $$

(5.5)

Equation (5.5)₂ can be integrated to yield

$$ \varphi =\frac{e_{33}}{\varepsilon_{33}}{u}_3+{B}_1{x}_3+{B}_2, $$

(5.6)

where B₁ and B₂ are integration constants, and B₂ is immaterial. Substituting Eq. (5.6) into the expressions for T₃₃, D₃, and Eq. (5.5)₁, we obtain

$$ {\displaystyle \begin{array}{l}{T}_{33}={\overline{c}}_{33}{u}_{3,3}+{e}_{33}{B}_1,\\ {}{D}_3=-{\varepsilon}_{33}{B}_1,\end{array}} $$

(5.7)

$$ {\overline{c}}_{33}{u}_{3,33}=-\rho\;{\omega}^2{u}_3, $$

(5.8)

where

$$ {\overline{c}}_{33}={c}_{33}\left(1+{k}_{33}^2\right),\kern1em {k}_{33}^2=\frac{e_{33}^2}{\varepsilon_{33}{c}_{33}}. $$

(5.9)

The general solution to Eq. (5.8) and the corresponding expression for the electric potential are

$$ {\displaystyle \begin{array}{l}{u}_3={A}_1\sin \xi {x}_3+{A}_2\cos \xi {x}_3,\\ {}\varphi =\frac{e_{33}}{\varepsilon_{33}}\left({A}_1\sin \xi {x}_3+{A}_2\cos \xi {x}_3\right)+{B}_1{x}_3+{B}_2,\end{array}} $$

(5.10)

where A₁ and A₂ are integration constants, and

$$ {\xi}^2=\frac{\rho }{{\overline{c}}_{33}}{\omega}^2. $$

(5.11)

The expression for stress is then

$$ {T}_{33}={\overline{c}}_{33}\left({A}_1\xi \cos \xi {x}_3-{A}_2\xi \sin \xi {x}_3\right)+{e}_{33}{B}_1. $$

(5.12)

The boundary conditions in Eq. (5.1) require that

$$ {\displaystyle \begin{array}{l}{\overline{c}}_{33}{A}_1\xi \cos \xi h-{\overline{c}}_{33}{A}_2\xi \sin \xi h+{e}_{33}{B}_1=0,\\ {}{\overline{c}}_{33}{A}_1\xi \cos \xi h+{\overline{c}}_{33}{A}_2\xi \sin \xi h+{e}_{33}{B}_1=0,\\ {}2\frac{e_{33}}{\varepsilon_{33}}{A}_1\sin \xi h+2{B}_1h=V.\end{array}} $$

(5.13)

Adding Eq. (5.13)_1,2 and subtracting them from each other, we can rewrite Eq. (5.13) as

$$ {\displaystyle \begin{array}{l}{\overline{c}}_{33}{A}_1\xi \cos \xi h+{e}_{33}{B}_1=0,\\ {}{\overline{c}}_{33}{A}_2\xi \sin \xi h=0,\\ {}2\frac{e_{33}}{\varepsilon_{33}}{A}_1\sin \xi h+2{B}_1h=V.\end{array}} $$

(5.14)

Equation (5.14) decouples into two sets of equations. One is Eq. (5.14)₂ alone. The other consists of Eq. (5.14)_1,3.

Consider free vibrations with V = 0 first. Equation (5.14)₂ determines what may be called antisymmetric modes for which

$$ {\overline{c}}_{33}{A}_2\xi \sin \xi h=0. $$

(5.15)

Nontrivial solutions may exist if

$$ \sin \xi h=0, $$

(5.16)

or

$$ {\xi}^{(n)}h=\frac{n\pi}{2},\kern1em n=0,2,4,6,\dots, $$

(5.17)

which determines the resonance frequencies as

$$ {\omega}^{(n)}=\frac{n\pi}{2h}\sqrt{\frac{{\overline{c}}_{33}}{\rho }},\kern1em n=0,\kern0.5em 2,\kern0.5em 4,\kern0.5em 6,\dots . $$

(5.18)

Equation (5.16) implies that B₁ = 0 and A₁ = 0. The corresponding modes are

$$ {u}_3^{(n)}=\cos {\xi}^{(n)}{x}_3,\kern1em {\varphi}^{(n)}=\frac{e_{33}}{\varepsilon_{33}}\cos {\xi}^{(n)}{x}_3, $$

(5.19)

where n = 0 is a rigid-body mode. Equation (5.14)_1,3 determines symmetric modes for which

$$ {\displaystyle \begin{array}{l}{\overline{c}}_{33}{A}_1\xi \cos \xi h+{e}_{33}{B}_1=0,\\ {}2\frac{e_{33}}{\varepsilon_{33}}{A}_1\sin \xi h+2{B}_1h=0.\end{array}} $$

(5.20)

The resonance frequencies are determined by

$$ \left|\begin{array}{cc}{\overline{c}}_{33}\xi \cos \xi h& {e}_{33}\\ {}\frac{e_{33}}{\varepsilon_{33}}\sin \xi h& h\end{array}\right|={\overline{c}}_{33}\xi h\cos \xi h-\frac{e_{33}^2}{\varepsilon_{33}}\sin \xi h=0, $$

(5.21)

or

$$ \tan \xi h=\frac{\xi h}{{\overline{k}}_{33}^2}, $$

(5.22)

where

$$ {\overline{k}}_{33}^2=\frac{e_{33}^2}{\varepsilon_{33}{\overline{c}}_{33}}=\frac{e_{33}^2}{\varepsilon_{33}{c}_{33}\left(1+{k}_{33}^2\right)}=\frac{k_{33}^2}{1+{k}_{33}^2}={\left({k}_{33}^t\right)}^2. $$

(5.23)

Equations (5.22) and (5.20) determine the resonance frequencies and modes. For symmetric modes , A₂ = 0.

Next consider forced vibrations. From Eq. (5.14)_1,3 we obtain

$$ {A}_1=\frac{\left|\begin{array}{cc}0& {e}_{33}\\ {}V& 2h\end{array}\right|}{\left|\begin{array}{cc}{\overline{c}}_{33}\xi \cos \xi h& {e}_{33}\\ {}2\frac{e_{33}}{\varepsilon_{33}}\sin \xi h& 2h\end{array}\right|}=\frac{-{e}_{33}V}{2{\overline{c}}_{33}\xi h\cos \xi h-2\frac{e_{33}^2}{\varepsilon_{33}}\sin \xi h}, $$

(5.24)

$$ {B}_1=\frac{\left|\begin{array}{cc}{\overline{c}}_{33}\xi \cos \xi h& 0\\ {}2\frac{e_{33}}{\varepsilon_{33}}\sin \xi h& V\end{array}\right|}{\left|\begin{array}{cc}{\overline{c}}_{33}\xi \cos \xi h& {e}_{33}\\ {}2\frac{e_{33}}{\varepsilon_{33}}\sin \xi h& 2h\end{array}\right|}=\frac{V{\overline{c}}_{33}\xi \cos \xi h}{2{\overline{c}}_{33}\xi h\cos \xi h-2\frac{e_{33}^2}{\varepsilon_{33}}\sin \xi h}. $$

(5.25)

Hence

$$ {D}_3=-{\varepsilon}_{33}{B}_1=-{\varepsilon}_{33}\frac{V}{2h}\frac{\xi h}{\xi h-{\overline{k}}_{33}^2\tan \xi h}=-{\sigma}^e, $$

(5.26)

where σ^e is the surface free charge per unit area on the electrode at x₃ = h. The frequency dependent capacitance per unit area is

$$ C=\frac{\sigma^e}{V}=\frac{\varepsilon_{33}}{2h}\frac{\xi h}{\xi h-{\overline{k}}_{33}^2\tan \xi h}. $$

(5.27)

The denominator of Eq. (5.27), if set to zero, gives Eq. (5.22). We note the following limits:

$$ {\displaystyle \begin{array}{l}\underset{e_{33}\to 0}{\lim }C=\frac{\varepsilon_{33}}{2h},\\ {}\underset{\omega \to 0}{\lim }C=\frac{\varepsilon_{33}}{2h}\frac{1}{1-\frac{k_{33}^2}{1+{k}_{33}^2}}=\frac{\varepsilon_{33}}{2h}\left(1+{k}_{33}^2\right)={C}_0,\end{array}} $$

(5.28)

where C₀ is the static capacitance.

5.2 Thickness Shear of a Ceramic Plate Through e₁₅

Consider an unbounded plate of ceramics poled along x₃ (see Fig. 5.2). The plate surfaces at x₂ = ±h are traction-free and are electroded. A time-harmonic voltage is applied across the plate thickness.

Fig. 5.2

An electroded ceramic plate with in-plane poling

For the specific material orientation in Fig. 5.2, the plate is driven into antiplane motion. From Sect. 2.9, the governing equations and boundary conditions are:

$$ {\displaystyle \begin{array}{l}c{\nabla}^2u+e{\nabla}^2\varphi =\rho \ddot{u},\\ {}e{\nabla}^2u-\varepsilon {\nabla}^2\varphi =0,\\ {}{T}_{23}=0,\kern1em {x}_2=\pm h,\\ {}\varphi \left({x}_2=h\right)-\varphi \left({x}_2=-h\right)=V\exp \left( i\omega \kern0.1em t\right).\end{array}} $$

(5.29)

By thickness vibration we mean fields in the following form:

$$ u=u\left({x}_2\right),\kern1em \varphi =\varphi \left({x}_2\right), $$

(5.30)

where the exp(iωt) factor has been dropped. The nontrivial components of the strain and electric fields are

$$ {S}_4=2{S}_{23}={u}_{,2},\kern1em {E}_2=-{\varphi}_{,2}. $$

(5.31)

The nontrivial stress and electric displacement components are

$$ {\displaystyle \begin{array}{l}{T}_4={cu}_{,2}+e{\varphi}_{,2},\\ {}{D}_2={eu}_{,2}-{\varepsilon \varphi}_{,2}.\end{array}} $$

(5.32)

The equations of motion and charge reduce to

$$ {\displaystyle \begin{array}{l}{cu}_{,22}+e{\varphi}_{,22}=-{\rho \omega}^2u,\\ {}{eu}_{,22}-{\varepsilon \varphi}_{,22}=0.\end{array}} $$

(5.33)

Equation (5.33)₂ can be integrated to yield

$$ \varphi =\frac{e}{\varepsilon }u+{B}_1{x}_2+{B}_2, $$

(5.34)

where B₁ and B₂ are integration constants, and B₂ is immaterial. Substituting Eq. (5.34) into the expressions for T₂₃, D₂, and Eq. (5.33)₁, we obtain

$$ {T}_{23}=\overline{c}{u}_{,2}+{eB}_1,\kern1em {D}_2=-\varepsilon {B}_1, $$

(5.35)

$$ \overline{c}{u}_{,22}=-\rho\;{\omega}^2u, $$

(5.36)

where

$$ \overline{c}=c\left(1+{k}^2\right),\kern0.5em {k}^2=\frac{e^2}{\varepsilon c}. $$

(5.37)

The general solution to Eq. (5.36) and the corresponding expression for the electric potential are

$$ {\displaystyle \begin{array}{l}u={A}_1\sin \xi {x}_2+{A}_2\cos \xi {x}_2,\\ {}\varphi =\frac{e}{\varepsilon}\left({A}_1\sin \xi {x}_2+{A}_2\cos \xi {x}_2\right)+{B}_1{x}_2+{B}_2,\end{array}} $$

(5.38)

where A₁ and A₂ are integration constants, and

$$ {\xi}^2=\frac{\rho }{\overline{c}}{\omega}^2. $$

(5.39)

The expression for stress is then

$$ {T}_{23}=\overline{c}\left({A}_1\xi \cos \xi {x}_2-{A}_2\xi \sin \xi {x}_2\right)+{eB}_1. $$

(5.40)

The boundary conditions require that

$$ {\displaystyle \begin{array}{l}\overline{c}{A}_1\xi \cos \xi h-\overline{c}{A}_2\xi \sin \xi h+{eB}_1=0,\\ {}\overline{c}{A}_1\xi \cos \xi h+\overline{c}{A}_2\xi \sin \xi h+{eB}_1=0,\\ {}2\frac{e}{\varepsilon }{A}_1\sin \xi h+2{B}_1h=V,\end{array}} $$

(5.41)

or, add the first two, and subtract the first two from each other:

$$ {\displaystyle \begin{array}{l}\overline{c}{A}_1\xi \cos \xi h+{eB}_1=0,\\ {}\overline{c}{A}_2\xi \sin \xi h=0,\\ {}2\frac{e}{\varepsilon }{A}_1\sin \xi h+2{B}_1h=V.\end{array}} $$

(5.42)

Equation (5.42) decouples into two sets of equations. One is Eq. (5.42)₂ alone. The other consists of Eq. (5.42)_1,3.

For free vibrations we have V = 0. Equation (5.42)₂ determines symmetric modes for which

$$ \overline{c}{A}_2\xi \sin \xi h=0. $$

(5.43)

Nontrivial solutions may exist if

$$ \sin \xi h=0, $$

(5.44)

or

$$ {\xi}^{(n)}h=\frac{n\pi}{2},\kern2em n=0,2,4,6,\dots, $$

(5.45)

which determines the resonance frequencies as

$$ {\omega}^{(n)}=\frac{2\pi }{2h}\sqrt{\frac{\overline{c}}{\rho }},\kern2em n=0,2,4,6,\dots . $$

(5.46)

Equation (5.44) implies that B₁=0 and A₁=0. The corresponding modes are

$$ {u}^{(n)}=\cos {\xi}^{(n)}{x}_2,\kern1em {\varphi}^{(n)}=\frac{e}{\varepsilon}\cos {\xi}^{(n)}{x}_2, $$

(5.47)

where n = 0 is a rigid-body mode. For antisymmetric modes

$$ {\displaystyle \begin{array}{l}\overline{c}{A}_1\xi \cos \xi h+{eB}_1=0,\\ {}2\frac{e}{\varepsilon }{A}_1\sin \xi h+2{B}_1h=0.\end{array}} $$

(5.48)

The resonance frequencies are determined by

$$ \left|\begin{array}{ll}\overline{c}\xi \cos \xi h& e\\ {}\frac{e}{\varepsilon}\sin \xi h& h\end{array}\right|=\overline{c}\xi h\cos \xi h-\frac{e^2}{\varepsilon}\sin \xi h=0, $$

(5.49)

or

$$ \tan \xi h=\frac{\xi h}{k^2}, $$

(5.50)

where

$$ {\overline{k}}^2=\frac{e^2}{\varepsilon \overline{c}}=\frac{e^2}{\varepsilon c\left(1+{k}^2\right)}=\frac{k^2}{1+{k}^2}. $$

(5.51)

Equations (5.50) and (5.48) determine the resonance frequencies and antisymmetric modes. For antisymmetric modes, A₂ = 0.

For forced vibrations, from Eq. (5.42)_1,3, we have

$$ {A}_1=\frac{\left|\begin{array}{cc}0& e\\ {}V& 2h\end{array}\right|}{\left|\begin{array}{ll}\overline{c}\xi \cos \xi h& e\\ {}2\frac{e}{\varepsilon}\sin \xi h& 2h\end{array}\right|}=\frac{- eV}{2\overline{c}\xi h\cos \xi h-2\frac{e^2}{\varepsilon}\sin \xi h}, $$

(5.52)

$$ {B}_1=\frac{\left|\begin{array}{ll}\overline{c}\xi \cos \xi h& 0\\ {}2\frac{e}{\varepsilon}\sin \xi h& V\end{array}\right|}{\left|\begin{array}{ll}\overline{c}\xi \cos \xi h& e\\ {}2\frac{e}{\varepsilon}\sin \xi h& 2h\end{array}\right|}=\frac{V\overline{c}\xi \cos \xi h}{2\overline{c}\xi h\cos \xi h-2\frac{e^2}{\varepsilon}\sin \xi h}. $$

(5.53)

Hence

$$ {D}_2=-\varepsilon \kern0.1em {B}_1=-\varepsilon \frac{V}{2h}\frac{\xi h}{\xi h-{\overline{k}}^2\tan \xi h}=-{\sigma}^e, $$

(5.54)

where σ^e is the surface free charge per unit area on the electrode at x₂ = h. The capacitance per unit area is

$$ C=\frac{\sigma^e}{V}=\frac{\varepsilon }{2h}\frac{\xi h}{\xi h-{\overline{k}}^2\tan \xi h}. $$

(5.55)

The denominator of Eq. (5.55), if set to zero, gives the frequency equation for the antisymmetric modes (see Eq. (5.50)). We note the following limits:

$$ {\displaystyle \begin{array}{l}\underset{e\to 0}{\lim }C=\frac{\varepsilon }{2h},\\ {}\underset{\omega \to 0}{\lim }C=\frac{\varepsilon }{2h}\frac{1}{1-\frac{k^2}{1+{k}^2}}=\frac{\varepsilon }{2h}\left(1+{k}^2\right).\end{array}} $$

(5.56)

Obviously, the problem treated in the previous section and this section are the same mathematically. However, separate and independent treatments are presented because of their importance in applications.

5.3 Thickness Shear of a Quartz Plate Through e₂₆

Consider an unbounded plate of rotated Y-cut quartz (see Fig. 5.3). The plate surfaces at x₂ = ±h are traction-free and are electroded. A time-harmonic voltage is applied across the plate thickness.

Fig. 5.3

An electroded plate of rotated Y-cut quartz

The governing equations for rotated Y-cut quartz can be found in Sect. 2.10. The boundary conditions are

$$ {\displaystyle \begin{array}{l}{T}_{2j}=0,\kern1em {x}_2=\pm h,\\ {}\varphi \left({x}_2=h\right)-\varphi \left({x}_2=-h\right)=V\exp \left( i\omega t\right).\end{array}} $$

(5.57)

For the specific material orientation of rotated Y-cut quartz, the plate is driven into thickness-shear vibration described by the following fields:

$$ {\displaystyle \begin{array}{l}{u}_1={u}_1\left({x}_2\right)\exp \left( i\omega\;t\right),\kern1em {u}_2={u}_3=0,\\ {}\varphi =\varphi \left({x}_2\right)\exp \left( i\omega\;t\right).\end{array}} $$

(5.58)

The nontrivial components of strain, electric field, stress, and electric displacement are

$$ 2{S}_{12}={u}_{1,2},\kern1em {E}_2=-{\varphi}_{,2}, $$

(5.59)

and

$$ {\displaystyle \begin{array}{l}{T}_{31}={c}_{56}{u}_{1,2}+{e}_{25}{\varphi}_{,2},\kern1em {T}_{12}={c}_{66}{u}_{1,2}+{e}_{26}{\varphi}_{,2},\\ {}{D}_2={e}_{26}{u}_{1,2}-{\varepsilon}_{22}{\varphi}_{,2},\kern1em {D}_3={e}_{36}{u}_{1,2}-{\varepsilon}_{23}{\varphi}_{,2},\end{array}} $$

(5.60)

where the time-harmonic factor has been dropped. The equations of motion and the charge equation reduce to

$$ {\displaystyle \begin{array}{l}{T}_{21,2}={c}_{66}{u}_{1,22}+{e}_{26}{\varphi}_{,22}=\rho {\ddot{u}}_1=-{\rho \omega}^2{u}_1,\\ {}{D}_{2,2}={e}_{26}{u}_{1,22}-{\varepsilon}_{22}{\varphi}_{,22}=0.\end{array}} $$

(5.61)

Equation (5.61)₂ can be integrated to yield

$$ \varphi =\frac{e_{26}}{\varepsilon_{22}}{u}_1+{B}_1{x}_2+{B}_2, $$

(5.62)

where B₁ and B₂ are integration constants, and B₂ is immaterial. Substituting Eq. (5.62) into the expressions for T₂₁, D₂, and Eq. (5.61)₁, we obtain

$$ {\displaystyle \begin{array}{l}{T}_{21}={\overline{c}}_{66}{u}_{1,2}+{e}_{26}{B}_1,\\ {}{D}_2=-{\varepsilon}_{22}{B}_1,\end{array}} $$

(5.63)

and

$$ {\overline{c}}_{66}{u}_{1,22}=-{\rho \omega}^2{u}_1, $$

(5.64)

where

$$ {\overline{c}}_{66}={c}_{66}\left(1+{k}_{26}^2\right),\kern2em {k}_{26}^2=\frac{e_{26}^2}{\varepsilon_{22}{c}_{66}}. $$

(5.65)

The general solution to Eq. (5.64) and the corresponding expression for the electric potential are

$$ {\displaystyle \begin{array}{l}{u}_1={A}_1\sin \xi {x}_2+{A}_2\cos \xi {x}_2,\\ {}\varphi =\frac{e_{26}}{\varepsilon_{22}}\left({A}_1\sin \xi {x}_2+{A}_2\cos \xi {x}_2\right)+{B}_1{x}_2+{B}_2,\end{array}} $$

(5.66)

where A₁ and A₂ are integration constants, and

$$ {\xi}^2=\frac{\rho }{{\overline{c}}_{66}}{\omega}^2. $$

(5.67)

Then the expression for the stress component relevant to boundary conditions becomes

$$ {T}_{21}={\overline{c}}_{66}\left({A}_1\xi \cos \xi {x}_2-{A}_2\xi \sin \xi {x}_2\right)+{e}_{26}{B}_1. $$

(5.68)

The boundary conditions in Eq. (5.57) require that

$$ {\displaystyle \begin{array}{l}{\overline{c}}_{66}{A}_1\xi \cos \xi h-{\overline{c}}_{66}{A}_2\xi \sin \xi h+{e}_{26}{B}_1=0,\\ {}{\overline{c}}_{66}{A}_1\xi \cos \xi h+{\overline{c}}_{66}{A}_2\xi \sin \xi h+{e}_{26}{B}_1=0,\\ {}2\frac{e_{26}}{\varepsilon_{22}}{A}_1\sin \xi h+2{B}_1h=V.\end{array}} $$

(5.69)

We add Eq. (5.69)_1,2, and also subtract them from each other. Then Eq. (5.69) becomes:

$$ {\displaystyle \begin{array}{l}{\overline{c}}_{66}{A}_1\xi \cos \xi h+{e}_{26}{B}_1=0,\\ {}{\overline{c}}_{66}{A}_2\xi \sin \xi h=0,\\ {}2\frac{e_{26}}{\varepsilon_{22}}{A}_1\sin \xi h+2{B}_1h=V.\end{array}} $$

(5.70)

Equation (5.70) decouples into two sets of equations. We treat free vibration and forced vibration separately below.

First, consider free vibration with V = 0. For symmetric modes, A₁ = B₁ = 0. Equation (5.70) reduces to

$$ {\overline{c}}_{66}{A}_2\xi \sin \xi h=0. $$

(5.71)

Nontrivial solutions may exist if

$$ \sin \xi h=0, $$

(5.72)

or

$$ {\xi}^{(n)}h=\frac{n\pi}{2},\kern2em n=2,4,6,\dots, $$

(5.73)

which determines the following resonance frequencies:

$$ {\omega}^{(n)}=\frac{n\pi}{2h}\sqrt{\frac{{\overline{c}}_{66}}{\rho }},\kern2em n=2,4,6,\dots . $$

(5.74)

The corresponding modes are

$$ {u}_1^{(n)}=\cos {\xi}^{(n)}{x}_2,\kern1em {\varphi}^{(n)}=\frac{e_{26}}{\varepsilon_{22}}\cos {\xi}^{(n)}{x}_2. $$

(5.75)

For antisymmetric modes, A₂=0. Equation (5.70) reduces to

$$ {\displaystyle \begin{array}{l}{\overline{c}}_{66}{A}_1\xi \cos \xi h+{e}_{26}{B}_1=0,\\ {}2\frac{e_{26}}{\varepsilon_{22}}{A}_1\sin \xi h+2{B}_1h=0.\end{array}} $$

(5.76)

The resonance frequencies are determined by

$$ \left|\begin{array}{ll}{\overline{c}}_{66}\xi \cos \xi h& {e}_{26}\\ {}\frac{e_{26}}{\varepsilon_{22}}\sin \xi h& h\end{array}\right|={\overline{c}}_{66}\xi h\;\cos \xi h-\frac{e_{26}^2}{\varepsilon_{22}}\sin \xi h=0, $$

(5.77)

or

$$ \cot \xi h=\frac{{\overline{k}}_{26}^2}{\xi h}, $$

(5.78)

where

$$ {\overline{k}}_{26}^2=\frac{e_{26}^2}{\varepsilon_{22}{\overline{c}}_{66}}=\frac{e_{26}^2}{\varepsilon_{22}{c}_{66}\left(1+{k}_{26}^2\right)}=\frac{k_{26}^2}{1+{k}_{26}^2}. $$

(5.79)

Equations (5.78) and (5.76) determine the resonance frequencies and modes.

For rotated Y-cut quartz, ${\overline{k}}_{26}^2$ is very small. If ${\overline{k}}_{26}^2$ is taken to be zero, the resonant frequencies determined by Eq. (5.78) are the same as those in Eq. (5.73) but n assumes 1, 3, 5, and so on. For a small and nonzero ${\overline{k}}_{26}^2$, Eq. (5.78) can be solved approximately. For example, for the most widely used fundamental mode with n = 1, we can write [2]

$$ \xi h=\frac{\pi }{2}-\Delta, $$

(5.80)

where Δ is a small number. Substituting Eq. (5.80) into Eq. (5.78), for small ${\overline{k}}_{26}^2$ and small Δ, we obtain

$$ \Delta \cong \frac{2}{\pi }{\overline{k}}_{26}^2, $$

(5.81)

or

$$ {\displaystyle \begin{array}{c}{\omega}^{(1)}\cong \frac{\pi }{2h}\sqrt{\frac{{\overline{c}}_{66}}{\rho }}\;\left(1-\frac{4}{\pi^2}{\overline{k}}_{26}^2\right)=\frac{\pi }{2h}\sqrt{\frac{c_{66}\left(1+{k}_{26}^2\right)}{\rho }}\;\left(1-\frac{4}{\pi^2}{\overline{k}}_{26}^2\right)\\ {}\cong \frac{\pi }{2h}\sqrt{\frac{c_{66}}{\rho }}\left(1+\frac{1}{2}{k}_{26}^2\right)\left(1-\frac{4}{\pi^2}{\overline{k}}_{26}^2\right)\;\\ {}\cong \frac{\pi }{2h}\sqrt{\frac{c_{66}}{\rho }}\left(1+\frac{1}{2}{k}_{26}^2-\frac{4}{\pi^2}{k}_{26}^2\right)>\frac{\pi }{2h}\sqrt{\frac{c_{66}}{\rho }}.\end{array}} $$

(5.82)

Static thickness-shear deformation and the first few thickness-shear modes in a rotated Y-cut quartz plate are shown in Fig. 5.4.

Fig. 5.4

Displacement distribution of thickness-shear deformation and modes along plate thickness

For forced vibration, from Eq. (5.70), we have A₂ = 0 and

$$ {A}_1=\frac{\left|\begin{array}{cc}0& {e}_{26}\\ {}V& 2h\end{array}\right|}{\left|\begin{array}{cc}{\overline{c}}_{66}\xi \cos \xi h& {e}_{26}\\ {}2\frac{e_{26}}{\varepsilon_{22}}\sin \xi h& 2h\end{array}\right|}=\frac{-{e}_{26}V}{2{\overline{c}}_{66}\xi h\cos \xi h-2\frac{e_{26}^2}{\varepsilon_{22}}\sin \xi h}, $$

(5.83)

$$ {B}_1=\frac{\left|\begin{array}{cc}{\overline{c}}_{66}\xi \cos \xi h& 0\\ {}2\frac{e_{26}}{\varepsilon_{22}}\sin \xi h& V\end{array}\right|}{\left|\begin{array}{cc}{\overline{c}}_{66}\xi \cos \xi h& {e}_{26}\\ {}2\frac{e_{26}}{\varepsilon_{22}}\sin \xi h& 2h\end{array}\right|}=\frac{V{\overline{c}}_{66}\xi \cos \xi h}{2{\overline{c}}_{66}\xi h\cos \xi h-2\frac{e_{26}^2}{\varepsilon_{22}}\sin \xi h}. $$

(5.84)

Hence

$$ {D}_2=-{\varepsilon}_{22}{B}_1=-{\varepsilon}_{22}\frac{V}{2h}\frac{\xi h}{\xi h-{\overline{k}}_{26}^2\tan \xi h}=-{\sigma}^e, $$

(5.85)

where σ^e is the surface free charge per unit area on the electrode at x₂ = h. The frequency dependent capacitance per unit area is then

$$ C=\frac{\sigma }{V}=\frac{\varepsilon_{22}}{2h}\frac{\xi h}{\xi h-{\overline{k}}_{26}^2\tan \xi h}. $$

(5.86)

The denominator of Eq. (5.86), if set to zero, gives Eq. (5.78). Note the following limits:

$$ {\displaystyle \begin{array}{l}\underset{e_{26}\to 0}{\lim }C=\frac{\varepsilon_{22}}{2h},\\ {}\underset{\omega \to 0}{\lim }C=\frac{\varepsilon_{22}}{2h}\frac{1}{1-\frac{k_{26}^2}{1+{k}_{26}^2}}=\frac{\varepsilon_{22}}{2h}\left(1+{k}_{26}^2\right).\end{array}} $$

(5.87)

5.4 Thickness Shear of a Quartz Plate Through e₃₆

Consider an unelectroded plate of rotated Y-cut quartz (see Fig. 5.5). The plate surfaces at x₂ = ±h are traction-free and are unelectroded. The free space electric field is neglected.

Fig. 5.5

An unelectroded plate of rotated Y-cut quartz

The plate in Fig. 5.5 can be excited into thickness-shear vibration by a lateral electric field [3] E₃ through the piezoelectric constant e₃₆. In the analysis below [4], we assume the presence of a spatially uniform driving electric field E₃ = E and its time-harmonic factor is dropped as usual. We begin with the following trial fields. They will be shown to satisfy all governing equations and boundary conditions of the theory of piezoelectricity:

$$ {\displaystyle \begin{array}{l}{u}_1={u}_1\left({x}_2\right),\kern1em {u}_2={u}_3=0,\\ {}\varphi =\phi \left({x}_2\right)-{Ex}_3.\end{array}} $$

(5.88)

The nontrivial components of the strain, electric field, stress, and electric displacement components are, correspondingly,

$$ 2{S}_{12}={u}_{1,2},\kern1em {E}_2=-{\phi}_{,2}\kern1em {E}_3=E, $$

(5.89)

$$ {\displaystyle \begin{array}{l}{T}_{31}={c}_{56}{u}_{1,2}+{e}_{25}{\phi}_{,2}-{e}_{35}E,\\ {}{T}_{21}={c}_{66}{u}_{1,2}+{e}_{26}{\phi}_{,2}-{e}_{36}E,\\ {}{D}_2={e}_{26}{u}_{1,2}-{\varepsilon}_{22}{\phi}_{,2}+{\varepsilon}_{23}E,\\ {}{D}_3={e}_{36}{u}_{1,2}-{\varepsilon}_{32}{\phi}_{,2}+{\varepsilon}_{33}E.\end{array}} $$

(5.90)

The equation of motion and the charge equation of electrostatics left to be satisfied by u₁ and ϕ are

$$ {\displaystyle \begin{array}{l}{T}_{21,2}={c}_{66}{u}_{1,22}+{e}_{26}{\phi}_{,22}=-{\rho \omega}^2{u}_1,\\ {}{D}_{2,2}={e}_{26}{u}_{1,22}-{\varepsilon}_{22}{\phi}_{,22}=0.\end{array}} $$

(5.91)

The displacement and potential fields determined from Eq. (5.91) are

$$ {\displaystyle \begin{array}{l}{u}_1={A}_1\sin \xi {x}_2+{A}_2\cos \xi {x}_2,\\ {}\phi =\frac{e_{26}}{\varepsilon_{22}}\left({A}_1\sin \xi {x}_2+{A}_2\cos \xi {x}_2\right)+{B}_1{x}_2+{B}_2,\end{array}} $$

(5.92)

where ${\xi}^2={\rho \omega}^2/{\overline{c}}_{66}$. Then the stress and electric displacement components needed for boundary conditions are

$$ {T}_{21}={\overline{c}}_{66}\left({A}_1\xi \cos \xi {x}_2-{A}_2\xi \sin \xi {x}_2\right)+{e}_{26}{B}_1-{e}_{36}E, $$

(5.93)

$$ {D}_2=-{\varepsilon}_{22}{B}_1+{\varepsilon}_{23}E. $$

(5.94)

For traction-free and unelectroded plate surfaces, we have

$$ {T}_{21}\left({x}_2=h\right)={\overline{c}}_{66}\left({A}_1\xi \cos \xi h-{A}_2\xi \sin \xi h\right)+{e}_{26}{B}_1-{e}_{36}E=0, $$

(5.95)

$$ {T}_{21}\left({x}_2=-h\right)={\overline{c}}_{66}\left({A}_1\xi \cos \xi h+{A}_2\xi \sin \xi h\right)+{e}_{26}{B}_1-{e}_{36}E=0, $$

(5.96)

$$ {D}_2\left({x}_2=\pm h\right)=-{\varepsilon}_{22}{B}_1+{\varepsilon}_{23}E=0. $$

(5.97)

Equations (5.95), (5.96), and (5.97) can be solved for A₁, A₂, and B₁. From these equations it can be seen that the lateral electric field E drives the plate as effective surface traction and charge.

5.5 Mass Sensitivity of Thickness-Shear Frequency

The electrodes on the crystal plate surfaces in Sects. 5.1, 5.2, and 5.3 were assumed to be very thin. Their mechanical effects such as inertia and stiffness were neglected. In this section we study the inertial effect of the electrodes only, which can also be viewed as the effect of additional mass layers on the surfaces of a plate with very thin electrodes with negligible mass (see Fig. 5.6).

Fig. 5.6

An electroded plate of rotated Y-cut quartz with mass layers

From Eq. (5.61), the governing equations are still

$$ {\displaystyle \begin{array}{l}{T}_{21,2}={c}_{66}{u}_{1,22}+{e}_{26}{\varphi}_{,22}=\rho {\ddot{u}}_1=-{\rho \omega}^2{u}_1,\\ {}{D}_{2,2}={e}_{26}{u}_{1,22}-{\varepsilon}_{22}{\varphi}_{,22}=0.\end{array}} $$

(5.98)

With consideration of the mass layer inertia, the boundary conditions are

$$ {\displaystyle \begin{array}{l}-{T}_{2j}=2{\rho}^{\prime }{h}^{\prime }{\ddot{u}}_j,\kern1em {x}_2=h,\\ {}{T}_{2j}=2{\rho}^{\prime }{h}^{{\prime\prime} }{\ddot{u}}_j,\kern1em {x}_2=-h,\\ {}\varphi \left({x}_2=h\right)=\varphi \left({x}_2=-h\right),\end{array}} $$

(5.99)

where ρ′ is the mass layer density. The general solution to Eq. (5.98) is still given by Eq. (5.66):

$$ {\displaystyle \begin{array}{l}{u}_1={A}_1\sin \xi {x}_2+{A}_2\cos \xi {x}_2,\\ {}\varphi =\frac{e_{26}}{\varepsilon_{22}}\left({A}_1\sin \xi {x}_2+{A}_2\cos \xi {x}_2\right)+{B}_1{x}_2+{B}_2,\end{array}} $$

(5.100)

where ${\xi}^2={\rho \omega}^2/{\overline{c}}_{66}$. For boundary conditions we also need the following field from Eq. (5.68):

$$ {T}_{21}={\overline{c}}_{66}\left({A}_1\xi \cos \xi {x}_2-{A}_2\xi \sin \xi {x}_2\right)+{e}_{26}{B}_1. $$

(5.101)

Substituting Eqs. (5.100) and (5.101) into the boundary conditions in Eq. (5.99), we obtain

$$ {\displaystyle \begin{array}{l}-{\overline{c}}_{66}\xi \left({A}_1\cos \xi h-{A}_2\sin \xi h\right)-{e}_{26}{B}_1\\ {}\kern1em =-{\omega}^22{\rho}^{\prime }{h}^{\prime}\left({A}_1\sin \xi h+{A}_2\cos \xi h\right),\\ {}{\overline{c}}_{66}\xi \left({A}_1\cos \xi h+{A}_2\sin \xi h\right)+{e}_{26}{B}_1\\ {}\kern1em =-{\omega}^22{\rho}^{\prime }{h}^{{\prime\prime}}\left(-{A}_1\sin \xi h+{A}_2\cos \xi h\right),\\ {}\frac{e_{26}}{\varepsilon_{22}}{A}_1\sin \xi h+{B}_1h=0.\end{array}} $$

(5.102)

For nontrivial solutions of the undetermined constants, the determinant of the coefficient matrix of Eq. (5.102) has to vanish. This results in the following frequency equation:

$$ \left|\begin{array}{ccc}-{\overline{c}}_{66}\xi \cos \xi h+{\omega}^22{\rho}^{\prime }{h}^{\prime}\sin \xi h& {\overline{c}}_{66}\xi \sin \xi h+{\omega}^22{\rho}^{\prime }{h}^{\prime}\cos \xi h& -{e}_{26}\\ {}{\overline{c}}_{66}\xi \cos \xi h-{\omega}^22{\rho}^{\prime }{h}^{{\prime\prime}}\sin \xi h& {\overline{c}}_{66}\xi \sin \xi h+{\omega}^22{\rho}^{\prime }{h}^{{\prime\prime}}\cos \xi h& {e}_{26}\\ {}\frac{e_{26}}{\varepsilon_{22}}\sin \xi h& 0& h\end{array}\right|=0. $$

(5.103)

We examine the special case of symmetric mass layers when h^″ = h^′. In this case the modes can be separated into symmetric and antisymmetric ones and Eq. (5.103) splits into two factors. For symmetric modes A₁ = 0 and B₁ = 0. The frequency equation is

$$ \tan \xi h=- R\xi h, $$

(5.104)

where R is the mass ratio defined by

$$ R=\frac{2{\rho}^{\prime }{h}^{\prime }}{\rho h}. $$

(5.105)

For antisymmetric modes A₂ = 0. The frequency equation is

$$ \cot \xi h-\frac{{\overline{k}}_{26}^2}{\xi h}= R\xi h. $$

(5.106)

For the fundamental antisymmetric mode, if we write [2]

$$ \xi h=\frac{\pi }{2}-\Delta, $$

(5.107)

we can obtain from Eq. (5.106), for small Δ, small ${\overline{k}}_{26}^2$, and small R,

$$ \Delta \cong \frac{2}{\pi }{\overline{k}}_{26}^2+\frac{\pi }{2}R, $$

(5.108)

which implies that

$$ {\omega}^{(1)}\cong \frac{\pi }{2h}\sqrt{\frac{{\overline{c}}_{66}}{\rho }}\kern0.24em \left(1-\frac{4}{\pi^2}{\overline{k}}_{26}^2-R\right). $$

(5.109)

Obviously, the mass layer inertia represented by R lowers the frequency as expected. When R = 0, Eq. (5.109) reduces to Eq. (5.82). The frequency effect of the inertia of the mass layer can be used to design sensors for measuring the mass layer density or thickness [5]. These mass sensors are called quartz crystal microbalances (QCMs).

5.6 Azimuthal Shear of a Ceramic Cylinder

Consider an infinite circular ceramic cylinder of inner radius a and outer radius b (see Fig. 5.7). It is made of ceramics with azimuthal poling. We arrange (2,3,1) along (r,θ,z) so that the poling direction is 3. The inner and outer surfaces are electroded and the electrodes may be open or shorted. There is no mechanical load applied.

Fig. 5.7

A ceramic cylinder with azimuthal poling

We study azimuthal or circumferential shear vibrations described by

$$ {\displaystyle \begin{array}{l}{u}_{\theta }={u}_{\theta }(r)\exp \left( i\omega \kern0.1em t\right),\kern1em {u}_r={u}_z=0,\\ {}\varphi =\varphi (r)\exp \left( i\omega \kern0.1em t\right).\end{array}} $$

(5.110)

For free vibrations, the boundary conditions are

$$ {\displaystyle \begin{array}{l}{n}_j{T}_{ji}=0,\kern1em r=a,\kern0.5em b,\\ {}\varphi \left(r=a\right)=\varphi \left(r=b\right),\kern1em \mathrm{if}\kern0.5em \mathrm{the}\kern0.5em \mathrm{electrodes}\kern0.5em \mathrm{are}\kern0.5em \mathrm{shorted},\\ {}\mathrm{or}\kern1em {D}_r=0,\kern1em r=a,\kern0.5em b,\kern1em \mathrm{if}\kern0.5em \mathrm{the}\kern0.5em \mathrm{electrodes}\kern0.5em \mathrm{are}\kern0.5em \mathrm{open}.\end{array}} $$

(5.111)

Corresponding to Eq. (5.110), the nontrivial components of strain, electric field, stress, and electric displacement are

$$ {\displaystyle \begin{array}{l}{S}_4=2{S}_{r\theta}=\frac{{d u}_{\theta }}{d r}-\frac{u_{\theta }}{r},\\ {}{E}_2={E}_r=-\frac{d\varphi}{d r},\end{array}} $$

(5.112)

$$ {\displaystyle \begin{array}{l}{T}_4={T}_{r\theta}={c}_{44}\left(\frac{{d u}_{\theta }}{d r}-\frac{u_{\theta }}{r}\right)+{e}_{15}\frac{d\varphi}{d r},\\ {}{D}_2={D}_r={e}_{15}\left(\frac{{d u}_{\theta }}{d r}-\frac{u_{\theta }}{r}\right)-{\varepsilon}_{11}\frac{d\varphi}{d r}.\end{array}} $$

(5.113)

Thus on the boundary surfaces at r = a and b there are no circumferential electric fields. The electric potential assumes constant values on the electrodes as required. The stress components T_rr and T_rz vanish everywhere, particularly on the lateral surfaces of the cylinder. The equation of motion and the charge equation to be satisfied are

$$ {\displaystyle \begin{array}{l}\frac{dT_{r\theta}}{dr}+\frac{2}{r}{T}_{r\theta}=-{\rho \omega}^2{u}_{\theta },\\ {}\frac{1}{r}{\left({r D}_r\right)}_{,r}=0.\end{array}} $$

(5.114)

Equation (5.114)₂ can be integrated as

$$ {D}_r={e}_{15}\frac{C_3}{r}, $$

(5.115)

where C₃ is an integration constant. Then, from Eq. (5.113)₂ we have

$$ {\varphi}_{,r}=\frac{e_{15}}{\varepsilon_{11}}\left(2{S}_{r\theta}-\frac{C_3}{r}\right). $$

(5.116)

The substitution of Eq. (5.116) into Eq. (5.113)₁ gives

$$ {\displaystyle \begin{array}{l}{T}_{r\theta}={c}_{44}\left(\frac{du_{\theta }}{dr}-\frac{u_{\theta }}{r}\right)+{e}_{15}\frac{e_{15}}{\varepsilon_{11}}\left(2{S}_{r\theta}-\frac{C_3}{r}\right)\\ {}\kern1em ={\overline{c}}_{44}\left(\frac{du_{\theta }}{dr}-\frac{u_{\theta }}{r}\right)-\frac{e_{15}^2}{\varepsilon_{11}}\frac{C_3}{r},\end{array}} $$

(5.117)

where

$$ {\overline{c}}_{44}={c}_{44}\left(1+{k}_{15}^2\right),\kern1em {k}_{15}^2=\frac{e_{15}^2}{\varepsilon_{11}{c}_{44}}. $$

(5.118)

Substituting Eq. (5.117) into Eq. (5.114)₁, we obtain

$$ {\displaystyle \begin{array}{l}{\overline{c}}_{44}\left(\frac{d^2{u}_{\theta }}{dr^2}-\frac{1}{r}\frac{du_{\theta }}{dr}+\frac{1}{r^2}{u}_{\theta}\right)+\frac{e_{15}^2}{\varepsilon_{11}}\frac{C_3}{r^2}\\ {}\kern1em +\frac{2}{r}{\overline{c}}_{44}\left(\frac{du_{\theta }}{dr}-\frac{u_{\theta }}{r}\right)-2\frac{e_{15}^2}{\varepsilon_{11}}\frac{C_3}{r^2}=-{\rho \omega}^2{u}_{\theta },\end{array}} $$

(5.119)

which can be written as

$$ {\overline{c}}_{44}\left(\frac{d^2{u}_{\theta }}{dr^2}+\frac{1}{r}\frac{du_{\theta }}{dr}-\frac{1}{r^2}{u}_{\theta}\right)+{\rho \omega}^2{u}_{\theta }=\frac{e_{15}^2}{\varepsilon_{11}}\frac{C_3}{r^2}, $$

(5.120)

or

$$ \frac{d^2{u}_{\theta }}{dr^2}+\frac{1}{r}\frac{du_{\theta }}{dr}-\frac{1}{r^2}{u}_{\theta }+{\xi}^2{u}_{\theta }={\overline{k}}_{15}^2\frac{C_3}{r^2}, $$

(5.121)

where

$$ {\xi}^2=\frac{{\rho \omega}^2}{{\overline{c}}_{44}},\kern1em {\overline{k}}_{15}^2=\frac{e_{15}^2}{\varepsilon_{11}{\overline{c}}_{44}}=\frac{e_{15}^2}{\varepsilon_{11}{c}_{44}\left(1+{k}_{15}^2\right)}=\frac{k_{15}^2}{1+{k}_{15}^2}. $$

(5.122)

We introduce a dimensionless radial coordinate R = ξ r. Equation (5.121) becomes

$$ \frac{d^2{u}_{\theta }}{dR^2}+\frac{1}{R}\frac{du_{\theta }}{dR}+\left(1-\frac{1}{R^2}\right){u}_{\theta }={\overline{k}}_{15}^2\frac{C_3}{R^2}, $$

(5.123)

which is Bessel’s equation of order one.

In the following we consider the case when the electrodes at r = a and b are open. The corresponding electrical boundary conditions in Eq. (5.111) imply that C₃ = 0. Then the general solution to Eq. (5.123) is

$$ {\displaystyle \begin{array}{c}{u}_{\theta }={C}_1{J}_1(R)+{C}_2{Y}_1(R)\\ {}={C}_1{J}_1\left(\xi r\right)+{C}_2{Y}_1\left(\xi r\right),\end{array}} $$

(5.124)

where J₁ and Y₁ are the first-order Bessel functions of the first and second kind, respectively. From Eq. (5.117) the shear stress is

$$ {\displaystyle \begin{array}{c}{T}_{r\theta}={\overline{c}}_{44}\left(\frac{du_{\theta }}{dr}-\frac{u_{\theta }}{r}\right)\\ {}={\overline{c}}_{44}\left[{C}_1{J}_1^{\prime}\left(\xi r\right)\xi +{C}_2{Y}_1^{\prime}\left(\xi r\right)\xi \right]\\ {}-{\overline{c}}_{44}\frac{1}{r}\left[{C}_1{J}_1\left(\xi r\right)+{C}_2{Y}_1\left(\xi\;r\right)\right]\\ {}={C}_1{\overline{c}}_{44}\xi \left[{J}_1^{\prime}\left(\xi r\right)-\frac{J_1\left(\xi\;r\right)}{\xi\;r}\right]+{C}_2{\overline{c}}_{44}\xi \left[{Y}_1^{\prime}\left(\xi r\right)-\frac{Y_1\left(\xi\;r\right)}{\xi\;r}\right].\end{array}} $$

(5.125)

The traction-free boundary conditions require that

$$ {\displaystyle \begin{array}{l}{C}_1{\overline{c}}_{44}\xi \left[{J}_1^{\prime}\left(\xi a\right)-\frac{J_1\left(\xi\;a\right)}{\xi\;a}\right]+{C}_2{\overline{c}}_{44}\xi \left[{Y}_1^{\prime}\left(\xi a\right)-\frac{Y_1\left(\xi\;a\right)}{\xi\;a}\right]=0,\\ {}{C}_1{\overline{c}}_{44}\xi \left[{J}_1^{\prime}\left(\xi b\right)-\frac{J_1\left(\xi\;b\right)}{\xi\;b}\right]+{C}_2{\overline{c}}_{44}\xi \left[{Y}_1^{\prime}\left(\xi b\right)-\frac{Y_1\left(\xi\;b\right)}{\xi\;b}\right]=0.\end{array}} $$

(5.126)

The frequency equation is given by that the determinant of the coefficient matrix of Eq. (5.126) has to vanish, that is,

$$ \left|\begin{array}{cc}{J}_1^{\prime}\left(\xi a\right)-\frac{J_1\left(\xi\;a\right)}{\xi\;a}& {Y}_1^{\prime}\left(\xi a\right)-\frac{Y_1\left(\xi\;a\right)}{\xi\;a}\\ {}{J}_1^{\prime}\left(\xi b\right)-\frac{J_1\left(\xi\;b\right)}{\xi\;b}& {Y}_1^{\prime}\left(\xi b\right)-\frac{Y_1\left(\xi\;b\right)}{\xi\;b}\end{array}\right|=0. $$

(5.127)

5.7 Axial Shear of a Ceramic Cylinder

Consider an infinite ceramic circular cylinder of inner radius a and outer radius b (see Fig. 5.8). We orient (1,2,3) along (r,θ,z) and the poling direction is 3. The inner and outer surfaces are electroded.

Fig. 5.8

A ceramic cylinder with axial poling

We study axial-shear vibrations which are antiplane motions. The notation for antiplane motions in Sect. 2.9 is employed. The governing equations are:

$$ {\displaystyle \begin{array}{l}\overline{c}{\nabla}^2u=\rho \ddot{u},\kern1em a<r<b,\\ {}{\nabla}^2\psi =0,\kern1em a<r<b,\\ {}\varphi =\psi +\frac{e}{\varepsilon }u,\kern1em a<r<b.\end{array}} $$

(5.128)

For boundary conditions we consider the following possibilities:

$$ {\displaystyle \begin{array}{l}u=0,\kern1em r=a,\kern0.5em b,\kern1em \mathrm{if}\kern0.5em \mathrm{the}\kern0.5em \mathrm{cylindical}\ \mathrm{surfaces}\kern0.5em \mathrm{are}\kern0.5em \mathrm{fixed},\\ {}\kern1em \mathrm{or}\kern1em {T}_{rz}=0,\kern1em r=a,\kern0.5em b,\kern1em \mathrm{if}\kern0.5em \mathrm{the}\kern0.5em \mathrm{cylindical}\ \mathrm{surfaces}\kern0.5em \mathrm{are}\kern0.5em \mathrm{free};\\ {}\varphi \left(r=a\right)=\varphi \left(r=b\right),\kern1em \mathrm{if}\kern0.5em \mathrm{the}\kern0.5em \mathrm{electrodes}\kern0.5em \mathrm{are}\kern0.5em \mathrm{shorted},\\ {}\kern1em \mathrm{or}\kern1em {D}_r=0,\kern1em r=a,\kern0.5em b,\kern1em \mathrm{if}\kern0.5em \mathrm{the}\kern0.5em \mathrm{electrodes}\kern0.5em \mathrm{are}\kern0.5em \mathrm{open}.\end{array}} $$

(5.129)

For axisymmetric motions [6] we look for solutions in the following form:

$$ {\displaystyle \begin{array}{l}u\left(r,t\right)=u(r)\exp \left( i\omega t\right),\\ {}\psi \left(r,t\right)=\psi (r)\exp \left( i\omega t\right),\end{array}} $$

(5.130)

and drop the exponential time factor. The equations for u and ψ reduce to

$$ {\displaystyle \begin{array}{l}{\nabla}^2u=\frac{\partial^2u}{\partial {r}^2}+\frac{1}{r}\frac{\partial u}{\partial r}=-\frac{{\rho \omega}^2}{\overline{c}}u,\\ {}{\nabla}^2\psi =\frac{\partial^2\psi }{\partial {r}^2}+\frac{1}{r}\frac{\partial \psi }{\partial r}=0.\end{array}} $$

(5.131)

The general solution to Eq. (5.131) is

$$ {\displaystyle \begin{array}{l}u={A}_1{J}_0\left(\xi r\right)+{A}_2{Y}_0\left(\xi r\right),\\ {}\psi ={A}_3\ln \frac{r}{b}+{A}_4,\end{array}} $$

(5.132)

where A₁, A₂, A₃, and A₄ are undetermined constants, J₀ and Y₀ are zero-order Bessel functions of the first and second kind, and

$$ {\xi}^2=\frac{{\rho \omega}^2}{\overline{c}}. $$

(5.133)

The stress and electric displacement components are

$$ {\displaystyle \begin{array}{l}\varphi =\frac{e}{\varepsilon}\left[{A}_1{J}_0\left(\xi r\right)+{A}_2{Y}_0\left(\xi r\right)\right]+{A}_3\ln \frac{r}{b}+{A}_4,\\ {}{T}_{rz}=\overline{c}{u}_{z,r}+e{\psi}_{,r}=-\overline{c}\xi \left[{A}_1{J}_1\left(\xi r\right)+{A}_2{Y}_1\left(\xi r\right)\right]+e\frac{A_3}{r},\\ {}{D}_r=-{\varepsilon \psi}_{,r}=-\varepsilon \frac{A_3}{r},\end{array}} $$

(5.134)

where ${J}_0^{\prime }=-{J}_1$ and ${Y}_0^{\prime }=-{Y}_1$ have been used.

In the case when the two cylindrical surfaces are fixed and the two electrodes are shorted and grounded, we have

$$ {\displaystyle \begin{array}{l}u=0,\kern1em r=a,b,\\ {}\varphi =0,\kern1em r=a,b,\end{array}} $$

(5.135)

which implies that

$$ \psi =0,\kern1em r=a,b. $$

(5.136)

Hence

$$ {A}_3=0,\kern1em {A}_4=0, $$

(5.137)

and the frequency equation is

$$ \left|\begin{array}{cc}{J}_0\left(\xi a\right)& {Y}_0\left(\xi a\right)\\ {}{J}_0\left(\xi b\right)& {Y}_0\left(\xi b\right)\end{array}\right|=0. $$

(5.138)

In the case when the cylindrical surfaces are traction-free and electrically open, we have

$$ {\displaystyle \begin{array}{l}{T}_{rz}=0,\kern1em r=a,\kern0.5em b,\\ {}{D}_r=0,\kern1em r=a,\kern0.5em b.\end{array}} $$

(5.139)

Then A₃ = 0 and the frequency equation is

$$ \left|\begin{array}{cc}{J}_1\left(\xi a\right)& {Y}_1\left(\xi a\right)\\ {}{J}_1\left(\xi b\right)& {Y}_1\left(\xi b\right)\end{array}\right|=0. $$

(5.140)

If the cylindrical surfaces are traction-free with shorted and grounded electrodes, we have

$$ {\displaystyle \begin{array}{l}{T}_{rz}=0,\kern1em r=a,\kern0.5em b,\\ {}\varphi =0,\kern1em r=a,\kern0.5em b.\end{array}} $$

(5.141)

In this case the frequency equation is

$$ {\displaystyle \begin{array}{l}\left|\begin{array}{cc}{J}_1\left(\xi a\right)& {Y}_1\left(\xi a\right)\\ {}{J}_1\left(\xi b\right)& {Y}_1\left(\xi b\right)\end{array}\right|\\ {}\kern1em =\frac{{\overline{k}}^2}{\xi b\ln \frac{a}{b}}\left|\begin{array}{cc}{J}_0\left(\xi a\right)-{J}_0\left(\xi b\right)& {J}_1\left(\xi a\right)-\frac{b}{a}{J}_1\left(\xi b\right)\\ {}{Y}_0\left(\xi a\right)-{Y}_0\left(\xi b\right)& {Y}_1\left(\xi a\right)-\frac{b}{a}{Y}_1\left(\xi b\right)\end{array}\right|.\end{array}} $$

(5.142)

For large x, Bessel functions can be approximated by

$$ {\displaystyle \begin{array}{l}{J}_v(x)\cong \sqrt{\frac{2}{\pi x}}\;\cos \left(x-\frac{\nu \pi}{2}-\frac{\pi }{4}\right),\\ {}{Y}_v(x)\cong \sqrt{\frac{2}{\pi x}}\;\sin \left(x-\frac{\nu \pi}{2}-\frac{\pi }{4}\right).\end{array}} $$

(5.143)

Then it can be shown that, for large a and b, Eq. (5.142) simplifies to

$$ \sin \xi \left(b-a\right)=\frac{{\overline{k}}^2}{\xi b\ln \frac{a}{b}}\left[\left(1+\frac{b}{a}\right)\cos \xi \left(b-a\right)-2\sqrt{\frac{b}{a}}\right]. $$

(5.144)

Setting 2h = b−a and letting a, b→∞, we have

$$ {\displaystyle \begin{array}{l}\xi b\ln \frac{a}{b}=\xi b\ln \frac{b-2h}{b}=\xi b\ln \left(1-\frac{2h}{b}\right)\cong \xi b\left(-\frac{2h}{b}\right)=-\xi 2h,\\ {}\sin \xi \left(b-a\right)=\sin \left(\xi 2h\right),\\ {}\left(1+\frac{b}{a}\right)\cos \xi \left(b-a\right)-2\sqrt{\frac{b}{a}}\\ {}\kern1em \cong 2\cos \left(\xi 2h\right)-2=-2\left[1-\cos \left(\xi 2h\right)\right]=-4{\sin}^2\xi h.\end{array}} $$

(5.145)

Then Eq. (5.144) reduces to

$$ \tan \xi h=\frac{\xi h}{{\overline{k}}^2}, $$

(5.146)

which is the frequency equation for thickness-shear vibrations of an electroded ceramic plate given by Eq. (5.50).

5.8 Extension of a Ceramic Rod Through e₃₃

Consider a cylindrical rod of length l made from polarized ceramics with axial poling (see Fig. 5.9). The shape of the cross section of the rod can be arbitrary but it has to be small compared to the length of the rod. The entire surface of the rod is traction-free. The cylindrical surface is unelectroded. The electric field in the surrounding free space is neglected. The two end faces are electroded. The rod can be excited into extensional vibration through e₃₃ if a voltage is applied across the two end electrodes. In this section we consider free vibrations with open electrodes. The wavelength of the vibration modes is much larger than the width and thickness of the rod.

Fig. 5.9

An axially poled ceramic rod

We consider a thin rod and make the following approximations of uniaxial stress and electric displacement:

$$ {\displaystyle \begin{array}{l}{T}_{33}={T}_{33}\left({x}_3,t\right),\kern1em \mathrm{all}\kern0.5em \mathrm{other}\kern0.5em {T}_{ij}=0,\\ {}{D}_3={D}_3\left({x}_3,t\right),\kern1em {D}_1={D}_2=0.\end{array}} $$

(5.147)

Thus the boundary conditions on the lateral surface are satisfied. The relevant equations of motion and charge reduce to

$$ {\displaystyle \begin{array}{l}{T}_{33,3}=\rho {\ddot{u}}_3,\\ {}{D}_{3,3}=0.\end{array}} $$

(5.148)

From Eq. (2.19), the relevant constitutive relations take the following form:

$$ {\displaystyle \begin{array}{l}{S}_{33}={s}_{33}{T}_{33}+{d}_{33}{E}_3,\kern1em \\ {}{D}_3={d}_{33}{T}_{33}+{\varepsilon}_{33}{E}_3,\end{array}} $$

(5.149)

where

$$ {\displaystyle \begin{array}{l}{S}_{33}={u}_{3,3},\kern1em {E}_3=-{\varphi}_{,3},\\ {}{s}_{33}={s}_{33}^E,\kern1em {\varepsilon}_{33}={\varepsilon}_{33}^T.\end{array}} $$

(5.150)

From Eq. (5.149) we obtain

$$ {\displaystyle \begin{array}{l}{T}_{33}=\frac{1}{s_{33}}\left({S}_{33}-{d}_{33}{E}_3\right)=\frac{1}{s_{33}}\left({u}_{3,3}+{d}_{33}{\varphi}_{,3}\right),\kern1em \\ {}{D}_3=\frac{d_{33}}{s_{33}}{S}_{33}+\left({\varepsilon}_{33}-\frac{d_{33}^2}{s_{33}}\right){E}_3=\frac{d_{33}}{s_{33}}{u}_{3,3}-{\varepsilon}_{33}\left[1-{\left({k}_{33}^l\right)}^2\right]{\varphi}_{,3},\end{array}} $$

(5.151)

where

$$ {\left({k}_{33}^l\right)}^2=\frac{d_{33}^2}{\varepsilon_{33}{s}_{33}}. $$

(5.152)

Substituting Eq. (5.151) into Eq. (5.148), we obtain

$$ {\displaystyle \begin{array}{l}\frac{1}{s_{33}}\left({u}_{3,33}+{d}_{33}{\varphi}_{,33}\right)=\rho {\ddot{u}}_3,\kern1em \\ {}\frac{d_{33}}{s_{33}}{u}_{3,33}-{\varepsilon}_{33}\left[1-{\left({k}_{33}^l\right)}^2\right]{\varphi}_{,33}=0.\end{array}} $$

(5.153)

Consider the case of open end electrodes. The electrical end conditions are

$$ {D}_3=0,\kern1em {x}_3=\pm l/2. $$

(5.154)

Equation (5.148)₂ implies that D₃ is independent of x₃. Hence, from Eq. (5.154), D₃ is identically zero, which implies that

$$ {\varphi}_{,3}=\frac{d_{33}}{\varepsilon_{33}{s}_{33}\left[1-{\left({k}_{33}^l\right)}^2\right]}{u}_{3,3}=\frac{1}{d_{33}}\frac{{\left({k}_{33}^l\right)}^2}{1-{\left({k}_{33}^l\right)}^2}{u}_{3,3}. $$

(5.155)

Then

$$ {T}_{33}=\frac{1}{s_{33}\left[1-{\left({k}_{33}^l\right)}^2\right]}{u}_{3,3}, $$

(5.156)

and Eq. (5.153)₁ becomes

$$ \frac{1}{s_{33}\left[1-{\left({k}_{33}^l\right)}^2\right]}{u}_{3,33}=\rho {\ddot{u}}_3. $$

(5.157)

For traction-free mechanical end conditions, in harmonic motions, the eigenvalue problem is

$$ {\displaystyle \begin{array}{l}{u}_{3,33}+\rho {s}_{33}\left[1-{\left({k}_{33}^l\right)}^2\right]{\omega}^2{u}_3=0,\kern1em -l/2<{x}_3<l/2,\\ {}{u}_{3,3}=0,\kern1em {x}_3=\pm l/2.\end{array}} $$

(5.158)

The solution of ω = 0 and u₃ = constant represents a rigid-body mode. For the rest of the modes we try u₃ = sinkx₁. Then, from Eq. (5.158)₁,

$$ k=\omega \sqrt{\rho {s}_{33}\left[1-{\left({k}_{33}^l\right)}^2\right]}. $$

(5.159)

To satisfy Eq. (5.158)₂ we must have

$$ \cos k\frac{l}{2}=0,\kern1em \Rightarrow \kern1em {k}^{(n)}\frac{l}{2}=\frac{n\pi}{2},\kern1em n=1,3,5,\dots, $$

(5.160)

or

$$ {\displaystyle \begin{array}{l}{\omega}^{(n)}\sqrt{\rho {s}_{33}\left[1-{\left({k}_{33}^l\right)}^2\right]}\frac{l}{2}=\frac{n\pi}{2},\\ {}{\omega}^{(n)}=\frac{n\pi}{l\sqrt{\rho {s}_{33}\left[1-{\left({k}_{33}^l\right)}^2\right]}},\kern1em n=1,3,5,\dots .\end{array}} $$

(5.161)

Similarly, by considering u₃ = coskx₁, the following frequencies can be determined:

$$ {\omega}^{(n)}=\frac{n\pi}{l\sqrt{\rho {s}_{33}\left[1-{\left({k}_{33}^l\right)}^2\right]}},\kern1em n=2,4,6,\dots . $$

(5.162)

Equations (5.161) and (5.162) show that ${k}_{33}^l$ raises the resonant frequencies, which is the so-called piezoelectric stiffening effect. The frequencies in Eqs. (5.161) and (5.162) are integral multiples of ω⁽¹⁾ and are called harmonics. ω⁽¹⁾ is called the fundamental and the rest are called the overtones.

5.9 Extension of a Ceramic Rod Through e₃₁

Consider a rectangular ceramic rod of length l, width w, and thickness t as shown in Fig. 5.10, where l >  > w >  > t. The poling direction is along x₃. We are interested in the extensional vibration of the rod [3].

Fig. 5.10

A ceramic rod with rectangular cross section

As an approximation, it is appropriate to take the vanishing boundary stresses on the surfaces bounding the two small dimensions to vanish everywhere. Consequently

$$ {T}_{11}={T}_1\left({x}_1,t\right),\kern0.5em \mathrm{and}\kern0.5em \mathrm{all}\kern0.5em \mathrm{other}\kern0.5em {T}_{ij}=0. $$

(5.163)

If the surfaces of the area lw are fully electroded with a driving voltage V across the electrodes, the appropriate electrical conditions are

$$ {E}_1={E}_2=0,\kern1em {E}_3=-\frac{V}{t}. $$

(5.164)

The pertinent constitutive relations are from Eq. (2.19):

$$ {\displaystyle \begin{array}{l}{S}_1={s}_{11}{T}_1+{d}_{31}{E}_3,\\ {}{D}_3={d}_{31}{T}_1+{\varepsilon}_{33}{E}_3,\end{array}} $$

(5.165)

where

$$ {s}_{33}={s}_{33}^E,\kern1em {\varepsilon}_{33}={\varepsilon}_{33}^T. $$

(5.166)

Equation (5.165)₁ can be inverted to give

$$ {T}_1=\frac{1}{s_{11}}{S}_1-\frac{d_{31}}{s_{11}}{E}_3. $$

(5.167)

Then the differential equation of motion and boundary conditions are

$$ {\displaystyle \begin{array}{l}\frac{1}{s_{11}}{u}_{1,11}=\rho {\ddot{u}}_1,\kern1em -l/2<{x}_1<l/2,\\ {}{T}_1=\frac{1}{s_{11}}{u}_{1,1}+\frac{d_{31}}{s_{11}}\frac{V}{t}=0,\kern1em {x}_1=\pm l/2.\end{array}} $$

(5.168)

Equation (5.168) shows that the applied voltage effectively acts like two extensional end forces on the rod. For free vibrations, the electrodes are shorted and V = 0. We look for a solution in the form

$$ {u}_1\left({x}_1,t\right)={u}_1\left({x}_1\right)\exp \left( i\omega \kern0.2em t\right). $$

(5.169)

Then the eigenvalue problem is

$$ {\displaystyle \begin{array}{l}{u}_{1,11}+\rho {s}_{11}{\omega}^2{u}_1,\kern1em -l/2<{x}_1<l/2,\\ {}{u}_{1,1}=0,\kern1em {x}_1=\pm l/2.\end{array}} $$

(5.170)

The solution of ω = 0 and u₁ = constant represents a rigid-body mode. For the rest of the modes we try u₁ = sinkx₁. Then, from Eq. (5.170)₁,

$$ k=\omega \sqrt{\rho {s}_{11}}. $$

(5.171)

To satisfy (5.170)₂ we must have

$$ \cos k\frac{l}{2}=0,\kern1em \Rightarrow \kern1em {k}_{(n)}\frac{l}{2}=\frac{n\pi}{2},\kern1em n=1,3,5,\dots, $$

(5.172)

or

$$ {\omega}_{(n)}\sqrt{\rho {s}_{11}}\frac{l}{2}=\frac{n\pi}{2},\kern1em {\omega}_{(n)}=\frac{n\pi}{l\sqrt{\rho {s}_{11}}},\kern1em n=1,3,5,\dots . $$

(5.173)

Similarly, by considering u₁ = coskx₁, the following frequencies can be determined:

$$ {\omega}_{(n)}=\frac{n\pi}{l\sqrt{\rho {s}_{11}}},\kern1em n=2,4,6,\dots . $$

(5.174)

The frequencies in Eqs. (5.173) and (5.174) do not show any piezoelectric stiffening effect because the electric field is zero in this case when the two electrodes are shorted.

If the surfaces of the area lt are fully electroded with a driving voltage V across the electrodes, the appropriate electrical conditions are

$$ {E}_1=0,\kern1em {D}_3=0,\kern1em {E}_2=-\frac{V}{w}. $$

(5.175)

The pertinent constitutive relations are

$$ {\displaystyle \begin{array}{l}{S}_1={s}_{11}{T}_1+{d}_{21}{E}_2+{d}_{31}{E}_3,\\ {}{D}_2={d}_{21}{T}_1+{\varepsilon}_{22}{E}_2+{\varepsilon}_{23}{E}_3.\end{array}} $$

(5.176)

From the boundary conditions on the areas of lw, we take the following to be approximately true everywhere:

$$ {D}_3={d}_{31}{T}_1+{\varepsilon}_{32}{E}_2+{\varepsilon}_{33}{E}_3=0. $$

(5.177)

With Eq. (5.177), we can write Eq. (5.176) as

$$ {\displaystyle \begin{array}{l}{S}_1={\tilde{s}}_{11}{T}_1+{\tilde{d}}_{21}{E}_2,\\ {}{D}_2={\tilde{d}}_{21}{T}_1+{\tilde{\varepsilon}}_{22}{E}_2,\end{array}} $$

(5.178)

where

$$ {\displaystyle \begin{array}{l}{\tilde{s}}_{11}={s}_{11}-{d}_{31}^2/{\varepsilon}_{33},\\ {}{\tilde{d}}_{21}={d}_{21}-{d}_{31}{\varepsilon}_{23}/{\varepsilon}_{33},\\ {}{\tilde{\varepsilon}}_{22}={\varepsilon}_{22}-{\varepsilon}_{23}^2/{\varepsilon}_{33}.\end{array}} $$

(5.179)

If the surfaces of areas lw and lt are not electroded, the appropriate electrical conditions are

$$ {D}_2={D}_3=0. $$

(5.180)

In this case, from Eq. (2.20), the pertinent constitutive relations are

$$ {\displaystyle \begin{array}{l}{S}_1={s}_{11}{T}_1+{g}_{11}{D}_1,\\ {}{E}_1=-{g}_{11}{T}_1+{\beta}_{11}{D}_1.\end{array}} $$

(5.181)

5.10 Radial Vibration of a Circular Ceramic Plate

A circular disk of ceramics poled in the thickness direction is positioned in a coordinate system as shown in Fig. 5.11. The top and bottom faces of the disk are traction-free and are completely coated with electrodes. The electrodes are connected to a voltage source of potential V exp (iω t). We consider axisymmetric radial modes [3].

Fig. 5.11

A circular ceramic plate with thickness poling

Under these circumstances, the boundary conditions at x₃ = ±b are

$$ {\displaystyle \begin{array}{l}{T}_{3j}=0,\kern1em {x}_3=\pm b,\\ {}\varphi =\pm \frac{V}{2}\exp \left( i\omega \kern0.1em t\right),\kern1em {x}_3=\pm b.\end{array}} $$

(5.182)

Since T₃, T₄, and T₅ vanish on both major surfaces of the plate and the plate is thin, these stresses cannot depart much from zero. Consequently they are assumed to vanish throughout. Thus we assume that

$$ {T}_3={T}_4={T}_5=0. $$

(5.183)

Furthermore, since the plate is thin and has conducting surfaces,

$$ {E}_1=0,\kern1em {E}_2=0,\kern1em {E}_3=-\frac{V}{2b}\exp \left( i\omega \kern0.1em t\right). $$

(5.184)

We consider radial modes with

$$ {u}_{\theta }=0,\kern1em \frac{\partial }{\partial \theta }=0. $$

(5.185)

The constitutive relations are

$$ {\displaystyle \begin{array}{l}{T}_{r r}={c}_{11}^p{u}_{r,r}+{c}_{12}^p{u}_r/r+{e}_{31}^p{\varphi}_{,3},\\ {}{T}_{\theta \theta}={c}_{11}^p{u}_r/r+{c}_{12}^p{u}_{r,r}+{e}_{31}^p{\varphi}_{,3},\\ {}{T}_{r\theta}=0,\\ {}{D}_3={e}_{31}^p\left({u}_{r,r}+{u}_r/r\right)-{\varepsilon}_{33}^p{\varphi}_{,3},\end{array}} $$

(5.186)

where

$$ {\displaystyle \begin{array}{l}\begin{array}{c}{c}_{11}^p={c}_{11}-{c}_{13}^2/{c}_{33},\\ {}{c}_{12}^p={c}_{12}-{c}_{13}^2/{c}_{33},\end{array}\kern1em \\ {}{e}_{31}^p={e}_{31}-{e}_{33}{c}_{13}/{c}_{33},\\ {}{\varepsilon}_{33}^p={\varepsilon}_{33}+{e}_{33}^2/{c}_{33},\end{array}} $$

(5.187)

are the effective material constants for a thin plate after the relaxation of the normal stress in the thickness direction. The one remaining equation of motion in cylindrical coordinates is

$$ \frac{\partial {T}_{rr}}{\partial r}+\frac{T_{rr}-{T}_{\theta \theta}}{r}=\rho {\ddot{u}}_r. $$

(5.188)

Substituting from Eq. (5.186) for the stress components, we obtain

$$ {c}_{11}^p\left({u}_{, rr}+{u}_{r,r}/r-{u}_r/{r}^2\right)=\rho {\ddot{u}}_r, $$

(5.189)

which, since we are assuming a steady-state problem with frequency ω, becomes

$$ {u}_{r, rr}+\frac{u_{r,r}}{r}+\left({\xi}^2-\frac{1}{r^2}\right){u}_r=0, $$

(5.190)

where

$$ {\xi}^2=\frac{\omega^2}{{\left({v}^p\right)}^2},\kern1em {\left({v}^p\right)}^2={c}_{11}^p/\rho . $$

(5.191)

Equation (5.190) can be written as Bessel’s equation of order one. For a solid disk, the motion at the origin is zero and the general solution is

$$ {u}_r={BJ}_1\left(\xi\;r\right)\exp \left( i\omega \kern0.1em t\right), $$

(5.192)

where J₁ is the first kind Bessel function of the first order. Equation (5.192) is subject to the boundary condition

$$ {T}_{rr}=0,\kern1em r=a, $$

(5.193)

hence Eq. (5.193) requires that

$$ {c}_{11}^pB{\left.\frac{dJ_1}{dr}\right|}_{r=a}+{c}_{12}^pB\frac{J_1}{a}=-{e}_{31}^p\frac{V}{2b}, $$

(5.194)

where, for convenience, the argument of the Bessel function is not written. From Eq. (5.194), B can be expressed in terms of V as follows:

$$ B={\left[\left(1-{\sigma}^p\right)\frac{J_1\left(\xi\;a\right)}{a}-\xi {J}_0\left(\xi\;a\right)\right]}^{-1}\frac{e_{31}^p}{c_{11}^p}\frac{V}{2b}, $$

(5.195)

where

$$ \frac{dJ_1(x)}{dx}={J}_0(x)-\frac{J_1(x)}{x}, $$

(5.196)

has been used and

$$ {\sigma}^p={c}_{12}^p/{c}_{11}^p, $$

(5.197)

which may be interpreted as a planar Poisson’s ratio, since the material is isotropic in the plane normal to x₃. The total charge on the electrode at the bottom of the plate is given by

$$ {Q}^e={\int}_A{D}_3 dA=2\pi {\int}_0^a{D}_3 rdr. $$

(5.198)

The substitution of Eq. (5.192) into Eq. (5.186)₄ and then into Eq. (5.198) yields

$$ {Q}^e=2\pi {e}_{31}^p{aBJ}_1\left(\xi\;a\right)-{\pi \varepsilon}_{33}^p{Va}^2/2b. $$

(5.199)

Hence we obtain for the current that flows to the bottom electrode of the plate as

$$ {\displaystyle \begin{array}{c}I=\frac{dQ^e}{dt}\\ {}= i\omega \left[\frac{2{\left({k}_{31}^p\right)}^2{J}_1\left(\xi\;a\right)}{\left(1-{\sigma}^p\right){J}_1\left(\xi\;a\right)-\xi\;{aJ}_0\left(\xi\;a\right)}-1\right]\frac{\varepsilon_{33}^p\pi {a}^2V}{2b},\end{array}} $$

(5.200)

where

$$ {\left({k}_{31}^p\right)}^2=\frac{{\left({e}_{31}^p\right)}^2}{\varepsilon_{33}^p{c}_{11}^p}. $$

(5.201)

For free vibrations, the applied voltage is zero. From Eq. (5.194),

$$ {\left.\frac{dJ_1}{dr}\right|}_{r=a}+{\sigma}^p\frac{J_1}{a}=0, $$

(5.202)

which determines resonance frequencies. Or, at resonances, the current goes to infinity. This condition is determined by setting the square bracketed factor in the denominator of Eq. (5.200) equal to zero. The resulting equation is

$$ \frac{\xi\;{aJ}_0\left(\xi\;a\right)}{J_1\left(\xi\;a\right)}=1-{\sigma}^p, $$

(5.203)

which can be brought into the same form as Eq. (5.202). The antiresonance frequencies result when the current goes to zero. The resulting equation is

$$ \frac{\xi\;{aJ}_0\left(\xi\;a\right)}{J_1\left(\xi\;a\right)}=1-{\sigma}^p-2{\left({k}_{31}^p\right)}^2. $$

(5.204)

5.11 Orthogonality of Modes

The free vibration of a piezoelectric body with frequency ω is governed by the differential equations

$$ {\displaystyle \begin{array}{l}{T}_{ji,j}\left(\mathbf{u},\varphi \right)=-{\rho \omega}^2{u}_i,\\ {}{D}_{i,i}\left(\mathbf{u},\varphi \right)=0,\end{array}} $$

(5.205)

where

$$ {\displaystyle \begin{array}{l}{T}_{ji}\left(\mathbf{u},\varphi \right)={c}_{ji kl}{S}_{kl}-{e}_{kji}{E}_k={c}_{ji kl}{u}_{k,l}+{e}_{kji}{\varphi}_{,k},\\ {}{D}_i\left(\mathbf{u},\varphi \right)={e}_{ik l}{S}_{kl}+{\varepsilon}_{ik}{E}_k={e}_{ik l}{u}_{k,l}-{\varepsilon}_{ik}{\varphi}_{,k}.\end{array}} $$

(5.206)

From Eqs. (2.32), (2.34), and (2.35), the homogeneous boundary conditions are

$$ {\displaystyle \begin{array}{l}{u}_i=0\kern1em \mathrm{on}\kern1em {S}_u,\\ {}{n}_j{T}_{ji}=0\kern1em \mathrm{on}\kern1em {S}_T,\\ {}\varphi =0\kern1em \mathrm{on}\kern1em {S}_{\varphi },\\ {}{n}_i{D}_i=0\kern1em \mathrm{on}\kern1em {S}_D.\end{array}} $$

(5.207)

For any two vectors u and v as well as two scalars φ and ψ, with the divergence theorem, we have

$$ {\displaystyle \begin{array}{l}{\int}_V\left[{T}_{ji,j}\left(\mathbf{u},\varphi \right){v}_i+{D}_{i,i}\Big(\mathbf{u},\varphi \Big)\psi \right] dV\\ {}\kern1em ={\int}_V\left\{\right.{\left[{T}_{ji}\left(\mathbf{u},\varphi \right){v}_i\right]}_{,j}-{T}_{ji}\left(\mathbf{u},\varphi \right){v}_{i,j}\\ {}\kern1em +{\left[{D}_i\left(\mathbf{u},\varphi \right)\psi \right]}_{,i}-{D}_i\left(\mathbf{u},\varphi \right){\psi}_{,i}\left.\right\} dV\\ {}\kern1em ={\int}_S\left[{n}_j{T}_{ji}\left(\mathbf{u},\varphi \right){v}_i+{n}_i{D}_i\Big(\mathbf{u},\varphi \Big)\psi \right] dS\\ {}\kern1em -{\int}_V\left({c}_{ji kl}{u}_{k,l}{v}_{i,j}+{e}_{kji}{\varphi}_{,k}{v}_{i,j}+{e}_{ik l}{u}_{k,l}{\psi}_{,i}-{\varepsilon}_{ik}{\varphi}_{,k}{\psi}_{,i}\right) dV.\end{array}} $$

(5.208)

Similarly,

$$ {\displaystyle \begin{array}{l}{\int}_V\left[{T}_{ji,j}\left(\mathbf{v},\psi \right){u}_i+{D}_{i,i}\Big(\mathbf{v},\psi \Big)\varphi \right] dV\\ {}\kern1em ={\int}_S\left[{n}_j{T}_{ji}\left(\mathbf{v},\psi \right){u}_i+{n}_i{D}_i\Big(\mathbf{v},\psi \Big)\varphi \right] dS\\ {}\kern1em -{\int}_V\left({c}_{ji kl}{v}_{k,l}{u}_{i,j}+{e}_{kji}{\psi}_{,k}{u}_{i,j}+{e}_{ik l}{v}_{k,l}{\varphi}_{,i}-{\varepsilon}_{ik}{\psi}_{,k}{\varphi}_{,i}\right) dV.\end{array}} $$

(5.209)

Subtracting Eqs. (5.208) and (5.209), using the symmetries of the material constants, we obtain the following identity [7] which corresponds to Green’s identity for the Laplacian operator and the reciprocal theorem in elasticity:

$$ {\displaystyle \begin{array}{l}{\int}_V\left[{T}_{ji,j}\left(\mathbf{u},\varphi \right){v}_i+{D}_{i,i}\Big(\mathbf{u},\varphi \Big)\psi \right] dV\\ {}\kern1em -{\int}_V\left[{T}_{ji,j}\left(\mathbf{v},\psi \right){u}_i+{D}_{i,i}\Big(\mathbf{v},\psi \Big)\varphi \right] dV\\ {}\kern1em ={\int}_S\left[{n}_j{T}_{ji}\left(\mathbf{u},\varphi \right){v}_i+{n}_i{D}_i\Big(\mathbf{u},\varphi \Big)\psi \right] dS\\ {}\kern1em -{\int}_S\left[{n}_j{T}_{ji}\left(\mathbf{v},\psi \right){u}_i+{n}_i{D}_i\Big(\mathbf{v},\psi \Big)\varphi \right] dS.\end{array}} $$

(5.210)

Now consider two vibration modes corresponding to two different resonance frequencies ω^(m) and ω⁽ⁿ⁾, that is,

$$ {\displaystyle \begin{array}{l}{T}_{ij,i}^{(m)}+\rho {\left({\omega}^{(m)}\right)}^2{u}_j^{(m)}=0,\\ {}{D}_{i,i}^{(m)}=0,\end{array}} $$

(5.211)

and

$$ {\displaystyle \begin{array}{l}{T}_{ij,i}^{(n)}+\rho {\left({\omega}^{(n)}\right)}^2{u}_j^{(n)}=0,\\ {}{D}_{i,i}^{(n)}=0.\end{array}} $$

(5.212)

The modes in Eqs. (5.211) and (5.212) also satisfy the boundary conditions in Eq. (5.207). In Eq. (5.210), letting

$$ {\displaystyle \begin{array}{l}\mathbf{u}={\mathbf{u}}^{(m)},\kern1em \varphi ={\varphi}^{(m)},\\ {}\mathbf{v}={\mathbf{u}}^{(n)},\kern1em \psi ={\varphi}^{(n)},\end{array}} $$

(5.213)

using Eqs. (5.207), (5.211) and (5.212), we obtain

$$ \left[{\left({\omega}^{(n)}\right)}^2-{\left({\omega}^{(m)}\right)}^2\right]{\int}_V\rho {u}_j^{(m)}{u}_j^{(n)} dV=0. $$

(5.214)

Since the two resonance frequencies are different, Eq. (5.214) implies that

$$ {\int}_V\rho {u}_j^{(m)}{u}_j^{(n)} dV=0, $$

(5.215)

which is called the orthogonality of modes [2].

In Eq. (5.208), if we choose

$$ {\displaystyle \begin{array}{l}\mathbf{u}={\mathbf{u}}^{(n)},\kern1em \varphi ={\varphi}^{(n)},\\ {}\mathbf{v}={\mathbf{u}}^{(n)},\kern1em \psi ={\varphi}^{(n)},\end{array}} $$

(5.216)

we obtain

$$ {\displaystyle \begin{array}{l}-{\left({\omega}^{(n)}\right)}^2{\int}_V\rho {u}_j^{(n)}{u}_j^{(n)} dV\\ {}\kern1em =-{\int}_V\left({c}_{jikl}{u}_{k,l}^{(n)}{u}_{i,j}^{(n)}+{e}_{kji}{\varphi}_{,k}^{(n)}{u}_{i,j}^{(n)}+{e}_{ik l}{u}_{k,l}^{(n)}{\varphi}_{,i}^{(n)}-{\varepsilon}_{ik}{\varphi}_{,k}^{(n)}{\varphi}_{,i}^{(n)}\right) dV,\end{array}} $$

(5.217)

which can be further written as

$$ {\left({\omega}^{(n)}\right)}^2=\frac{\int_V\left({c}_{jikl}{u}_{k,l}^{(n)}{u}_{i,j}^{(n)}+2{e}_{kji}{\varphi}_{,k}^{(n)}{u}_{i,j}^{(n)}-{\varepsilon}_{ik}{\varphi}_{,k}^{(n)}{\varphi}_{,i}^{(n)}\right) dV}{\int_V\rho {u}_j^{(n)}{u}_j^{(n)} dV}. $$

(5.218)

Equation (5.218) can be viewed as the ratio between the electric enthalpy and the kinetic energy corresponding to the mode. More generally, we can define the following functional for the variational formulation of the free vibration eigenvalue problem:

$$ R\left(\mathbf{u},\varphi \right)=\frac{\int_V\left({c}_{jikl}{u}_{k,l}{u}_{i,j}+2{e}_{kji}{\varphi}_{,k}{u}_{i,j}-{\varepsilon}_{ik}{\varphi}_{,k}{\varphi}_{,i}\right) dV}{\int_V\rho {u}_j{u}_j dV}, $$

(5.219)

where the admissible u and φ satisfy the boundary conditions in Eq. (5.207). Then it can be shown [7] that R assumes stationary values when u = u⁽ⁿ⁾ and φ = φ⁽ⁿ⁾, and the stationary value of R is (ω⁽ⁿ⁾)². R is called the Rayleigh quotient.