Theory of elasticity

A solid subjected to external forces gets deformed and returns to its original form when the forces are removed, the body is said to be elastic. Various types of forces like, mechanical, thermal, electrical and magnetic can be employed for the generation of elastic waves in an elastic medium, however it is more suitable to choose materials in which the stress varies linearly with applied electric field. Elasticity of a solid is concerned with the internal forces within the solid and its particle displacement from equilibrium position [38]. The forces are expressed as stress T, while the displacements are expressed in terms of strain S. The particle is an elementary region of a material much larger than the inter-atomic distance and much smaller than any characteristic elastic dimensions such as wavelength. Assume the equilibrium state of solid with a particle

located at point x = (x1, x2, x3) and is displaced by an amount u = (u1, u2, u3) under the application of external force. Where, the components u1, u2 and u3 are the general displacement components along coordinates x1, x2 and x3, respectively. Thus the particle has been displaced to a new position x+ u as shown in Fig. 2.1. If u is independent of x there will be no internal force, since this simply shows a displacement of a material as a whole, and also

x2

x1

x3 x = (x1, x2, x3)

x +u u x

Fig. 2.1. Particle position in equilibrium and deformed states of solid body. Note:

The deformation of the solid is shown in broken lines [38].

TH-1583_09610201 17

Chapter 2 Modeling and Simulation of SAW Devices there will be no force if the material is rotated. The strain at each point can thus be defined as

𝑆𝑆𝑖𝑖𝑖𝑖�𝑥𝑥1,𝑥𝑥2,𝑥𝑥3�=1 2�𝜕𝜕𝑢𝑢_𝑖𝑖

𝜕𝜕𝑥𝑥𝑖𝑖+𝜕𝜕𝑢𝑢_𝑖𝑖

𝜕𝜕𝑥𝑥𝑖𝑖�; 𝑖𝑖 ,𝑗𝑗= 1,2,3. (2.1) Thus the strain is related to the internal forces. The strain is a second rank tensor and symmetrical as Sij = Sji. The stress T is defined as internal stress in the material i.e. force per unit area. The second-rank stress tensor is defined as Tij (x1, x2, x3) and is symmetric, Tij = Tji. According to Hooke’s law, each components of stress is given by the linear combination of the strain components and is expressed as

1 2 3

=

∑∑

=

( ) , , , , , , .

ij mech ijkl kl

k l

T c S i j k l _(2.2)

where, Cijkl is the fourth-rank stiffness tensor. These elements are the physical properties of the materials under consideration [3].

2.1.1 Piezoelectricity and constitutive equations

Piezoelectricity is a phenomenon which couples the elastic stress and strain to the electric field and displacement that occurs only in anisotropic materials, whose internal lattice structure lacks a center of symmetry [38]. The stress, Tij (elec) produced by the piezoelectric effect is given by

( ) =

∑

, , , , =1,2,3.

ij elec k kij k i j k l

T e E (2.3)

where, electric field 𝐸𝐸_𝑘𝑘 =− 𝜕𝜕𝜕𝜕 𝜕𝜕𝑥𝑥⁄ _𝑘𝑘, Total stress Tij is the sum of stresses due to the electric field Ek and mechanical strain and can be expressed as [3], [38-40]

∂

=

∑∑

−

∑

=

∑∑

+

∑

∂

k l k k l k k

E E

ij ijkl kl kij k ijkl kl kij V

T c S e E c S e x (2.4)

When an electric field is applied to the dielectric medium, the electric displacement can be expressed by electric field and permittivity tensor εij of the dielectric medium. In case of piezoelectric material, the additional electric field displacement caused by the strain is developed due to piezoelectric effect. The total electric displacement Di can then be given as

ε

=

∑

^s +

∑∑

i ij j ijk jk

j j k

D

E

e S ^(2.5)

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Chapter 2 Modeling and Simulation of SAW Devices where, 𝐶𝐶_{𝑖𝑖𝑖𝑖𝑘𝑘𝑖𝑖}^𝐸𝐸 is the stiffness tensor for constant electric field (N/m²), Sjk is the strain components, eijk is the piezoelectric tensor relating elastic to electric fields (C/m²), 𝜀𝜀_{𝑖𝑖𝑖𝑖}^𝑠𝑠 is the permittivity tensor for constant stress (F/m), and Ej is the electric field vector (V/m) [38].

2.1.2 Equation of motion

If the stress and strain are functions of time and position then the above equations can be expressed in the terms of equation of motion using Newton’s laws [38]. Theoretically, the electric field distribution can be found by solving the equations of Newton and Maxwell simultaneously. For an elementary cube centered at X’ = (𝑥𝑥′₁, 𝑥𝑥′2, 𝑥𝑥′3) with edge length of δ, density ρ, and the mass ρδ³, the total force in terms of stress along xi direction is given by

3

x

δ

′

 ∂ 

 ∂ 

 



∑

^ij

j j

T

x ^(2.6)

2

2 , , 1,2,3.

ρ^∂ = ^∂ =

∂ ⁱ

∑

∂ ^ij

j j

u T i j

t x ^(2.7)

In case of piezoelectric medium elastic waves travel much slowly than the electromagnetic waves. Hence the magnetic field due to these waves is negligible and electric field can be expressed using quasi static approximation. Thus Maxwell equation can be approximated as

curlE , gradV

t

= −∂ = = −

∂B 0 E (2.8)

where, B is the magnetic field, E is the electric field derived from the scalar electric potential V [3].

i i

E V

x

= −∂

∂ (2.9)

From equations (2.1), (2.9) and (2.4) the equation of motion for piezoelectric material can be expressed as [3]

i E k

kij ijkl

j k j k l j l

u e V c u

x x x x

ρ^∂∂²t2 =

∑∑

^ ∂ ∂^∂2 +

∑

∂ ∂^{∂2 } (2.10) In addition, the material is taken to be an insulator there are no free charges. Thus div D = 0. From equation (2.5)

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Chapter 2 Modeling and Simulation of SAW Devices

S j

ij ijk

i j i j k i k

V e u

x x x x



ε

∂ ∂ 

 − =

 ∂ ∂ ∂ ∂ 

 

 

∑∑

²

∑

^{2 0 (2.11)}

The degrees of freedom (dependent variables) are the global displacements u1, u2, and u3

in the global x1, x2, and x3 directions, respectively, and the electric potential V can be obtained by solving the equations (2.10) and (2.11) using appropriate boundary conditions [38].

2.1.3 Solution of surface wave in piezoelectric media

The detailed procedure for finding the solution of surface waves in isotropic and anisotropic materials is explained in [38]. In this section, boundary conditions and solutions of surface waves are briefly elaborated. The boundary conditions at the surface should be satisfied in order to determine the phase, velocity and amplitude of the waves. The appropriate boundary conditions should be applied in order to determine the phase velocity and amplitude of the wave [38]. In general, two cases are considered; the first case is called as free surface case where the space above the surface of piezoelectric material is a vacuum and no conductor is present, so that there are no free charges. The second case is called as metalized case where the surface of piezoelectric material is covered with thin metal layer with infinite conductivity and short the horizontal components of the electric field E. These two cases give different surface wave velocity and measure the coupling between wave and electrical perturbation at the surface. For free surface case, if the wave number of the surface wave is β then the potential in vacuum

(

^x₃^≥0

)

using Laplace’s equation can be expressed as

( ^x ) j t( x)

V V e= ₀ ^{−β 3}e ^{ω β− 1} (2.12) where, V₀ is a constant. For metalized surface case the potential at x3 = 0 is zero. The mechanical boundary conditions at the surface of the substrate is stress-free condition as expressed in equation below

T13 = T23 = T33 = 0 at x3 = 0 (2.13) The electrical boundary is imposed by Maxwell’s equation, the normal component of D being continuous across the air/substrate interface (at x3 = 0). To find the surface wave solutions, we first consider the partial waves in which the displacements and potential, denoted by u’ and V’, take the form

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Chapter 2 Modeling and Simulation of SAW Devices

0 3 0 1

'= ' exp(j xτ ) exp[ (j t K xω − )],

u u (2.14)

0 3 0 1

' ' exp( )exp[ ( )]

V =V j xτ j t K xω − (2.15)

where K0 is the wavenumber of the surface wave, assumed to be real. These expressions are substituted to equation (2.10) and (2.11). Then these equations are solved for τ numerically, giving eight solutions for the same. Accordingly for every solution of τ, relative solutions of u^’and V^’ are obtained. There are four values of τ in general denoted by τ1, τ2

τ3 ,and τ4 . The partial waves are therefore,

' '

0

exp(

3

)exp[ (

0 1

)],

i

=

i

j x τ

i

j t K x ω −

u u

(2.16)

' '

0

exp(

3

)exp[ (

0 1

)], 1,2,3,4

i i i

V V = j x τ j t K x ω − i =

_(2.17)

In the half-space it is assumed that the solutions is a linear sum of these partial waves, so that

4 '

1 i i

i

A

=

=

∑

u u (2.18)

4 '

1 i i

i

V AV

=

=

∑

(2.19)

The coefficients Ai are to be such that the solution satisfies the electrical and mechanical boundary conditions. These conditions give a determinant which must be zero for a valid solution for a right choice of K0. The velocity than can be determined by ω/K0 and displacement and potential are given by equation (2.18) and (2.19).

2.1.4 Acoustic waves in isotropic media

In an isotropic material the stiffness tensor cijkl is given by [3]

cijkl = λδijδkl + μ(δikδjl+δilδjk) (2.20) Where, the Kronecker delta is defined by δij = 1 for i=j and δij = 0 for i≠j. The constants λ and μ are known as Lamé constants and in practice these are always positive. μ is also called rigidity. Substituting cijkl into (2.2) and the stress is written in form of

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Chapter 2 Modeling and Simulation of SAW Devices

Tij = λδijΔ+2μSij (2.21)

ii i

i i i

S u

x

∆ = = ∂

∑ ∑

∂ (2.22)

The equation of motion (2.7) rewritten as

2 2

2^j ( ) j

j

u u

t x

ρ∂ = +λ µ ∂∆+ ∇µ

∂

∂ (2.23)

Where ∇ = ∂

∑

∂ ²

2

i xi2

Considering an infinite isotropic medium supporting plane waves, with frequency ω and the displacement u can be given by

u = u0 exp [j(ωt – k.x )] (2.24) The wave vector k = (k1, k2, k3), gives the direction of propagation. Wave fronts are solutions of k.x = constant, and they are perpendicular to k. The phase velocity of the wave is V=ω/|k|. From eq (2.24), ∂u/∂xj = –jkju, and substituting into eq (2.23) and we obtain

ω²ρuj = (λ+µ) (k.u) kj+µ|k|²uj, j = 1,2,3, (2.25) where |k|² = 𝑘𝑘₁²+𝑘𝑘₂²+𝑘𝑘₃² , substituting for u^j in eq (2.25), and results in vector form, gives

ω²ρu0 = (λ+µ) (k.u0) k+µ|k|² u0 (2.26) Here, there are two terms parallel to u0 and the other term is parallel to k, with the latter including the scalar product k.u0.Therefore two cases are to be considered, firstly if u0 is perpendicular to k and the scalar product is zero, and the remaining terms in the equation are parallel. Secondly, if u0 is not perpendicular to k the scalar product is non-zero. Hence, for non-trivial solutions we must have u0 parallel to k. These two cases give shear wave and longitudinal wave solutions respectively.

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Chapter 2 Modeling and Simulation of SAW Devices Considering shear wave case, u0 is perpendicular to k gives shear waves and wave vector denoted by kt given by |kt|² = ω²ρ/µ and the phase velocity of shear waves is denoted by Vt, equal to ω/|kt|, and then transvers wave

𝜕𝜕_𝑡𝑡=�𝜇𝜇 𝜌𝜌� (2.27)

For longitudinal wave, u0 is parallel to k then k = ±u0 (|k|/|u0|), using this relation we find (k.u0)k = u0|k|², and substituting in eq (2.26) gives |kl|² = ω²ρ/(λ+2µ), and then longitudinal phase velocity is given by

𝜕𝜕_𝑖𝑖 =�(𝜆𝜆+2𝜇𝜇)

�𝜌𝜌 (2.28)