PSI - Issue 80
Akihide Saimoto et al. / Procedia Structural Integrity 80 (2026) 352–367 A. Saimoto et al. / Structural Integrity Procedia 00 (2023) 000–000
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s E i jk ℓ is a compliance under constant electric field, d ki j is a piezoelectric constant and ϵ σ ik is a permittivity under constant stress field. ε i j is a component of strain, σ i j is a component of stress, E i is a component of electric field and D i is a component of electric displacement, they are listed in the table of nomenclature with units. Based upon the assumption of infinitesimal deformation, the strain components are expressed by a 1st-order gradient of elastic displacements u i as ε i j = ( u i , j + u j , i ) / 2 and the component of electric field can be obtained as E i = − ϕ , i , where comma with subsequent index i denotes the partial di ff erentiation with respect to coordinate variable x i and ϕ is the electrostatic potential. In the following, y axis coincides with the poling direction and plane strain condition in z direction is assumed. The existence of electrostatic potential automatically satisfies the relation ∂ E x ∂ y − ∂ E y ∂ x = − ∂ 2 ϕ ∂ x ∂ y + ∂ 2 ϕ ∂ x ∂ y = 0 (2) This equation is one of the Maxwell’s equations in the electrostatics. As in a same manner, the equilibrium conditions with no body force σ i j , j = 0 is satisfied if Airy’s stress function F ( x , y ) exists. The stress components are expressed in terms of stress function as
∂ 2 F ∂ y 2
∂ 2 F ∂ x 2
∂ 2 F ∂ x ∂ y
(3)
, τ xy = −
σ x =
, σ y =
The extended constitutive equation in the matrix form can be expressed as follows. ε x ε y γ xy D x D y = s 11 s 12 0 0 d 12 s 12 s 22 0 0 d 22 0 0 s 66 d 61 0 0 0 d 61 ϵ 11 0 d 12 d 22 0 0 ϵ 22 ∂ 2 F /∂ y 2 ∂ 2 F /∂ x 2 − ∂ 2 F /∂ x ∂ y − ∂ϕ/∂ x − ∂ϕ/∂ y
(4)
In this expression, d 12 , d 22 and d 61 relates to lateral, longitudinal and shear piezoelectric strains, respectively. If these constants are all zero, the elastic and electric fields are not coupled and can be analyzed separately as an orthotropic elasticity and electrostatics. The Saint-Venant compatibility condition becomes ∂ 2 ε x ∂ y 2 + ∂ 2 ε y ∂ x 2 − ∂ 2 γ xy ∂ x ∂ y = s 11 ∂ 4 F ∂ y 4 + (2 s 12 + s 66 ) ∂ 4 F ∂ x 2 ∂ y 2 + s 22 ∂ 4 F ∂ x 4 − d 12 ∂ 3 ϕ ∂ y 3 + ( d 61 − d 22 ) ∂ 3 ϕ ∂ x 2 ∂ y = 0 (5) and the Gauss equation with no electric charge becomes ∂ D x ∂ x + ∂ D y ∂ y = d 12 ∂ 3 F ∂ y 3 − ( d 61 − d 22 ) ∂ 3 F ∂ x 2 ∂ y − ϵ 11 ∂ 2 ϕ ∂ x 2 − ϵ 22 ∂ 2 ϕ ∂ y 2 = 0 (6) Equations (5) and (6) must be satisfied simultaneously. In order to simplify the notation, the following operates are introduced.
∂ 4 ∂ y 4
∂ 4 ∂ x 2 ∂ y 2
∂ 4 ∂ x 4
∂ 3 ∂ y 3 −
∂ 3 ∂ x 2 ∂ y
∂ 2 ∂ x 2
∂ 2 ∂ y 2
L 4 = s 11
+ (2 s 12 + s 66 )
+ s 22
, L 3 = d 12
( d 61 − d 22 )
(7)
, L 2 = ϵ 11
+ ϵ 22
Then equations (5) and (6) can be rewritten as L 4 F = L 3 ϕ, L 3 F = L 2 ϕ
(8)
Second of Eq.(8) is automatically satisfied if following extended potential Ψ ( x , y ) exists F ( x , y ) = L 2 Ψ ( x , y ) , ϕ ( x , y ) = L 3 Ψ ( x , y ) Substituting Eqs.(9) into 1st of Eq.(8), a sixth order partial di ff erential equation for Ψ ( x , y ) is obtained. ( L 4 L 2 − L 3 L 3 ) Ψ ( x , y ) = 0
(9)
(10)
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