Free vibration of piezoelectric semiconductor composite structure with fractional viscoelastic layer

doi:10.1007/s10483-025-3237-8

摘要/Abstract

Abstract:

In this study, the free vibration of a piezoelectric semiconductor (PS) composite structure composed of a PS layer, a fractional viscoelastic layer, and an elastic substrate with simply-supported boundary conditions is investigated. The fractional derivative Zener model is used to establish the constitutive relation of the viscoelastic layer. The first-order shear deformation theory and Hamilton's principle are used to derive the motion equations of the present problem. The frequency parameter is numerically resolved with the Newton-Raphson method through the eigenvalue equation. The effects of either geometric parameters, carrier density, and electric voltage applied on the surface of the composite structure or the fractional order of the Zener model on both the natural frequency and loss factor are discussed, and some interesting conclusions are drawn. This work will be helpful for designing and manufacturing PS materials and structures.

Key words: fractional viscoelastic material, free vibration, composite structure, piezoelectric semiconductor (PS)

中图分类号:

O327

. [J]. Applied Mathematics and Mechanics (English Edition), 2025, 46(4): 683-698.

Yansong LI, Wenjie FENG, Lei WEN. Free vibration of piezoelectric semiconductor composite structure with fractional viscoelastic layer[J]. Applied Mathematics and Mechanics (English Edition), 2025, 46(4): 683-698.

图/表 13

$ω (λ 1, λ 2)$	$n 0 = 10 16$ m $− 3$		$n 0 = 10 23$ m $− 3$
$ω (λ 1, λ 2)$	Guo et al.^[32]	Present	Guo et al.^[32]	Present
$ω$ (1, 1)	0.991 1	0.991 3	0.983 3	0.983 3
$ω$ (1, 2)	2.157 1	2.157 2	2.148 8	2.148 9
$ω$ (2, 2)	3.071 3	3.071 5	3.065 0	3.065 4

ISD-110	ZnO	ZnO
$a ∗ = 0.028 7$ MPa	$c 11 = c 22 = 209.7$ GPa	$e 15 = e 24 = − 0.48$ C/m $2$
$b ∗ = 1.035$ MPa	$c 12 = 121.1$ GPa	$ε 11 = ε 22 = 7.57 × 10 − 11$ F/m
$c ∗ = 0.011$	$c 13 = c 23 = 105.1$ GPa	$ε 33 = 9.03 × 10 − 11$ F/m
$α = 0.5$	$c 33 = 210.9$ GPa	$μ 11 = μ 22 = 0.02$ m $2$ / $(V ⋅ s)$
$ρ c = 965$ kg/m $3$	$c 44 = 42.47$ GPa	$μ 33 = 0.02$ m $2$ / $(V ⋅ s)$
	$c 55 = 42.47$ GPa	$d 11 = d 22 = 0.000 52$ m $2$ /s
	$c 66 = 44.3$ GPa	$d 33 = 0.000 52$ m $2$ /s
	$e 31 = e 32 = − 0.573$ C/m $2$	$ρ t = 5 700$ kg/m $3$
	$e 33 = 1.32$ C/m $2$

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