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Dynamic Stability of a Thin Film Bonded to a Compliant Substrate Subjected to a Step Load with Damping

  • Zhicheng Ou , Xiaohu Yao and Xiaoqing Zhang EMAIL logo
Published/Copyright: May 9, 2017

Abstract

The flexible electronic structure is based on the buckling of a thin film on a compliant substrate. This paper studies the dynamic stability of this structure subjected to a uniaxial step load with damping. The equation of motion is derived by Hamilton’s principle and Euler-Lagrange equation. Through the qualitative analysis of the phase portraits of the dynamic equation, as well as the quantitative analysis of the responses according to Budiansky-Roth criterion, the critical dynamic load is determined. It is the same as that in static buckling. Affected by damping, at the stage of pre-buckling, the amplitude of the film vibrates and attenuates, where the maximum response is the initial amplitude. The structural damping can be derived by the logarithm of the ratio between two adjacent peaks of the vibration. At the stage of post-buckling, the amplitude of the film vibrates and tends to a stable response, which is the amplitude in static buckling. The upper and lower bounds of the post-buckling response are asymmetric and solved by modified Krylov-Bogoliubov method.

Appendix

The approximate upper and lower bounds of eq. (40) when μ>1 can be solved analytically by using modified Krylov-Bogoliubov method [35, 36, 37].

When 0\ltf02<μ1, the solution of eq. (40) without the damping term is given by

(1)f(τ)=aκ1nd(mτϕ,κ)

where κ12=1κ2. Under the initial condition eq. (38), the coefficients in eq. (1) are

(2)ω2=12a2,κ2=2+2(1μ)a2a=2(μ1)f02,ϕ=0

The upper and lower bounds of solution (1) are

(3)flow=aκ1=2(μ1)a2,fup=a

The time derivative of eq. (1) is

(4)f(τ)=amκ1κ2cd(mτϕ,κ)sd(mτϕ,k)

The trial solution of eq. (40) with the damping term is assumed in terms of Jacobi elliptic functions as eq. (1), but the parameters are all time-dependent,

(5)f(τ)=a(τ)κ1(τ)nd(M(τ),κ(τ))

with

M(τ)=0τm(s)dsϕ(τ)

The initial conditions are given by

(6)f(0)=a(0)κ1(0)=a01κ02=f0κ(0)=κ0,a(0)=a0,ϕ(0)=0

The trial solution eq. (5) should satisfy three additional constraints. The first two constraints are obtained from equation (2) where the parameters are time-dependent,

(7)m(τ)2=12a(τ)2,κ(τ)2=2+2(1μ)a(τ)2

The third constraint is usual in Krylov-Bogoliubov method: the time derivative of the trial solution must have the same form as that of the generating solution. The third constraint can be rewritten from eq. (4),

(8)f(τ)=a(τ)m(τ)κ1(τ)κ(τ)2cd(M(τ),κ(τ))sd(M(τ),k(τ))

The time derivative of trial solution eq. (5) is not dependent on the time derivative of the parameters.

For convenience, the Italic symbol pq denotes corresponding Jacobi elliptic function, such as pq=pq(M(τ),μ(τ)). Then eq. (8) can be rewritten as

(9)f=amκ1κ2cdsd

Differentiating eq. (5) with respect to τ and using the constraint eq. (9) gives

(10)aκ12aκκcn(cn+Isd)aκ2κ12ϕcnsd=0

with

I=E(M,κ)κ12M

where E(M,κ) is elliptic integral of the second kind. Differentiating eq. (9) with respect to τ gives

(11)f′′=κ1ma+amcdsd+aκ1m(ϕm)(12cd2)nd+amκκ1Iκ1112cd2nd+2κ1cdsd

From eqs (5) and (7), we obtain

(12)f3+(1μ)f=12a3κ1nd[κ12(2nd21)1]

Differentiating the constraints eq. (7) gives

(13)mm=κκ2κ2

Submitting eqs (9), (11) and (12) into eq. (40), and using the constraints eqs (7), (10) and (13), we obtain

(14)κ12cn4κ212cn2κ12aacn222cn2κ2mm=2ξκ2κ12cn2sn2

The coefficients of a/a, m/m and ξ in eq. (14) are roughly periodic. So eq. (14) can be reduced to a simple form by using the averaging principle method [38],

(15)aaR1(κ)mm=R2(κ)ξ

with

(16)R1(κ)=Q1Q0,R2(κ)=Q2Q0

and

(17a)Q0=κ12cn4κ212cn2κ12
(17b)Q1=cn222cn2κ2
(17c)Q2=2κ2κ12cn2sn2

The averaging operator is given by

(18)Qi(M,κ)=14K(κ)04K(κ)Qi(M,κ)dM
K(κ) here and E(κ) in the following text are the complete elliptic integrals of the first and second kind. The coefficients in eq. (15) are integrated according to equations (16)–(18),
(19)R1(κ)=47κ2+3κ4K(κ)+22κ2E(κ)21κ22κ2E(κ)21κ2K(κ)R2(κ)=1
R1 can be expanded as a series form
(20)R1(κ)1+4κ2+35κ232+O(κ)

Submitting eqs (13), (19) and (20) into eq. (15) gives

(21)aa35κ4+32κ2+128κ32κ2κ2=ξ

Setting a(τ)=b(τ)eξτ, eq. (21) can be simplified as,

(22)bb=35κ4+32κ2+128κ32κ2κ2

Integrating both sides of eq. (22),

b0bdbb=κ0κ35κ4+32κ2+12832κ2κ2dκ

we have

(23)lnbb0=3564κ02κ2+8332ln2κ022κ2+2lnκκ0

Submitting eq. (7) and a=beξτ, eq. (23) can be rewritten as,

(24)ξτ=35(μ1)321a21a02+3532lna02a2+lna02+1μa2+1μ

From initial condition (6) and constraint (7), we have

(25)a02=2μ2f02

By submitting eq. (25) into eq. (24), the upper bound of eq. (40), fup=a, is satisfied,

(26)ξτ=35(μ1)3212μ2f021fup2+358ln2μ2f02fup2+lnμ1f02fup2+1μ

According to eq. (3), the lower bound of eq. (40), flow, is satisfied,

(27)ξτ=35(μ1)3212μ2f0212μ2flow2+358ln2μ2f022μ2flow2+lnμ1f02μ1flow2

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Received: 2015-7-10
Accepted: 2017-3-1
Published Online: 2017-5-9
Published in Print: 2017-5-24

© 2017 Walter de Gruyter GmbH, Berlin/Boston

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