ABSTRACT

An infinitely long circular cylinder, consisting generally of a finite number of coaxial viscoelastic layers, surrounded by a deformable medium is considered. The dynamic stress—the deformed state of a piecewise-homogeneous cylindrical layer from a harmonic wave is investigated. The numerical results of stress, depending on the wavelength are obtained.

Keywords:

Viscoelasticity, Fluid, Frequency, Longitudinal and Transverse Wave, Shell, Plane Strain

1. Introduction

During seismic impacts, modern underground pipelines operate under conditions of not only static but also dynamic loads, which are accompanied by large damage and even failure of the whole system [1] - [6] . In the case of a sufficiently long cavity, the impact perpendicular to its longitudinal axis, the environment surrounding the cavity and the lining are in conditions of plane deformation, and the task of determining the stress state of the array and lining reduces to a flat problem of the dynamic theory of elasticity (and whether visco-elasticity) [7] [8] [9] [10] . In [11] , the problem of stress concentration in an infinite linearly elastic cavity near a circular cavity with the propagation of longitudinal harmonic waves was solved. The solution of the diffraction problem for a harmonic transverse wave was obtained in [12] . This paper investigates the interaction of cylindrical stress waves with a cylinder, which in the general case consists of a finite number of coaxial viscoelastic layers. Due to the fact that long-term seismic waves, as a rule, exceed the characteristic dimensions of the cross section designs (for example, diameter D), therefore, when solving diffraction problems,

it is necessary to consider long-wave effects ( $\frac{D}{\lambda }<1$ , $\lambda$ is the wavelength). With longer wavelengths ( $\frac{D}{\lambda }=0.04÷0.16$ ) maximum coefficients of dynamic concentrations turned out to be 5% - 10% more than with the corresponding biaxial static loaded ( $\lambda \to \infty$ ) [13] . At $\frac{D}{\lambda }>0.16$ dynamic stress concentrations are significantly lower than static. In [14] , it was shown that the difference in

dynamic stress concentrations in the case of a rigid inclusion and cavity can be attributed to the possibility of propagation of generalized Rayleigh-type waves on a concave free cylindrical surface of the cavity. A significant contribution to the calculation of flexible pipelines was made in [15] [16] , which investigated such important issues as accounting without a backing zone and determining the stability of underground pipelines. There are a large number of authors who have applications in other branches of technology who can be successfully applied to the calculation of underground pipelines. These works are devoted to the study of the stress distribution in plastic, weakened by a reinforced bore, operating under plane strain conditions. The most significant research in this area can be attributed to the work that solved the problem in stretching a plate in which a ring is embedded or soldered.

2. Problem Statement and Basic Relations

In this paper, the interaction of a cylindrical stress wave by parallel-layered elastic layers with a liquid is investigated. It is assumed that the linear source in Figure 1 is a continuous source of dilatation (or transverse) stress waves with an angular velocity ω and amplitude ${\phi }_0$ (or ${\psi }_0$ ), and the layered package is a thick-walled and thin-walled layers of the cylinder. In describing the movement of thin-walled elements, the equations of the theory of such shells are used, which are based on the Kirchhoff-Love hypotheses. For thin-walled layers, the original equations are the linear theory of elasticity. The numbering of the layers is the product in ascending order of their radii from k = 1 to k = N (Figure 1). The value characterizing the properties and the state of the elements correspond to the values $j=1,2,\cdots ,N$ , where K is the elastic layer enclosed between K-1 and K-m layers. The environmental parameters are denoted by the indices K = N (Figure 1). Under the assumption of a generalized plane-deformed state, the equation of motion in displacements has the form [1]

$\left({\tilde{\lambda }}_j+2{\tilde{\mu }}_j\right)grad\left(div\left(u_j\right)\right)-\tilde{\mu }rot\left(rot\left(u_j\right)\right)+b_j={\rho }_j\frac{{\partial }^2{u}_j}{\partial t^2}.$ (1)

where ${\lambda }_j$ and ${\mu }_j$ ( $j=1,2,\cdots ,N$ , $j=N$ —relate to the environment, $j=1,2,\cdots ,N-1$ —to layer) operator moduli of elasticity [13]

$\begin{array}{l}{\tilde{\lambda }}_{j}f\left(t\right)={\lambda }_{0j}\left[f\left(t\right)-{\displaystyle \underset{-\infty }{\overset{t}{\int }}{R}_{\lambda }^{\left(i\right)}\left(t-\tau \right)f\left(\tau \right)\text{d}\tau }\right]\\ {\tilde{\mu }}_{j}f\left(t\right)={\mu }_{0j}\left[f\left(t\right)-{\displaystyle \underset{-\infty }{\overset{t}{\int }}{R}_{\mu }^{\left(i\right)}\left(t-\tau \right)f\left(\tau \right)\text{d}\tau }\right];\end{array}$

$b_j$ —Volume force density vector ( $b_j=0$ ); $f\left(t\right)$ —some function; ${\rho }_j$ —density of materials, $R_{\mu }^{\left(i\right)}\left(t-\tau \right)$ and $R_{\lambda }^{\left(i\right)}\left(t-\tau \right)$ —relaxation core, ${\lambda }_{oj},{\mu }_{oj}$ —instantaneous elastic moduli of viscoelastic material, $u_j\left(u_{rj},u_{\theta j}\right)$ —displacement vector which depends on $r,\theta ,t$ . At pressures up to 100 MPa, the movement of the fluid in is satisfactorily described by the wave velocity of the particle particles [14]

$\Delta {\phi }_{\text{0}}=\frac{1}{{С}_{о}^2}\frac{{\partial }^2{\phi }_{\text{0}}}{\partial t^2}$ . (2)

where $\Delta =\frac{{\text{d}}^2}{\text{d}{r}^2}+\frac{\text{d}}{r\text{d}r}+\frac{{\text{d}}^2}{r^2\text{d}{\theta }^2}$ ; С_о—acoustic velocity of sound in a fluid. Potential ${\phi }_0$ and the fluid velocity vector are dependent $V=grad{\phi }_0$ . Fluid pressure $r=R_0$ is determined using the linearized Cauchy-Lagrange integral $Р=-{\rho }_{о}{С}_0\frac{\partial {\phi }_0}{\partial t}$ —fluid pressure on the wall of the cylindrical layer and ρ_о-fluid density Under the condition of continuous flow of fluid, the normal component of the velocity of the fluid and the layer on the surface of their contact r = R₀ must be equal

${\frac{\partial {\phi }_0}{\partial r}|}_{r=R_0}={\frac{\partial u_{r1}}{\partial t}|}_{r=R_0}$ . (3)

where $u_{r1}$ —moving the layer along the normal. On the contact of two bodies r = R_j equal displacements and stresses (fixed contact condition)

$u_{rj}=u_{r\left(j+1\right)};{\sigma }_{rrj}={\sigma }_{rr\left(j+1\right)};u_{\theta j}=u_{\theta \left(j+1\right)};{\sigma }_{r\theta j}={\sigma }_{r\theta \left(j+1\right)}$ (4)

Note that in the case of sliding ground contact over the pipe surface, the last Equation in (4) takes the form [16] :

${\sigma }_{r\theta j}=0$ .

where ${\sigma }_{nn}^{\left(j\right)}$ and ${\sigma }_{n{s}_1}^{\left(j\right)}$ —radial and tangential stresses in the j-th viscoelastic body; $u_n^{\left(j\right)}$ and $u_{s_1}^{\left(j\right)}$ —radial and tangential displacement of the j-th body. The solution of the wave Equation (1) in the displacement potentials satisfies at infinity $r\to \infty$ the Summerfield radiation condition [14] :

$\underset{r\to \infty }{\lim }{\phi }_N=0,\underset{r\to \infty }{\lim }{\left(\sqrt{r}\right)}^{к}\left(\frac{\partial {\phi }_N}{\partial r}+i{\alpha }_N{\phi }_N\right)=0$ , (5)

$\underset{r\to \infty }{\lim }{\psi }_N=0,\underset{r\to \infty }{\lim }{\left(\sqrt{r}\right)}^{к}\left(\frac{\partial {\psi }_N}{\partial r}+i{\beta }_N{\psi }_N\right)=0$ .

As $r\to \infty$ , natural oscillations by the first and third conditions (5) are not fulfilled, therefore, the shortened Sommerfeld conditions at infinity are set, which in detail in [14] was considered.

If in place the equation of motion (1) is used cylindrical shells, then the equation of motion of shells in a flat formulation is:

$\begin{array}{l}\frac{{\partial }^{2}u}{\partial {\theta }^2}+\frac{\partial W}{\partial \theta }=-\frac{R^2}{B}{X}_1\\ \frac{\partial u}{\partial \theta }+b^2\left(\frac{{\partial }^{4}W}{\partial {\theta }^4}+2\frac{{\partial }^{2}V}{\partial {\theta }^2}+W\right)+W=\frac{R^2}{B}{X}_2\end{array}$ (6)

where u and W- longitudinal and transverse displacements respectively:

$X_1=-{{\sigma }_{r\theta }|}_{r=R_0+h_c/2}-{\rho }_c{h}_c\frac{{\partial }^{2}u}{\partial t^2};X_2=-{{\sigma }_{rr}|}_{r=R_0+h_c/2}-{\rho }_c{h}_c\frac{{\partial }^{2}W}{\partial t^2};$

$b^2=\frac{h_c^2}{12R_0^2},B=\frac{E_c{h}_c}{1-v_c^2}.$

$b^2=\frac{h_c^2}{12R_0^2},B=\frac{E_c{h}_c}{1-v_c^2}.$

Shell radius $R_0$ , ${\rho }_c$ —shell density, ${\nu }_c$ —shells Puasson ratio, $h_c$ —shell thickness, $E_c$ —shell elastic modulus, ${\sigma }_{rr}$ and ${\sigma }_{r\theta }$ —normal and tangent components of the reaction from the environment.

The contact between the shell and the environment can be hard or sliding:

${u|}_{r=R_0+h_c/2}={u_{\theta }|}_{r=R_0+h_c/2},{W|}_{r=R_0+h_c/2}={u_r|}_{r=R_0+h_c/2}$ (7)

Consider a longitudinal wave generated by a longitudinal source of expansion waves located at $\overline{О}$ (Figure 1). The displacement potentials of the incident expansion wave can be represented as [15]

${\phi }_N^{Р}={\phi }_{NO}i\text{π}{H}_0^{\left(1\right)}\left({\alpha }_N\bar{r}\right){\text{e}}^{-i\omega t}$ . (8)

where $H_0^{\left(1\right)}$ —is the divergent functions of Henkel (the first kind of zero order); ${\phi }_{NO}$ —expansion wave amplitude; ${\alpha }_N$ —compression wave number; ${\alpha }_N^2={\omega }^2/c_{\alpha }^2$ , $c_{\alpha }^2=\left({\lambda }_N+2{\mu }_N\right)/{\rho }_N$ , w—circular frequency. In the absence of an incident wave (6), the natural oscillations of a reinforced bore located in a viscoelastic medium are considered.

3. Solution Methods

The problem is solved in displacement potentials, for this we present the displacement vector in the form:

$u_j=grad\left({\phi }_j\right)+rot\left({\psi }_j\right),\left(j=1,2,\cdots ,N\right)$

where ${\phi }_j$ —longitudinal wave potential; ${\psi }_j\left({\psi }_{rj},{\psi }_{\theta j}\right)$ —vector potential of transverse waves.

The basic equations of the theory of visco elasticity (1) for this problem of plane strain are reduced to the following equation

$\begin{array}{l}\left({\lambda }_{oj}+2{\mu }_{oj}\right){\nabla }^2{\phi }_j-{\lambda }_{oj}{\displaystyle \underset{-\infty }{\overset{t}{\int }}{R}_{\lambda }^{\left(j\right)}\left(t-\tau \right){\nabla }^2{\phi }_j\text{d}\tau }\\ -2{\mu }_{oj}{\displaystyle \underset{-\infty }{\overset{t}{\int }}{R}_{\mu }^{\left(j\right)}\left(t-\tau \right){\nabla }^2{\phi }_j\text{d}\tau }={\rho }_j\frac{{\partial }^2{\phi }_j}{\partial t^2}\end{array}$

${\mu }_{oj}{\nabla }^2{\psi }_j-{\mu }_{oj}{\displaystyle \underset{-\infty }{\overset{t}{\int }}{R}_{\mu }^{\left(j\right)}\left(t-\tau \right){\nabla }^2{\psi }_j\text{d}\tau }={\rho }_j\frac{{\partial }^2{\psi }_j}{\partial t^2}$ (9)

where ${\nabla }^2=\frac{{\partial }^2}{\partial r^2}+\frac{1}{r}\frac{\partial }{\partial r}+\frac{1}{r^2}+\frac{{\partial }^2}{\partial {\theta }^2}$ —differential operators in cylindrical coordinates and $v_j$ — Poisson’s ratio.

At infinity, $r\to \infty$ potentials of longitudinal and transverse waves with $j=N$ satisfy the Summerfield radiation condition (5).

The solution of Equation (9) can be sought in the form:

${\phi }_j\left(r,\theta ,t\right)={\displaystyle \underset{k=1}{\overset{\infty }{\sum }}{q}_{kj}^{\left(\phi \right)}\left(r,\theta \right){\text{e}}^{-i\omega t};}{\psi }_j\left(r,\theta ,t\right)={\displaystyle \underset{k=1}{\overset{\infty }{\sum }}{q}_{kj}^{\left(\psi \right)}\left(r,\theta \right){\text{e}}^{-i\omega t}.}$ (10)

where $q_{kj}^{\left(\phi \right)}\left(r,\theta \right)$ and $q_{kj}^{\left(\psi \right)}\left(r,\theta \right)$ —complex function which is solving the following Equations (10)

${\nabla }^2{q}_{kj}^{\left(\phi \right)}\left(r,\theta \right)+{\alpha }_j^2{q}_{kj}^{\left(\phi \right)}=0,{\nabla }^2{q}_{kj}^{\left(\psi \right)}\left(r,\theta \right)+{\beta }_j^2{q}_{kj}^{\left(\psi \right)}=0,$

${\nabla }^2{q}_{k0}^{\left(\phi \right)}\left(r,\theta \right)+{\alpha }_0^2{q}_{k0}^{\left(\phi \right)}=0,j=1,2,\cdots ,N$ (11)

where ${\alpha }_j^2=\frac{\rho {\omega }^2}{{\lambda }_{oj}\left(1-{\bar{\lambda }}_{oj}\right)+2{\mu }_{oj}\left(1-{\bar{\mu }}_{oj}\right)}$ , ${\beta }_j^2=\frac{\rho {\omega }^2}{{\mu }_{oj}\left(1-{\bar{\mu }}_{oj}\right)}$ , ${\alpha }_0^2=\frac{{\omega }^2}{{С}_0^2}$

${\bar{\lambda }}_{oj}=a_{\lambda j}\left(\omega \right)+i{b}_{\lambda j}\left(\omega \right)$ , ${\bar{\mu }}_{oj}=a_{\lambda \mu }\left(\omega \right)+i{b}_{\mu j}\left(\omega \right)$ ,

$a_{\lambda j}\left(\omega \right)={\displaystyle \underset{0}{\overset{\infty }{\int }}{R}_{\lambda j}\left(\tau \right)\sin \left(\omega \tau \right)\text{d}\tau }$ , $b_{\lambda j}\left(\omega \right)={\displaystyle \underset{0}{\overset{\infty }{\int }}{R}_{\mu j}\left(\tau \right)\cos \left(\omega \tau \right)\text{d}\tau }$ .

The solution of Equation (9) with regard to (11) is expressed in terms of Henkel functions of the 1st and 2-nd kind of the nth order:

${\varphi }_j={\displaystyle \underset{n=0}{\overset{\infty }{\sum }}\left[A_{nj}{H}_n^{\left(1\right)}\left({\alpha }_{j}r\right)+{A^{\prime }}_{nj}{H}_n^{\left(2\right)}\left({\alpha }_{j}r\right)\right]\cos \left(n\theta \right){\text{e}}^{-i\omega t}}$

${\psi }_j={\displaystyle \underset{n=0}{\overset{\infty }{\sum }}\left[B_{nj}{H}_n^{\left(1\right)}\left({\beta }_{j}r\right)+{B^{\prime }}_{nj}{H}_n^{\left(2\right)}\left({\beta }_{j}r\right)\right]}\sin \left(n\theta \right){\text{e}}^{-i\omega t};j=1,2,\cdots ,N-1$

${\varphi }_N={\displaystyle \underset{n=0}{\overset{\infty }{\sum }}\left[C_{nN}{H}_n^{\left(1\right)}\left({\alpha }_{N}r\right)+D_{nN}{H}_n^{\left(2\right)}\left({\alpha }_{N}r\right)\right]}\cos \left(n\theta \right){\text{e}}^{-i\omega t}$ (12)

${\psi }_N={\displaystyle \underset{n=0}{\overset{\infty }{\sum }}\left[M_{nN}{H}_n^{\left(1\right)}\left({\beta }_{N}r\right)+L_{nN}{H}_n^{\left(2\right)}\left({\beta }_{N}r\right)\right]}\sin \left(n\theta \right){\text{e}}^{-i\omega t}$

${\varphi }_{\text{0}}={\displaystyle \underset{n=0}{\overset{\infty }{\sum }}\left[K_{n0}{J}_n\left({\alpha }_{0}r\right)+{K^{\prime }}_{n0}{N}_n\left({\alpha }_{0}r\right)\right]}\cos \left(n\theta \right){\text{e}}^{-i\omega t}$

where $A_{nj},{A^{\prime }}_{nj},B_{nj},{B^{\prime }}_{nj},C_{nj},D_{nj},L_{nN},M_{nN},K_{nN}$ and ${K^{\prime }}_{nN}$ —decomposition coefficients, which are determined by the corresponding boundary conditions; $H_n^{\left(1\right)}\left({\alpha }_{j}r\right)$ and $H_n^{\left(2\right)}\left({\alpha }_{j}r\right)$ —respectively, the Henkel function of the 1st and 2nd kind of the nth order $H_n^{\left(1\right),\left(2\right)}\left(\alpha r\right)=J_n\left(\alpha r\right)\pm i{N}_n\left(\alpha r\right)$ .

Solution (12) with $j=N$ satisfies at infinity $r\to \infty$ the Somerfield radiation condition (5) and is represented as:

$\begin{array}{l}{\varphi }_N={\displaystyle \underset{т=0}{\overset{\infty }{\sum }}{C}_{nN}{Н}_{п}^{\left(1\right)}\left({\alpha }_{N}r\right)\cos \left(m\theta \right){\text{e}}^{-i\omega t}};\\ {\psi }_N={\displaystyle \underset{т=0}{\overset{\infty }{\sum }}{M}_{nN}{H}_{п}^{\left(1\right)}\left({\beta }_{N}r\right)\sin \left(n\theta \right){\text{e}}^{-i\omega t}}.\end{array}$

Solving problem (2) with $r\to 0$ satisfies the condition of limiting the force factors [10] and it follows that ${K^{\prime }}_n=0$

${\varphi }_{\text{0}}={\displaystyle \underset{n=0}{\overset{\infty }{\sum }}{K}_{n0}{J}_n\left({\alpha }_{0}r\right)\cos \left(n\theta \right){\text{e}}^{-i\omega t}}$

The full potential can be determined by imposing the potentials of the incident and reflected waves. Thus, the displacement potentials will be

${\varphi }_N={\phi }_N^{\left(p\right)}+{\phi }_N,{\Psi }_N={\psi }_N,{\varphi }_j={\phi }_j,{\Psi }_j={\psi }_j,{\varphi }_0={\phi }_{\text{0}}$ (13)

To determine the stress-strain state, it is first necessary to express the incident wave through wave functions (13). Using the geometric construction in Figure 1 and moving from the coordinates $\bar{r},\bar{\theta }$ to coordinates $r,\theta$ in the area of $r\le r_N$

${\varphi }_N^{\left(p\right)}={\varphi }_{0}i\text{π}{\displaystyle \underset{n=1}{\overset{\infty }{\sum }}\left[{\left(-1\right)}^n{Е}_n{J}_n\left({\alpha }_{N}r\right)H_n^{\left(1\right)}\left({\alpha }_{N}z\right)\right]\cos \left(n\theta \right){\text{e}}^{-i\omega t}}$ .

where $E_n=\{\begin{array}{l}1,n=0\\ 2,n\ge 1\end{array}$ , $J_n$ —cylindrical Bessel function of the first kind.

It follows that voltages and offsets can easily be expressed in terms of displacement potentials [15]

$u_{rj}=\frac{\partial {\phi }_j}{\partial r}+\frac{1}{r}\frac{\partial {\psi }_j}{\partial \theta };u_{\theta j}=\frac{1}{r}\frac{\partial {\phi }_j}{\partial \theta }-\frac{\partial {\psi }_j}{\partial r},$ (14)

${\sigma }_{rrj}=\bar{\lambda }{\nabla }^2{\varphi }_j+2{\bar{\mu }}_j[\frac{{\partial }^2{\varphi }_j}{\partial r^2}+\frac{\partial }{\partial r}\left(\frac{1}{r}\frac{\partial {\psi }_j}{\partial \theta }\right)];$

$\begin{array}{l}{\sigma }_{\theta \theta j}=\bar{\lambda }{\nabla }^2{\varphi }_j+2{\bar{\mu }}_j[\frac{1}{r}\left(\frac{\partial {\varphi }_j}{\partial r}+\frac{1}{r}\frac{{\partial }^2{\varphi }_j}{\partial {\theta }^2}\right)+\frac{1}{r}\left(\frac{1}{r}\frac{\partial {\psi }_j}{\partial \theta }-\frac{{\partial }^2{\psi }_j}{\partial r\partial \theta }\right)];\\ {\sigma }_{r\theta j}=\bar{\mu }\left\{2\left[\frac{1}{r}\frac{{\partial }^2{\varphi }_j}{\partial \theta \partial r}-\frac{1}{r^2}\frac{\partial {\varphi }_j}{\partial \theta }\right]+\left[\frac{1}{r^2}\frac{{\partial }^2{\psi }_j}{\partial {\theta }^2}-r\frac{\partial }{\partial r}\left(\frac{1}{r}\frac{\partial {\psi }_j}{\partial r}\right)\right]\right\}\end{array}$

Displacement and stresses for the case of a compression wave falling on a layer $\psi$ it turns out.

Substituting (13) into (14) after considering (11), we obtain the following expression for displacement and stress:

$\begin{array}{c}{u}_{rN}=r^{-1}{\displaystyle \underset{n=0}{\overset{\infty }{\sum }}[{\varphi }_0{E}_n{i}^n{E}_{51}^{\left(1N\right)}\left({\alpha }_{N}r\right)+C_{nN}{E}_{51}^{\left(3N\right)}\left({\alpha }_{N}r\right)}\\ +M_{nN}{E}_{52}^{\left(3N\right)}\left({\beta }_{N}r\right)]\cos \left(n\theta \right){\text{e}}^{i\omega t}\end{array}$

$\begin{array}{c}{u}_{\theta N}=r^{-1}{\displaystyle \underset{n=0}{\overset{\infty }{\sum }}[{\varphi }_0{E}_n{i}^n{E}_{61}^{\left(1N\right)}\left({\alpha }_{N}r\right)+C_{nN}{E}_{61}^{\left(3N\right)}\left({\alpha }_{N}r\right)}\\ +M_{nN}{E}_{62}^{\left(3N\right)}\left({\beta }_{N}r\right)]\cos \left(n\theta \right){\text{e}}^{-i\omega t}\end{array}$

$\begin{array}{c}{u}_{rj}=r^{-1}{\displaystyle \underset{n=0}{\overset{\infty }{\sum }}[A_{nj}{E}_{51}^{\left(3j\right)}\left({\alpha }_{j}r\right)+A_{nj}^1{E}_{61}^{\left(4j\right)}\left({\alpha }_{j}r\right)}\\ +B_{nj}{E}_{52}^{\left(3j\right)}\left({\beta }_{j}r\right)+B_n^1{E}_{52}^{\left(4j\right)}]\cos \left(n\theta \right){\text{e}}^{-i\omega t}\end{array}$

$\begin{array}{c}{u}_{\theta j}=r^{-1}{\displaystyle \underset{n=0}{\overset{\infty }{\sum }}[A_{nj}{E}_{61}^{\left(3j\right)}\left({\alpha }_{j}r\right)+A_{nj}^1{E}_{61}^{\left(4j\right)}\left({\alpha }_{j}r\right)+B_{nj}{E}_{62}^{\left(3j\right)}\left({\beta }_{j}r\right)}\\ +B_n^1{}_j{E}_{62}^{\left(4j\right)}\left({\beta }_{j}r\right)]\sin \left(n\theta \right){\text{e}}^{-i\omega t}\end{array}$ (15)

$\begin{array}{c}{\sigma }_{rrN}=2{\mu }_{0N}\left(1-M_{kN}\right)r^{-2}{\displaystyle \underset{n=0}{\overset{\infty }{\sum }}{\varphi }_0{E}_n{i}^n{E}_{11}^{\left(1N\right)}\left({\alpha }_{N}r\right)+C_{nN}{E}_{11}^{\left(3N\right)}\left({\alpha }_{N}r\right)}\\ +M_{nN}{E}_{12}^{\left(3N\right)}\left({\beta }_{N}r\right)]\cos \left(n\theta \right){\text{e}}^{-i\omega t}\end{array}$

$\begin{array}{c}{\sigma }_{\theta \theta N}=2{\mu }_{0N}\left(1-M_{kN}\right)r^{-2}{\displaystyle \underset{n=0}{\overset{\infty }{\sum }}[{\varphi }_0{E}_n{i}^n{E}_{21}^{\left(1N\right)}\left({\alpha }_{N}r\right)+C_{nN}{E}_{21}^{\left(3N\right)}\left({\alpha }_{N}r\right)}\\ +M_{nN}{E}_{22}^{\left(3N\right)}\left({\beta }_{N}r\right)]\cos \left(n\theta \right){\text{e}}^{-i\omega t}\end{array}$

$\begin{array}{c}{\sigma }_{r\theta N}=2{\mu }_{0N}\left(1-M_{kN}\right)r^{-2}{\displaystyle \underset{n=0}{\overset{\infty }{\sum }}[{\varphi }_0{E}_n{i}^n{E}_{41}^{\left(1N\right)}\left({\alpha }_{N}r\right)+C_{nN}{E}_{41}^{\left(3N\right)}\left({\alpha }_{N}r\right)}\\ +M_{nN}{E}_{42}^{\left(3N\right)}\left({\beta }_{N}r\right)]\sin \left(n\theta \right){\text{e}}^{-i\omega t}\end{array}$

$\begin{array}{c}{\sigma }_{rrj}=2{\mu }_{0j}\left(1-M_{kj}\right)r^{-2}{\displaystyle \underset{n=0}{\overset{\infty }{\sum }}[A_{nj}{E}_{11}^{\left(3j\right)}\left({\alpha }_{j}r\right)+A_n^1{}_j{E}_{11}^{\left(4j\right)}\left({\alpha }_{j}r\right)}\\ +B_{nj}{E}_{12}^{\left(3j\right)}\left({\beta }_{j}r\right)+B_{nj}^1{E}_{11}^{\left(4j\right)}\left({\beta }_{j}r\right)]\cos \left(n\theta \right){\text{e}}^{-i\omega t}\end{array}$

$\begin{array}{c}{\sigma }_{\theta \theta j}=2{\mu }_{0j}\left(1-M_{kj}\right)r^{-2}{\displaystyle \underset{n=0}{\overset{\infty }{\sum }}[A_{nj}{E}_{21}^{\left(3j\right)}\left({\alpha }_{j}r\right)+A_{nj}^1{E}_{21}^{\left(4j\right)}\left({\alpha }_{j}r\right)}\\ +B_{nj}{E}_{22}^{\left(3j\right)}\left({\beta }_{j}r\right)+B_{nj}^1{E}_{22}^{\left(4j\right)}\left({\beta }_{j}r\right)]\cos \left(n\theta \right){\text{e}}^{-i\omega t}\end{array}$

$\begin{array}{c}{\sigma }_{r\theta j}=2{\mu }_{0j}\left(1-M_{kj}\right)r^{-2}{\displaystyle \underset{n=0}{\overset{\infty }{\sum }}[A_{nj}{E}_{51}^{\left(3j\right)}\left({\alpha }_{j}r\right)+A_{nj}^1{E}_{61}^{\left(4j\right)}\left({\alpha }_{j}r\right)}\\ +B_{nj}{E}_{52}^{\left(3j\right)}\left({\beta }_{j}r\right)+B_{nj}^1{E}_{52}^{\left(4j\right)}\left({\beta }_{j}r\right)]\sin \left(n\theta \right){\text{e}}^{-i\omega t}\end{array}$

where

$E_{11}^{\left(kj\right)}=\left(n^2+n-\frac{{\beta }_j^2{r}^2}{2}\right)Y_n^{\left(kj\right)}\left({\alpha }_{j}r\right)-{\alpha }_{j}r{Y}_{n-1}^{\left(kj\right)}\left({\alpha }_{j}r\right)$

$E_{12}^{\left(kj\right)}=n\left[\left(n+1\right)Y_n^{\left(kj\right)}\left({\beta }_{j}r\right)+{\beta }_{j}r{Y}_{n-1}^{\left(kj\right)}\left({\beta }_{j}r\right)\right]$

$E_{21}^{\left(kj\right)}=-\left(n^2+n+\frac{{\beta }_j^2{r}^2}{2}-{\alpha }_j^2{r}^2\right)Y_n^{\left(kj\right)}\left({\alpha }_{j}r\right)+{\alpha }_{j}r{Y}_{n-1}^{\left(kj\right)}\left({\alpha }_{j}r\right)$

$E_{22}^{\left(kj\right)}=n\left[{\beta }_{j}r{Y}_n^{\left(kj\right)}\left({\beta }_{j}r\right)-\left(n+1\right)Y_n^{\left(kj\right)}\left({\beta }_{j}r\right)\right]$

$E_{31}^{\left(kj\right)}=\left({\alpha }_j^2{r}^2-\frac{{\beta }_j^2{r}^2}{2}\right)Y_n^{\left(kj\right)}\left({\alpha }_{j}r\right)$

$E_{41}^{\left(kj\right)}=n\left[\left(n+1\right)Y_n^{\left(kj\right)}\left({\alpha }_{j}r\right)-{\alpha }_{j}r{Y}_{n-1}^{\left(kj\right)}\left({\alpha }_{j}r\right)\right]$

$E_{42}^{\left(kj\right)}=-\left(n^2+n-\frac{{\beta }_j^2{r}^2}{2}\right)Y_n^{\left(kj\right)}\left({\beta }_{j}r\right)+{\beta }_{j}r{H}_{n-1}^{\left(kj\right)}\left({\beta }_{j}r\right)$

$E_{51}^{\left(kj\right)}=\left[{\alpha }_{j}r{Y}_{n-1}^{\left(kj\right)}\left({\alpha }_{j}r\right)-n{Y}_n^{\left(kj\right)}\left({\alpha }_{j}r\right)\right]$

$E_{52}^{\left(kj\right)}=-n{Y}_n^{\left(kj\right)}\left({\beta }_{j}r\right)$

$E_{61}^{\left(kj\right)}=-n{Y}_n^{\left(kj\right)}\left({\alpha }_{j}r\right)$

$E_{62}^{\left(kj\right)}=\left[n{Y}_n^{\left(kj\right)}\left({\beta }_{j}r\right)-{\beta }_{j}r{Y}_{n-1}^{\left(kj\right)}\left({\beta }_{j}r\right)\right],k=1,2,3,4$

where $Y_n^{\left(1j\right)}=J_n,Y_n^{\left(2j\right)}=N_n,Y_n^{\left(3j\right)}=H_n^{\left(1\right)},Y_n^{\left(4j\right)}=H_n^{\left(2\right)}$ .

The construction of a formal solution does not meet fundamental difficulties, but the study of such a solution requires a huge amount of computation. The tasks are reduced to solving inhomogeneous algebraic equations with complex coefficients

$\left[C\right]\left\{q\right\}=\left\{p\right\}$ . (16)

where {q} is a vector column containing arbitrary constants; {F}—vector column of external loads; [C] is a square matrix, the elements of which are expressed through the functions of Bessel and Henkel. Equation (13) is solved by the Gauss method with the selection of the main element. In the work of movement and stress is reduced in dimensionless types

$\begin{array}{l}{u}_{rj}^{*}=\frac{u_{rj}}{i\alpha {\phi }_A};u_{\theta j}^{*}=\frac{u_{\theta j}}{i\alpha {\phi }_A};{\sigma }_{rrj}^{*}=\frac{{\sigma }_{rrj}}{{\sigma }_0};\\ {\sigma }_{r\theta j}^{*}=\frac{{\sigma }_{r\theta j}}{{\sigma }_0};{\sigma }_0=-\bar{\mu }{\beta }^2{\phi }_A.\end{array}$

In the case when $E_1=E_2=\cdots =E_N$ , ${\rho }_1={\rho }_2=\cdots ={\rho }_N$ and ${\nu }_1={\nu }_2=\cdots ={\nu }_N$ , we get holes (r = R₀) that are in infinitely elastic medium ( $a_{\lambda N}\left(\omega \right)=0$ , $b_{\lambda N}\left(\omega \right)=0$ ). The boundary (r = R₀) is stress free, i.e. no liquid. In this case, the circumferential voltage on the surface of the cavity is reduced to the following:

${\sigma }_{\theta \theta }{}_N\left(R,\theta ,t\right)=\frac{-4}{\text{π}}{\beta }_N^2{\mu }_N{\varphi }_0\left(1-\frac{1}{k^2}\right){\displaystyle \underset{n=0}{\overset{\infty }{\sum }}{\left(-1\right)}^n}{\in }_n{H}_n^{\left(1\right)}\left({\alpha }_{N}Z\right){\bar{T}}_n{}_N\cos \left(n\theta \right){\text{e}}^{-i\omega t},$ (17)

where

$\begin{array}{l}{\overline{Т}}_{nN}={Т}_{nN}\left(R_N\right)={\left[{\alpha }_N{R}_N{H}_{n-1}\left({\alpha }_N{R}_N\right)-H_{n-1}\left({\alpha }_N{R}_N\right)+Q_n{}_N\left({\beta }_N{R}_N\right)\right]}^{-1}\\ Q_n{}_N\left({\beta }_N{R}_N\right)=\frac{\left(n^3-n+\frac{1}{2}{\beta }_N^2{R}_N^2\right){\beta }_N{R}_N{H}_{n-1}\left({\beta }_N{R}_N\right)-\left(n^2+n-\frac{1}{4}{\beta }_N^2{R}_N^2\right){\beta }_N^2{R}_N^2{H}_n^{\left(1\right)}\left({\beta }_N{R}_N\right)}{\left(n^2-1\right){\beta }_N{R}_N{H}_{n-1}^{\left(1\right)}\left({\beta }_N{R}_N\right)-\left(n^2-n+\frac{1}{2}{\beta }_N^2{R}_N^2\right)H_n^{\left(1\right)}\left({\beta }_N{R}_N\right)}\\ k_N^2=C_{pN}^2/C_{SN}^2.\end{array}$

Now consider some limiting cases.

With $r\to 0$ :

$\begin{array}{l}{H_0^{\left(1\right),\left(2\right)}\left(z\right)|}_{r\to 0}=\pm \frac{2i}{\text{π}}\ln z-i{\left(\frac{z}{2}\right)}^2\left(1-\frac{2}{\text{π}}\ln z\right)+0\left(z^4\ln z\right),\\ {H_{т}^{\left(1\right),\left(2\right)}\left(z\right)|}_{r\to 0}=\mp \frac{i}{\text{π}}{\left(\frac{z}{2}\right)}^2\left\{\left(n-1\right)!+n!z^2+0\left(z^4\right)\right\}.\end{array}$

And $r\to \infty$ : $H_{т}^{\left(1\right),\left(2\right)}\left(z\right)={\left(\frac{2}{\text{π}z}\right)}^{1/2}{\text{e}}^{\pm i\left(kz-\text{π}/4\right)}$ , the asymptotic formulas of Hankel of the 1st and 2nd kind were used [17] . If in expression (17) Z tends to infinity, then we can use the asymptotic expansions of the Henkel function for large values of the argument [17] (α- is a finite)

$\underset{Z\to \infty }{\lim }{{\sigma }_{\theta \theta }^{\ast }|}_{r=R_N}\approx \frac{4}{4}\left[1-\frac{1}{k_N^2}\right]{\displaystyle \underset{n=0}{\overset{\infty }{\sum }}{i}^{n+1}{E}_n{T}_{nN}\cos \left(n\theta \right){\text{e}}^{-i\omega t}}$ (18)

This expression completely coincides with the expressions obtained by [17] for a plane incident wave. If the wave number tends to zero, then the limiting process describes static solutions for long waves. This limiting process allows us to use approximating expressions for the Henkel functions for small values of the argument (Z is finite)

$\begin{array}{c}\underset{\alpha \to 0}{\lim }{{\sigma }_{\theta \theta }|}_{r=R_1}\approx 4\left[1+{\left(\frac{R_N}{Z}\right)}^2+\frac{4}{Z}\cos \theta \right]\\ \ast {\displaystyle \underset{m=2}{\overset{\infty }{\sum }}{\left(-1\right)}^{m-2}{\left(-\frac{R_N}{Z}\right)}^{m-2}\left(m-1\right)\cos \left(m\theta \right)}\end{array}$ (19)

This solution exactly coincides with the solution of the static problem obtained in [18] . If the cylindrical cavity contains an ideal fluid, then the ring stresses take the form

$\begin{array}{c}{\sigma }_{\theta \theta }^{\ast }=-\frac{4}{\text{π}}{\displaystyle \underset{n=0}{\overset{\infty }{\sum }}\frac{{\in }_n{i}^{n+1}}{{\Delta }_n}}\{[\left(\frac{1}{{\chi }_N^2}-1\right)\left(\frac{n{\beta }_N^2{R}^2}{2}\left(n-1\right)-n^2\left(n^2-1\right)-n\frac{{\beta }_N^4{R}^4}{4}\right)\\ +\frac{{\beta }_N^2{R}^3}{4}\eta \left(n\left(n+1\right)-\frac{1}{2}{\beta }_N^2{R}^2\right)]I_n\left({\alpha }_{0}R\right)H_n\left({\beta }_{N}R\right)\\ +\left[\left(\frac{1}{{\chi }_N^2}-1\right)\left(n^3-n^2-n\right)-\frac{1}{4}n{\beta }_N^2{R}^2\right]{\beta }_{N}R{I}_n\left({\alpha }_{0}R\right)H_{n-1}\left({\beta }_{N}R\right)\\ +\left(\frac{1}{{\chi }_N^2}-1\right)\left(n^3-n+\frac{1}{2}{\beta }_N^2{R}^2\right){\alpha }_{0}R{I}_{n-1}\left({\alpha }_{0}R\right)H_n\left({\beta }_{N}R\right)\\ +\left(\frac{1}{{\chi }_N^2}-1\right)\left(1-n^2\right){\alpha }_0{\beta }_N^2{R}^2{I}_{n-1}\left({\alpha }_{0}R\right)H_{n-1}\left({\beta }_{N}R\right)\}\cos \left(n\theta \right){\text{e}}^{-i\omega t}\end{array}$

where ${\chi }_N^2=\frac{2\left(1-v_N\right)}{1-2v_N}=\frac{{\beta }_N^2}{{\alpha }_N^2};\eta =\frac{{\rho }_1}{{\rho }_N}$ . At ${\alpha }_{1}R\to 0$ , then it turns out the solution of a static problem

${\sigma }_{rr}^{\ast }=\frac{{\lambda }_1}{{\lambda }_1+{\mu }_N};{\sigma }_{\theta \theta }^{\ast }=\frac{1}{1-v_N}\left[\left(1-\frac{2{\lambda }_1\left(1-v_N\right)}{{\lambda }_1+{\mu }_N}\right)-\left(2-4v_N\right)\cos \left(2\theta \right)\right].$

In the limiting processes in expressions (15) and (16) is described by physical results, is given in Table 1. The stress concentration coefficient ${\sigma }_{\theta \theta }^{\ast }{}_N$ is determined by the following formulas

${{\sigma }_{\theta \theta }^{\ast }|}_{r=R_N}=\frac{{{\sigma }_{\theta \theta }|}_{r=R_N}}{{\sigma }_{\theta \theta }^{\left(p\right)}}$ . (20)

where ${\sigma }_{\theta \theta }^{\left(p\right)}=i\pi {\phi }_0\mu {\alpha }^2\left[\pi \phi \mu H_2^{\left(1\right)}\left(\alpha \bar{r}\right)+\left(1-k^2\right)H_0^{\left(1\right)}\left(\alpha \bar{r}\right)\right]{\text{e}}^{-i\omega t}$ .

If we take the fluid into account, then using (2), (3) and (15) we can determine the corresponding stresses ${\sigma }_{rr}^{\ast }{}_N$ and ${\sigma }_{\theta \theta }^{\ast }{}_N$ . In the absence of an incident wave (6), the natural oscillations of an unsupported (or reinforced) hole located in the medium are considered.

At the boundary r = R, we set a voltage-free condition, i.e.

${{\sigma }_{rr}|}_{r=a}={{\sigma }_{r\theta }|}_{r=a}=0$ . (21)

Substituting (12) into (21), we obtain the frequency equation

$Z_{1n}{X}_{2n}+Z_{2n}{X}_{1n}=0$ . (22)

where

$X_{1n}={\Omega }_0{H}_{n+1}^{\left(1\right)}\left({\Omega }_o\right)+\left(a_{n2}^1-d_1{\Omega }_0^2\right)H_0^{\left(1\right)}\left({\Omega }_o\right);$

$X_{2n}=n\left[\left(n-1\right)H_n^{\left(1\right)}\left({\Omega }_1\right)-{\Omega }_1{H}_{n+1}^{\left(1\right)}\left({\Omega }_1\right)\right];$

$Z_{1n}=n\left[\left(1-n\right)H_n^{\left(1\right)}\left({\Omega }_o\right)-{\Omega }_o{H}_{n+1}^{\left(1\right)}\left({\Omega }_o\right)\right];$

$Z_{2n}=\left(a_{n2}^1-{\Omega }_1^2/2\right)H_n^{\left(1\right)}\left({\Omega }_1\right)+{\Omega }_1{H}_{n+1}^{\left(1\right)}\left({\Omega }_1\right);$

$d_1=\left(1-v_1\right)/\left(1-2v_1\right);\text{}{a}_{n2}=n^2;a_{n2}^1=n^2-n;{\Omega }_o={\Omega }_1{L}_1;$

$l_1=\left(1-2v_1\right)/\left(2\left(1-v_1\right)\right);{\Omega }_1=\omega a/C_{p1}$