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In this paper, the case of the interaction of a flat compression pulse with a layered cylindrical body in an infinite homogeneous and isotropic elastic medium is studied. The problem by the methods of integral Fourier transforms is solved. The inverse transform numerically by the Romberg method is calculated. With a time of toast and a decrease in momentum, the accuracy is not less than 2%. Taking into account the diffracted waves the results are obtained.

Various issues related to the interaction of bodies with a continuous medium (the creation of effective mathematical models is, theoretical and experimental methods for the study of non-stationary problems of dynamics) are described in monographs [

The problems of the no stationary dynamics of a homogeneous isotropic linearly elastic medium in cylindrical coordinates are given in the work of C. Chree [

Some issues related to the diffraction of no stationary waves on cavities and absolutely rigid obstacles are considered in the works of A.N. Guzz, V.D. Kubenko and M.A. Cherevko [

The influence of various factors on the behavior of a smooth infinitely long thin cylindrical shell during the diffraction of a plane shock wave on it (a plane problem) was studied by many authors [

The problem of the action of non-stationary waves on layered cylindrical bodies with radius R_{k} is considered. The motion vector of the medium is connected with the potentials φ N and ψ k by means of the formulas

u k = g r a d φ k + r o t ( ψ k ) ( k = 1 , 2 , ⋯ , N ) .

Suppose that the elastic medium is in plane strain conditions in the plane. In polar coordinates r , θ , the basic ratios of the plane problem are

u r k = ∂ φ k ∂ r + 1 r ∂ ψ k ∂ θ , u θ k = 1 r ∂ φ k ∂ θ − ∂ ψ k ∂ r

σ r r k = λ k c 1 k ∂ 2 φ k ∂ t 2 + 2 μ k ( ∂ 2 φ k ∂ r 2 + 1 r ∂ 2 ψ k ∂ r ∂ θ − 1 r 2 ∂ ψ k ∂ θ ) σ θ θ k = ρ k ∂ 2 φ k ∂ t 2 − 2 μ k ( ∂ 2 φ k ∂ r 2 + 1 r ∂ 2 ψ k ∂ r ∂ θ − 1 r 2 ∂ ψ k ∂ θ ) σ r θ k = ρ k ∂ 2 ψ k ∂ t 2 + 2 μ k ( 1 r ∂ 2 φ k ∂ r ∂ θ − 1 r 2 ∂ φ k ∂ θ − ∂ 2 ψ k ∂ r 2 )

here λ k and μ k the Lame elastic constants of the k-th layer; ρ k —density of the material of the k-th layer; σ r r k , σ θ θ k , σ r θ k —components of the stress tensor of the k-th layer.

Non-stationary stress waves σ x x ( i ) and σ x y ( i ) , whose front is parallel to the longitudinal axis of the cylinder, fall on a layered cylinder (

The basic equations of the theory of elasticity for this problem of plane strain in displacement potentials are reduced to the following:

∇ 2 φ j = 1 c p j 2 ∂ 2 φ j к ∂ t 2 ; ( j = 1 , 2 , ⋯ , N ) ∇ 2 ψ j = 1 c ρ j 2 ∂ 2 ψ j ∂ t 2 . (1)

where U_{j} and ψ j are the displacement potentials of the j-th layer, с_{рj} and с β j —are the phase velocities of the extension and shear waves of the j-th layer.

Suppose that time t is counted from the moment when the incident pulse touches the surface of the external (N − 1)-th cylinder at point r = r N , θ = 0 . Until that moment, peace remains. In accordance with the foregoing, the task of finding the field of diffracted waves and the stress-strain state caused by the incident pulse [

σ x x ( i ) = σ 0 H ( t ^ ) , σ x y ( i ) = σ 0 ν N 1 − ν N H ( t ^ ) , t ^ = t − ( x + r N ) / C P N (2)

σ 0 —the amplitude of the incident waves; H ( t ^ ) —the unit Heaviside function, reduces to solving differential Equations (1). Boundary conditions on the contact of two cylindrical surfaces should be equal to displacement and tension

r = а к : σ r r к = σ r r ( к + 1 ) ; σ r θ к = σ r θ ( к + 1 ) ; σ r z к = σ r z ( к + 1 ) ; u к = u к + 1 ; ϑ к = ϑ к + 1 ; w к = w к + 1 .

At infinity ( r → ∞ ), the perturbations must die out. If φ N and ψ N —diverging waves, then

φ N → 0 , ψ N → 0 at x 2 + y 2 + z 2 → ∞ .

The problem is solved under the following zero initial conditions [

∂ ϕ j ∂ r + 1 r ∂ ψ j ∂ θ | t = 0 = ∂ ∂ t ( ∂ ϕ j ∂ r + 1 r ∂ ψ j ∂ θ ) | t = 0 = 0 , 1 r ∂ ψ j ∂ θ − ∂ ϕ j ∂ r = ∂ ∂ t ( 1 r ∂ ψ j ∂ θ − ∂ ϕ j ∂ r ) | t = 0 = 0 , (3)

where j = 1 , 2 , ⋯ , N ; Nt = j—is the number of cylindrical layers; j = N —environment.

It is required to determine the dynamic stress-strain state of the cylinder and its environment caused by the incident voltage pulse (2).

To solve the plane problem, the integral Laplace transform (or Fourier transform) over time t is often used. When applying the integral Laplace transform for a function f ( t ) that is integral in the sense of Lebegue on any open interval 0 < t < T , is expressed by the formula

f L ( s ) = ∫ 0 ∞ e − s t f ( t ) d t = L [ f ( t ) ]

The function f L ( s ) is usually called the image (transform ant), the function of the f ( t ) —original. The inversion of the Laplace transform is determined by the formula

f ( t ) = 1 2 π i ∫ γ − i ∞ γ + i ∞ e s t f L ( s ) d s = L − 1 [ f L ( s ) ] ,

where the integral is taken along the path to the right of the singularities of the integrand. Using the Laplace transform problem, the interaction of non-stationary waves with a layered cylindrical body is a time-consuming task. Under the integral function is complex and has a complex form. Therefore, to find the exact expression of the original and bring to the numerical calculation is almost impossible. This method is applied in the work of V.D. Kubenko [

Integral Fourier transform. The stress field caused by the forces (2) satisfies the wave Equation (1), i.e. every cylindrical layer satisfies it. To solve the above problem, apply the t-integral Fourier transform with respect to time

ϕ F ( ξ ) = 1 2 π ∫ − ∞ + ∞ ϕ ( Ω ) l − i ξ Ω d Ω ; ϕ ( Ω ) = 1 2 π ∫ − ∞ + ∞ ϕ F ( ξ ) l i ξ Ω d ξ (4)

Using zero initial conditions, we obtain the depicted problem

1 r ∂ ∂ r ( r ∂ ϕ j F ∂ r ) + 1 r 2 ∂ 2 ϕ j F ∂ θ 2 + Ω 2 C P j 2 ϕ ¯ j F = 0 , 1 r ∂ ∂ r ( r ∂ ψ ¯ j F ∂ r ) + 1 r 2 ∂ 2 ψ ¯ j F ∂ θ 2 + Ω 2 C P j 2 ψ ¯ j F = 0 , (5)

where Ω —Fourier transforms parameter; φ j F , ψ j F —image of the Fourier transform of functions φ j ( t ) and ψ j ( t ) respectively. Then the solution of Equations (4) and (5) will be

ϕ j F ( r , θ , Ω ) = ϕ ¯ j F ( r , Ω ) cos n θ ; ψ j F ( r , θ , Ω ) = ψ ¯ j F ( r , Ω ) sin n θ (6)

here

ϕ ¯ j F ( r , Ω ) = { A n H n ( 1 ) ( Ω r / C P N ) , r ≥ r N , A n j H n ( 1 ) ( Ω r / C P j ) + B n j H ( 2 ) ( Ω r / C P j ) , r 0 ≤ 2 ≤ r N ( j = 1 , 2 , ⋯ , N − 1 ) , A n 0 I n ( Ω r / C S N ) , 0 ≤ r ≤ r 0 ; (7)

ψ ¯ j F ( r , Ω ) = { C n J H n ( 1 ) ( Ω r / C S j ) + L n J H n ( 2 ) ( Ω r / C S j ) , r 0 ≤ r ≤ r N , C n H n ( 1 ) ( Ω r / C S N ) , r ≥ r N , C n 0 I n ( Ω r / C S 0 ) , r n ≤ r ≤ r . (8)

Coefficients A n 0 , A n j , A n N , B n j , С n j , C n N —determined from the boundary conditions (7)-(8), which are placed on the contact of two cylindrical surfaces. Boundary conditions at r = R n taking into account the incident waves (1) take the form

а) σ r r N F + σ r r N ( i ) F = σ r r ( N − 1 ) F ,

b) σ r θ N F + σ r θ N ( i ) F = σ r θ ( N − 1 ) F ,

v) u r N F + u r N ( i ) F = u r ( N − 1 ) F ,

g) u θ N F + u θ N ( i ) F = u θ ( N − 1 ) F ,

where

а) σ r r N ( i ) F ( Ω ) = σ 01 ( P ) ∑ n = 0 ∞ ( − 1 ) n ∈ n I n ( Ω r / C P N ) cos n θ ;

b) σ r r N F ( Ω ) = σ r r N F ( cos 2 θ + ∈ N sin 2 θ ) ;

v) σ r θ N F = − σ r r F [ ( 1 − E N ) / 2 ] sin 2 θ ;

g) u r N F = u r N F cos θ ;

d) u θ N F = u θ N F sin θ ;

σ 01 ( P ) = σ 0 e − N Ω / C P N .

Substituting (5) and (6) into the boundary conditions (7) and (8), we obtain a system of complex algebraic equations with ( 4 j + 3 ) unknowns in the form

[ Z ] { g } = { P } , (9)

[ Z ] = ( [ Z 1 ] 0 [ Z 2 ] [ Z ( N − 1 ) ] 0 [ Z N ] )

[ Z j ] —4 × 4 matrix, the elements of which are of the nth order first and second kind Bessel and Henkel functions; { g } —Vector columns of unknown coefficients; { P } − { 0 , 0 , ⋯ , 0 , P 1 N , P 2 N , P 3 N , P 4 N } T —vector columns characterizing the falling loads, where P 1 N , P 2 N , P 3 N , P 4 N corresponds to σ r r N ( i ) F , σ r θ N ( i ) F , u r N ( i ) F , u θ N ( i ) F . Let the stepped waves interact with a cylindrical hole when r = r 0 and a stress-free hole ( σ r r | r = a = σ r θ | r = a = 0 ) . The only voltage that does not vanish at r = r 0 , is the ring voltage σ θ θ n / σ 0 . Applying the Fourier transform to the equation of motion and the boundary conditions [

σ θ θ n * = σ θ θ n ( r 01 θ , t ) σ = 1 2 π ∫ − ∞ ∞ Δ 1 ( r 0 Ω ) e i Ω t Ω 1 [ Δ 2 Δ 3 + Δ 4 Δ 5 ] d Ω , (10)

Δ 1 ( r 01 Ω ) = ( Δ 3 + τ 0 E ) [ 2 Ω H n − 1 ( 1 ) ( Ω ) − ( ( 2 n 2 + 2 n ) + Ω 2 ) H n ( 1 ) ( Ω ) ] + [ τ 0 Δ 2 − Δ 4 ] [ 2 n ( n + 1 ) H n ( 1 ) ( ( C P 1 / C S 1 ) Ω ) + 2 C P n Ω C S 1 H n − 1 ( 1 ) ( C P C S Ω ) ] .

Expression Δ k ( k = 1 , 2 , 3 , 4 , 5 ) given in [

σ θ θ n * = 1 2 π ∫ ω a ω b Δ 1 ( r 01 Ω 1 ) Ω 1 [ Δ 2 Δ 3 + Δ 4 Δ 5 ] e − i Ω t d Ω . (11)

Values of the limits of integration ω a , ω b are selected depending on the type of incident pulse. Numerical values of spectral density σ r r ( i ) F ( Ω ) from (9) of the final incident pulse; only in a small frequency range is significantly different from zero. Limits of integration ω a , ω b should be selected in accordance with this range and taking into account the required accuracy. At the same time, the question remains open as to what error the neglect of the contribution of integrals of the type (10), within the limits of integration of − ∞ to ω a and from ω b to ∞ . The numerical summation of the infinite sum (10) is, of course, also impossible. However, it was shown in [

Magnitude σ θ θ n / σ 0 from (11) is calculated on a computer as follows. All numeric parameters required for calculations are specified. The following notation is introduced: x 1 = Ω , x 2 = n 1 Ω , where n 1 = C P 1 / C S 1 ; Ω = ω α / C P 1 . For two values x k ( k = 1 , 2 ) Bessel function is determined I n ( ξ ) и N n ( ξ ) ( n = 1 , 2 , ⋯ , 10 ) . These arrays are calculated by the formula

u n ( ξ ) = 2 ( n − 1 ) ξ u n − 1 ( ξ ) − u n − 2 ( ξ ) , u n ( ξ ) = I n ( ξ ) , N n ( ξ ) (12)

As shown in [

I ¯ n ( ξ ) = 2 ( n − 1 ) ξ I ¯ n + 1 ( ξ ) − I ¯ n + 2 ( ξ ) (13)

in the direction of decreasing index (from n = N to n = 0), an auxiliary function is calculated I ¯ n ( ξ ) . To calculate the integral (11) of the integrand function

χ 1 ( r 0 , Ω , t ) = ( Δ 1 ( r 0 , Ω 1 ) / Ω 1 ( Δ 2 Δ 3 + Δ 4 Δ 5 ) ) e i Ω t

can be integrated numerically by writing it in the form

χ 1 ( r 0 , Ω , t ) = x 1 ( r 0 , Ω , t ) − i x 2 ( r 0 , Ω , t ) .

The falling pulse σ x x ( i ) ( Ω ) [

σ x x ( i ) ( Ω ) = f 1 ( Ω , t ) − i f 2 ( Ω , t ) ,

where f 1 ( Ω , t ) , f 2 ( Ω , t ) —real functions. Using Euler’s formula for l x p ( i Ω t ) , dividing (18) into real and imaginary (19) parts, after some transformations we get

σ θ θ n ∗ = 1 2 π ∫ − ∞ ∞ [ x 1 ( Ω , t ) − i x 2 ( Ω , t ) ] d Ω (14)

Dividing the integral (14) into two terms

σ θ θ n = 1 2 π ∫ − ∞ 0 [ x 1 ( Ω , t ) − i x 2 ( Ω , t ) ] d Ω + 1 2 π ∫ 0 ∞ [ x 1 ( Ω , t ) − i x 2 ( Ω , t ) ] d Ω . (15)

And replacing the variable in the first integral Ω on − Ω , will have

σ θ θ n = 1 2 π ∫ 0 ∞ [ x 1 ( Ω , t ) − x 1 ( − Ω , t ) ] − i [ x 2 ( Ω , t ) − x 2 ( − Ω , t ) ] d Ω . (16)

Since (16) is the inverse Fourier transform and contains the real value in the left-hand side [

x 1 ( Ω , t ) = − x 1 ( − Ω , t ) ; x 2 ( Ω , t ) = − x 2 ( − Ω , t ) . (17)

Considering it, from (17) we finally get

σ θ θ n ∗ = 2 π ∫ ω a ω b [ x 1 ( Ω , t ) + i x 2 ( Ω , t ) ] d Ω .

The value of integral (17) can be found numerically using the Romberg method [

Numerical results are presented for the ring voltage at r = r 0 , caused by the incident flat shock wave with a stepped distribution of voltage over time. Numerical results were obtained for ν = 0.25 : С S 1 / С P 1 = 0.5 ; θ = 0 ∘ и 90 ∘ . To determine the integral (17) of the boundary of the integral ω a and ω b have chosen [ 10 − 4 − N ] , N = 1 , 2 , 3 , 4 , 5 , а step h = 0.1 , 0.01 , 0.001 . At N = 5 and N = 6 the value of the ring voltage differs from the previous one by the fifth decimal place. Change σ θ θ * depending on the τ at various n = 0 , 1 , 2 , 3 , 4 , 5 shown in

Let the inner boundary ( r = r 0 ) free from voltage, and on contact with the environment, the condition of equality of displacements and stresses (7) [

σ r r 2 = 2 μ 2 r − 2 ∑ k = 1 2 ∑ n = 0 ∞ ∫ − ∞ + ∞ [ C n k ε 1 n ( k ) + D n k ε 2 n ( k ) ] e i Ω τ d Ω , σ θ θ 2 = 2 μ 2 r − 2 ∑ k = 1 2 ∑ n = 0 ∞ ∫ − ∞ + ∞ [ C n k ε 3 n ( k ) + D n k ε 4 n ( k ) ] e i Ω τ d Ω , σ r θ 2 = 2 μ 2 r − 2 ∑ k = 1 2 ∑ n = 0 ∞ ∫ − ∞ + ∞ [ C n k ε 5 n ( k ) + D n k ε 6 n ( k ) ] e i Ω τ d Ω , σ r r 1 = 2 μ 1 r − 2 ∑ k = 1 2 ∑ n = 1 ∞ ∫ − ∞ + ∞ [ A n δ n ( 1 ) + B n δ n ( 2 ) ] e i Ω τ d Ω ,

where C n k , D n k , A n , B n —arbitrary constants: C n k = σ k n ( c ) / Δ n , D n k = σ k n ( D ) / Δ n , A n = δ n ( A ) / Δ n , B n = σ n ( B ) / Δ n ; σ k n ( k ) and Δ n —square complex matrices (6 × 6). The remaining elements of the stress tensor are written similarly (17)

C n k = Re C n k + i Im C n k , D n k = Re D n k + i Im D n k , A n = Re A n + i Im A n , B n = Re B n + i Im B n , δ n ( e ) = Re δ n ( e ) + i Im δ n ( e ) , e = 1 , 2 , ε m n ( k ) = Re ε m n ( k ) + i Im ε m n ( k ) , e i Ω t = cos Ω t + i sin Ω t , m = 1 , 2 , 3 , 4 , 5 (18)

Substituting (18) into (17), after some transformations, we obtain the stress tensor

σ j i = ∑ k = 1 2 ∑ n = 0 ∞ ∫ ω a ω b Re σ ′ i j d Ω . (19)

All these procedures are stored in the memory of the machine. A universal algorithm for calculating integrals of type (19) has been developed. The results of the calculations are shown in

θ = 90 ∘ ( ν 1 = 0.2 ; ν 2 = 0.25 ; r 0 / r 1 = 0.5 ; E 1 / E 2 = 0.1 ; η = 0.1 )

The obtained data are compared with known results [

The difference between the stresses on the outer and inner surfaces reaches ≈15% - 20%, and the difference between the stresses on the middle and inner surfaces ≈10% ( r 0 / r 1 = 0.5 ) . Calculations show that when τ = 12 α / C P 1 the results of this study are approaching the exact static value σ θ θ * = 8.13 . The dependence of the circumferential voltage on τ presented in

Let a non-stationary step load (1) fall on an elastic two-layer cylindrical body for t > 0. A hard contact condition is set at the borders of the contact. The stress tensor in each layer is written as

σ i j ( k ) ( r 1 θ 1 t ) = 1 π ∑ n = 0 ∞ ∫ ω a ω b Re σ ′ n i j ( k ) ( r 1 θ 1 Ω ) d Ω , k = 1 , 2 , 3. (20)

Stress tensor σ i j ( k ) represents the functions of Bessel and Hankel of the first and second kind of the n-th order. Integral (20) is calculated according to the developed algorithm of the first chapter. The decision was limited to five members of the series (20), since the retention of the next members of the series has almost no effect on the results. For example, holding ten members (20) changes the voltage value by less than 2% - 3%. The following parameters were used in the calculations.: r 0 / r 2 = 0.2 ; r 1 / r 2 = 0.6 ; ν 1 = 0.2 ; ν 2 = 0.25 ; ν 3 = 0.2 ; E 1 / E 2 = 0.3 ; E 3 / E 2 = 0.1 ; ρ 1 / ρ 2 = 0.3 ; ρ 3 / ρ 2 = 0.1 .

1) In this paper, a method and algorithm are proposed for solving the problem of no stationary interaction of elastic waves on multilayer cylindrical bodies.

2) A new approach to solving dynamic problems of bodies interacting with the environment, based on the methods of Fourier and the Romberg method, is proposed.

3) It has been established that with the same loading characteristics in the material of the outer layer of a two-layer body, stress waves with the same parameters are formed at the initial moments of time.

The authors declare no conflicts of interest regarding the publication of this paper.

Ibrohimovich, S.I., Rakhimovich, K.N., Khudoyberdiyevich, T.M. and Urinovich, K.N. (2019) Interaction of Nonstationary Waves on Cylindrical Body. Applied Mathematics, 10, 435-447. https://doi.org/10.4236/am.2019.106031