Open access peer-reviewed chapter - ONLINE FIRST

Cylindrical Surface Wave: Revisiting the Classical Biot’s Problem

By Jeremiah Rushchitsky

Submitted: January 25th 2019Reviewed: May 17th 2019Published: September 27th 2019

DOI: 10.5772/intechopen.86910

Downloaded: 43

Abstract

The problem on a surface harmonic elastic wave propagating along the free surface of cylindrical cavity in the direction of cavity axis is considered. In the case of isotropic medium, this is the classical Biot’s problem of 1952. First, the Biot pioneer work is revisited: the analytical part of Biot’s findings is shown in the main fragments. The features are using two potentials and representation of solution by Macdonald functions of different indexes. Then the new direct generalization of Biot’s problem on the case of transversely isotropic medium within the framework of linear theory of elasticity is proposed. Transition to the transverse isotropy needs some novelty—necessity of using the more complex representations of displacements through two potentials. Finally, a generalization of Biot’s problem on the case of isotropic and transversely isotropic media in the framework of linearized theory of elasticity with allowance for initial stresses is stated. This part repeats briefly the results of A.N. Guz with co-authors of 1974. The main features are using the linearized theory of elasticity and one only potential. All three parts are shown as analytical study up to the level when the numerical methods have to be used.

Keywords

  • surface harmonic cylindrical wave
  • classical Biot’s problem
  • generalization to the case of transversely isotropic medium

1. Introduction

Note first that the seismic waves include mainly the primary and secondary body waves and different kinds of surface waves. This chapter is devoted to one kind of surface waves. The problem is stated as follows: the infinite medium with cylindrical circular cavity having the symmetry axis Ozand constant radius is analyzed. An attenuating in depth of medium surface harmonic wave propagates along the cavity surface in directionOz. In this case, the problem becomes mathematically the axisymmetric one. This problem is solved by Biot in 1952 [1] with assumption that the medium is isotropic. The context of this chapter includes four parts. The subchapter 1 “Introduction” is the standard one. The subchapter 2 is named: “Main Stages of Solving the Classical Biot’s Problem on Surface Wave along Cylindrical Cavity.” Here, the analytical part of Biot’s findings is shown in the main fragments. The features are using two potentials and representation of solution by Macdonald functions of different indexes. The subchapter 3 “Direct Generalization of Biot’s Problem on the Case of Transversely Isotropic Media within the Framework of Linear Theory of Elasticity” contains the new approach to the classical Biot’s problem and represents the direct generalization of this problem that uses the Biot’s scheme of analysis. Transition to the case of transverse isotropy needs some novelty—necessity of using the more complex representations of displacements through two potentials. The subchapter 4 “Genera-lization of Biot’s Problem on the case of Isotropic and Transversely Isotropic Media within the framework of Linearized Theory of Elasticity with Allowance for Initial Stresses” repeats briefly the results of A.N. Guz with co-authors (1974). They considered a generalization of the Biot’s problem on the case of elastic media with allowance for the initial stresses. The main features are using the linearized theory of elasticity, one only potential, and Macdonald function of one index.

2. Main stages of solving the classical Biot’s problem on surface wave along a cylindrical cavity

2.1 Statement of problem and main equations in potentials

A cylindrical system of coordinates Orϑzis chosen, and a harmonic wave is considered that has the phase variable σ=kzvt, unknown wave number k=ω/v, unknown phase velocity v, and arbitrary (but given) frequency ωand ampitude A. It is supposed that the wave propagates in an infinite medium with cylindrical cavity of constant radius roin the direction of vertical coordinate zand possibly attenuates in the direction of radial coordinate r. In this linear statement and in assumption that deformations are small, the problem is axisymmetric, and deformations are described by two displacements urrztuφrzt=0and uzrztand two Lame equations of the form

C11C122Δrzur1r2ur+C11+C122ur,r+1rur+uz,z,r=ρur,tt,E1
12C11C12Δrzuz+12C11+C12ur,r+1rur+uz,z,z=ρuz,tt,E2

or

λ+2μur,rr+1rur,r1r2ur+uz,rz+μur,zzuz,rz=ρur,ttE3
λ+2μur,rz+1rur,z+uz,zzμ1rur,zuz,r+ur,rzuz,rr=ρuz,tt.E4

Further the potentials Φrzt,Ψrztare introduced

ur=Φ,rΨ,z,uz=Φ,z+Ψ,r+1/rΨ.E5

When Eq. (5) is substituted into Eqs. (3) and (4), then two uncoupled linear wave equations are obtained:

ΔrzΦ1/vL2Φ,tt=0,E6
ΔrzΨ1/r2Ψ1/vT2Ψ,tt=0.E7

Here the standard notations of Laplace operator Δrzand velocities of longitudinal and transverse waves in isotropic elastic medium vL=λ+2μ/ρ,vT=μ/ρare used.

2.2 Solving the wave equations in the form of Macdonald functions

The solution of Eqs. (6) and (7) is found in the form of harmonic waves in the direction of vertical coordinate:

Φrzt=Φreikzωt,Ψrzt=Ψreikzωt,E8

Φrzt=Φrcoskzvt,Ψrzt=Ψrsinkzvt.

A substitution of representations (8) into the wave Eqs. (6) and (7) gives the equations relative to the unknown amplitudes Φr,ΨrΦ,rr+1/rΦ,rk2kL2Φ=0.

Φ,rr+1/rΦ,rk21v/vL2Φ=0,E9
Ψ,rr1/rΨ,rk2kT2+1/r2Ψ=0Ψ,rr1/rΨ,rk21v/vT2+1/r2Ψ=0E10

These equations correspond to the Bessel equation for Macdonald functions Kλx(modified Bessel functions of the second kind [2, 3, 4])

y+1/xy1+λ2/x2y=0E11

More exactly, Eqs. (9) and (10) have the solutions in the form of Macdonald functions, if the conditions.

k2kL2>0,k2kT2>0k21v/vL2>0k21v/vT2>0E12

are fulfilled. According to conditions (12), the wave number of cylindrical wave must be real, and the wave velocity must be less of the velocities of classical longitudinal and transverse plane waves.

Further the wave Eqs. (9) and (10) are considered separately. The first equation is written in the form

Φ,rr+1/rΦ,rmL2Φ=0mL=k1v/vL2E13

This equation has the solution in the form of Macdonald function:

Φr=AΦK0mLrE14

of zeroth order and unknown argument x=mLr, which includes the unknown phase velocity of wave.

The second equation can be written in the form

Ψ,rr1/rΨ,rmT2+1/r2Ψ=0mT=k1v/vT2.E15

The corresponding solution under conditions (12) is expressed by the Macdonald function K1rk2kT2

Ψr=AΨK1mTrE16

of the first order and unknown argument x=mTr, which includes the unknown wave velocity. The amplitude coefficient AΨis assumed to be constant and arbitrary.

Note that the Macdonald functions have the property of attenuation with increasing arguments which is shown in Figure 1. Therefore, the propagation along the vertical coordinate zwaves (15) and (16) can be considered as the waves with amplitudes Φr,Ψr, which attenuate with increasing the radial coordinate r.

Figure 1.

Plots of the first five Macdonald functions.

This means that amplitudes can decrease essentially with increasing the distance from the surface of cylindrical cavity. In this sense, the waves (15) and (16) are the surface ones. This forms also the sense of conditions (12). The same conditions are used in the analysis of classical Rayleigh surface wave which propagates along the plane surface of isotropic elastic medium [5, 6, 7, 8, 9]. But the Rayleigh wave attenuates as an exponential function when being moved from the free surface, whereas the cylindrical surface Biot’s wave attenuates as the Macdonald functions. At that, the arguments in exponential function and Macdonald functions are identical and depend on the wave velocity.

2.3 Boundary conditions: equations for unknown wave number

The boundary conditions correspond to the absence of stresses on surface r=ro

σrrr=rozt=0,σrzr=rozt=0.E17

The stresses

σrr=2μur,r+λur/r+ur,r+uz,z,σrz=μur,z+uz,rE18

are written through the potentials

σrr=λ+2μΦ,rrΨ,rz+λ1/rΦ,rΨ,z+Φ,zz+Ψ,rz+1/rΨ,z,E19
σrz=μΦ,rzΨ,zz+Φ,zr+Ψ,rr+1/rΨ,r1/r2Ψ.E20

Then the boundary conditions (17) can be written in the form.

2μΦ,rrΨ,rz+λΔΦr=ro=0,μ2Φ,rzΨ,zz+ΔΨ1/r2Ψr=ro=0E21

In the work [1], Biot has used the expressions.

ΔΦ1/vL2Φ,tt=0, ΔΨ1/r2Ψ1/vT2Ψ,tt=0and rewrite Eq. (21) in such a way Φ,rrΨ,rz+λ/2μ1/vL2Φ,ttr=ro=0, 2Φ,rzΨ,zz+1/vT2Ψ,ttr=ro=0.

Then the substitution of solutions (14) and (16) into the boundary conditions (21) gives two homogeneous algebraic equations relative to the unknown constant amplitude coefficients

1v/vL2λμv/vL2K0mLroK0mLro+K2mLroAΦ1v/vL2AΨ=0,E22
21v/vL2AΦ+2v/vT2K1mTroK1mLroAΨ=0.E23

An analysis of these equations that describe the cylindrical surface wave is very similar to the analysis that has been carried out by Rayleigh for the classical wave propagating along the plane surface. Some novelty in analysis of systems (22) and (23) is consideration of the system relative to quantities K1mLroAΦand K1mSroAΨ

1v/vL2K0mLroK1mLro+1mLroλ2μv/vL2K0mLroK1mLroK1mLroAΦ+1v/vS2K0mTroK1mTro+1mTroK1mTroAΨ=0,E24
21v/vL2K1mLroAΦ+2v/vT2K1mTroAΨ=0.E25

Solving of systems (24) and (25) gives two results. First, the solution is found accurate within one amplitude factor. Second, an equation for determination of phase velocity of cylindrical surface wave can be obtained in an explicit form.

The work of Biot (1952) has demonstrated some art in handling the Macdonald functions and has written Eq. (24) through only functions of the zeroth and first orders. For that, the known formulas

K0x=K1x,K1x=K0x,K0x+1/xK0x=K0x,K0x=1/xK1x+K0xE26

have been used [3]. As a result, the equation for determination of phase velocity of cylindrical wave has the form

2v/vT22v/vT2K0mLroK1mLro+1v/vL2mLro41v/vL21v/vT2K0mTroK1mTro+1mTro=0.E27

Let us write the corresponding equation for the Rayleigh wave [5, 6, 7, 8, 9] as

41v/vL21v/vS22v/vS22=0.E28

Thus, a presence of Macdonald functions in Eq. (27) complicates essentially an analysis of this equation because according to relations mL=k1v/vL2, mS=k1v/vT2these functions have the unknown velocity in argument.

If the cavity radius is not small, then the Macdonald functions can be represented by the simple formula K0r=K1r=erπ/2r, and Eq. (27) is reduced to the Rayleigh Eq. (28).

Strictly speaking, the analytical part of analysis is ended by obtaining Eq. (27). Further analysis can be continued with the aim of the numerical methods. Biot in [1] has shown some comments and conclusions based on resources of the 1950s.

A possibility of analytical approach is still saved in the problem on existence of the appropriate wave velocity. First of all, Eq. (27) depends on the elastic constants, and this dependence can be shown in the form of dependence on the ratio of known velocities vL/vT. If the notation v2/vT2=zis used, then Eq. (27) can be written in the form

2zvL/vT22zK0rok1zvL/vT2K1rok1zvL/vT2+1zvL/vT2rok1zvL/vT241z1zvL/vT2K0rok1zK1rok1z+1rok1z=0.E29

It seems appropriate to recall here the most known ways of proving the existence of velocity of the classical Rayleigh wave. An initial equation is always Eq. (28). Two different notations v2/vT2=zand v=1/θare used, which generate two different representations of Eq. (28)

zz38z1z21vT2/vL2=0,E30
2θ21/vT224θ2θ21/vT2θ21/vL2=0.E31

Finding the real root of Eq. (30) is the key step in the analysis of the Rayleigh wave [5, 6, 7, 8, 9] . For more than 100 years of analysis of this wave, many methods of proving the existence of real velocity of wave were elaborated.

First of all, the sufficiently useful and exact empirical Viktorov’s formula [5].

v/vT=z0.87+1.12υ1+υυthe Poisson ratioE32

should be shown.

Let us show further briefly some phenomenological methods. Note that the restriction on the Rayleigh wave velocity is already obtained from a statement of the problem—it is less of the velocity of plane transverse wave. This restriction can be written in the form z<1or θ>1/cT.

Method 1 (graphical method [10, 11]). Eq. (30) is considered as a sum of two summands Z1+Z2=0. The first summand Z1 = z3 describes a cubic parabola; the lower branch of which lies in the first quadrant of the plane zOZ1. The second summand describes a quadratic parabola Z2=8z1z21cT2/cL2, which is concave in the direction of coordinate axis OZ2. Further the ratio cT2/cL2=μ/λ+2μcan be estimated from below and top 0cT2/cL21/2with allowance for the shear modulus μthat is positive. These parabolas are intersected on the interval 01. More exactly, one of the roots z=zCof Eq. (30) can be estimated 0.764z=c/cT20.912.Here, the minimal value corresponds to the case when the parabola is tangent to the abscissa axis, and the maximal value corresponds to the case when the parabola is moved partially into the fourth quadrant. Thus, the velocity of Rayleigh wave is close to the velocity of plane transverse wave, but always less of its 0.874 ≤ (c/cT) ≤ 0.955.

Method 2 (method of finding the interval, on ends of which the equation possesses the different by sign values [2, 11]). This method is based on the analysis of Eq. (30). The value of equation that corresponds to the point cR=cTis positive and equal to 1. The second point is chosen as cR=εcT, where εis assumed as the small quantity (this point is close to 0). When this value is substituted into Eq. (30), then expression 2ε21cT2/cL2is always negative. Hence, at least one root of equation lies in the interval εcTcT.

Method 3 (another method of finding the interval, on ends of which the equation possesses the different by sign values [5]). This method is based on the analysis of Eq. (31). The right point is chosen as θ=1/cT(similar to method 2). Then Eq. (31) possesses the positive value. The left point corresponds to θ. Further an expression (31) is expanded into the power series near the point at infinity. This series starts with the term 2θ21/cT21/cL2, which is always negative. So this equation possesses in the chosen points the different sign values. Thus, at least one root of the equation lies in the interval 1/cT.

Method 4 (method based on assumption relative to the Poisson ratio [7]). This assumption consists in the choice of value of Poisson ratio that is often used in the analysis of seismic waves in Earth’s crust ν=λ/2λ+μ=1/4λ=μ. Then cubic Eq. (31) (the zeroth root θ1=0is ignored from a physical considerations) can be solved exactly, and the roots possess the values θ2=4,θ3=2+2/3,θ4=22/3. Since the condition θ<1has been fulfilled, then the corresponding root is equal to θ4=0.8453.

The main conclusion from the shown above methods is that they really allow to establish an existence of real root of Rayleigh equation (the real value of velocity of harmonic Rayleigh wave). They give the positive answer on the question whether the Rayleigh wave exists. In the case of other surface waves including the cylindrical wave under consideration, the experience of the classical Rayleigh wave analysis can be quite useful.

3. Cylindrical wave propagating along the surface of the cylindrical cavity in the direction of vertical axis: The case of transversal isotropy of medium

Let us return to the initial statement of problem and consider an infinite medium with cylindrical circular cavity that has the symmetry axis Ozand radius ro. The medium is assumed to be the transversely isotropic elastic one. It is assumed further that the wave is harmonic in time, and attenuating deep into medium wave propagates in the direction of axis Ozalong the cavity surface. Such a problem can be considered as some generalization of Biot’s [1] problem that is solved in the assumption of isotropy of medium on the case of transversal isotropy of medium. Therefore, it seems expedient to recall some facts from the theory of elasticity of transversally isotropic medium.

3.1 Some information on transversally isotropic medium

Let us consider the case when Ox3is the axis of symmetry and Ox1x2is the plane of isotropy. This symmetry corresponds to the hexagonal crystalline system. The matrix of elastic properties is characterized by 5 independent elastic constants C11,C12,C13,C33,C44and 12 non-zero components [11, 12, 13]:

CIK=C11C12C13000C12C11C13000C13C13C33000000C44000000C440000001/2C11C12.E33

Then the constitutive relations σεhave the form [12, 14].

σ11=C11klεkl=C11ε11+C12ε22+C13ε33,σ22=C22klεkl=C12ε11+C22ε22+C13ε33,σ33=C33klεkl=C13ε11+C13ε22+C33ε33,σ12=C11C12ε12,σ13=2C44ε13,σ23=2C44ε23,E34

or in notations σu[12, 14].

σ11=C11u1,1+C12u2,2+C13u3,3,σ22=C12u1,1+C11u2,2+C13u3,3,σ33=C13u1,1+C13u2,2+C11u3,3,σ12=1/2C11C12u1,2+u2,1,σ13=C44u1,3+u3,1,σ23=1/2C44u2,3+u3,2.E35

Also, five independent elastic technical constants are often used.

Ex=Ey,Ex=Ey,Ez,Gxy,Gxz=Gyz,υxy,υxz=υyz,Gxy=Ex/1+2υxy. They are evaluated through CNMby the following formulas:

The longitudinal Young modulus that corresponds to tension along the symmetry axis Oz

Ez=C332C132/C11+C12.E36

The transverse Young modulus that corresponds to tension in the isotropy plane Oxy

Ex=C11C12C11+C12C332C132/C11C33+C132.E37

The shear modulus that corresponds to the shear along the isotropy plane Oxy

Gxy=C66=1/2C11C12.E38

The shear modulus that corresponds to the shear along the symmetry axis Oz

Gxz=C44.E39

The Poisson ratio that corresponds to the shear along the symmetry axis Ozunder tension in the isotropy plane and characterizes the shortening in this plane

υxz=C13/C11+C12.E40

Sometimes, the corresponding Lame moduli are used.

λxy+2μxy=C11,λxy=C12,μxy=1/2C11C12λxz+2μxz=C33,λxz=C13,μxz=C44.E41

The Poisson ratio (40) is determined by the known formula of isotropic theory υxz=λxz/2λxy+μxy.

The Poisson ratio υxythat corresponds to the shear along the symmetry axis Ozunder tension along the isotropy plane is determined also by the classical formula υxy=λxy/λxy+μxy.

The constants C11,C12,C13,C33,C44are represented through the technical constants E,E,ν,ν,Gby the formulas.

C11=1ν2E/E1ν2+1+2νν2E/EE,C12=vν2E/E1ν2+1+2νν2E/EE,C13=ν1ν1ν2+1+2νν2E/EE,C33=1ν21ν2+1+2νν2E/EE,C44=G.E42

Let us comment briefly some features of transversally isotropic materials. They can be divided on the natural and artificial ones. An example of the classical natural material is the rock. An example of the modern material is a family of fibers “Kevlar®.” Kevlar® KM2 [15] is characterized by elastic constants Ex=1.34GPa,Ez=84.62GPa,Gxz=24.40GPa,υxy=0.24,υxz=0.60.

An example of composite materials can be four fibrous composites of micro- and nanolevels, which are described in [15]. The corresponding elastic constants for some variants of these materials are as follows [15]:

10% of carbon microfibers Ex=3.59GPa,Ez=25.22GPa,Gxz=1.17GPa,υxy=0.39,υxz=0.58.

10% of graphite microwhiskers Ex=3.69GPa,Ez=102.4GPa,Gxz=1.14GPa,υxy=0.39,υxz=0.62.

10% of zig-zag carbon nanotubes Ex=3.70GPa,Ez=67.21GPa,Gxz=1.14GPa,υxy=0.39,υxz=0.62.

10% of chiral carbon nanotubes Ex=3.67GPa,Ez=126.4GPa,Gxz=1.14GPa,υxy=0.39,υxz=0.62.

The shown above values are typical for the transversally isotropic materials, and therefore they are briefly commented below.

Comment 1. The Young modulus in the direction along the symmetry axis Ezexceeds essentially the Young modulus in the isotropy plane Ex(from 6 to 34 times in examples above but can in some cases exceed 100 times).

Comment 2. The Lame moduli λxand λzrepeat the relations between Exand Ez.

Comment 3. The Poisson ratio υxzalong the symmetry axis Ozexceeds the classical red line in 0.5 for values of this ratio.

Comment 4. The shear moduli Gxyand Gxzare differed quite moderately.

3.2 The basic formulas for elastic transversely isotropic medium with axial symmetry

Let us write the basic formulas for the case of symmetry axis Oz. Then displacements are characterized by two components urrzt,uzrzt. The motion equations in stresses have the form.

σrr,r+σrz,z+1/rσrrσφφ=0,σrz,r+1/rσφz,φ+σzz,z+1/rσrz=0.E43

The substitution of constitutive equations.

σrr=C11ur,r+C121/rur+C13uz,z,σzz=C13ur,r+C131/rur+C33uz,z,σrz=1/2C44uz,r+ur,z,σφz=σ=0E44

into the motion Eqs. (43) gives the motion equations in displacements

C11ur,rr+1/rur,r1/r2ur+C44ur,zz+C13+C44uz,rz=ρur,tt,E45
C44uz,rr+1/ruz,r+C33uz,zz+C13+C44ur,rz+1/rur,z=ρuz,tt.E46

Note that Eqs. (45) and (46) include only four constants (the constant C12is not represented in these equations). This means that displacements and strains are described by only four constants. But the stress state is already described by all five constants.

3.3 Three classical ways of introducing the potentials in transversely isotropic elasticity

The basic equations of the theory of transversely isotropic elasticity are frequently analyzed by the use of potentials. The potentials are introduced in theory of elasticity mainly for static problems. Transition to the dynamic problems is associated with complications that are sometimes impassable. Because the problem on waves is related to the dynamic ones, let us show further the possible complications with introducing the potentials.

Way 1 [12]. It is proposed for the axisymmetric problems of equilibrium (not motion) and is based on introducing one only potential φrzas the function of stresses. The formulas for stresses include four unknown parameters a,b,c,d, which is characteristic for representations in the transversely isotropic elasticity.

σrr=φ,rr+b1/rφ,r+aφ,zz,z,σθθ=bφ,rr+1/rφ,r+aφ,zz,z,E47
σzz=cφ,rr+c1/rφ,r+dφ,zz,z,σrz=φ,rr+1/rφ,r+aφ,zz,r.E48

The next step consists in substitution of formulas (47) and (48) into the first equation of equilibrium and the equations that are obtained from the Cauchy relations and formulas for the strain tensor. This permits to determine the unknown parameters through the elastic constants represented in the equilibrium equations. Further, the second equation of equilibrium gives the biharmonic equation for finding the potentials

Δs1Δs2φ=0,E49

where ΔsNφ=φ,rr+1/rφ,r+1/sN2φ,zzN=1;2are some “complicated” copies of classical expressions Δφ=φ,rr+1/rφ,r+φ,zzassociated with the Laplace operator. Two constants sNare determined from the algebraic equations

s4a+c/ds2+1/d=0,s1,3=±a+c+a+c24d2d,s2,4=±a+ca+c24d2d.E50

Thus, a transition from the isotropic case to the transversally isotropic one complicates the procedure of solving the static problems. Here a necessity of solving the classical biharmonic equation is changed on necessity of solving some generalization of this equation in the form (49).

Way 2 [12, 16]. This way is also proposed for the static problems. Here, two potentials are introduced which are linked immediately with displacements

ur=ϕ1,r+ϕ2,r,uz=k1ϕ1,z+k2ϕ2,z.E51

A substitution of representations (51) into equations of equilibrium (45), (46)

C11ur,rr+1/rur,r1/r2ur+C44ur,zz+C13+C44uz,rz=0,
C44uz,rr+1/ruz,r+C33uz,zz+C13+C44ur,rz+1/rur,z=0

allows to determine the unknown constants k1,k2. An idea consists in that both equations must be transformed in identical equations relative to the potentials by comparing some coefficients

k12C13+C44+C44C11=kC33k12C44+C13+C44=V.

This expression gives the quadratic equation for k12and V

V2+C132C44+C33C11C33C11C44V+C33C11=0.E52

Note that the simple link VN=1/sNexists between constants VNand sN, which makes the ways 1 and 2 very similar. Then the potentials fulfill the equations

ΔrzNφN=φN,rr+1/rφN,r+1/VN2φN,zz.E53

The stresses are expressed through new potentials in such a way

σrr=C11C121/rϕ1,rr+ϕ2,rrC441+k1ϕ1,zz+1+k2ϕ2,zz,σθθ=C11C121/rϕ1,rr+ϕ2,rrC13k1C12V1ϕ1,zz+C13k2C12V2ϕ2,zz,σzz=C33k1C13V1ϕ1,zz+C33k2C13V2ϕ2,zz,σrz=C441+k1ϕ1,rz+1+k2ϕ2,rz.E54

Way 3 [1, 16]. This way is proposed for equations of motion, but only for the isotropic theory of elasticity. It can be used for the static problems of transversely isotropic theory of elasticity. The initial equations here are the equations of motion (43) without inertial summands

C11ur,rr+1/rur,r1/r2ur+C44ur,zz+C13+C44uz,rz=0,E55
C44uz,rr+1/ruz,r+C33uz,zz+C13+C44ur,rz+1/rur,z=0.E56

The potentials are introduced like (51), but the representations are complicated by necessity of introducing two new unknown parameters:

ur=Φ,rΨ,z,uz=nΦ,z+mΨ,r+m1/rΨ,E57

A substitution of representations (57) into equations of motion (45) and (46) gives equations relative to the potentials. Eq. (45) gives two equations:

Φ,rr+1/rΦ,r+C44+nC13+C44C11Φ,zz=0,E58
Ψ,rr+1/rΨ,r1/r2Ψ+C44C11mC13+C44Ψ,zz=0,E59

whereas Eq. (46) gives three equations:

Φ,rr+1/rΦ,r+nC33nC44+C13+C44Φ,zz=0,E60
Ψ,rrz+1/rΨ,rz1/r2Ψ,z+C33mC13+C44C44mΨ,zzz=0,E61
Ψ,rr+1/rΨ,r1/r2Ψ+C33mC13+C44C44mΨ,zz=0.E62

The last two equations are identical. Eqs. (58) and (60) and (59) and (62) have to be identical. This means that the coefficients in these equations have to be identical. As a result, two equations can be obtained for the determination of unknown constants n,m.

C44+nC13+C44C11=nC33nC44+C13+C44n2nC11C33C442C13+C442C44C13+C44+1=0,E63
n1,2=C11C33C442C13+C4422C44C13+C44×1±14C44C13+C44C11C33C442C13+C4422,E64
C44C11mC13+C44=C33mC13+C44C44mm2+mC442+C11C33+C13+C442C33C13+C44+C11C33=0.E65
m1,2=C442+C11C33+C13+C4422C33C13+C44×1±14C11C33C33C13+C44C442+C11C33+C13+C4422.E66

The unknown potentials Φrzand Ψrzhave to be determined from the simple Eqs. (63) and (65) which are the classical Bessel equations of orders 0 and 1 and arguments depending on some rational combination of elastic constants.

Thus, three ways of introduction of potentials in the static problems of transversely isotropic theory of elasticity are shown. The different attempts to transfer these ways into the dynamic problems meet some troubles—the presence of inertial summands generates new additional conditions for the unknown constants in representations of potentials. Introducing the new constants does not help—the number of conditions is still more than the number of all constants.

3.4 Solving the problem on the propagation in the direction of vertical axis surface cylindrical wave for the case of transversal isotropy of medium

Consider now equations of motion (45) and (46) and introduce the potentials by the formula (57). A substitution of formula (57) into equations of motion gives five equations relative to the potentials. Eq. (57) gives two equations:

Φ,rr+1/rΦ,r+C44+nC13+C44C11Φ,zz=ρC11Φ,tt,E67
Ψ,rr+1/rΨ,r1/r2Ψ+C44C11mC13+C44Ψ,zz=ρC11mC13+C44Ψ,tt.E68

Eq. (46) gives three equations:

Φ,rr+1/rΦ,r+nC33nC44+C13+C44Φ,zz=nC44+C13+C44Φ,tt,E69
Ψ,rrz+1/rΨ,rz1/r2Ψ,z+C33mC13+C44C44mΨ,zzz=ρC44Ψ,ztt,E70
Ψ,rr+1/rΨ,r1/r2Ψ+C33mC13+C44C44mΨ,zz=ρC44Ψ,tt.E71

Two last equations are identical. Also the equations for potential Φmust be identical as well as the equations for potential Ψmust be identical. Let us assume additionally that the problem in hand considering the solution in the form of harmonic in time cylindrical wave with unknown wave number kand known frequency ω:

Φrzt=Φreikzωt,Ψrzt=Ψreikzωt.E72

Note that characterization of an attenuation of wave depth down functions Φr,Ψris unknown. They must be found from equations, which are obtained by substitution of representations (72) into Eqs. (67) and (71):

Φ,rr+1/rΦ,rC44+nC13+C44C11k2kL112Φ=0,E73

kL11=ω/vL11, vL11=C11/ρ,

Φ,rr+1/rΦ,rnnC44+C13+C44C33k2C11kL112Φ=0,E74
Ψ,rr+1/rΨ,r1/r2ΨC44C11mC13+C44k2kT442Ψ=0,E75

kT44=ω/vL44, vL44=C44/ρ,

Ψ,rr+1/rΨ,r1/r2ΨC33mC13+C44C44mk2kT442Ψ=0.E76

As a result, two equations can be obtained that permit to determine the constants n,m

n22N1n+N2=0,m2+2M1m+M2=0,E77
N±M±=N1M1±N1M12N2M2,E78
2N1=C11C33C13+C442k2C11C11C44kL112C442C44C13+C44k2,N2=C44C11kL112C44k2=0E79
2M1=C442C11C33C13+C442k2C442C11C44kT442C13+C44C33k2C44kT442,M2=C11C33k2C44kT442k2.E80

Note that restriction on the kind of solution (it has to be a wave) allows to unite two different conditions into one—conditions for equaling coefficients in summands with the second derivative by time tand vertical coordinate z. In this case, the number of unknown constants coincides with the number of conditions which are necessary for the determination of potentials. As a result, the wave attenuation-transformed potentials can be determined from the equations of Bessel type:

Φ,rr+1/rΦ,rML112Φ=0,ML11=C44+nC13+C44C11k2kL112,E81
Ψ,rr+1/rΨ,r1/r2+MT442Ψ=0,MT44=C44C11mC13+C44k2kT442,E82

A success in the determination of transformed potentials is accompanied by a complication of conditions which provide the wave attenuation. They have the form.

C44+nC13+C44C11k2kL112>0,C33mC13+C44C44mk2kT442>0.E83

Let us recall that the similar conditions for the case of isotropic medium k2kL2>0, k2kT2>0are slightly simpler and coincide with the corresponding conditions of classical Rayleigh surface wave [5, 6, 7, 8, 9, 17]. A complexity of conditions (83) is increased by the complex form of dependence of constants n,mon the wave number k.

If the conditions (83) are fulfilled, then the solution of wave equations for potentials can be written in the form.

Φr=AΦK0ML11r,Ψr=AΨK1MT44r.E84

With allowance for formulas (84), the representations of potentials becomes more definite

Φrzt=AΦK0ML11reikzωt,Ψrzt=AΨK1MT44reikzωt.E85

The formula (85) completes the first analytical part of solving the problem on cylindrical surface wave.

3.5 Boundary conditions: equations for the unknown wave number

This part of analysis can be treated as the second analytical part. The boundary conditions have the form identical for all kinds of symmetry of properties. That is, they have the form (17) or (21). The formulas for stresses depend already on the symmetry of medium. The expressions for stresses through the potential reflect the features of introducing the potentials. In this case, they have the form

σrr=λ+2μΦ,rrΨ,rz+λ1/rΦ,rΨ,z++nΦ,zz+mΨ,rz+m1/rΨ,z,E86
σrz=μΦ,rzΨ,zz+nΦ,zr+mΨ,rr+m1/rΨ,rm1/r2Ψ.E87

Further, the representations (86) and (87) should be substituted into the boundary conditions, and the formulas on differentiation of Macdonald functions [3] should be taken into account:

dK0ML11rx/dr=ML11K1ML11rx,
d2K0ML11r/dr2=ML111/rK1ML11r+ML112K0ML11r,
dK1MT44r/dr=1/rK1MT44rMT44K0MT44r.

Then the boundary conditions are transformed into the algebraic equations relative to quantities K1ML11roAΦ, K1MT44roAΨ

ML111ro+vL2vT2ML112vL2vT2vL2nk2K0ML11roK1ML11roAΦK1ML11roikvL2vT2vT2×21m+vT2vL2vT21ro++1m+vT2vL2vT2MT44K0MT44roK1MT44roAΨK1MT44ro=0,E88
1+nikK0ML11roK1ML11r0K1ML11roAΦ+mMT442+k2K1MT44rAΨ=0.E89

When the determinant of linear homogeneous system of Eqs. (88) and (89) is equaled to zero, then the equations for the unknown wave number can be obtained:

1+nk2vL2vT2vT2K0ML11roK1ML11r021m+vT2vL2vT21/ro+1m+vT2vL2vT2MT44K0MT44roK1MT44romMT442+k2ML111ro+vL2vT2ML112vL2vT2vL2nk2K0ML11roK1ML11ro=0.E90

Note that the sufficiently complex expression relative to the wave number is hidden coefficients ML11,MT44of Macdonald’s functions K0ML11roK1ML11r0, K0MT44roK1MT44ro. Therefore, the analytical part of analysis is finished on these formulas. Further, the numerical approaches have to be utilized.

Note also that the simple and convenient condition from analysis of classical surface Rayleigh wave [6, 7, 8, 9, 10, 17], when the wave number depends only on ratio vL2/vT2, does not exist in the analysis of cylindrical surface wave. Here, the parameters ML11,MT44depend on the complicated form on all elastic constants. Of course, the Macdonald functions can be represented approximately through their arguments. But only the numerical methods can give the final result—the value of wave number or phase velocity.

4. Solving the problem on propagating in the direction of symmetry axis surface wave within the framework of linearized theory of elasticity with allowance for initial stresses

Note that analysis of cylindrical surface wave in isotropic medium was first carried out by Biot [1] in 1952 and the transversally isotropic medium with initial stresses was first carried out by Guz et al. in 1974 [18].

Let us show below an analysis of the problem in hand that is carried out in Subchapter “Longitudinal Waves” of Chapter 4 “Waves in Cylindrical Media” of volume 2 of edition [19]. Here, the cylinder of circular cross-section is considered, and the longitudinal wave is defined as the wave propagating in the direction of cylinder axis Oy3. The problem is assumed to be axisymmetric and is described within the framework of linearized theory of elasticity for bodies with initial stresses. The cylindrical coordinates r'θy3are introduced, and displacements are taken in the form

ur'=ur'r'y3t,uθ=0,uy3=uy3r'y3tE91

The medium is assumed isotropic or transversally isotropic. The main relations for transversal isotropy are described by independent constant

ω1111,ω1122,ω1133,ω1221,ω1313,ω1331,ω3113,ω3333.E92

Note that as shown in (92), eight constants are necessary in the linearized theory, but in the framework of linear theory, they have the form (33), and their number is five.

Further, the general solutions of basic equations in displacements are utilized. These equations have the form (3.174) [19]

ωlmαβ2uα/xkxβ=ρδ2uα/τ2E93

where only eight independent constants (92) must be taken into account.

The corresponding equations of linear theory of elasticity for the case of transversally isotropic medium without of initial stresses are written above as Eqs. (45) and (46).

The general solutions for the case of axial symmetry are expressed through one potential in the form (4.13) [19]

ur'=2/ry3X,E94
u3=ω'1111+ω'13131ω'1111Δ'1+ω'31132/y32ρ'2/τ2X',Δ'=2/r'2+1/r'/r'.

Note that in Section 3 of this chapter, two potentials Φ,Ψare introduced by formula (57), which corresponds and generalizes the procedure used in Biot’s analysis [1].

The longitudinal harmonic wave is described analytically through the potential in the form (101) [19]

X'r'y3τ=X'1r'eiky3ωτ,E95

where the unknown amplitude X'1r'has to be determined by substitution of solution (4.13) [19] into the second Eq. (3.362) [19] (for potential X'). This gives Eq. (4.16) [19]:

ω'1111ω'1331Δ'1k2ξ'22Δ'1k2ξ'32k2ρ'Ccp2ω'1111+ω'1331Δ'1k2ω'1111+ω'3113+ρ'2Ccp2X'1=0,E96
Ccp=ω/k,ξ2,3'2=c'±c'2ω'3333ω'3113/ω'1111ω'1331,c'=1/2ω'3333/ω'1331+ω'3113/ω'1111ω'1111+ω'13312/ω'1111ω'1331,

which further is written in the form

Δ'1ζ2'2Δ'1ζ3'2=0E97

The unknown quantities ζ'2,3must be found from the linear algebraic equation of the fourth degree (4.20) [19].

ω'1111ω'1331ζ'4+k4ρ'Ccp2ω'3333ρ'Ccp2ω'3113+k2ω'1111ρ'Ccp2ω'3333+ω'1331ρ'Ccp2ω'3113+ω'1111+ω'31132ζ'2=0,E98

The solution (95) describes the surface wave, if amplitude X'1r'attenuates with increasing the radius. This is provided by the condition that quantities ζ'2,3is unequal and pure imaginary. Then the potential gains the form (4.22) [19].

X'1r'=B10J0ζ'2r'+B20K0ζ'2r'+B30J0ζ'3r'+B40K0ζ'3r',E99

The shown part of analysis from introducing the potential by formula (94) to representation of solution by formula (99) inclusive can be compared with analogous part of analysis from Section 3 of this chapter (from introducing the potentials by formula (57) to the solution in the form of (85)). It is easy to see a difference in representations (99) and (85): formula (99) uses the Bessel functions and in particular the Macdonald function of zero index, whereas formula (85) uses (like the Biot’s solution (14)) the Macdonald functions (16) of the zero and first indexes.

The next part of analysis of cylindrical wave consists in substitution of solution into boundary conditions of the form (99) [19]

Q'r'r'=0,Q'r'3=0whenr'=R'1,R'2.E100

The case of oscillatory behavior of wave in the direction of radius is considered with pointing that the case of surface wave is the same type. A substitution of solution (99) into conditions (4.79) [19] gives the dependence of velocity of surface wave or its wave number on frequency—a dispersion equation in the form of determinant of the fourth order in the form (4.26) [19].

detαijΔωk=0;i,j=1,2,3,4.E101

This finishes the analytical part of analysis shown in [19]. It corresponds to the part of Section 3.5 of this chapter, where the explicit form of dispersive equations is proposed in the form (90) that includes the Macdonald functions of the zero and first orders which represent some generalization of dispersion Eq. (27) obtained by Biot.

5. Conclusions

This chapter proposes three fragments of analytical analysis of the cylindrical surface wave propagating in the vertical direction of circular cylindrical cavity. The first fragment shows the analytical part of pioneer work of Biot. It represents the classicism of mathematical procedures and physical comments of Biot. Properly speaking, the clear and understandable Rayleigh’s scheme is saved, but it is complemented by some findings reflecting the features of cylindrical waves. Two next fragments show the more late development of the Biot’s problem. They are different by influence of the Biot’s procedure. The approach shown in Section 3 is more close to the Biot’s analytical scheme, whereas Section 4 proposes as an independent scheme that is more close to the Rayleigh scheme. Nevertheless, all fragments testify the mathematical complexity in solving the problem on the cylindrical surface waves. Thus, revisiting the old Biot’s problem shows that it still generates new scientific and practical problems.

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Jeremiah Rushchitsky (September 27th 2019). Cylindrical Surface Wave: Revisiting the Classical Biot’s Problem [Online First], IntechOpen, DOI: 10.5772/intechopen.86910. Available from:

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