where d is the vibration direction of the displacement and c is the wave velocity. For a fixed time t,

Transcription

1 3 Plane waves 3.1 Plane waves in unbounded solid Consider a plane wave propagating in the direction with the unit vector p. The displacement of the plane wave is assumed to have the form ( u i (x, t) = d i f t p x ) (190) c where d is the vibration direction of the displacement and c is the wave velocity. For a fixed time t, t p x/c = const (191) represents a plane with normal p, and as t increases, this plane moves in the direction p with velocity c. Substitution eq.(190) into the equation of motion C ijl u,lj ρ 2 u i / t 2 = 0 (192) yields [ ] Cijl p j p l d ρd i f (ξ) = 0. (193) c 2 If f (ξ) 0, it then follows that [ Cijl p j p l ρc 2 δ i ] d = 0. (194) For d 0, furthermore, the following characteristic equation has to be satisfied. C ijl p j p l ρc 2 δ i = 0. (195) Since C ijl p j p l are components of a real symmetric matrix, all eigenvalues ρc 2 are real. In general, eq.(195) gives three solutions for ρc 2, thus we have three inds of waves propagating in a solid. Isotropic solid For an isotropic solid, we have only two independent elastic constants, λ and µ. Hence C ijl p j p l = {λδ ij δ i + µ(δ i δ jl + δ il δ j )}p j p l = (λ + µ)p i p + δ i µ. (196) Substituting eq.(196) into (195), we obtain (λ + µ)p i p + δ i (µ ρc 2 ) = 0 (197) which can be reduced to the equation (µ ρc 2 ) 2 (λ + 2µ ρc 2 ) = 0. (198) Thus there are only two possible wave speeds, given by c = (λ + 2µ)/ρ = c p and c = µ/ρ = c s (199) Eq.(194) can be written for an isotropic solid as follows: (ρc 2 µ)d i = (λ + µ)p i p d or For ρc 2 = µ, and ρc 2 = λ + 2µ, we have (ρc 2 µ)d = (λ + µ)(p d)p. (200) p d = 0, and d = (p d)p, (201) respectively. The former case shows the trasnverse (S) wave with the vibration direction d perpendicular to the propagation direction p, and the latter one shows the longitudinal (P) wave with the the vibration direction d as the propagation vector p. The vibration direction of S wave is often divided into the horizontal and vertical directions, and the horizontal and vertical components are called an SH wave and an SV wave, respectively. 24

2 3.2 Reflection of a plane wave in a half space time harmonic plane wave with the displacement where = ω/c. u i (x, t) = Ad i cos ω(t p x/c) = Re {Ad i exp[ iω(t p x/c)]} = Re {Ad i exp(ip x) exp( iωt)} (202) The time factor exp[ iωt] appears as a common term contained in all wave fields and the real part can be taen after calculation in a complex form. In the sequel, therefore, the following wave fields are considered. displacement stress u i (x) = Ad i exp(ip x) (203) τ ij (x) = C ijl u,l (x) = iac ijl d p l exp(ip x) (204) Assume the wave fields, u i = u in i incident wave + u ref i and τ i = τ in i + τ ref i in a half space as follows:. u in i (x) = A (0) d (0) i exp(i (0) p (0) x), τij in (x) = ia (0) (0) C ijl d (0) exp(i (0) p (0) x) (205) reflected wave u ref i = 3 m=1 A (m) d (m) i exp(i (m) p (m) x), τ ref ij (x) = 3 m=1 ia (m) (m) C ijl d (m) p (m) l The boundary conditions τ 22 = τ 21 = τ 23 = 0 are satisfied on the free surface at x 2 = 0: τ 2j = τ2j in + τ ref 2j { 3 exp(i (m) p (m) x) (206) = i A (0) (0) C 2jl d (0) exp(i (0) p (0) x) + A (m) (m) C 2jl d (m) p (m) l exp(i (m) p (m) x)} m=1 x2 =0 = 0 (j = 1, 2, 3) (207) Eq.(207) must be valid for all values of x 1 and x 3 and hence all exponentials in eq.(207) have common arguments, i.e., (0) p (0) 1 = (1) p (1) 1 = (2) p (2) 1 = (3) p (3) 1, (0) p (0) 3 = (1) p (1) 3 = (2) p (2) 3 = (3) p (3) 3 (208) Then eq.(207) is rewritten in a matrix form: (1) C 21l d (1) p(1) l (2) C 21l d (2) p(2) l (3) C 21l d (3) (1) C 22l d (1) p(1) l (2) C 22l d (2) p(2) l (3) C 22l d (3) (1) C 23l d (1) p(1) l (2) C 23l d (2) p(2) l (3) C 23l d (3) p(3) l p(3) l p(3) l A (1) /A (0) A (2) /A (0) A (3) /A (0) = (0) C 21l d (0) (0) C 22l d (0) (0) C 23l d (0) (209) Reflection of a plane SV wave in an isotropic half space As shown in Fig. 8, the incident wave is assumed to be a plane SV wave with parameters of (0) = s ω/c s, (210) p (0) = (sin θ (0), cos θ (0), 0), d (0) = ( cos θ (0), sin θ (0), 0). (211) 25

3 x 2 x 1 x 2 u (0) u (2) (1) u x 3 ƒæ (1) ƒæ (0) ƒæ (2,3) d (2) d (0) p (2,3), u (3) d (3) p (0) x 1 p (1) = d (1) Figure 8: Reflection of a plane SV wave in an isotropic half space. From eq.(199), there are two wave speeds, c p and c s, in an isotropic solid, where c s is obtained as double roots. So the reflected waves consist of three types of elastic waves with the wave numbers: (1) = p = ω/c p, (2) = (3) = s = ω/c s. (212) The propagation vectors p (m) can be determined from eq.(208) and the direction of vibration d (1) is obtained using (201.a). The directions of vibration d (2) and d (3) for S waves can, however, be any orthogonal vectors in the plane perpendicular to the vector p, satisfying the relation (201.b). To determine unique directions, therefore, d (2) is taen in the x 1 -x 2 plane and d (3) is defined by d (3) = p (3) d (2). Then the results are shown as follows: p (1) = (sin θ (1), cos θ (1), 0) = (κ sin θ (0), 1 κ 2 sin 2 θ (0), 0), d (1) = p (1) (213) p (2) = (sin θ (2), cos θ (2), 0) = (sin θ (0), cos θ (0), 0), d (2) = (cos θ (0), sin θ (0), 0) (214) p (3) = (sin θ (3), cos θ (3), 0) = (sin θ (0), cos θ (0), 0), d (3) = (0, 0, 1) (215) where κ = c p /c s. The amplitudes of the reflected waves are obtained substituting eqs.(210)-(215) into eq.(209). 2 p µ cos θ (1) sin θ (1) s µ(sin 2 θ (0) cos 2 θ (0) ) 0 A (1) /A (0) p (λ + 2µ cos 2 θ (1) ) 2 s µ cos θ (0) sin θ (0) 0 A (2) /A (0) 0 0 s µ cos θ (0) A (3) /A (0) s µ(sin 2 θ (0) cos 2 θ (0) ) = 2 s µ cos θ (0) sin θ (0) 0 (216) Note that in an isotropic solid, the inplane wave motions in the x 1 -x 2 plane are separated from the antiplane wave motions as seen in the matrix on the LHS of eq.(216). For the incident angle θ (0) > θ c sin 1 (1/κ), the component p (1) 2 becomes p (1) 2 = cos θ (1) = 1 κ 2 sin 2 θ (0) = i κ 2 sin 2 θ (0) 1 (217) The reflected P-wave may then be written as u ref:p = A (1) d (1) exp[± p κ 2 sin 2 θ (0) 1x 2 ] exp[i s sin θ (0) x 1 ] (218) The positive sign in the first exponential term in eq.(218), must be chosen so that the amplitude of the reflected wave is finite in a half space x 2 0. The reflected P wave is a wave propagating in the x 1 -direction with wavenumber s sin θ (0) and the amplitude of the reflected P wave decays with the depth into the half space. This type of wave is called a surface wave. 26

4 3.3 Reflection and transmission of antiplane SH waves in an isotropic layered half space Let us consider antiplane wave motions in a two-dimensional isotropic layered half space as shown in Fig. 9. The layer D with the thicness h is perfectly bonded to the lower half space D, where a time harmonic plane SH wave is incident from the far field at the bottom. The material properties in the domains D and D are given by ( µ, ρ) and (µ, ρ), respectively, where µ is the shear modulus and ρ is the mass density. D _ D x =h 2 (2) x 2 (3) (0) (1) x 1 ƒæ (0) Figure 9: Reflection and transmission of antiplane SH waves in an isotropic layered half space. In a 2-D isotropic solid, the antiplane wave motion has a displacement component with the vibration direction d = (0, 0, 1), which is independent of inplane wave motions in the x 1 -x 2 plane. Therefore the displacment components in the layered media can be expressed as u (m) 1 = u (m) 2 = 0, u (m) 3 = A (m) exp(i (m) p (m) x) (m = 0,..., 3) (219) where the incident wave is denoted by m = 0 and the reflected and refracted waves are represented by m = 1, 2, 3 as shown in Fig. 9. The parameters of the incident wave are given by (0) = ω/c s = ω/ µ/ρ, p (0) = (sin θ (0), cos θ (0), 0) and the prescribed amplitude A (0). Since all antiplane waves propagate in the x 1 -x 2 plane, we have p (1) 3 = p (2) 3 = p (3) 3 = 0. Also the displacements in eq.(219) satisfies the equation of motion (174). So the wavenumbers (m) can be obtained as (1) = (0) = s = ω/c s = ω/ µ/ρ, (220) (2) = (3) = s = ω/ c s = ω/ µ/ ρ. (221) Now unnown terms in eq.(219) are the amplitude A (m) (m = 1, 2, 3) and the components p (m) 1 and p (m) 2 (m = 1, 2, 3) of the propagation vectors. To determine these unnowns, we use the boundary conditions at x 2 = 0 and x 2 = h given by u (0) 3 + u (1) 3 = u (2) 3 + u (3) 3, τ (0) 23 + τ (1) 23 = τ (2) 23 + τ (3) 23 at x 2 = 0 (222) τ (2) 23 + τ (3) 23 = 0 at x 2 = h (223) where τ (m) 23 is a shear stress defined as { τ (m) µ 23 = µ } u (m) 3 x 2 { µ = i µ } (m) p (m) 2 A (m) exp(i (m) p (m) x). (224) Substitution of eq.(219) into the boundary conditions (222) and (223) yields A (0) e i(0) p (0) 1 x1 + A (1) e i(1) p (1) 1 x1 = A (2 e i(2) p (2) 1 x1 + A (3) e i(3) p (3) 1 x1 (225) iµ (0) p (0) 2 A(0) e i(0) p (0) 1 x1 + iµ (1) p (1) 2 A(1) e i(1) p (1) 1 x1 = i µ (2) p (2) 2 A(2) e i(2) p (2) 1 x1 + i µ (3) p (3) 2 A(3) e i(3) p (3) 1 x1 (226) i µ (2) p (2) 2 A(2) e i(2) (p (2) 1 x1+p(2) 2 h) + i µ (3) p (3) 2 A(3) e i(3) (p (3) 1 x1+p(3) 2 h) = 0 (227) 27

5 These equations are satisfied for all x 1 coordinates. For this sae, all exponentials must be common. Thus, we have (0) p (0) 1 = (1) p (1) 1 = (2) p (2) 1 = (3) p (3) 1 (228) From eqs.(220), (221) and (228), the propagation vectors are obtained as follows: p (1) = (sin θ (0), cos θ (0), 0), p (2) = (β sin θ (0), q θ, 0), p (3) = (β sin θ (0), q θ, 0) (229) where β = c s /c s and q θ = 1 β 2 sin 2 θ (0). With the aid of eqs.(220), (221), the algebraic equations (225)-(227) can be simplified to A (1) /A (0) 1 cos θ (0) µ s µ s q θ µ s µ s q θ A 0 q θ e i sq θ h q θ e i sq θ (2) /A (0) h A (3) /A (0) = cos θ (0). (230) 0 Fig. 10 shows the displacement amplifications /A (0) = (u (2) 3 +u(3) 3 )/A(0) on the free surface x 2 = h as a function of h s for the incident waves with the incident angles θ (0) = 0 and π/4. The material constants in the layer are µ/µ = 1/10 and ρ/ρ = 1. We can see large amplification at h s = (2n 1)π/2 (n = 1, 2,...), due to the resonance in the soft top layer. 6 u3 / A (0) (0) 0 / _ h s Figure 10: Displacement amplifications on the free surface. Problem 3-1 As seen in eq.(195), there are three types of waves for an anisotropic solid, and their velocities (or slownesses; reciprocal numbers of velocities) change according to the direction of wave propagation, p. Draw the figure showing slownesses of three waves versus propagation directions, which is called a slowness diagram in the x 1 x 3 plane for the anisotropic solid with elastic constants: C 1111 = 235.0, C 1122 = 3.69, C 1133 = 3.69, C 2222 = 26.0, C 2233 = 3.32, C 3333 = 26.0, C 2323 = 5.52, C 3131 = 28.2, C 1212 = 28.2, [ dynes/cm 2 ] and ρ = 1.00 [g/cm 3 ] where other components are zero. Problem 3-2 Consider the reflection of an incident plane SV wave in an isotropic half space with the Poisson s ratio ν = Show the amplitude ratio A (1) /A (0) and A (2) /A (0) as a function of the incident angle θ (0). Problem 3-3 Determine the reflection coefficient A (3) of a reflected SH wave in a plain half space if the incident wave is an SH wave with the wavenumber s, the propagation direction p (0) = (sin θ (0), cos θ (0), 0), the vibration direction d (0) = (0, 0, 1), and the amplitude A (0). 28

6 Problem 3-4 For general anisotropic media, reflection and transmission problems can be solved in a numerical manner with iterative calculation. Show the flow chart of the calculation by using eqs.(208) and (209). Problem 3-5 Obtain a propagation matrix for two dimensional inplane motions in a multi-connected region as seen in eq.(64) for a one dimensional problem. 29