Body Waves from a Shearing Motion

Chapter 1 Introduction

3.2 Body Waves from a Shearing Motion

coordinates,

U_r(t) = d(t) sinφ (r≤r₀)

U_φ(t) =d(t) cosφ (r≤r₀) (3.13) σ_rz =σ_zφ = 0 (r≥r₀)

where d(t) is the horizontal displacement of the rigid plate, r₀ is the radius of the disk, andU_r,U_φ, andU_z are the displacements of a point atr, φ, zin these coordinate directions (radial, tangential, and depth). Assuming a particular solution similar to that given in Equations 3.28 through 3.33, in conjunction with the boundary conditions given by Equation 3.13, and forcing the disk to remain horizontal when a force is applied, one can solve for A(ζ) and C(ζ). The procedure is used in the next section. Using this method, Bycroft (1956) and Cherry (1962) find the stress field under the plate to be

σ_rz = P sinφ 2πr₀p

r₀²−r² (r≤r₀) σ_zφ = P cosφ

2πr₀p

r₀²−r² (r≤r₀)

whereP is the maximum applied shear force. The sinusoidal nature of the excitation force (sin(ωt)) has been neglected for convenience, and will be brought in once the displacement field is calculated. The displacement field calculated byCherry (1962) is provided in Equations 3.22 through 3.24 in the proceeding analysis. The displacement field is given in cylindrical coordinates, wherer andφ give the coordinates of a point on the free surface (z = 0) of the half-space. After the solution is achieved, it is transformed to spherical coordinates to simplify the integration of the displacement fields. The following integrals have been expanded and re-arranged to match the integral format presented in Appendix D in order to easily visualize the required

integrations. By utilizing the Bessel function identity

∂

∂rJ₁(ζr) = ζJ₀(ζr)−J₁(ζr)

r (3.14)

to expand the integral form of the displacement field given by Cherry (1962) in Equations 17 to 19 to integrals of the form solved in Appendix D, the displacement field is as follows,

U_r = P sinφ 2πµr₀

·

− Z _∞

0

2ζp

ζ²−k²

F(ζ) sin(ζr₀)ζJ₀(ζr)e^−αz dζ+ Z _∞

0

2p

ζ²−k²

ζrF(ζ) sin(ζr₀)ζ²J₁(ζr)e^−αzdζ+ Z _∞

0

sin(ζr0) rζ³p

ζ²−k²ζJ₁(ζr)e^−βzdζ+ (3.15)

Z _∞

0

(2ζ²−k²)p

ζ² −k²

ζF(ζ) sin(ζr₀)ζJ₀(ζr)e^−βzdζ+

− Z _∞

0

(2ζ²−k²)p

ζ²−k²

ζ³rF(ζ) sin(ζr0)ζ²J1(ζr)e^−βzdζ

¸

Uφ = P cosφ 2πµr₀

·

− Z _∞

0

2p

ζ²−k²

ζrF(ζ) sin(ζr0)ζ²J1(ζr)e^−αz dζ+ Z _∞

0

sin(ζr0) ζp

ζ²−k²ζJ0(ζr)e^−βzdζ+

(3.16)

− Z _∞

0

sin(ζr₀) ζ³rp

ζ²−k²ζ²J₁(ζr)e^−βzdζ+ Z _∞

0

(2ζ²−k²)p

ζ² −k²

ζ³rF(ζ) sin(ζr₀)ζ²J₁(ζr)e^−βzdζ

¸

U_z = P sinφ 2πµr0

· Z _∞

0

2p

ζ²−h²p

ζ²−k²

ζF(ζ) sin(ζr₀)ζ²J₁(ζr)e^−αzdζ+

(3.17)

− Z _∞

0

(2ζ²−k²)

ζF(ζ) sin(ζr0)ζ²J1(ζr)e^−βzdζ

¸

where,

α = p

ζ²−h² and β =p

ζ²−k²

the symbols commonly used for the P and S wave velocities, α and β, are here only used in the exponentials to distinguish which integrals are associated with compres- sional waves (α) and shear waves (β). Instead, the symbols V_P and V_S specify the wave velocities. Also,

h = ω

V_p and k= ω

V_s and Λ = k h = V_P

V_S

where h and k are wavenumbers, and F(ζ) is the Rayleigh frequency equation (By- croft, 1956) given by,

F(ζ) = (k²−2ζ²)²−4p

ζ² −h²p

ζ²−k²ζ²

(3.18)

= (k²−2ζ²)²−4αβζ²

where ζ = −msinθ (the saddle point) and m can take on the values of either h or k, depending on the type of waves, P or S, being studied, respectively. Expanding F(ζ) for both saddle points, it is found that the Rayleigh frequency equation becomes either

F(−hsinθ) = h⁴ h

(Λ² −2 sin²θ)²−4 sin²θcosθp

Λ²−sin²θ i

(3.19)

or

F(−ksinθ) = k⁴

·

(1−2 sin²θ)² −4 sin²θcosθ r 1

Λ² −sin²θ

¸

(3.20)

After performing the necessary integrations and transforming the resulting dis- placement fields to obtain a spherical coordinate set by utilizing the following trans- formation,

U_R=U_rsinθ+U_zcosθ U_θ =U_rcosθ−U_zsinθ (3.21)

As illustrated in Appendix E.2, the first-order terms (_R¹) of the displacement field are as follows

U_R = P 2πRµ

·h⁴p

Λ²−sin²θ

F(−hsinθ) sin 2θsinφ

¸

sin(ωt−hR) (3.22)

U_θ = P 2πRµ

·k⁴(1−2 sin²θ)

F(−ksinθ) cosθsinφ

¸

sin(ωt−kR) (3.23)

U_φ = P

2πRµ [ cosφ] sin(ωt−kR) (3.24)

where the sinusoidal time variant has been incorporated into the solution, and equal signs are used for the displacement field as only the first-order terms will be used in a later section to estimate the radiated energy of the model. This field is in agreement with the asymptotic field computed by Cherry (1962). The higher order terms are omitted for brevity, as for most of the integrals, they include several dozen terms and would be too cumbersome to read. If the reader wishes to obtain these terms, he/she can use the method given in Appendix E to compute them for the integrals presented here or ones with similar structure. Figures 3.3 through 3.5 provide plots of the displacement fields given by Equations 3.22 through 3.24, minus the sinusoidal time term. It should be noted here that the reason that Figure 3.4a has the large peak associated with it is that the square root term in Equation 3.20 becomes imaginary when sinθ= _Λ¹, and the behavior of the amplitude of U_θ changes dramatically at this point. This is not the case for Equation 3.19, and therefore terms containing it do not show this behavior.

Taking the time derivative of the displacement field to obtain the velocity field to

Phase (radians)

θ (radians) 0.2 0.4 0.6 0.8 1 1.2 1.4 -3

-2 -1 0 1 2 y 3

z

0.05 0.15 0.25

0.05

0.15

0.25

x

y -0.1 -0.05 0.05 0.1

-0.1 -0.05 0.05 0.1

A) B) C)

Figure 3.3 Displacement field forU_R. The gray arrows through the origin show the excitation direction for a horizontal force.

All three figures correspond to the first term of the expansion (1/Rterm). Figure A shows a vertical polar cross section (φ= π/2), figure B the corresponding phase argument, and figure C the corresponding polar map views of the radiation patterns (θ=π/8,z = 0). The harmonic source for the left and right plots is located at the plot’s origin, and it oscillates along the x axis.

Phase (radians)

θ (radians) 0.2 0.4 0.6 0.8 1 1.2 1.4 -3

-2 -1 0 1 2 y 3

z

0.5 1 1.5 2 2.5

0.5 1 1.5 2 2.5

x

y

-1 -0.5 0.5 1

-1 -0.5 0.5

1

A) B) C)

Figure 3.4 Displacement field forU_θ. The gray arrows through the origin show the excitation direction for a horizontal force.

All three figures correspond to the first term of the expansion (1/Rterm). Figure A shows a vertical polar cross section (φ= π/2), figure B the corresponding phase argument, and figure C the corresponding polar map views of the radiation patterns (θ = π/8, z = 0). The large peak present at approximately 35^o from the vertical axis is due to the asymptotic expansion.

The harmonic source for the left and right plots is located at the plot’s origin, and it oscillates along the x axis.

0.2 0.4 0.6 0.8 1 1.2 1.4 -3

-2 -1 0 1 2 3

Phase (radians)

θ (radians)

0.2 0.6 1 1.4

0.2

0.6

1

1.4

y

z

-1 -0.5 0.5 1

-1 -0.5 0.5 1

x

y

A) B) C)

Figure 3.5 Displacement field for U_φ. The gray arrows through the origin show the excitation direction for a horizontal force.

All three figures correspond to the first term of the expansion (1/Rterm). Figure A shows a vertical polar cross section (φ= π/2), figure B the corresponding phase argument, and figure C the corresponding polar map views of the radiation patterns (θ=π/8,z = 0). The harmonic source for the left and right plots is located at the plot’s origin, and it oscillates along the x axis.

be integrated at a later section,

V_R = P ω 2πRµ

·h⁴p

Λ²−sin²θ

F(−hsinθ) sin 2θsinφ

¸

cos(ωt−hR) (3.25)

V_θ = P ω 2πRµ

·k⁴(1−2 sin²θ)

F(−ksinθ) cosθsinφ

¸

cos(ωt−kR) (3.26)

V_φ = P ω

2πRµ [ cosφ] cos(ωt−kR) (3.27)

the final result is achieved.