Determining anti-plane responses of symmetric canyon embedded within single-layer half-space

Abstract

This paper proposes a modified hybrid method combining finite element method and a Lamb series, to derive the displacement amplitude of a symmetric canyon embedded within a single-layer half-space subjected to shear horizontal (SH) waves. Four canyon shapes and various site effects were also examined under various incident angles and dimensionless frequencies. The site effects included the canyon-decay effects and canyon-area effects (due to the existing canyon) as well as the thickness effect (due to the soft layer). We discuss the resonance frequencies of the single-layer half-space under various shifts in incident angle imposed by the existed canyon. We also compare the responses obtained in this paper with those of a canyon embedded in a half-space. Finally, we employed the fast Fourier transform to obtain responses in the time domain with series responses in the frequency domain. These results support our discovery that site effects dominated responses at the free surface. This paper provides a valuable reference furthering our understanding of site effects associated with surface irregularities in a single-layer half-space.

Appendix 1

The single-layer half-space in Fig. 

Fig. 19

Schematic diagram showing single-layer half-space under excitation via surface loading

19 was disturbed by a virtual load acting at the original point:

$$ \left( {\sigma_{zy1}^{s} } \right)_{m} = - \mu_{1} \delta^{\left( m \right)} \left( x \right),\;{\text{at}}\;\left( {x,z} \right) = \left( {0,0} \right) $$

(A-1)

where $\delta \left( x \right)$ is the Dirac delta function. Symbol $\delta^{\left( m \right)}$ represents the m-th derivative of $\delta \left( x \right)$ with respect to x. The virtual load in Eq. (A-1) is one type of Lamb load.

The displacements generated by the disturbance satisfied the governing equation as well as the corresponding boundary conditions as follows:

$$ \frac{{\partial^{2} u_{yj}^{s} }}{{\partial x^{2} }} + \frac{{\partial^{2} u_{yj}^{s} }}{{\partial z^{2} }} + k_{sj}^{2} u_{yj}^{s} = 0;\;j = 1\;,\;2 $$

(A-2)

$$ \left( {\sigma_{zy1}^{s} } \right)_{m} = \left( {\sigma_{zy2}^{s} } \right)_{m} ,\;{\text{at}}\;z = H $$

(A-3)

$$ \left( {u_{y1}^{s} } \right)_{m} = \left( {u_{y2}^{s} } \right)_{m} ,\;{\text{at}}\;z = H $$

(A-4)

where superscript s refers to displacements or stresses in the scattering field. The solutions to Eq. (A-2) are series functions, based on the given value of m, which can be obtained using the following boundary conditions: (1) traction free at the free surface except at the original point, as shown in Eqs. (1) and (2) continuous displacements and tractions along interface L, as shown in Eqs. (A-3) and (A-4). The series functions also satisfied the radiation condition, which means that the energy of waves decayed after propagating over an extended distance. This makes it possible to express the displacements $\left( {u_{yj}^{s} } \right)_{m}^{{}}$, stresses including $\left( {\sigma_{xyj}^{s} } \right)_{m}^{{}}$ and $\left( {\sigma_{zyj}^{s} } \right)_{m}^{{}}$, as well as tractions $\left( {t_{yj}^{s} } \right)_{m}$ in the scattered field, as follows:

$$ \left( {u_{y1}^{s} } \right)_{m}^{{}} = \frac{1}{2\pi }\int_{ - \infty }^{\infty } {A_{U}^{1} e^{{ - \nu^{\prime}_{1} (H - z) - ikx}} dk} + \frac{1}{2\pi }\int_{ - \infty }^{\infty } {A_{D}^{1} e^{{ - \nu^{\prime}_{1} z - ikx}} dk} $$

(A=5a)

$$ \left( {u_{y2}^{s} } \right)_{m}^{{}} = \frac{1}{2\pi }\int_{ - \infty }^{\infty } {A_{D}^{2} e^{{ - \nu^{\prime}_{2} (z - H) - ikx}} dk} $$

(A-5b)

$$ \nu^{\prime}_{j} = \sqrt {k^{2} - k_{sj}^{2} } ,\;\;\;\;\;\;j = 1\;,\;2 $$

(A-5c)

$$ A_{D}^{1} = \frac{1}{{1 - R_{D}^{12} e^{{ - 2\nu^{\prime}_{1} H}} }}\frac{{\left( { - {\text{i}}k} \right)^{n} }}{{\nu^{\prime}_{1} }};\;A_{U}^{1} = R_{D}^{12} A_{D}^{1} ;\;A_{D}^{2} = T_{D}^{12} A_{D}^{1} $$

(A=5d)

$$ R_{D}^{12} = \frac{{\mu_{1} \nu^{\prime}_{1} - \mu_{2} \nu^{\prime}_{2} }}{{\mu_{1} \nu^{\prime}_{1} + \mu_{2} \nu^{\prime}_{2} }}e^{{ - \nu^{\prime}_{1} H}} ;\;T_{D}^{12} = \frac{{2\mu_{1} \nu^{\prime}_{1} }}{{\mu_{1} \nu^{\prime}_{1} + \mu_{2} \nu^{\prime}_{2} }}e^{{ - \nu^{\prime}_{1} H}} $$

(A-5e)

and

$$ \left( {\sigma_{xyj}^{s} } \right)_{m}^{{}} = \mu_{j} \frac{\partial }{\partial x}\left( {u_{yj}^{s} } \right)_{m}^{{}} ;\;\left( {\sigma_{zyj}^{s} } \right)_{m}^{{}} = \mu_{j} \frac{\partial }{\partial z}\left( {u_{yj}^{s} } \right)_{m}^{{}} $$

(A-6a)

$$ \left( {t_{yj}^{s} } \right)_{m} = \left( {\sigma_{xyj}^{s} } \right)_{m} n_{x}^{{}} + \left( {\sigma_{zyj}^{s} } \right)_{m} n_{z} ;\;j = 1,\;2. $$

(A-6b)

Note that $\left( {u_{yj}^{s} } \right)_{m}$ is expressed using the integral form, as shown in Eq. (A-5a) or Eq. (A-5b). It can be obtained using the modified steepest descent method [37].