Prediction of Horizontal and Torsional Vibrations of Foundations Resting on Saturated Porous Media Using Cone Model

Abstract

Seismic analysis of machinery foundations on saturated porous media and soil–structure interaction can be performed by rigorous and approximate methods. The boundary element method, indirect finite element method, and scaled boundary finite element method are among the rigorous methods, and the cone model is known as one of the most widely used approximate methods. In the latter case, the soil medium is modeled with incomplete cones and the propagation of waves is followed until the wave is sufficiently damped. In this paper, the dynamic analysis of foundations rested on saturated porous media under horizontal and torsional vibrations is investigated via the approximate cone model method. In doing so, an efficient framework based on the cone model method has been coded in MATLAB and validated with previous research studies. Careful consideration is spent to obtain the system of differential equations governing horizontal and torsional vibrations in a porous medium by considering the effect of soil dilatancy. Furthermore, the effect of different parameters including shear modulus, porosity, density, Poisson’s ratio has been investigated on the response of foundations under shear and torsional vibrations. The results show that the cone model formulations can offer a good level of accuracy and high calculation efficiency for the prediction of horizontal and torsional vibrations of foundations resting on saturated porous media.

Appendix A

Differential Equations for Vertical Displacement Under Shear Vibration

Due to the dilation conditions for the soil, by applying vibrational shear stress to the disk, both horizontal and vertical displacements are created. According to the relationships related to soil dilation for soil strain, the following relationship can be obtained:

$$A{\text{d}}\varepsilon = \left( {\varepsilon + \frac{\partial \varepsilon }{{\partial x}}{\text{d}}x} \right)\left( {A + \frac{\partial A}{{\partial x}}{\text{dx}}} \right) - \varepsilon A$$

(23)

Simplifying Eq. (23) gives:

$$A{\text{d}}\varepsilon = A\frac{\partial \varepsilon }{{\partial x}}{\text{d}}x + \varepsilon \frac{\partial A}{{\partial x}}{\text{d}}x$$

(24)

With similar operations for shear angle:

$$A{\text{d}}\gamma = A\frac{\partial \gamma }{{\partial x}}{\text{d}}x + \gamma \frac{\partial A}{{\partial x}}{\text{d}}x$$

(25)

By dividing Eqs. 24 by 25, we have

$$\frac{{A{\text{d}}\varepsilon }}{{A{\text{d}}\gamma }} = \frac{{A\frac{\partial \varepsilon }{{\partial x}}{\text{d}}x + \varepsilon \frac{\partial A}{{\partial x}}{\text{d}}x}}{{A\frac{\partial \gamma }{{\partial x}}{\text{d}}x + \gamma \frac{\partial A}{{\partial x}}{\text{d}}x}}$$

(26)

Differentiating disk area with respect to x and substituting to Eq. (26) gives:

$$\frac{{{\text{d}}\varepsilon }}{{{\text{d}}\gamma }} = \frac{{\frac{\partial \varepsilon }{{\partial x}}{\text{d}}x + \varepsilon \frac{2}{x}{\text{d}}x}}{{\frac{\partial \gamma }{{\partial x}}{\text{d}}x + \gamma \frac{2}{x}{\text{d}}x}}$$

(27)

Knowing that the strain is equal to the change in vertical displacement along the length and the shear angle is equal to the change in horizontal displacement to length, Eq. (27) is rewritten as follows.

$$\frac{{{\text{d}}\varepsilon }}{{{\text{d}}\gamma }} = \frac{{\frac{{\partial^{2} u}}{{\partial x^{2} }}{\text{d}}x + \frac{\partial u}{{\partial x}}\frac{2}{x}{\text{d}}x}}{{\frac{{\partial^{2} v}}{{\partial x^{2} }}{\text{d}}x + \frac{\partial v}{{\partial x}}\frac{2}{x}{\text{d}}x}}$$

(28)

where u and v are total vertical and horizontal displacements of the soil, respectively. By simplifying the numerator and denominator of the above fraction, the following equation is obtained which is equal to the sine function of dilation degree ($\psi$) as follows:

$$\frac{{{\text{d}}\varepsilon }}{{{\text{d}}\gamma }} = \frac{{\frac{{\partial^{2} \left( {xu} \right)}}{{\partial x^{2} }}}}{{\frac{{\partial^{2} \left( {xv} \right)}}{{\partial x^{2} }}}} = - \sin \psi$$

(29)

Equation (29) can be rewritten in the following form:

$$\frac{{\partial^{2} \left( {xu} \right)}}{{\partial x^{2} }} = - \frac{{\partial^{2} \left( {xv} \right)}}{{\partial x^{2} }}\sin \psi$$

(30)

The total vertical displacement due to shear vibration is equal to the sum of the vertical displacement of the fluid ($u_{f}$) and solid ($u_{s}$) phases as:

$$u = u_{f} + u_{s}$$

(31)

By substituting Eqs. 31 to 30, the following expression is developed:

$$\frac{{\partial^{2} \left( {x\left( {u_{f} + u_{s} } \right)} \right)}}{{\partial x^{2} }} = - \sin \psi \frac{{\partial^{2} \left( {xv} \right)}}{{\partial x^{2} }}$$

(32)

Separating the left part of Eq. 32 gives:

$$\frac{1}{ - \sin \psi }\left( {\frac{{\partial^{2} \left( {xu_{s} } \right)}}{{\partial x^{2} }} + \frac{{\partial^{2} \left( {xu_{f} } \right)}}{{\partial x^{2} }}} \right) = \frac{{\partial^{2} \left( {xv} \right)}}{{\partial x^{2} }}$$

(33)

On the other hand, the differential equation of fluid conservation that was proved by Amini [43] (which is based on the motion of fluid following the Darcy law) is defined as follows:

$$D_{f} \frac{1 - n}{n}\frac{{\partial^{2} }}{{\partial x^{2} }}\left( {xu_{s} } \right) + D_{f} \frac{{\partial^{2} }}{{\partial x^{2} }}\left( {xu_{f} } \right) = \rho_{f} \frac{{\partial^{2} \left( {xu_{f} } \right)}}{{\partial t^{2} }} + \frac{{n\rho_{f} g}}{k}\left( {\frac{{\partial \left( {xu_{f} } \right)}}{\partial t} - \frac{{\partial \left( {xu_{s} } \right)}}{\partial t}} \right)$$

(34)

In this relation, k and g are coefficient of permeability and the acceleration of gravity, respectively. $D_{f}$ is the fluid bulk modulus and it is defined as:

$$D_{f} = \frac{1}{{\beta_{f} + \frac{1 - s}{p}}}$$

(35)

where $\beta_{f}$ is fluid compressibility and p is absolute pressure equal to one atmosphere. Due to the fact that fluid contains air bubbles$,{ }D_{f} { }$ is strongly depended on the degree of soil saturation. In this study, $1/\beta_{f}$ is assumed as 2140 MPa [43]. Putting Eqs. 34 and 35 in a system of equations and solving equations simultaneously, the unknowns that are the vertical displacements of solid and fluid phases due to shear vibration can be obtained:

$$\left\{ {\begin{array}{*{20}l} {\frac{1}{ - \sin \psi }\left( {\frac{{\partial^{2} \left( {xu_{s} } \right)}}{{\partial x^{2} }} + \frac{{\partial^{2} \left( {xu_{f} } \right)}}{{\partial x^{2} }}} \right) = \frac{{\partial^{2} \left( {xv} \right)}}{{\partial x^{2} }}} \hfill \\ {D_{f} \frac{1 - n}{n}\frac{{\partial^{2} \left( {xu_{s} } \right)}}{{\partial x^{2} }} + D_{f} \frac{{\partial^{2} \left( {xu_{f} } \right)}}{{\partial x^{2} }} = \rho_{f} \frac{{\partial^{2} \left( {xu_{f} } \right)}}{{\partial t^{2} }} + \frac{{n\rho_{f} g}}{k}\left( {\frac{{\partial \left( {xu_{f} } \right)}}{\partial t} - \frac{{\partial \left( {xu_{s} } \right)}}{\partial t}} \right)} \hfill \\ \end{array} } \right.$$

(36)

To solve Eq. 36, the following equations should be used to convert Eq. 36 from the time to frequency domain.

$$\left\{ {\begin{array}{*{20}c} {xv\left( {x,\omega ,t} \right) = V\exp \left( {\gamma x} \right)\exp \left( {i\omega t} \right)} \\ {xu_{s} \left( {x,\omega ,t} \right) = U_{s} \exp \left( {\gamma x} \right)\exp \left( {i\omega t} \right)} \\ {xu_{f} \left( {x,\omega ,t} \right) = U_{f} exp\left( {\gamma x} \right)exp\left( {i\omega t} \right)} \\ \end{array} } \right.$$

(37)

Substituting the right-hand sides of Eqs. (37) to (36) gives:

$$\left\{ {\begin{array}{*{20}l} {\frac{1}{ - \sin \psi }\left( {U_{s} + U_{f} } \right) = V} \hfill \\ {\left( {\gamma^{2} D_{f} \frac{1 - n}{n} + i\omega \frac{{n\rho_{f} g}}{k}} \right)U_{s} + \left( {\gamma^{2} D_{f} + \omega^{2} \rho_{f} - i\omega \frac{{n\rho_{f} g}}{k}} \right)U_{f} = 0} \hfill \\ \end{array} } \right.$$

(38)

where V is the horizontal displacement. As can be seen from Eq. 37 in the shear vibration, the vertical displacements of solid and fluid phases for the soil under the massless rigid disk are a function of horizontal displacement of the soil. Further simplification gives:

$$\left\{ {\begin{array}{*{20}l} {{ }U_{f} = \frac{{A_{1} A_{2} }}{{A_{3} }}V} \hfill \\ {{ }U_{s} = \left( { - A_{2} - \frac{{A_{1} A_{2} }}{{A_{3} }}} \right)V} \hfill \\ \end{array} } \right.$$

(39)

where $U_{f}$ and $U_{s}$ represent the vertical displacements of the fluid and solid phase, respectively, so that both are expressed in terms of V. Parameters A₁, A₂ and A₃ are defined through Eqs. 40 to 42.

$$A_{1} = \gamma^{2} D_{f} \frac{1 - n}{n} + i\omega \frac{{n\rho_{f} g}}{k}$$

(40)

$$A_{2} = \sin \psi$$

(41)

$$A_{3} = \left( {\gamma^{2} D_{f} + \omega^{2} \rho_{f} - 2i\omega \frac{{n\rho_{f} g}}{k} - \gamma^{2} D_{f} \frac{1 - n}{n}} \right)$$

(42)

Vertical Dynamic Stiffness Under the Simultaneous Effect of Shear and Vertical Vibrations

To calculate the vertical dynamic stiffness, the vertical displacement due to the application of normal and shear stresses to the disk located on half-space is considered. It should be noted that the former case was studied by Amini [43], and the latter will be discussed in this section. For the case where the foundation is only subjected to vertical vibration, it can be written as:

$$P_{0} \left( \omega \right) = S\left( \omega \right)U_{0} \left( \omega \right)$$

(43)

when the foundation is under the effect of both vertical and shear or vertical and torsional vibration, the force–displacement relationship is expressed as:

$$P_{0} \left( \omega \right) = S^{\prime}\left( \omega \right)\left[ {U_{0} \left( \omega \right) + U_{{0_{s} }} \left( \omega \right)} \right]$$

(44)

Combining Eqs. 43 and 44 yields:

$$S\left( \omega \right)U_{0} \left( \omega \right) = S^{\prime}\left( \omega \right)\left[ {U_{0} \left( \omega \right) + U_{{0_{s} }} \left( \omega \right)} \right]$$

(45)

Vertical stiffness under both shear and vertical vibrations is obtained as Eq. 46, which is a coefficient of vertical stiffness resulting from vertical vibration. This coefficient is a function of both types of vertical displacement.

$$S^{\prime}\left( \omega \right) = S\left( \omega \right)\frac{{U_{0} \left( \omega \right)}}{{\left[ {U_{0} \left( \omega \right) + U_{{0_{s} }} \left( \omega \right)} \right]}}$$

(46)

Differential Equations for Displacement Under Torsional Vibration

Consider the disk shown in Fig. 17, which is modeled on the surface of a homogeneous half-space with shear modulus G and mass density , as a one-sided incomplete semi-infinite circular cone with the same material properties as the half-space. A torsional moment (torque) with an amplitude M is applied to the disk, resulting in a torsional rotation (twisting) of the disk. The downward wave propagation with amplitude $\vartheta \left( {x,\omega } \right)$ and the cone’s opening angle of the torsional cone are to be determined. For transition cones, the area increases with increasing depth, but in rotary cones, increasing the depth increases the polar moment of inertia, so it is:

Fig. 17

Cone model for rotational degree of freedom

$$\frac{I\left( x \right)}{{I_{0} }} = \left( {\frac{x}{{x_{0} }}} \right)^{4}$$

(47)

where $I_{0} = \pi r_{0}^{4} /2$. According to Fig. 17, the equation of motion is expressed in the following form for an infinitesimally small element:

$$- T\left( {x,\omega } \right) + T\left( {x,\omega } \right) + T\left( {x,\omega } \right)_{{{^{\prime}}x}} {\text{d}}x + \omega^{2} I\left( x \right){\text{d}}x\left( {\left( {1 - n} \right)\rho_{s} + n\rho_{f} } \right)\vartheta = 0$$

(48)

which can be also expressed in the following form:

$$\frac{\partial T}{{\partial x}} + \omega^{2} I\left( x \right)\left( {\left( {1 - n} \right)\rho_{s} + n\rho_{f} } \right)\vartheta = 0$$

(49)

Substituting ${\text{I}}\left( {\text{x}} \right) = \frac{{{\pi r}_{0}^{4} }}{2}\left( {\frac{{\text{x}}}{{{\text{x}}_{0} }}} \right)^{4}$ and ${\text{T}} = {\text{G}}\vartheta {\text{I}}\left( {\text{x}} \right)$ to Eq. (49) gives:

$$\frac{{\partial^{2} \vartheta }}{{\partial x^{2} }} + \frac{4}{x}\frac{\partial \vartheta }{{\partial x}} + \frac{{\omega^{2} }}{{c_{ss}^{2} }}\vartheta = 0$$

(50)

where c_ss is shear wave velocity. The answer to this equation of motion can be calculated as:

$$\vartheta \left( {x,\omega } \right) = c_{2} \left( {\frac{1}{{x^{3} }} + i\frac{\omega }{{c_{ss} }}\frac{1}{{x^{2} }}} \right)e^{{ - i\frac{\omega }{{c_{ss} }}x}}$$

(51)

where c₂ is the integration constant. The term with $e^{{ - i\omega x/c_{ss} }}$ corresponds to a wave propagating in the positive x-direction with the velocity c_ss. The other term with $e^{{i\omega x/c_{ss} }}$ (not included in Eq. 51) corresponds to a wave propagating in the negative x-direction, i.e., towards the disk, which is physically impossible. Enforcing the boundary condition $\vartheta \left( {x = x_{0} ,\omega } \right) = \vartheta_{0} \left( \omega \right)$ gives the final form of displacement for torsional vibration as follows:

$$\vartheta \left( {x,\omega } \right) = \frac{{\left( {\frac{{x_{0}^{3} }}{{x^{3} }}} \right) + i\left( {\frac{\omega }{{c_{ss} }}} \right)\left( {\frac{{x_{0}^{3} }}{{x^{2} }}} \right)}}{{1 + i\left( {\frac{\omega }{{c_{ss} }}} \right)x_{0} }}e^{{ - i\frac{\omega }{{c_{ss} }}\left( {x - x_{0} } \right)}} \vartheta_{0} \left( \omega \right)$$

(52)

Torsional Dynamic Stiffness Under Torsional Vibration

In the case of torsional vibration, the equilibrium is established for the common boundary of disk and homogeneous half-space, i.e., $x_{0} = x$, to obtain the soil stiffness. At this point, the amplitude of the torsional moment applied to the disk is equal to the sub-disk moment of inertia.

$$M\left( {x = x_{0} ,\omega } \right) = - T\left( {x = x_{0} ,\omega } \right)$$

(53)

The torsion–rotation relationship is defined as:

$$T\left( \omega \right) = {\text{GI}}\left( x \right)\vartheta \left( {x,\omega } \right)_{{,_{x} }}$$

(54)

Differentiating $\vartheta \left( {{\text{x}},{\upomega }} \right)$ with respect to x and for $x_{0} = x$ gives:

$$M_{0} \left( x \right) = {\text{GI}}_{0} \left[ {\frac{{3\left( {\frac{1}{{x_{0} }}} \right) + 2i\left( {\frac{\omega }{{c_{ss} }}} \right)}}{{1 + i\left( {\frac{\omega }{{c_{ss} }}} \right)x_{0} }} - \frac{i\omega }{{c_{ss} }}} \right]\vartheta_{0} \left( \omega \right)$$

(55)

The dynamic stiffness of a homogeneous porous half-space with respect to the torsion–rotation relationship is expressed as follows:

$$M_{0} \left( \omega \right) = S\left( \omega \right)\vartheta_{0} \left( \omega \right)$$

(56)

Finally, torsional stiffness is equal to:

$$S\left( \omega \right) = {\text{GI}}_{0} \left[ {\frac{{3\left( {\frac{1}{{x_{0} }}} \right) + 2i\left( {\frac{\omega }{{c_{ss} }}} \right)}}{{1 + i\left( {\frac{\omega }{{c_{ss} }}} \right)x_{0} }} - \frac{i\omega }{{c_{ss} }}} \right]$$

(57)