**2. Proposed Calculation Method**

#### *2.1. Model Establishment*

As shown in Figure 2, the pile group supported 3.6 MW offshore wind turbine can be divided into two parts. The first part is the pile group foundation, which is partially embedded in the soil and partially submerged in the seawater. The second part is the superstructure, which is under distributed and thrust wind load.

**Figure 2.** Pile group supported 3.6 MW offshore wind turbine.

### *2.2. Pile Submerged in the Seawater*

The pile diameter of the pile group is relatively small compared with the wavelength (*D*/*Lwl* ≤ 0.2). Therefore, the wave load applied to the pile can be calculated by the Morison equation. For linear waves, at height z of the cylinder, the horizontal wave load *qG*(*z*) can be calculated according to Equation (1) [28]:

$$q\_{G}(z) = \frac{1}{2} \mathcal{C}\_{D} \rho\_{w} D \left( \frac{\pi H\_{W}}{T} \frac{\cosh(KL\_{w})}{\sinh(KL\_{w})} \right)^{2} \cos \theta |\cos \theta| dz + \\ \mathcal{C}\_{M} \frac{\pi D^{2}}{4} \frac{2 \pi^{2} H\_{W}}{T^{2}} \frac{\cosh(KL\_{w})}{\sinh(KL\_{w})} \sin \theta dz$$

where *C*<sup>D</sup> is the drag coefficient, *CM* is the inertia coefficient, *θ* = −*ωt*, *ρ<sup>w</sup>* is the density of the seawater, *D* is the pile diameter, *Hw* is the wave height, *T* is the wave period, and *K* = <sup>2</sup>*<sup>π</sup> Lwl* , where *Lwl* is the wavelength. This paper mainly discusses the dynamic response of the pile under different frequencies; therefore, it is needed to calculate the maximum horizontal wave load applied to the whole pile body. The wave load applied to the whole pile body can be calculated according to Equation (2) [28]:

$$\begin{array}{l} F\_T = \int\_0^{L\_W} q\_G(z) = F\_{HD} \cos \theta |\cos \theta| + F\_{HI} \sin \theta\\ F\_{HD} = \mathcal{C}\_\mathcal{D} \frac{\gamma D H\_W^2}{2} K\_1, F\_{HI} = \mathcal{C}\_M \frac{\gamma \pi D^2 H\_W}{8} K\_2 \end{array} \tag{2}$$

where *K*<sup>1</sup> = <sup>2</sup>*KLw*+sinh2*KLw* 8sinh2*KLw* and *K*<sup>2</sup> = tanh*KLw*. As shown in Equation (2), the value of *θ* to determine the maximum of *FHD* and *FH I* is not the same. Therefore, it is needed to determine the value of *θ* when the total wave force is the largest. According to the calculation, the maximum total wave force depends on the value of *FHD* and *FH I*.


In engineering, *FH I* < 2*FHD* is rarely seen. Therefore, in this paper, we consider *FH I* ≥ 2*FHD*, which means *FT*max = *FH I*. Then, we can calculate the wave load applied to the cylinder at height z when the maximum total wave load happens [28]:

$$\eta\_G(z) = \mathcal{C}\_M \frac{\pi D^2}{4} \frac{2\pi^2 H\_w}{T^2} \frac{\cosh(KZ\_w)}{\sinh(KL\_w)}\tag{3}$$

Equation (3) is used for single piles in the seawater. However, the wave load applied to the pile group can be complex due to the pile–pile interaction and pile–cap interaction. According to the research [29], considering the pile–pile interaction, this paper considers the wave load applied to each pile within the pile group as 0.8 times the wave load applied to the single pile, and Equation (3) can be modified as:

$$q\_G(z) = 0.8C\_M \frac{\pi D^2}{4} \frac{2\pi^2 H\_w}{T^2} \frac{\cosh(KZ\_w)}{\sinh(KL\_w)}\tag{4}$$

Based on the water–soil interface, the local coordinate system is established, and the *z* axis is towards the seabed. The pile submerged in the seawater can be divided into n parts, and the length of each part is *hi*. According to dynamic equilibrium conditions, the dynamic equation of each pile section can be obtained:

$$\frac{d^4 w\_i^{a1}}{dz^4} + 4\gamma^4 w\_i^{a1}(z) = \frac{q\_G(z)}{E\_p I\_p} \tag{5}$$

where *w* is the lateral displacement of the pile section, superscript *a*1 denotes the first part of the active pile, and *γ* = <sup>−</sup>*mpω*<sup>2</sup> 4*Ep Ip* 1/4 , where *ω* is the load frequency, *mp* is the unit mass of the pile body, *Ep* is the modulus of elasticity of the pile, and *Ip* is the cross-section inertia moment of the pile. The solution of Equation (5) is the superposition of general solution and particular solution, as shown in Equation (6):

$$w\_i^{a1}(z) = e^{(-1+i)\gamma z} A\_{1i}^{a1} + e^{(-1-i)\gamma z} A\_{2i}^{a1} + e^{(1-i)\gamma z} A\_{3i}^{a1} + e^{(1+i)\gamma z} A\_{4i}^{a1} + q\_{\;\;G}'(z) \tag{6}$$

where *q <sup>G</sup>* = *apqG*, *ap* = <sup>1</sup> *Ep Ip*(*K*4+4*γ*4) , *Aa*<sup>1</sup> <sup>1</sup>*<sup>i</sup>* , *<sup>A</sup>a*<sup>1</sup> <sup>2</sup>*<sup>i</sup>* , *<sup>A</sup>a*<sup>1</sup> <sup>3</sup>*<sup>i</sup>* , *<sup>A</sup>a*<sup>1</sup> <sup>4</sup>*<sup>i</sup>* are undetermined coefficients, which can be obtained according to boundary conditions.

According to differential relations, we can obtain the lateral displacement *w*, rotation angle *θ*, bending moment *M*, and shearing force *Q* of the pile section. For simplicity, we define *M* = *M*/*Ep Ip* and *Q* = *Q*/*Ep Ip* to represent the equivalent bending moment and shearing force. Written in matrix form:

$$\begin{Bmatrix} w\_i^{a1} \\ \phi\_i^{a1} \\ M\_i^{a1} \\ Q\_i^{a1} \\ 1 \end{Bmatrix} = \left\{ t\_i^{a1} \right\} \begin{Bmatrix} A\_{1i}^{a1} \\ A\_{2i}^{a1} \\ A\_{3i}^{a1} \\ A\_{4i}^{a1} \\ 1 \end{Bmatrix} \tag{7}$$

where the matrix *ta*1 *i* is shown in Appendix A.

Equation (7) can be used to calculate each pile section of the pile body. The transfer matrix between the pile top and pile tip of each pile section can then be obtained, as shown in Equation (8):

$$\begin{Bmatrix} w^{a1}(h\_i) \\ \Phi^{a1}(h\_i) \\ M^{a1}(h\_i) \\ Q^{a1}(h\_i) \\ 1 \end{Bmatrix} = \left\{ t\_i^{a1}(h\_i) \right\} \left\{ t\_i^{a1}(0) \right\}^{-1} \begin{Bmatrix} w^{a1}(0) \\ \Phi^{a1}(0) \\ M^{a1}(0) \\ Q^{a1}(0) \\ 1 \end{Bmatrix} \tag{8}$$

Let {*Ta*<sup>1</sup> *<sup>i</sup>* } <sup>=</sup> *ta*1 *<sup>i</sup>* (*hi*) *ta*<sup>1</sup> *<sup>i</sup>* (0) −<sup>1</sup> and *<sup>T</sup>a*<sup>1</sup> = *<sup>T</sup>a*<sup>1</sup> *<sup>n</sup>* ··· *<sup>T</sup>a*<sup>1</sup> *<sup>i</sup>* ··· *<sup>T</sup>a*<sup>1</sup> <sup>2</sup> *<sup>T</sup>a*<sup>1</sup> <sup>1</sup> ; the transfer matrix between the pile tip (water–soil interface) and the pile top (water–air interface) can then be obtained, as shown in Equation (9):

$$\begin{Bmatrix} w^{a1}(L\_w) \\ \Phi^{a1}(L\_w) \\ M^{a1}(L\_w) \\ Q^{a1}(L\_w) \\ 1 \end{Bmatrix} = \{T^{a1}\} \begin{Bmatrix} w^{a1}(0) \\ \Phi^{a1}(0) \\ M^{a1}(0) \\ Q^{a1}(0) \\ 1 \end{Bmatrix} \tag{9}$$

Thus, we obtain the transfer matrix of the active pile submerged in the seawater. The solution can also be used for the passive pile and the active pile under the influence of the passive pile. Let {*Tp*1} <sup>=</sup> {*Ta*1} <sup>=</sup> {*Ta*<sup>1</sup> }, written in matrix form:

$$\begin{Bmatrix} w^{p1}(L\_w) \\ \Phi^{p1}(L\_w) \\ M^{p1}(L\_w) \\ Q^{p1}(L\_w) \\ 1 \end{Bmatrix} = \{T^{p1}\} \begin{Bmatrix} w^{p1}(0) \\ \Phi^{p1}(0) \\ M^{p1}(0) \\ Q^{p1}(0) \\ 1 \end{Bmatrix}$$

$$\left\{ \begin{array}{l} \overline{w}^{q1}(L\_w) \\ \overline{\Phi}^{q1}(L\_w) \\ \overline{M}^{q1}(L\_w) \\ \overline{Q}^{q1}(L\_w) \\ 1 \end{array} \right\} = \{T^{a1}\} \begin{Bmatrix} \overline{w}^{1}(0) \\ \overline{\Phi}^{q1}(0) \\ \overline{M}^{q1}(0) \\ \overline{Q}^{q1}(0) \\ 1 \end{Bmatrix} \tag{10}$$

where *wp*<sup>1</sup> is the lateral displacement of the passive pile and *wa*<sup>1</sup> is the lateral displacement of the active pile under the influence of the passive pile.
