2.4.1. Active Pile

For the active pile, the continuity condition is:

$$\begin{Bmatrix} w^{a2}(0) \\ \phi^{a2}(0) \\ M^{a2}(0) \\ Q^{a2}(0) \\ 1 \end{Bmatrix} = \begin{Bmatrix} w^{a1}(L\_w) \\ \phi^{a1}(L\_w) \\ M^{a1}(L\_w) \\ Q^{a1}(L\_w) \\ 1 \end{Bmatrix} \tag{35}$$

Substituting Equations (10) and (35) into Equation (16), the overall transfer matrix can be obtained:

$$\begin{Bmatrix} w^a(L) \\ \phi^a(L) \\ M^a(L) \\ Q^a(L) \\ 1 \end{Bmatrix} = \{T^{a2}\} \{T^{a1}\} \begin{Bmatrix} w^a(0) \\ \phi^a(0) \\ M^a(0) \\ Q^a(0) \\ 1 \end{Bmatrix} \tag{36}$$

Substituting boundary conditions:

$$\begin{array}{l} M\_{z=0} = \frac{d^2 w(z)}{dz^2} \Big|\_{z=0} = 0\\ Q\_{z=0} = \frac{d^3 w(z)}{dz^3} \Big|\_{z=0} = 0\\ w\_{z=L} = w(z)|\_{z=L} = 0\\ \theta\_{z=L} = \frac{dw(z)}{dz} \Big|\_{z=L} = 0 \end{array} \tag{37}$$

Equation (36) can then be solved, and the displacement, rotation angle, bending moment and the shearing force of the active pile top can be obtained; the value of single pile swaying impedance *K*∗ *hh*, rocking impedance *K*<sup>∗</sup> *rr*, swaying-rocking impedance *K*<sup>∗</sup> *hr*, and rocking-swaying impedance *K*∗ *rh* can then be obtained.

#### 2.4.2. Passive Pile

For the passive pile, the continuity condition is:

$$\begin{Bmatrix} w^{p^2}(0) \\ \Phi^{p^2}(0) \\ M^{p^2}(0) \\ Q^{p^2}(0) \\ 1 \end{Bmatrix} = \begin{Bmatrix} w^{p^1}(L\_w) \\ \Phi^{p^1}(L\_w) \\ M^{p^1}(L\_w) \\ Q^{p^1}(L\_w) \\ 1 \end{Bmatrix} \tag{38}$$

Substituting Equations (10), (11) and (38) into Equation (25), the overall transfer matrix can be obtained:

$$\begin{Bmatrix} w^p(L) \\ \Phi^p(L) \\ M^p(L) \\ Q^p(L) \\ 1 \end{Bmatrix} = \left\{ T\_1^{p2} \right\} \left\{ T^{p1} \right\} \begin{Bmatrix} w^p(0) \\ \Phi^p(0) \\ M^p(0) \\ Q^p(0) \\ 1 \end{Bmatrix} + \left\{ T\_2^{p2} \right\} \left\{ T^{a1} \right\} \begin{Bmatrix} w^a(0) \\ \Phi^a(0) \\ M^a(0) \\ Q^a(0) \\ 1 \end{Bmatrix} \tag{39}$$

Substituting the solution to Equation (36) and boundary conditions (37), the equation can then be solved, and the displacement, rotation angle, bending moment, and the shearing force of the passive pile top can be obtained.

#### 2.4.3. Active Pile under Secondary Wave

For the active pile under secondary wave:

$$\begin{Bmatrix} \overline{w}^{a2}(0) \\ \overline{\phi}^{a2}(0) \\ \overline{M}^{a2}(0) \\ \overline{Q}^{a2}(0) \\ 1 \end{Bmatrix} = \begin{Bmatrix} \overline{w}^{1}(L\_{w}) \\ \overline{\phi}^{a1}(L\_{w}) \\ \overline{M}^{11}(L\_{w}) \\ \overline{Q}^{11}(L\_{w}) \\ 1 \end{Bmatrix} \tag{40}$$

Substituting Equations (10), (11) and (40) into Equation (34), the overall transfer matrix can be obtained:

$$\begin{Bmatrix} \overline{w}^{a}(L) \\ \overline{\Phi}^{a}(L) \\ \overline{M}^{a}(L) \\ \overline{Q}^{a}(L) \\ 1 \end{Bmatrix} = T\_{1} \{ \overline{T}^{a1} \} \left\{ \begin{matrix} \overline{w}^{a}(0) \\ \overline{\Phi}^{a}(0) \\ \overline{M}^{a}(0) \\ \overline{Q}^{a}(0) \\ 1 \end{matrix} \right\} + T\_{2} \{ T^{p1} \} \left\{ \begin{matrix} w^{p}(0) \\ \Phi^{p}(0) \\ M^{p}(0) \\ Q^{p}(0) \\ 1 \end{matrix} \right\} + T\_{3} \{ T^{d1} \} \left\{ \begin{matrix} w^{a}(0) \\ \Phi^{a}(0) \\ M^{a}(0) \\ Q^{a}(0) \\ 1 \end{matrix} \right\} \tag{41}$$

Substituting the solutions to Equations (36) and (39), and boundary conditions (37), the equation can be solved. The displacement, rotation angle, bending moment, and the shearing force of the active pile top under the secondary wave can be obtained.

2.4.4. Pile–Pile Interaction and Pile Group Dynamic Response

The pile–pile interaction factor is defined as [31]:

$$
\pi^G = \frac{w^p}{w^a} \tag{42}
$$

where *wa* is the displacement of the active pile and *wp* is the displacement of the passive pile. To reflect the influence of the passive pile to the active pile, the definition of the pile–pile interaction factor is modified as [32]:

$$\begin{array}{l} \kappa^{G} = \frac{w^{p}}{\overline{w^{t}} - \overline{w}^{t}} = \frac{w^{p}}{(1 - \kappa)w^{t}}\\ \kappa = \frac{\overline{w^{t}}}{\overline{w^{t}}} \end{array} \tag{43}$$

where *wa* is the displacement of active pile under secondary wave.

Suppose the number of piles of the pile group is n, the cap is rigid, and the mass of the cap is ignored. Under the lateral harmonic load *Peiω<sup>t</sup>* , the lateral displacement of the pile group *w<sup>G</sup>* can be considered equal to the displacement of each pile *w<sup>i</sup>* , which means *w*<sup>1</sup> = *w<sup>i</sup>* = *wG*. In the pile group, each pile plays the role of both active pile and passive pile, so the single pile displacement can be calculated:

$$w^i = \left(1 - \sum\_{j=1, j \neq i}^n \kappa\_{ij}\right) w^i + \sum\_{j=1, j \neq i}^n a\_{ij}^G w^j \tag{44}$$

The dynamic equation of the pile group can be written in matrix form:

$$\frac{1}{K\_{h\hbar}^{\*}} \begin{bmatrix} 1 & \sum\_{j=1, j\neq 1}^{n} \kappa\_{1j} & a\_{12}^{G} & \cdots & a\_{1n}^{G} \\ & a\_{21}^{G} & 1 - \sum\_{j=1, j\neq 2}^{n} \kappa\_{2j} & \cdots & a\_{2n}^{G} \\ & \vdots & \vdots & \vdots & \vdots \\ & & a\_{n1}^{G} & & \cdots & 1 - \sum\_{j=1, j\neq n}^{n} \kappa\_{nj} \end{bmatrix} \begin{Bmatrix} P\_{1} \\ P\_{2} \\ \vdots \\ P\_{n} \end{Bmatrix} = \begin{Bmatrix} w^{1} \\ w^{2} \\ \vdots \\ w^{n} \end{Bmatrix} \tag{45}$$

where *K*∗ *hh* is the single pile impedance, which can be obtained by solving the Equation (36), *Pj* is the load distributed to pile *j*, and *a<sup>G</sup> ij* is the interaction factor between pile *i* and *j*, *i* = *j*. The load applied to the pile group is the sum of the load applied to each pile:

$$P = \sum\_{j=1}^{n} P\_j \tag{46}$$

Then, Equation (45) can be solved, and the horizontal pile group impedance can be calculated:

$$K\_{\mathbb{G}}^{\*} = P/w\_{\mathbb{G}} = K\_{\mathbb{G}} + i\mathbb{C}\_{\mathbb{G}} \tag{47}$$

According to the previous research [31], the swaying interaction factor, the swayrocking interaction factor, and the rock-swaying interaction factor between piles can be ignored. Therefore, *K*∗ *rr*, *K*<sup>∗</sup> *hr K*<sup>∗</sup> *rh* are the sum of single pile calculation results. The pile group impedance matrix can then be obtained:

$$K^\* = \begin{Bmatrix} K\_G^\* & K\_{lr}^\* \\ K\_{rh}^\* & K\_{rr}^\* \end{Bmatrix} \tag{48}$$

#### *2.5. Superstructure Dynamic Response*

#### 2.5.1. Distributed Wind Load

The distributed wind load *Fd*(*z*) can be calculated according to the equation below [33]:

$$F\_d(z) = 1/2\rho\_a \mathbb{C}\_D D\_T(z) V^2(z) \tag{49}$$

where *ρ<sup>d</sup>* is the density of the air, valued 1.225 kg/m<sup>3</sup> in this paper, *C*<sup>D</sup> is the drag coefficient, valued according to the Reynolds number and structure surface roughness, i.e., 1.2 in this paper, *V* (m/s) is the average wind speed, and *DT* (m) is the tower diameter, the value of which changes with increasing tower height.

The wind profile [34] can be calculated according to the following equation:

$$V(z) = V\_d \left(\frac{z}{H\_a}\right)^{a\_{iv}}\tag{50}$$

where *Va* is the wind speed at height *Ha* and *α<sup>w</sup>* is the power law coefficient, valued 0.12 in open seas with waves.
