4. Free vibration of power conductor

The self-damping model of the power conductor provided the basis to analyze free vibration experienced by the conductor. Free vibration occurs when the forcing function causing the power conductor to vibrate become zero. Hence the equation of motion is expressed in Eq. (28) as [19]:

$$EI\frac{\partial^4 y(\mathbf{x},t)}{\partial \mathbf{x}^4} - S\frac{\partial^2 y(\mathbf{x},t)}{\partial \mathbf{x}^2} + \beta I \frac{\partial^5 y(\mathbf{x},t)}{\partial \mathbf{x}^4 \partial t} + C\frac{\partial y(\mathbf{x},t)}{\partial t} + \rho A \frac{\partial^2 y(\mathbf{x},t)}{\partial t^2} = 0 \tag{28}$$

Applying the principle of separation of variable to the equation of motion yields Eq. (29):

$$\stackrel{\text{III}}{\text{EI}}\stackrel{\text{III}}{\text{X}}(\mathbf{x})\mathbf{T}(\mathbf{t}) - \stackrel{\text{III}}{\text{S}}(\mathbf{x})\mathbf{T}(\mathbf{t}) + \beta \stackrel{\text{III}}{\text{X}}^{\text{III}}(\mathbf{x})\dot{\mathbf{T}}(\mathbf{t}) + \text{CX}(\mathbf{x})\dot{\mathbf{T}}(\mathbf{t}) + \rho A \mathbf{T}(\mathbf{t})\mathbf{X}(\mathbf{x}) = \mathbf{0} \tag{29}$$

� � <sup>n</sup>π<sup>x</sup> Integrating the eigenfunction Xn <sup>x</sup> <sup>l</sup> <sup>ð</sup> Þ ¼ sin in the model yields Eqs. (30)–(33): High Voltage Transmission Line Vibration: Using MATLAB to Implement the Finite Element Model of a Wind… 47 http://dx.doi.org/10.5772/intechopen.75186

$$\begin{split} EI\left[\left(\frac{n\pi}{l}\right)^4 \sin\left(\frac{n\pi\chi}{l}\right)\right] T(t) - S\left[\left(\frac{-n\pi}{l}\right) \sin\left(\frac{n\pi\chi}{l}\right)\right] T(t) \\ + \beta I\left[\left(\frac{n\pi}{l}\right)^4 \sin\left(\frac{n\pi\chi}{l}\right)\right] \dot{T}(t) + \mathbb{C}\left[\sin\left(\frac{n\pi\chi}{l}\right)\right] \dot{T}(t) + \rho A\left[\sin\left(\frac{n\pi\chi}{l}\right)\right] \ddot{T}(t) = 0 \end{split} \tag{30}$$

$$\begin{split} \sin\left(\frac{n\pi x}{l}\right) \begin{bmatrix} EI\left(\frac{n\pi}{l}\right)^4 T(t) + S\left(\frac{n\pi}{l}\right)^2 \dot{T}(t) \\ + \beta I\left(\frac{n\pi}{l}\right)^4 T(t) + \mathcal{C}\dot{T}(t) + \rho A\ddot{T}(t) \end{bmatrix} = 0 \end{split} \tag{31}$$

$$
\rho A \ddot{T}(t) + \left[ \beta I \left( \frac{n \pi}{l} \right)^4 + \mathbb{C} \right] \dot{T}(t) + \left[ S \left( \frac{n \pi}{l} \right)^2 + EI \left( \frac{n \pi}{l} \right) \right] T(t) = 0 \tag{32}
$$

$$\ddot{T}(t) + \left[\frac{\beta I}{\rho A} \left(\frac{n\pi}{l}\right)^4 + \frac{\mathbb{C}}{\rho A}\right] \dot{T}(t) + \left[\frac{\mathbb{S}}{\rho A} \left(\frac{n\pi}{l}\right)^2 + \frac{EI}{\rho A} \left(\frac{n\pi}{l}\right)^4\right] T(t) = 0\tag{33}$$

Considering that the vibration model represents a multi-degree vibration system. The natural frequency of the power conductor is determined and expressed in Eqs. (34) and (35):

$$
\omega\_n^2 = \frac{S}{\rho A} \left(\frac{n\pi}{l}\right)^2 + \frac{EI}{\rho A} \left(\frac{n\pi}{l}\right)^4 \tag{34}
$$

$$\left\|2\xi\omega\_n\right\|^2 = \left[\frac{\beta I}{\rho A}\left(\frac{n\pi}{l}\right)^4 + \frac{\mathcal{C}}{\rho A}\right] \tag{35}$$

The temporal solution to the free vibration model is expressed in Eq. (36) as:

$$T\_n = A\_1 e^{-\xi\_n \omega\_n t} \sin \left(\omega\_d t + \phi \right) \tag{36}$$

The solution can also be represented in Eq. (37) and expressed as:

$$T\_n = e^{-\xi\_n \omega\_n t} (B\_1 \sin \omega\_d t + B\_2 \cos \omega\_d t) \tag{37}$$

Where the damped frequency of the power conductor is expressed in Eq. (38) as:

$$
\omega\_d = \omega\_n \sqrt{1 - \xi^2} \tag{38}
$$

The system response is expressed in Eq. (39) as:

$$y(\mathbf{x},t) = \sum\_{n=1}^{\infty} A\_1 e^{-\xi\_n \omega\_n t} \sin\left(\omega\_d t + \varphi\right) \sin\frac{n\pi\mathbf{x}}{l} \tag{39}$$

The response can also be represented in Eq. (40) and expressed as:

$$y(\mathbf{x}, t) = \sum\_{n=1}^{\infty} \left[ e^{-\xi\_n \omega\_d t} (B\_1 \sin \omega\_d t + B\_2 \cos \omega\_d) \right] \sin \frac{n \pi \mathbf{x}}{l} \tag{40}$$
