Nonlinear vibration analysis of vortex-induced vibrations in overhead power lines with nonlinear vibration absorbers

Abstract

Vortex-induced vibrations are one of the major factors in fatigue failure of power transmission lines and can be mitigated using vibration absorbers in the form of Stockbridge dampers. Since power transmission lines play an important role in modern infrastructure, a thorough understanding of the nonlinear dynamical interactions between conductors, dampers, and wind forces is crucial. Although different nonlinear models exist for conductor vibration with attached dampers or under wind force, no work combines all these nonlinearities in a single model and examines the dynamics of the conductor along with dampers. In an attempt to fill this gap, this work combines the nonlinearities from the mid-plane stretching of the conductor, equivalent cubic stiffness of the Stockbridge damper, and fluctuating lift force modeled as a Van der Pol oscillator in a single model to investigate the nonlinear vortex-induced vibrations. In this work, the conductor is modeled as a simply supported beam and the Stockbridge damper as a mass–spring–damper–mass system with a combination of cubic and linear stiffness. The governing equations of motion are solved analytically using the method of multiple scales for the case of primary resonance between the fluctuating lift-force and conductor. Analytical findings are further validated by comparing against the numerical integration of a reduced-order system, and the results show an excellent match. The analysis is extended by conducting a parametric study to investigate the effect of different system parameters on the frequency response curves. These findings are promising and further provide a direction to design an optimal vibration absorber.

Appendix A: Expressions used in Eq. (34)

For the sake of simplicity, slow flow equations (Eq (33)), governing the amplitude and phase, can be written in a more compact form as

$$\begin{aligned}&D_2 a_y=A_{11}+B_{11}\sin (\Gamma )\,, \end{aligned}$$

(39a)

$$\begin{aligned}&D_2 q_y=A_{12}+B_{12}\cos (\Gamma )\,, \end{aligned}$$

(39b)

$$\begin{aligned}&D_2 \Gamma =\sigma +B_{14}\sin (\Gamma )-A_{13}-B_{13}\cos (\Gamma )\,, \end{aligned}$$

(39c)

where $A_{ij}$ (for $i=1,\,2,\,3$ and $j=1,\,2,\,3$) and $B_{ij}$ (for $i=1,\,2,\,3$ and $j=1,\,2,\,3,\,4$) are the function of system parameters, excitation frequency, and amplitudes $a_y$ and $q_y$. These are given by

$$\begin{aligned}&A_{11} =\frac{\sum _{p=1}^{n}\Big [{a_y} {c_{dp}}({\Psi _p}-1){Y_p}( {\xi _p}) ^{2}+{a_y} {k_p}{\Psi _{4,p}}(1-\Psi _p) {Y_p}( {\xi _p})\Big ]}{\sum _{p=1}^{n}\Big [{k_p}{Y_p}( {\xi _p}) {\Psi _{3,p}}+2{Y_p}( {\xi _p})^{2}{\alpha _{1p}}\Big ]+2b_1}\nonumber \\&\qquad -\,\frac{2{\omega _y}{a_y} {f^*_L}{\bar{\alpha }}{b_1} +2{\omega _y}{a_y} {\bar{\mu }}{ b_1}{\omega _s}}{\sum _{p=1}^{n}\Big [{\omega _s}{\omega _y}{k_p}{Y_p}( {\xi _p}) {\Psi _{3,p}}+2{\omega _s}{ \omega _y}{Y_p}( {\xi _p})^{2}{\alpha _{1p}}\Big ]+2\omega _y\omega _s b_1} \nonumber \\\end{aligned}$$

(40a)

$$\begin{aligned}&B_{11} =\frac{{f^*_L}{q_y} { b_1}{\omega _s}}{\sum _{p=1}^{n}\Big [{\omega _s}{\omega _y}{k_p}{Y_p}( {\xi _p}) {\Psi _{3,p}}+2{\omega _s}{ \omega _y}{Y_p}( {\xi _p})^{2}{\alpha _{1p}}\Big ]+2\omega _y\omega _s b_1} \nonumber \\\end{aligned}$$

(40b)

$$\begin{aligned}&A_{12}\nonumber \\&\quad ={ \omega _s}G{{C_{L0}^*}}^{2}{q_y} {b_1}- {q_y}^{3}G{\omega _s} {b_{12}},\quad \,B_{12}=\frac{1}{2b_1}{a_y}{F^*}{\omega _y} {b_1}\,, \end{aligned}$$

(40c)

$$\begin{aligned}&A_{13}={\frac{\sum _{p=1}^{n} \left( 6{a_y}^{3}{ k_p}{Y_p}( {\xi _p}) {\Psi _{2,p}}-6{a_y}^{3}{\gamma _p}{Y_p}( {\xi _p})^{4}{ \Psi _{1,p}}\right) }{\sum _{p=1}^{n} {a_y}{\omega _y} \Big [16{b_1}+8{k_p}{Y_p}( {\xi _p}) { \Psi _{3,p}}+16{Y_p}( {\xi _p})^{2}{\alpha _{1p}}\Big ]}}\nonumber \\&\qquad -\,{\frac{3{a_y} ^{3}\lambda {b_{2}}{b_{3}}}{\sum _{p=1}^{n} {a_y}{\omega _y} \Big [16{b_1}+8{k_p}{Y_p}( {\xi _p}) { \Psi _{3,p}}+16{Y_p}( {\xi _p})^{2}{\alpha _{1p}}\Big ]}} \end{aligned}$$

(40d)

$$\begin{aligned}&B_{13}={\frac{-8{f^*_L}{q_y} {b_1}}{\sum _{p=1}^{n} {a_y}{\omega _y} \Big [16{b_1}+8{k_p}{Y_p}( {\xi _p}) { \Psi _{3,p}}+16{Y_p}( {\xi _p})^{2}{\alpha _{1p}}\Big ]}},\nonumber \\&B_{14}=-\frac{1}{2q_y}{a_y} {F^*}{\omega _y }\,. \end{aligned}$$

(40e)

As mentioned in the main text, the steady state amplitudes and phase can be obtained by setting $D_2 a_y=D_2 q_y=D_2\Gamma =0$, which further leads to

$$\begin{aligned}&A^{*}_{11}+B^{*}_{11}\sin (\Gamma ^{*})=0\,, \end{aligned}$$

(41a)

$$\begin{aligned}&A^{*}_{12}+B^{*}_{12}\cos (\Gamma ^{*})=0\,, \end{aligned}$$

(41b)

$$\begin{aligned}&\sigma +B^{*}_{14}\sin (\Gamma ^{*})-A^{*}_{13}-B^{*}_{13}\cos (\Gamma ^{*})=0\,. \end{aligned}$$

(41c)

In the above equation, superscript $^{*}$ refers to steady state quantities. Equations (41a) and (41b) can be solved for $\sin (\Gamma ^{*})$ and $\cos (\Gamma ^{*})$ to get

$$\begin{aligned} \sin (\Gamma ^{*})=-\frac{A^{*}_{11}}{B^{*}_{11]}},\quad \cos (\Gamma ^{*})=-\frac{A^{*}_{12}}{B^{*}_{12}}\,. \end{aligned}$$

(42)

In the next step, by using trigonometric identity, and substituting Eq. (42) in Eq. (41c), we get two algebraic equations in the form of

$$\begin{aligned}&\left( \frac{A^{*}_{11}}{B^{*}_{11]}}\right) ^2+\left( \frac{A^{*}_{12}}{B^{*}_{12}}\right) ^2-1=0\,, \end{aligned}$$

(43a)

$$\begin{aligned}&\sigma -B^{*}_{14}\frac{A^{*}_{11}}{B^{*}_{11]}}-A^{*}_{13}+B^{*}_{13}\frac{A^{*}_{12}}{B^{*}_{12}}=0\,, \end{aligned}$$

(43b)

Note that in the above equations except $a^{*}_y$ and $q^{*}_y$ all the system and excitation parameters are known, hence, these simultaneous algebraic equations can be used to govern the steady state amplitudes $a^{*}_y$ and $q^{*}_y$ and can be written in a compact form as

$$\begin{aligned} \begin{aligned} G_1(a^{*}_y,\,q^{*}_y)&=\left( \frac{A^{*}_{11}}{B^{*}_{11]}}\right) ^2+\left( \frac{A^{*}_{12}}{B^{*}_{12}}\right) ^2-1=0\\ G_2(a^{*}_y,\,q^{*}_y)&=\sigma -B^{*}_{14}\frac{A^{*}_{11}}{B^{*}_{11]}}-A^{*}_{13}+B^{*}_{13}\frac{A^{*}_{12}}{B^{*}_{12}}=0. \end{aligned} \end{aligned}$$

(44)