Active Vibration Control of Flexible Beams Based on Infinite-Dimensional Lyapunov Stability Theory: An Experimental Study

Abstract

Flexible beams are parts of lighter structures and flexible manipulators, which possess numerous advantages. However, the flexibility leads to a vibration problem. This paper presents an experimental study of active vibration control of flexible beams. The beam assumed to behave according to the Euler–Bernoulli theory is modeled by a partial differential equation. The control objective is to suppress the vibration of the beam. Based on the Lyapunov stability theory of infinite-dimension systems, a control law is directly derived without using any approximated finite-dimensional model of the beam. The success of the controller is experimentally demonstrated on a cantilever aluminum beam with a laser displacement sensor and a piezoelectric actuator. Experimental results show that the controller effectively suppresses the vibration.

Appendices

Appendix 1

The following inequalities are used in the subsequent derivation:

$$\begin{aligned}&\frac{u^{2}}{\alpha }+\alpha v^{2}\ge \;|uv|,\quad \forall u,v,\alpha \in R,\quad \forall \alpha >0\end{aligned}$$

(24)

$$\begin{aligned}&\quad \int _0^l {{u}'^{2}(x,t)\mathrm{d}x} \le l^{2}\int _0^l {{u}''^{2}(x,t)\mathrm{d}x} \end{aligned}$$

(25)

First, we derive the condition for $\beta _0$ such that $V(t)>0$. By substituting

$$\begin{aligned} \rho (x)=b_\mathrm{b} \rho _\mathrm{b} t_\mathrm{b} +2b_\mathrm{p} \rho _\mathrm{p} t_\mathrm{p} S(x) \ge b_\mathrm{b} \rho _\mathrm{b} t_\mathrm{b} \end{aligned}$$

(26)

and

$$\begin{aligned} EI(x)=EI_\mathrm{b} +EI_\mathrm{p} S(x)\;\ge EI_\mathrm{b} \end{aligned}$$

(27)

into Eq. (17), it gives

$$\begin{aligned} V_{1} (t) \ge \frac{1}{2}b_\mathrm{b} \rho _\mathrm{b} t_\mathrm{b} \int _{0}^{l} \dot{w}^{2} (x,t)\mathrm{d}x + \frac{1}{2}EI_\mathrm{b} \int _{0}^{l} w^{\prime \prime 2} (x,t)\mathrm{d}x\nonumber \\ \end{aligned}$$

(28)

By using Eqs. (24) and (26), $V_2 (t)$ which is given in Eq. (19) can be derived as

$$\begin{aligned}&V_{2} (t) \ge - b_\mathrm{b} \rho _\mathrm{b} t_\mathrm{b} \int _{0}^{l} \left( {\alpha _{1} w^{2} (x,t) + \frac{{\dot{w}^{2} (x,t)}}{{\alpha _{1} }}} \right) \mathrm{d}x \nonumber \\&\quad +\frac{B}{2}\int _{0}^{l} w^{2} (x,t)\mathrm{d}x - C\int _{0}^{l} \left( {\alpha _{2} w^{2} (x,t) + \frac{{\dot{w}^{2} (x,t)}}{{\alpha _{2} }}} \right) \mathrm{d}x\nonumber \\ \end{aligned}$$

(29)

where $\alpha _1$ and $\alpha _2$ are positive constants. Then, by substituting Eqs. (28) and (29) into Eq. (21) and using Eq. (25), it yields

$$\begin{aligned} V(t)&\ge b_\mathrm{b} \rho _\mathrm{b} t_\mathrm{b} \left( {1 - 2\frac{{\beta _{0}}}{{\alpha _{1} }}} \right) \int _{0}^{l} \dot{w}^{2} (x,t)\mathrm{d}x \nonumber \\&+ \beta _{0} \left( {\frac{B}{2} - \alpha _{1} b_\mathrm{b} \rho _\mathrm{b} t_\mathrm{b} - C\alpha _{2} } \right) \int _{0}^{l} w^{2} (x,t)\mathrm{d}x \nonumber \\&+ \left( {\frac{{EI_\mathrm{b} }}{2} - \frac{{\beta _{0} Cl^{2}}}{{\alpha _{2} }}} \right) \int _{0}^{l} w^{\prime \prime 2} (x,t)\mathrm{d}x \end{aligned}$$

(30)

By setting

$$\begin{aligned}&1-2\frac{\beta _0 }{\alpha _1 }>0,\quad \;\frac{B}{2}-\alpha _1 b_\mathrm{b} \rho _\mathrm{b} t_\mathrm{b} -C\delta _2 >0,\; \nonumber \\&\frac{EI_\mathrm{b} }{2}-\frac{\beta _0 Cl^{2}}{\alpha _2 }>0, \end{aligned}$$

(31)

we obtain $V(t)>0$. From Eq. (31), we have

$$\begin{aligned}&2\beta _0 <\alpha _1 <\frac{B}{4b_\mathrm{b} \rho _\mathrm{b} t_\mathrm{b} },\nonumber \\&\frac{2\beta _0 Cl^{2}}{EI_\mathrm{b} }<\alpha _2 <\frac{B}{4C}. \end{aligned}$$

(32)

To assure Eq. (32) which gives $V(t)>0$, $\beta _0$ should be chosen as

$$\begin{aligned} \beta _0 <\min \left\{ {\frac{B}{8b_\mathrm{b} \rho _\mathrm{b} t_\mathrm{b} },\frac{EI_\mathrm{b} B}{8C^{2}l^{2}}} \right\} . \end{aligned}$$

(33)

Next, the condition which yields $\dot{V}(t)<0$is examined. By substituting Eq. (23) into Eq. (22), we have

$$\begin{aligned} \dot{V}(t)&\le - \int _{0}^{l} (B - \beta _{0} \rho (x))\dot{w}^{2} (x,t)\mathrm{d}x \nonumber \\&-\, \beta _{0}\int _{0}^{l} EI(x)w^{\prime \prime 2} (x,t)\mathrm{d}x \nonumber \\&+\, C\beta _{0} \int _{0}^{l} \dot{w}(x,t)w^{\prime }(x,t)\mathrm{d}x \end{aligned}$$

(34)

Define

$$\begin{aligned} \rho _{\max } =b_\mathrm{b} \rho _\mathrm{b} t_\mathrm{b} +2b_\mathrm{p} \rho _\mathrm{p} t_\mathrm{p} \ge \rho (x). \end{aligned}$$

(35)

By using Eqs. (24), (25) and (35), it yields

$$\begin{aligned} \dot{V}(t)&\le \int _{0}^{l} \left( {\beta _{0} \rho _{{\max }} + \frac{{C\beta _{0} }}{{\alpha _{3} }} - B} \right) \dot{w}^{2} (x,t)\mathrm{d}x \nonumber \\&+\, \beta _{0} \int \int _{0}^{l} \left( {\alpha _{3} Cl^{2} - EI(x)} \right) w^{\prime \prime 2} (x,t)\mathrm{d}x \end{aligned}$$

(36)

where $\alpha _3$ is a positive constant. By setting

$$\begin{aligned} \beta _0 \rho _{\max } +\frac{C\beta _0 }{\alpha _3 }-B<0,\quad \alpha _3 Cl^{2}-EI(x)<0, \end{aligned}$$

(37)

we obtain $\dot{V}(t)<0$. Thus, $\beta _0$ and $\alpha _3$ should be chosen as

$$\begin{aligned}&\beta _0 <\frac{B}{\rho _{\max } +\frac{C}{\alpha _3 }},\end{aligned}$$

(38)

$$\begin{aligned}&\alpha _3 <\frac{EI(x)}{Cl^{2}}<\frac{EI_\mathrm{b} }{Cl^{2}} \end{aligned}$$

(39)

to guarantee $\dot{V}(t)<0$. Therefore, according to Eqs. (33) and (38), $\beta _0$ should be chosen as

$$\begin{aligned} \beta _0 <\min \left\{ {\frac{B}{8b_\mathrm{b} \rho _\mathrm{b} t_\mathrm{b} },\frac{EI_\mathrm{b} B}{8C^{2}l^{2}},\frac{B}{\rho _{\max } +\frac{C}{\alpha _3 }}} \right\} . \end{aligned}$$

(40)

where $\alpha _3$ satisfies Eq. (39).

Note that the conditions (39) and (40) are achievable. For our experimental system (using numerical values form Table 1), we have $\frac{EI_\mathrm{b} }{Cl^{2}}=9.8$. Thus, we can simply choose $\alpha _3 =1$ to satisfy Eq. (39). Similarly, we have $\frac{B}{8b_\mathrm{b} \rho _\mathrm{b} t_\mathrm{b} }=386$, $\frac{EI_\mathrm{b} B}{8C^{2}l^{2}}=3,070$ and $\frac{B}{\rho _{\max } +C}=665$. Thus, $\beta _0 <386$ satisfies Eq. (40).

Appendix 2

To estimate $g(t)$ from $w(l,t)$, we considered only the first vibration mode of the beam. The deflection equation can be expressed as (Beer et al. 2014)

$$\begin{aligned} w(x,t)=\frac{F}{EI_\mathrm{b} }\left( {\frac{lx^{2}}{2}-\frac{x^{3}}{6}} \right) . \end{aligned}$$

(41)

Taking its derivative with respect to $x$ yields

$$\begin{aligned} {w}'(x,t)=\frac{F}{EI_\mathrm{b} }\left( {lx-\frac{x^{2}}{2}} \right) . \end{aligned}$$

(42)

From Eq. (41),

$$\begin{aligned} \frac{F}{EI_\mathrm{b} }=3\frac{w(l,t)}{l^{3}}. \end{aligned}$$

(43)

Substituting Eq. (43) into Eq. (42) yields

$$\begin{aligned} {w}'(x,t)=3\frac{w(l,t)}{l^{3}}\left( {lx-\frac{x^{2}}{2}} \right) . \end{aligned}$$

(44)

Thus, we have

$$\begin{aligned} g(t)=3\frac{w(l,t)}{l^{3}}\left( {l(l_2 -l_1 )-\frac{(l_2^2 -l_1^2)}{2}} \right) . \end{aligned}$$

(45)