**2. Active beam vibration reduction with additional elements**

In this problem, the additional elements make the concentrated masses and actuators and all constitute the mechanical set beam-actuators-masses. Adding actuators (and the glue at the same time) is the technical necessity but they introduce to the mechanical set the additional dynamics effects namely, local stiffness and concentrated masses. As far as concentrated masses are concerned, adding them is substantiated as follows. The proposed optimal distribution of the actuators needs asymmetrical beam vibrations and these ones may be ensured by at least one concentrated mass.

### **2.1 Uniform beam vibration with damping**

There are four theories (models) for the transversely vibrating uniform beam (Han et al., 1999): Euler-Bernoulli, Rayleigh, shear and Timoshenko. The first of them, called the

An Optimal Distribution of Actuators in

they are arranged as depicted in Fig. 3.

For simplicity, let P E,J,h, ,S, = {

1

Fig. 3. Distribution of actuators and glue layers

b b J (bh )/12 = , 3 2

Fig. 4. Cross-sections of the set beam−actuator−glue

The parameters of the set beam-actuators-glue may be written as

*ha*

*hg*

where *<sup>s</sup> s n* <sup>=</sup> 1,2,..., , PPP s a = + <sup>g</sup> , <sup>0</sup>

0

s

bb ss

4

μ

The Eq. (6) may be written down quite similar like Eq. (1), namely

μ

x x2

ρ μ

= = { b,a,g} {[b]eam,[a]ctuator,[g]lue} , for example S bh

0

ϑ

Active Beam Vibration – Some Aspects, Theoretical Considerations 401

Furthermore, the dynamic effects of the actuators and glue on the beam vibration are introduced. The location and length of separate actuators, and the glue layers simultaneously, are denoted commonly with coordinates { x and s} { s} respectively and

<sup>1</sup> ... <sup>s</sup> ...

2

beam, actuators and glue, i.e. {Young's modulus, surface moment of inertia, thickness, mass density, surface of the rectangular cross-section, inner damping factor} respectively. Furthermore all parameters are supplemented with following index

rectangular cross-section, b − beam / glue layer width. Moments of inertia are calculated relatively of y -axis, see Fig. 4, where the neutral axis displacement d is neglected , hence 3

*b*

( <sup>+</sup> ) <sup>+</sup> ( JEuDHJEJE <sup>+</sup> ) ( ) ( SuDDHJE ++ ) uDHS <sup>+</sup> <sup>2</sup>

( ) ( ) 4 4 <sup>2</sup> EJD u EJD D u S D u f + ++ =

Heaviside step function in point x1s and so on, { x ,x x 2,x 2 1s 2s s s s s } =− + { } . For ns aktuators ( ns glue layers) and nr concentrated masses, Eq. (1) takes the form

s

δ

μ

r rt + − =− <sup>r</sup> m (x x )D u f

bbb sss

*hb d*

g g gb g J (bh ) /12 S (h 2 h 2) = ++ , 3 2

x

xs

} means the physical and geometrical parameters of the

 ϑ

a a ab g a J (bh ) /12 S (h 2 h h 2) = + ++ .

0

t bb ss

ρ

(6)

t s 0

ρ

<sup>t</sup> − (7)

H H(x x ) H(x x ) H(x x ) = −= −− − 1s 2s 1s 2s , H(x x ) − 1s −

b s1s 2sb s s s P P P H(x x ) P P H =+ − = + (5)

4

α

0

2

 ρt r *y*

= means the surface of the

ϑ

classical beam theory, is applied here. It is simple and provides reasonable results for formulated problem.

Fig. 1. The geometry of the simple supported beam

Let be the beam as depicted in Fig. 1. The Bernoulli-Euler equation governs transverse vibration (or bending or lateral vibration) of the beam has a following standard form (Kaliski, 1986; Pietrzakowski, 2004),

$$\text{EJD}^4\text{u} + \text{\tiny\mu\text{EJD}^4(\text{D}\_\text{\text}\text{u})} + \text{\tiny\mu\text{SD}\_\text{\text}^2\text{u}} = -\text{f} \tag{1}$$

where u u(x,t) <sup>=</sup> – beam deflection at the point x and the time t , f f(x,t) <sup>=</sup> – load force, 44 4 D (.) (.) / x =∂ ∂ , D (.) (.) / t <sup>t</sup> =∂ ∂ ; hereafter the rest symbols are jointly explained.

To solve Eq. (1) explicitly, four boundary conditions, at the ends of the beam, are needed. In general, boundary conditions represent displacement, slope, moment and shear respectively. Here, it is assumed that the beam is simple supported, then both displacement and the bending moment equal zero

$$\mathbf{u}(0,\mathbf{t})=0,\qquad\mathbf{D}^2\mathbf{u}(0,\mathbf{t})=0\tag{2}$$

$$\mathbf{u}(\ell, \mathbf{t}) = \mathbf{0}, \qquad \mathbf{D}^2 \mathbf{u}(\ell, \mathbf{t}) = \mathbf{0} \tag{3}$$

To solve over determined problem, one needs to know initial conditions. But here, the harmonic steady state plays a major part, so that the initial conditions are omitted.

### **2.2 Beam vibration with concentrated masses**

To solve the intended problem, Eq. (1) must be rounded out. First of all, to obtain asymmetric modes and consistently asymmetric general vibration, a few concentrated masses are added to the beam (Low & Naguleswaran, 1998; Majkut, 2010; Naguleswaran, 1999). They are marked by { mr} , and their distribution is described with set of coordinates { xr} , see Fig. 2, hence

$$\sum\_{\mathbf{r}} \mathbf{m}\_{\mathbf{r}} \delta(\mathbf{x} - \mathbf{x}\_{\mathbf{r}}) = \mathbf{m}\_1 \delta(\mathbf{x} - \mathbf{x}\_1) + \mathbf{m}\_2 \delta(\mathbf{x} - \mathbf{x}\_2) + \dots + \mathbf{m}\_r \delta(\mathbf{x} - \mathbf{x}\_r) + \dots \tag{4}$$

where r r 1,2,...,n = , δ(.) − Dirac's delta function.

$$\begin{array}{cccccccccc} & & & \mathbf{m\_1} & & & \mathbf{m\_2} & & \dots & & \mathbf{m\_r} & & \dots & & & & & \mathbf{x\_r} \\ \hline \hline 0 & & & & \mathbf{x\_1} & & & & & \mathbf{x\_2} & & & & & & \mathbf{x\_r} \\ \hline 0 & & & & & & & \mathbf{x\_2} & & & & & & & \mathbf{x\_r} \\ \end{array}$$

Fig. 2. Distribution of the concentrated masses

Furthermore, the dynamic effects of the actuators and glue on the beam vibration are introduced. The location and length of separate actuators, and the glue layers simultaneously, are denoted commonly with coordinates { x and s} { s} respectively and they are arranged as depicted in Fig. 3.

### Fig. 3. Distribution of actuators and glue layers

400 Acoustic Waves – From Microdevices to Helioseismology

classical beam theory, is applied here. It is simple and provides reasonable results for

Let be the beam as depicted in Fig. 1. The Bernoulli-Euler equation governs transverse vibration (or bending or lateral vibration) of the beam has a following standard form

where u u(x,t) <sup>=</sup> – beam deflection at the point x and the time t , f f(x,t) <sup>=</sup> – load force, 44 4 D (.) (.) / x =∂ ∂ , D (.) (.) / t <sup>t</sup> =∂ ∂ ; hereafter the rest symbols are jointly explained. To solve Eq. (1) explicitly, four boundary conditions, at the ends of the beam, are needed. In general, boundary conditions represent displacement, slope, moment and shear respectively. Here, it is assumed that the beam is simple supported, then both displacement

To solve over determined problem, one needs to know initial conditions. But here, the

To solve the intended problem, Eq. (1) must be rounded out. First of all, to obtain asymmetric modes and consistently asymmetric general vibration, a few concentrated masses are added to the beam (Low & Naguleswaran, 1998; Majkut, 2010; Naguleswaran, 1999). They are marked by { mr} , and their distribution is described with set of coordinates

> r r1 12 2 r r <sup>r</sup> m (x x ) m (x x ) m (x x ) ... m (x x ) ...

0

δδδ

(.) − Dirac's delta function.

x2 xr x1

m1 m2 mr ... ...

harmonic steady state plays a major part, so that the initial conditions are omitted.

f(x)

( ) 44 2 EJD u + μEJD D ut t + =− ρSD u f (1)

u(0,t) 0, = <sup>2</sup> D u(0,t) 0 = (2)

u( ,t) 0, = <sup>2</sup> D u( ,t) 0 = (3)

 δ

x

− = − + − ++ − + (4)

x

formulated problem.

u

(Kaliski, 1986; Pietrzakowski, 2004),

and the bending moment equal zero

{ xr} , see Fig. 2, hence

where r r 1,2,...,n = ,

**2.2 Beam vibration with concentrated masses** 

δ

Fig. 2. Distribution of the concentrated masses

Fig. 1. The geometry of the simple supported beam

0

For simplicity, let P E,J,h, ,S, = { ρ μ } means the physical and geometrical parameters of the beam, actuators and glue, i.e. {Young's modulus, surface moment of inertia, thickness, mass density, surface of the rectangular cross-section, inner damping factor} respectively. Furthermore all parameters are supplemented with following index ϑ = = { b,a,g} {[b]eam,[a]ctuator,[g]lue} , for example S bh ϑ ϑ = means the surface of the rectangular cross-section, b − beam / glue layer width. Moments of inertia are calculated relatively of y -axis, see Fig. 4, where the neutral axis displacement d is neglected , hence 3 b b J (bh )/12 = , 3 2 g g gb g J (bh ) /12 S (h 2 h 2) = ++ , 3 2 a a ab g a J (bh ) /12 S (h 2 h h 2) = + ++ .

Fig. 4. Cross-sections of the set beam−actuator−glue

The parameters of the set beam-actuators-glue may be written as

$$\mathbf{P} = \mathbf{P\_b} + \sum\_{s} \mathbf{P\_s} \cdot \mathbf{H(x\_{1s} - x\_{2s})} = \mathbf{P\_b} + \sum\_{s} \mathbf{P\_s} \{\mathbf{H}\}^0 \tag{5}$$

where *<sup>s</sup> s n* <sup>=</sup> 1,2,..., , PPP s a = + <sup>g</sup> , <sup>0</sup> H H(x x ) H(x x ) H(x x ) = −= −− − 1s 2s 1s 2s , H(x x ) − 1s − Heaviside step function in point x1s and so on, { x ,x x 2,x 2 1s 2s s s s s } =− + { } . For ns aktuators ( ns glue layers) and nr concentrated masses, Eq. (1) takes the form

( <sup>+</sup> ) <sup>+</sup> ( JEuDHJEJE <sup>+</sup> ) ( ) ( SuDDHJE ++ ) uDHS <sup>+</sup> <sup>2</sup> t s 0 t bb ss 4 s 0 bbb sss 4 s 0 bb ss μ μ ρ ρ 2 r rt + − =− <sup>r</sup> m (x x )D u f δ(6)

The Eq. (6) may be written down quite similar like Eq. (1), namely

$$\rm E/D^4u + \mu E/D^4(D\_iu) + \left(\rho S + \alpha\_r\right)D\_iu = -f \tag{7}$$

An Optimal Distribution of Actuators in

Substituting (14) into (11) gives

where the dispersion relationship is given by

The Eq. (17) is very important and the solution to it is

where Krylov functions are defined as, (Kaliski, 1986),

omitted in Eq. (11).

or

hence

0

form, Fig. 6,

Active Beam Vibration – Some Aspects, Theoretical Considerations 403

considered in coupling conditions between third and forth elements and therefore it is

4 2 EJ DXT S X DT 0 jj j j j j t + = ρ

<sup>4</sup> <sup>2</sup> jj j <sup>t</sup> <sup>2</sup>

4 4 DX X 0 j jj − = λ

2 2 DT T 0 <sup>t</sup> + = ω

4 2 j j

j

λ ω ω

2

K (z) ch(z) cos(z) 2 , <sup>3</sup> = − ( ) K (z) sh(z) sin(z) 2 <sup>4</sup> = + ( ) (21)

<sup>x</sup> u2 u4 u1 u3

ω

γ

jj j

S E J ρ

X (x) A K ( x) B K ( x) C K ( x) D K ( x) j j1 j j2 j j3 j j4 j = +++ λλλ

K (z) ch(z) cos(z) 2 , <sup>1</sup> = + ( ) K (z) sh(z) sin(z) 2 <sup>2</sup> = − ( ) ,

e <sup>2</sup> e <sup>4</sup> e <sup>3</sup> e 00 0 <sup>1</sup> 0

The boundary conditions in local coordinates, j x [0,e ] ∈ , to the separate j -element have the

Fig. 6. Geometry of the set beam-one actuator-one mass in local coordinates

• boundary conditions at the left end of the 1st−element

EJ DX D T SX T

u (x,t) X (x) T(t) j j = (14)

=− = (16)

= = (19)

 λ

mr

X (0) 0, <sup>1</sup> = <sup>2</sup> D X (0) 0 <sup>1</sup> = (22)

(20)

(15)

(17)

(18)

Let the solution be represented by a product of spatial and temporal functions

jj j

ρ

where hereafter

$$\mathbf{E}\mathbf{J} = \mathbf{E}\_{\mathrm{b}}\mathbf{J}\_{\mathrm{b}} + \sum\_{s} \mathbf{E}\_{s}\mathbf{J}\_{s} \left\{ \mathbf{H} \right\}^{0}, \qquad \mu \, \mathbf{E}\mathbf{J} = \mu\_{\mathrm{b}} \, \mathbf{E}\_{\mathrm{b}} \, \mathbf{J}\_{\mathrm{b}} + \sum\_{s} \mu\_{s} \mathbf{E}\_{s} \mathbf{J}\_{s} \left\{ \mathbf{H} \right\}^{0},$$

$$\rho \, \mathbf{S} = \rho\_{\mathrm{b}} \mathbf{S}\_{\mathrm{b}} + \sum\_{s} \rho\_{s} \mathbf{S}\_{s} \left\{ \mathbf{H} \right\}^{0}, \qquad \alpha\_{\mathrm{r}} = \sum\_{\mathrm{r}} \mathbf{m}\_{\mathrm{r}} \, \delta(\mathbf{x} - \mathbf{x}\_{\mathrm{r}}) \tag{8}$$

On the ground of the EJ , ρ S and αr form, Eq. (7) can not be understood in a classical manner. To solve it, some methods may be applied. One of them is presented in (Ercoli & Laura, 1987; Kasprzyk & Wiciak, 2007; Majkut, 2010); another attitude may be found in (C.N. Bapat & C. Bapat, 1987) and it is applied here.

$$\begin{array}{ccccc} \hline \begin{array}{c} \ell\_{\star} \\ \hline 0 \end{array} & \begin{array}{c} \ell\_{\star} \\ \hline \begin{array}{c} \mathbf{x}\_{1s} \\ \hline \end{array} & \begin{array}{c} \mathbf{x}\_{2s} \\ \hline \end{array} & \begin{array}{c} \mathbf{m}\_{\star} \\ \hline \end{array} & \begin{array}{c} \mathbf{x}\_{1s} \\ \hline \end{array} \end{array} \end{array} \end{array}$$

Fig. 5. Geometry of the set beam−one actuator−one mass

At the latter attitude, the beam is divided into some uniform elements. The division may not be coincidental. To clearly explain this problem, for simplicity consider a set beam-one actuator (and glue)-one concentrated mass, Fig. 5. The division is imposed out of the change of physical properties namely, properties of the actuators (and glue) and concentrated masses. So, the beam is divided into j j = = 1,2,...,n 4 elements. All elements may be considered separately and the solution to Eq. (7) can be expressed as

$$\mathbf{u}(\mathbf{x}, \mathbf{t}) = \sum\_{\perp} \mathbf{u}\_{\parallel}(\mathbf{x}, \mathbf{t}) \tag{9}$$

where u (x,t) is the solution on j j -element and it is fulfilled the following equation

$$\left(E\_{\rangle}\mathbf{I}\_{\rangle}\mathbf{D}^{4}\boldsymbol{u}\_{\rangle} + \boldsymbol{\mu}\_{\rangle}\mathbf{E}\_{\rangle}\mathbf{I}\_{\rangle}\mathbf{D}^{4}\left(\mathbf{D}\_{i}\boldsymbol{u}\_{\rangle}\right) + \left(\boldsymbol{\rho}\_{\rangle}\mathbf{S}\_{\rangle} + \boldsymbol{\alpha}\_{r}\right)\mathbf{D}\_{i}^{2}\boldsymbol{u}\_{\rangle} = -\boldsymbol{f}\_{\rangle} \tag{10}$$

To find u (x,t) with the separation of variables method, the eigenvalues and eigenfunctions j for each element are needed.

### **2.3 Eigenvalues and eigenfunctions problem**

In this problem it is assumed that ( , ) 0 *<sup>j</sup> f xt* = and j μ= 0 , hence based on Eq. (10) one obtains

$$\mathbf{E}\_{\parallel} \mathbf{J}\_{\perp} \mathbf{D}^{4} \mathbf{u}\_{\parallel} + \rho \mathbf{S}\_{\parallel} \mathbf{S}\_{\parallel} \mathbf{D}^{2}\_{\text{t}} \mathbf{u}\_{\parallel} = \mathbf{0} \tag{11}$$

where E J and j j j j ρ S may be different on the separate elements, but here, as depicted in Fig. 5, is

$$\mathbf{E\_1J\_1 = E\_3J\_3 = E\_4J\_4 = E\_bJ\_{b'}} \qquad \mathbf{E\_2J\_2 = E\_bJ\_b + E\_aJ\_a + E\_gJ\_g} \tag{12}$$

$$
\rho\_1 \mathbf{S}\_1 = \rho\_3 \mathbf{S}\_3 = \rho\_4 \mathbf{S}\_4 = \rho\_b \mathbf{S}\_{b\prime} \qquad \rho\_2 \mathbf{S}\_2 = \rho\_b \mathbf{S}\_b + \rho\_a \mathbf{S}\_a + \rho\_g \mathbf{S}\_g \tag{13}
$$

The boundary conditions for the *j* −element consist of boundary conditions of the problem and coupling conditions between neighboring elements. The concentrated mass mr is considered in coupling conditions between third and forth elements and therefore it is omitted in Eq. (11).

Let the solution be represented by a product of spatial and temporal functions

$$\mathbf{u}\_{\rangle}(\mathbf{x}, \mathbf{t}) = \mathcal{X}\_{\rangle}(\mathbf{x}) \, \mathbf{T}(\mathbf{t}) \tag{14}$$

Substituting (14) into (11) gives

$$\mathbf{E}\_{\flat} \mathbf{J}\_{\flat} \mathbf{D}^{4} \mathbf{X}\_{\flat} \mathbf{T} + \rho \mathbf{S}\_{\flat} \mathbf{X}\_{\flat} \mathbf{D}\_{\flat}^{2} \mathbf{T} = \mathbf{0} \tag{15}$$

or

402 Acoustic Waves – From Microdevices to Helioseismology

bb ss <sup>s</sup> EJ E J E J H , = + <sup>0</sup>

0

S S SH, = + r rr

manner. To solve it, some methods may be applied. One of them is presented in (Ercoli & Laura, 1987; Kasprzyk & Wiciak, 2007; Majkut, 2010); another attitude may be found in

x xs x <sup>r</sup> 1s

At the latter attitude, the beam is divided into some uniform elements. The division may not be coincidental. To clearly explain this problem, for simplicity consider a set beam-one actuator (and glue)-one concentrated mass, Fig. 5. The division is imposed out of the change of physical properties namely, properties of the actuators (and glue) and concentrated masses. So, the beam is divided into j j = = 1,2,...,n 4 elements. All elements may be

x2s

where u (x,t) is the solution on j j -element and it is fulfilled the following equation

( )( ) 4 4 <sup>2</sup> *EJ D u EJ D Du S D u jj j j jj <sup>t</sup> j jj r t j j* + ++ =

To find u (x,t) with the separation of variables method, the eigenvalues and eigenfunctions j

4 2 EJ Du S D u 0 jj j j j t j + = ρ

11 33 44 bb S S S S, == = 22 bb aa g g

The boundary conditions for the *j* −element consist of boundary conditions of the problem and coupling conditions between neighboring elements. The concentrated mass mr is

 ρ

μ

S may be different on the separate elements, but here, as depicted in

EJ EJ EJ EJ , 11 33 44 bb == = EJ EJ EJ EJ 22 bb aa =++ g g (12)

ρρρρ

α

μμ

α

b b b s ss s

 δ= − <sup>r</sup>

 μEJ E J E J H = + ,

r form, Eq. (7) can not be understood in a classical

<sup>j</sup> <sup>j</sup> u(x,t) u (x,t) <sup>=</sup> (9)

− *f* (10)

(11)

= 0 , hence based on Eq. (10) one obtains

SSSS =++ (13)

mr

m (x x ) (8)

x

0

α

 ρ

bb ss s

ρρ

(C.N. Bapat & C. Bapat, 1987) and it is applied here.

ρS and

s

considered separately and the solution to Eq. (7) can be expressed as

μ

Fig. 5. Geometry of the set beam−one actuator−one mass

where hereafter

0

On the ground of the EJ ,

for each element are needed.

where E J and j j j j

Fig. 5, is

ρ

**2.3 Eigenvalues and eigenfunctions problem**  In this problem it is assumed that ( , ) 0 *<sup>j</sup> f xt* = and j

ρρ

 ρ  ρ

$$\frac{\mathbf{E}\_{\parallel}\mathbf{J}\_{\parallel}}{\rho\_{\parallel}\mathbf{S}\_{\parallel}}\frac{\mathbf{D}^{4}\mathbf{X}\_{\parallel}}{\mathbf{X}\_{\parallel}} = -\frac{\mathbf{D}\_{\text{t}}^{2}\mathbf{T}}{\mathbf{T}} = o^{2} \tag{16}$$

hence

$$\mathbf{D}^4 \mathbf{X}\_{\parallel} - \lambda\_{\parallel}^4 \mathbf{X}\_{\parallel} = \mathbf{0} \tag{17}$$

$$\mathbf{D}\_t^2 \mathbf{T} + \rho \mathbf{o}^2 \mathbf{T} = \mathbf{0} \tag{18}$$

where the dispersion relationship is given by

$$\mathcal{A}\_{\parallel}^{4} = \alpha^{2} \frac{\rho\_{\parallel} \mathbf{S}\_{\parallel}}{\mathbf{E}\_{\parallel} \mathbf{J}\_{\parallel}} = \frac{\alpha^{2}}{\mathcal{Y}\_{\parallel}} \tag{19}$$

The Eq. (17) is very important and the solution to it is

$$\mathbf{X}\_{\uparrow}(\mathbf{x}) = \mathbf{A}\_{\uparrow}\mathbf{K}\_{1}(\boldsymbol{\lambda}\_{\uparrow}\mathbf{x}) + \mathbf{B}\_{\uparrow}\mathbf{K}\_{2}(\boldsymbol{\lambda}\_{\uparrow}\mathbf{x}) + \mathbf{C}\_{\uparrow}\mathbf{K}\_{3}(\boldsymbol{\lambda}\_{\uparrow}\mathbf{x}) + \mathbf{D}\_{\uparrow}\mathbf{K}\_{4}(\boldsymbol{\lambda}\_{\uparrow}\mathbf{x}) \tag{20}$$

where Krylov functions are defined as, (Kaliski, 1986),

$$\mathbf{K}\_{1}(\mathbf{z}) = \left(\operatorname{ch}(\mathbf{z}) + \cos(\mathbf{z})\right) \Big/ 2, \qquad \mathbf{K}\_{2}(\mathbf{z}) = \left(\operatorname{sh}(\mathbf{z}) - \sin(\mathbf{z})\right) \Big/ 2.$$

$$\mathbf{K}\_{3}(\mathbf{z}) = \left(\operatorname{ch}(\mathbf{z}) - \cos(\mathbf{z})\right) \Big/ 2, \qquad \mathbf{K}\_{4}(\mathbf{z}) = \left(\operatorname{sh}(\mathbf{z}) + \sin(\mathbf{z})\right) \Big/ 2. \tag{21}$$

$$\begin{array}{ccccc} \mathbf{u}\_1 & \mathbf{u}\_2 & \mathbf{u}\_3 & \mathbf{u}\_4 & \mathbf{u}\_4 & \mathbf{x}\_4 \\ \hline \mathbf{e}\_1 \mathbf{0} & \mathbf{e}\_1 \mathbf{0} & \mathbf{e}\_2 \mathbf{0} & \mathbf{e}\_2 \mathbf{0} & \mathbf{e}\_3 \mathbf{0} & \mathbf{e}\_4 \mathbf{0} \\ \end{array}$$

Fig. 6. Geometry of the set beam-one actuator-one mass in local coordinates

The boundary conditions in local coordinates, j x [0,e ] ∈ , to the separate j -element have the form, Fig. 6,

• boundary conditions at the left end of the 1st−element

$$\mathcal{X}\_1(0) = 0, \qquad \mathcal{D}^2 \mathcal{X}\_1(0) = 0 \tag{22}$$

An Optimal Distribution of Actuators in

λ

λ

λ

**1 2 1**

**B B 0 C**

**1 2 1**

**A A 0 B**

Active Beam Vibration – Some Aspects, Theoretical Considerations 405

2 2 2 1 11 4 1 11 2 2 22

<sup>=</sup> ′′ ′′ ′′ ′′ ′′

K K 00 0

′ ′ −

3 2 2

λ

−

E

3 3 33

<sup>−</sup> ′′ <sup>−</sup>

λ

<sup>=</sup> ′′′ ′′′ ′′′ ′′′ ′′′

00 0

3 3 3 1 11 1 1 11 3 2 22

′ ′ −

1 3 1 1 2

′ ′ −

′′ ′′ ′′ ′′

EJ K EJ K 0 0 EJ 0 EJ K EJ K 0 EJ 0 0

K K 10 0 0

1234 2 2 2 3 2 4 2 1 2222 2 22 3 2 22 4 2 22 1 2 22 2

′′ ′′ ′′ ′′ ′′ ′′ ′′ ′′

333 2 4 2 22 1 2 22 2 2 22 3 J K EJ K EJ K EJ K λλλ

3

3 3 K

4

λ

3 3 ′′′ = K K( e) υ υ λ4 4 ′′′′= (34)

λ

λ

λ

(31)

(32)

(33)

λ

λλλ 234 ν , , ν ν} and

<sup>1</sup> = and it

KKKK KKKK EJ K EJ K EJ K EJ K

λλλλ

λλλλ

2 3 33

λ

1234 3 2 3 3 3 4 3 1 2222 3 33 3 3 33 4 3 33 1 3 33 2 3333 3 33 4 3 33 1 3 33 2 3 3

′′′ ′′′ ′′′ ′′′ ′′′ ′′′ ′′′ ′′′

1 0 00

1234 3412

= B ,D ,A ,B ,C ,D ,A ,B ,C ,D ,A ,B ,C ,D 1 1 22 2 2 33 3 3 44 4 4 **x** (35)

} of the system beam-actuator-mass.

0 0 EJ 0 m EJ 0 0 KKKK KKKK

2 4 44

λ

λλλλ

′′′

′′′ ′′′ ′′′

λλλλ

λλλλ

00 0

2 3 r 4 44

<sup>−</sup> <sup>−</sup> ′′′ ′′′ ′′′′ <sup>−</sup> <sup>=</sup> ′′′′ − − ′′′′ ′′′′ ′′′′ ′′′′ ′′′′ ′′′′ ′′′′ ′′′′

 λ

ν

{K K ,K ,K ,K

To solve of the homogeneous matrix equation (29), one assumes that det ( ) 0 **A**

ων

} = { <sup>1234</sup>} ,

2 2 ′′ = K K ( e ), υ υ λ

[ ]<sup>T</sup>

= 1,2,...,n . Based on Eq. (28) one can calculate {

ω

**1**

**C**

**0**

υ

10 0 0

0 0 EJ 0 0 EJ 0 0 KKKK KKKK EJ K EJ K EJ K EJ K EJ K EJ K EJ K EJ

λ

In the current boundary problem, the separate blocks take the form

 λ

2 4

 λ

 λ

**0**

′′ ′′′ −

**0**

**1 2 1**

1 1 ′ = K K ( e ), υ υ λ

**C C 0 D**

where the symbols in matrices are given by

The unknowns are collected in column matrix

K K ( e ),

λ1ν } , ν

finally, based on Eq. (19), the frequency {

υ υ λ

gives the set {

**1**

**B**

′ ′′ ′ ′ −

• coupling conditions between 1st and 2nd−elements

$$\mathbf{X}\_1(\mathcal{A}\_1 \mathbf{e}\_1) = \mathbf{X}\_2(\mathcal{A}\_2 \mathbf{0}), \qquad \mathbf{D} \mathbf{X}\_1(\mathcal{A}\_1 \mathbf{e}\_1) = \mathbf{D} \mathbf{X}\_2(\mathcal{A}\_2 \mathbf{0}),$$

$$\mathbf{E}\_1 \mathbf{J}\_1 \mathbf{D}^2 \mathbf{X}\_1(\mathcal{A}\_1 \mathbf{e}\_1) = \mathbf{E}\_2 \mathbf{J}\_2 \mathbf{D}^2 \mathbf{X}\_2(\mathcal{A}\_2 \mathbf{0}), \qquad \mathbf{E}\_1 \mathbf{J}\_1 \mathbf{D}^3 \mathbf{X}\_1(\mathcal{A}\_1 \mathbf{e}\_1) = \mathbf{E}\_2 \mathbf{J}\_2 \mathbf{D}^3 \mathbf{X}\_2(\mathcal{A}\_2 \mathbf{0})\tag{23}$$

• coupling conditions between 2nd and 3rd−elements

$$\mathbf{X}\_{2}(\nexists \mathbf{z}\_{2}\mathbf{e}\_{2}) = \mathbf{X}\_{3}(\nexists \mathbf{0}\_{3}\mathbf{0}), \qquad \mathbf{D}\mathbf{X}\_{2}(\not\mathbf{z}\_{2}\mathbf{e}\_{2}) = \mathbf{D}\mathbf{X}\_{3}(\not\mathbf{z}\_{3}\mathbf{0}),$$

$$\mathbf{E}\_{2}\mathbf{J}\_{2}\,\mathbf{D}^{2}\mathbf{X}\_{2}(\not\mathbf{z}\_{2}\mathbf{e}\_{2}) = \mathbf{E}\_{3}\mathbf{J}\_{3}\,\mathbf{D}^{2}\mathbf{X}\_{3}(\not\mathbf{z}\_{3}\mathbf{0}), \qquad \mathbf{E}\_{2}\mathbf{J}\_{2}\,\mathbf{D}^{3}\mathbf{X}\_{2}(\not\mathbf{z}\_{2}\mathbf{e}\_{2}) = \mathbf{E}\_{3}\mathbf{J}\_{3}\,\mathbf{D}^{3}\mathbf{X}\_{3}(\not\mathbf{z}\_{3}\mathbf{0})\tag{24}$$

• coupling conditions between 3rd and 4th−elements

$$\mathbf{X\_3(\mathcal{J}\_3\mathbf{e\_5}) = X\_4(\mathcal{J}\_4 \mathbf{0})}, \qquad \mathbf{D}\mathbf{X\_3(\mathcal{J}\_5\mathbf{e\_5}) = D X\_4(\mathcal{J}\_4 \mathbf{0})}, \qquad \mathbf{E\_3\mathcal{J}\_3 D^2 X\_3(\mathcal{J}\_5\mathbf{e\_5}) = E\_4 \mathbf{J}\_4 D^2 X\_4(\mathcal{J}\_4 \mathbf{0})}$$

and

$$\mathbf{E\_{3}J\_{3}D^{3}X\_{3}(\mathcal{J}\_{3}\mathbf{e\_{3}}) + m\_{r}}\boldsymbol{\rho}^{2}\mathbf{X\_{3}(\mathcal{J}\_{3}\mathbf{e\_{3}}) = E\_{4}J\_{4}D^{3}\mathbf{X\_{4}(\mathcal{J}\_{4}\mathbf{0})}}$$

or

$$\rm E\_3 I\_3 \, D^3 \mathcal{X}\_3(\mathcal{Z}\_3 \mathbf{e}\_3) = m\_r \rho^2 \mathcal{X}\_4(\mathcal{Z}\_3 \mathbf{0}) + E\_4 \mathcal{I}\_4 \, D^3 \mathcal{X}\_4(\mathcal{Z}\_4 \mathbf{0}) \tag{25}$$

• boundary conditions at the right end of the 4th−element

$$X\_4(\mathcal{J}\_4 e\_4) = 0, \qquad D^2 X\_4(\mathcal{J}\_4 e\_4) = 0 \tag{26}$$

Since λλλλ <sup>1234</sup> ≠≠≠ then, to calculate them, the Eq. (19) must be used. It is convenient to express { λλλ <sup>234</sup> , , } as a function λ<sup>1</sup> , hence

$$
\lambda\_1^4 \,\, \chi\_1 = \lambda\_2^4 \,\, \chi\_2 = \lambda\_3^4 \,\, \chi\_3 = \lambda\_4^4 \,\, \chi\_4 = \,\, \phi^2 \tag{27}
$$

or

$$
\lambda\_2^4 = \lambda\_1^4 \left(\boldsymbol{\gamma}\_1/\boldsymbol{\gamma}\_2\right), \ \lambda\_3^4 = \lambda\_1^4 \left(\boldsymbol{\gamma}\_1/\boldsymbol{\gamma}\_3\right), \ \lambda\_4^4 = \lambda\_1^4 \left(\boldsymbol{\gamma}\_1/\boldsymbol{\gamma}\_4\right) \tag{28}
$$

Substituting Eq. (20) into boundary conditions (22) it appears that A 0 <sup>1</sup> = , C 0 <sup>1</sup> = . In the same way, the rest of conditions given by Eqs. (23) − (26) lead to the set of algebraic equations and it may be written in the matrix form

$$\mathbf{A}\,\mathbf{x} = \mathbf{0} \tag{29}$$

The matrix **A** is too large, to presented it in explicit form. Hence, its elements fall into blocks so that the matrix **A** can be written as

$$\mathbf{A} = \begin{bmatrix} \mathbf{A}\_1' & \mathbf{A}\_2'' & \mathbf{0} & \mathbf{0} \\ \mathbf{0} & \mathbf{B}\_1'' & \mathbf{B}\_2'' & \mathbf{0} \\ \mathbf{0} & \mathbf{0} & \mathbf{C}\_1'' & \mathbf{C}\_2'' \\ \mathbf{0} & \mathbf{0} & \mathbf{0} & \mathbf{D}\_1'' \end{bmatrix} \tag{30}$$

 = DX ( e ) DX ( 0) 1 11 2 2 λ

 = DX ( e ) DX ( 0) 2 22 3 3 λ

> λ

32 3 E J D X ( e ) m X ( e ) E J D X ( 0) 33 3 3 3 r 3 3 3 44 4 4

32 3 E J D X ( e ) m X ( 0) E J D X ( 0) 33 3 3 3 r 4 3 44 4 4

 ωλ+ =

 ωλ= +

= 3 3 E J D X ( e ) E J D X ( 0) 11 1 1 1 22 2 2

= 3 3 E J D X ( e ) E J D X ( 0) 22 2 2 2 33 3 3

λ= <sup>2</sup>

> λ γ

ω

 , ( ) 4 4 λ λ 4 1 14 = γ γ

<sup>1234</sup> ≠≠≠ then, to calculate them, the Eq. (19) must be used. It is convenient to

4444 2

Substituting Eq. (20) into boundary conditions (22) it appears that A 0 <sup>1</sup> = , C 0 <sup>1</sup> = . In the same way, the rest of conditions given by Eqs. (23) − (26) lead to the set of algebraic

The matrix **A** is too large, to presented it in explicit form. Hence, its elements fall into

**1 2**

**A**

′ ′′ ′′ ′′′ <sup>=</sup> ′′′ ′′′′ ′′′′

**0 0CC 0 0 0D**

**AA 0 0 0BB 0**

**1 2**

**1 2 1**

====

λ γ

11 22 33 44

 , ( ) 4 4 λ λ 3 1 13 = γ γ

 λ= ,

 λ= ,

 = 2 2 E J D X ( e ) E J D X ( 0) 33 3 3 3 44 4 4 λ

λ

λ

4 44 *DX e* ( )0 (26)

**Ax 0** = (29)

 λ= (23)

 λ= (24)

=

(25)

(27)

(30)

(28)

 λ

λ

λ

• coupling conditions between 1st and 2nd−elements

λ

2 2 E J D X ( e ) E J D X ( 0), 11 1 1 1 22 2 2

• coupling conditions between 2nd and 3rd−elements

λ

2 2 E J D X ( e ) E J D X ( 0), 22 2 2 2 33 3 3

• coupling conditions between 3rd and 4th−elements

λ

λ

X ( e ) X ( 0), 3 33 4 4

λλλλ

λλλ

<sup>234</sup> , , } as a function

λ λ 2 1 12 = γ γ

equations and it may be written in the matrix form

blocks so that the matrix **A** can be written as

 λ

λ

and

or

Since

or

express {

X ( e ) X ( 0), 1 11 2 2

X ( e ) X ( 0), 2 22 3 3

 = DX ( e ) DX ( 0), 3 33 4 4 λ

λ

λ

λ<sup>1</sup> , hence

*X e* 4 44 ( ) 0, λ=

> λ γ

• boundary conditions at the right end of the 4th−element

λ

( ) 4 4

γ

 λ

 λ  λ

 λ In the current boundary problem, the separate blocks take the form

2 4 1 3 1 1 2 2 2 2 1 11 4 1 11 2 2 22 3 3 3 1 11 1 1 11 3 2 22 1234 2 2 2 3 2 4 2 1 2222 2 22 3 2 22 4 2 22 1 2 22 2 3 2 2 K K 10 0 0 K K 00 0 EJ K EJ K 0 0 EJ 0 EJ K EJ K 0 EJ 0 0 KKKK KKKK EJ K EJ K EJ K EJ K E λ λ λ λ λ λ λ λ λ λλλλ λλλλ λ ′ ′ − ′ ′ − ′ ′ − ′ ′′ ′ ′ − <sup>=</sup> ′′ ′′ ′′ ′′ ′′ ′′ ′′ ′′ ′′ ′′ ′′ ′′ ′′ **1 2 1 A A 0 B 0** 333 2 4 2 22 1 2 22 2 2 22 3 J K EJ K EJ K EJ K λλλ ′′ ′′ ′′ ′′ (31) 3 2 3 33 3 3 33 1234 3 2 3 3 3 4 3 1 2222 3 33 3 3 33 4 3 33 1 3 33 2 3333 3 33 4 3 33 1 3 33 2 3 3 10 0 0 00 0 0 0 EJ 0 0 EJ 0 0 KKKK KKKK EJ K EJ K EJ K EJ K EJ K EJ K EJ K EJ λ λ λ λλλλ λλλλ λλλλ − <sup>−</sup> ′′ <sup>−</sup> ′′ ′′′ − <sup>=</sup> ′′′ ′′′ ′′′ ′′′ ′′′ ′′′ ′′′ ′′′ ′′′ ′′′ ′′′ ′′′ ′′′ ′′′ ′′′ ′′′ **1 1 2 1 B B B 0 C 0** 3 3 K ′′′ (32) 4 2 4 44 2 3 r 4 44 1234 3412 1 0 00 00 0 0 0 EJ 0 m EJ 0 0 KKKK KKKK ν λ λ ω λ <sup>−</sup> <sup>−</sup> ′′′ ′′′ ′′′′ <sup>−</sup> <sup>=</sup> ′′′′ − − ′′′′ ′′′′ ′′′′ ′′′′ ′′′′ ′′′′ ′′′′ ′′′′ **1 1 2 1 C C C 0 D 0** (33)

where the symbols in matrices are given by

$$\left\{\mathbf{K}\_{\boldsymbol{\nu}}\right\} = \left\{\mathbf{K}\_{1}, \mathbf{K}\_{2}, \mathbf{K}\_{3}, \mathbf{K}\_{4}\right\},$$

$$\mathbf{K}\_{\boldsymbol{\nu}}^{\prime} = \mathbf{K}\_{\boldsymbol{\nu}}(\boldsymbol{\lambda}\_{1}\mathbf{e}\_{1}), \qquad \mathbf{K}\_{\boldsymbol{\nu}}^{\prime} = \mathbf{K}\_{\boldsymbol{\nu}}(\boldsymbol{\lambda}\_{2}\mathbf{e}\_{2}), \qquad \mathbf{K}\_{\boldsymbol{\nu}}^{\prime\prime} = \mathbf{K}\_{\boldsymbol{\nu}}(\boldsymbol{\lambda}\_{3}\mathbf{e}\_{3}), \qquad \mathbf{K}\_{\boldsymbol{\nu}}^{\prime\prime\prime} = \mathbf{K}\_{\boldsymbol{\nu}}(\boldsymbol{\lambda}\_{4}\mathbf{e}\_{4})\tag{34}$$

The unknowns are collected in column matrix

$$\mathbf{x} = \begin{bmatrix} \mathbf{B}\_{1\prime} \mathbf{D}\_{1\prime} \mathbf{A}\_{2\prime} \mathbf{B}\_{2\prime} \mathbf{C}\_{2\prime} \mathbf{D}\_{2\prime} \mathbf{A}\_{3\prime} \mathbf{B}\_{3\prime} \mathbf{C}\_{3\prime} \mathbf{D}\_{3\prime} \mathbf{A}\_{4\prime} \mathbf{B}\_{4\prime} \mathbf{C}\_{4\prime} \mathbf{D}\_{4\prime} \end{bmatrix}^{\mathrm{T}} \tag{35}$$

To solve of the homogeneous matrix equation (29), one assumes that det ( ) 0 **A** λ<sup>1</sup> = and it gives the set { λ1ν } , ν = 1,2,...,n . Based on Eq. (28) one can calculate { λλλ 234 ν , , ν ν } and finally, based on Eq. (19), the frequency {ων} of the system beam-actuator-mass.

An Optimal Distribution of Actuators in

ν μ

ν μ

0 j = 2 j = 3

Fig. 7. Geometry of set with nj −elements

 ρ

m1 m2 m3

2 2

ωω

Since the term 2 2

ρ

problem.

where 1 2 i ( 1) = − ,

ω ω ν μ

> ν μ

**2.5 Forced vibrations with damping** 

ω

ν μ

ωωρ

ωωρ

ν μ

μν

ν μ

Active Beam Vibration – Some Aspects, Theoretical Considerations 407

+ = m X (0)X (0) E J r4 4 ν

( ) ( ) 32 32 X (e )D X (e ) DX (e )D X (e ) X (e )D X (e ) DX (e )D X (e ) 4 4 44 4 4 44 44 4 4 44 4 4

 μ

⋅ − −− (44)

 ν

− ++++

 μ

+ = m X (0)X (0) 0 r4 4 ν

The orthogonality condition, Eq. (45), may be generalized in a simple way. Let the system beam-actuators-masses be divided into nj −elements as depicted in Fig. 7. In this case one has

<sup>j</sup> jj j j jj j n1 n n n n <sup>j</sup> S X X dx m X (0)X (0) m X (e )X (e ) 0

− ++ = (46)

+

ν

 ρ

ω

ω

ν

, the general orthogonality condition is given by

<sup>+</sup>

<sup>t</sup> j jj <sup>t</sup> j j − (48)

 μ

 μ

++ = <sup>=</sup> (47)

The Eq. (47) in particular case is used in deriving the solution to the forced vibration

( ) 44 2 EJ Du EJ D Du S D u f jj j j jj + +=

The solution to Eq. (48) is forced vibrations with damping. Let be the load force in the form

j j <sup>f</sup> f (x,t) f (x)exp(i t) =

u (x,t) X (x)exp(i t) j j = f f

( ) ( ) j jjj j

( ) j jjj j <sup>j</sup> jj j j jj j n1 n n n n <sup>2</sup> <sup>j</sup> 0, S X X dx m X (0)X (0) m X (e )X (e ) ,

μ =ν

> μ

A point departure for further consideration is Eq. (7); for j −element one has

Applying separation of variables method, the solution to Eq. (48) is assumed as

μ

f – excited frequency.

Substituting Eqs. (49) and (50) to Eq. (48) one obtains

ν

ν

11 1 1 22 2 2 33 3 3 44 4 4 <sup>0000</sup> S X X dx S X X dx S X X dx S X X dx

11 1 1 22 2 2 33 3 3 44 4 4 <sup>0000</sup> S X X dx S X X dx S X X dx S X X dx

ρ

ρ

) 4 4 ⋅

ν μ

ν μ

 μ ρ

ρ

...

 μ

ν

(49)

(50)

β ν μ

≠

ν μ

) (45)

 ν ν μ

ν μ

 μ

− ++++

( )( 12 34 eeee 2 2

( )( 12 34 eeee 2 2

...

ν μ

ν μ

ρ

μν

ρ

Because of Eq. (26), the right-hand-side is zero, hence

j = 1 ...

ν μ

− is canceled for

Now, the unknowns put down in column matrix, Eq. (35), should be determined. Let the main matrix elements **A** be written as two suffix quantities Aαβ , where α and β label the rows and columns respectively. Let Mαβ be the minor of the Aαβ element. The general solution to Eq. (29) is

$$\mathbf{B}\_1 \colon \mathbf{D}\_1 \colon \mathbf{A}\_2 \colon \dots = (-1)^{a+1} \mathbf{M}\_{a1} \colon (-1)^{a+2} \mathbf{M}\_{a2} \colon (-1)^{a+3} \mathbf{M}\_{a3} \colon \dots \tag{36}$$

Substituting { λjν } and unknowns **x** to Eq. (20), the ν −eigenfunctions (ν −modes) assigned to the j −element are obtained. The solution to Eq. (10) is given by

$$X(\mathbf{x}) = \sum\_{\mathbf{j}} X\_{\mathbf{j}}(\mathbf{x}) = \sum\_{\mathbf{j}\nu} X\_{\mathbf{j}}(\mathcal{J}\_{\mathbb{p}}\mathbf{x}) = \sum\_{\mathbf{j}\nu} X\_{\mathbf{j}\nu}(\mathbf{x}) = \sum\_{\nu} X\_{\nu}(\mathbf{x}) \tag{37}$$

where j j (...) (...) ν ν<sup>=</sup> and the separate modes are equal

$$\mathbf{X}\_{\left[\nu\right]}\left(\mathbf{x}\right) = \mathbf{A}\_{\left]}\mathbf{K}\_{1}\left(\boldsymbol{\uplambda}\_{\left]\nu}\mathbf{x}\right) + \mathbf{B}\_{\left]}\mathbf{K}\_{2}\left(\boldsymbol{\uplambda}\_{\left]\nu}\mathbf{x}\right) + \mathbf{C}\_{\left]}\mathbf{K}\_{3}\left(\boldsymbol{\uplambda}\_{\left]\nu}\mathbf{x}\right) + \mathbf{D}\_{\left]}\mathbf{K}\_{4}\left(\boldsymbol{\uplambda}\_{\left]\nu}\mathbf{x}\right) \right.\tag{38}$$

### **2.4 Orthogonality of modes**

Orthogonality condition of the uniform beam modes may be found in (Kaliski, 1986; de Silva, 2000). First of all, based on twice integration by parts, one has

$$\int\_{0}^{\prime} \mathbf{X}\_{\nu}(\mathbf{x}) \mathbf{D}^{4} \mathbf{X}\_{\mu}(\mathbf{x}) d\mathbf{x} = \left( \mathbf{X}\_{\nu}(\mathbf{x}) \mathbf{D}^{3} \mathbf{X}\_{\mu}(\mathbf{x}) - \mathbf{D} \mathbf{X}\_{\nu}(\mathbf{x}) \mathbf{D}^{2} \mathbf{X}\_{\mu}(\mathbf{x}) \right) \Big|\_{0}^{\prime} + \int\_{0}^{\prime} \mathbf{D}^{2} \mathbf{X}\_{\nu}(\mathbf{x}) \mathbf{D}^{2} \mathbf{X}\_{\mu}(\mathbf{x}) d\mathbf{x} \tag{39}$$

The separate modes X (x) μ , X (x) ν, fulfill the following modal equations

$$\mathbb{E}\mathbb{I}\mathcal{D}^4 \mathcal{X}\_{\mu}(\mathbf{x}) = \mathcal{O}\_{\mu}^2 \rho \mathbb{S} \mathcal{X}\_{\mu}(\mathbf{x}) \tag{40}$$

$$\text{EJE}^4 \mathcal{X}\_\nu(\mathbf{x}) = a\_\nu^2 \,\rho \,\text{S} \mathcal{X}\_\nu(\mathbf{x}) \tag{41}$$

Multiplying above equations by X (x) ν and X (x) μ respectively, integrate both in range o integration x [0, ] ∈ , use Eq. (39), subtract the second result from the first one, one obtains (for simplicity an argument (x) is omitted)

$$\mathbb{E}\left(a\_{\nu}^{2} - a\_{\mu}^{2}\right)\rho\mathbb{S}\left[\int\_{0}^{\ell} \mathbb{X}\_{\nu} \mathbb{X}\_{\mu} \mathrm{d}\mathbf{x} = \mathbb{E}\right] \left[\left(\mathbb{X}\_{\mu} \mathrm{D}^{3} \mathbb{X}\_{\nu} - \mathrm{D} \mathbb{X}\_{\mu} \mathrm{D}^{2} \mathbb{X}\_{\nu}\right) - \left(\mathbb{X}\_{\nu} \mathrm{D}^{3} \mathbb{X}\_{\mu} - \mathrm{D} \mathbb{X}\_{\nu} \mathrm{D}^{2} \mathbb{X}\_{\mu}\right)\right]\_{0}^{\ell} \tag{42}$$

For standard boundary conditions, the right-hand-side equals zero.

The procedure outlined above can be used to the problem presented in Fig. 6, but Eq. (39) must be applied to the separate j −element, namely

$$\begin{aligned} \left\| \int\_{0}^{\boldsymbol{\alpha}\_{\parallel}} \mathcal{X}\_{\boldsymbol{\mu}\prime}(\mathbf{x}) \mathcal{D}^{4} \mathcal{X}\_{\boldsymbol{\mu}\prime}(\mathbf{x}) d\mathbf{x} &= \left( \mathcal{X}\_{\boldsymbol{\mu}\prime}(\mathbf{x}) \mathcal{D}^{3} \mathcal{X}\_{\boldsymbol{\mu}\prime}(\mathbf{x}) - \mathcal{D} \mathcal{X}\_{\boldsymbol{\mu}\prime}(\mathbf{x}) \mathcal{D}^{2} \mathcal{X}\_{\boldsymbol{\mu}\prime}(\mathbf{x}) \right) \right\|\_{0}^{\mathbb{H}} + \\ &+ \int\_{0}^{\boldsymbol{\alpha}\_{\parallel}} \mathcal{D}^{2} \mathcal{X}\_{\boldsymbol{\mu}\prime}(\mathbf{x}) \mathcal{D}^{2} \mathcal{X}\_{\boldsymbol{\mu}\prime}(\mathbf{x}) d\mathbf{x} \end{aligned} \tag{43}$$

Considering both boundary conditions of the problem and coupling conditions between neighboring elements, Eqs. (22)−(26), instead of Eq. (42) one has

Now, the unknowns put down in column matrix, Eq. (35), should be determined. Let the

<sup>123</sup> B : D : A :... ( 1) M : ( 1) M :( 1) M : ... <sup>112</sup> <sup>1</sup> <sup>2</sup> <sup>3</sup>

j j j j jj j X(x) X (x) X ( x) X (x) X (x) ν

X (x) A K ( x) B K ( x) C K ( x) D K ( x) j j

Orthogonality condition of the uniform beam modes may be found in (Kaliski, 1986; de

( ) 4 3 22 <sup>2</sup> <sup>0</sup> <sup>0</sup> <sup>0</sup> X (x)D X (x)dx X (x)D X (x) DX (x)D X (x) D X (x)D X (x)dx

4 2 EJD X (x) S X (x)

4 2 EJD X (x) S X (x)

 and X (x) μ

( )() ( ) <sup>2</sup> <sup>2</sup> 32 32

μν

j

 νμ

D X (x)D X (x)dx

+

neighboring elements, Eqs. (22)−(26), instead of Eq. (42) one has

integration x [0, ] ∈ , use Eq. (39), subtract the second result from the first one, one obtains

 μμ = ω ρ

 νν = ω ρ

μ

ν

ν

For standard boundary conditions, the right-hand-side equals zero.

=− +

 νμ

, fulfill the following modal equations

0 0

 μ

 ν

 νμ

=− +

 μ

 μν

S X X dx EJ X D X DX D X X D X DX D X

− = − −−

The procedure outlined above can be used to the problem presented in Fig. 6, but Eq. (39)

( ) <sup>j</sup> <sup>j</sup>

<sup>e</sup> <sup>e</sup> 4 32 jj jj jj <sup>0</sup> <sup>0</sup> <sup>e</sup> 2 2 j j <sup>0</sup>

X (x)D X (x)dx X (x)D X (x) DX (x)D X (x)

ν

Considering both boundary conditions of the problem and coupling conditions between

 = +++ 1j j λλλ

 == = = λ

αβ

ν

νν

αβ

++ + =− − − (36)

−eigenfunctions (

νν

νν

be the minor of the A

ααα

 ν

ααα

ν

, where

αβ

α and β

ν

 ν

ν

 μ

(40)

(41)

 ν  μ

(43)

respectively, integrate both in range o

(39)

(42)

 λ2j j 3j j 4 j (38)

label the

element. The general

−modes) assigned

(37)

main matrix elements **A** be written as two suffix quantities A

} and unknowns **x** to Eq. (20), the

<sup>=</sup> and the separate modes are equal

Silva, 2000). First of all, based on twice integration by parts, one has

 νμ

 , X (x) ν

to the j −element are obtained. The solution to Eq. (10) is given by

rows and columns respectively. Let M

solution to Eq. (29) is

ν

λjν

j j (...) (...)

**2.4 Orthogonality of modes** 

νμ

The separate modes X (x)

ω ωρ

ν μ ν

 ν

μ

Multiplying above equations by X (x)

(for simplicity an argument (x) is omitted)

 νμ

must be applied to the separate j −element, namely

νμ

Substituting {

where

( )( 12 34 eeee 2 2 11 1 1 22 2 2 33 3 3 44 4 4 <sup>0000</sup> S X X dx S X X dx S X X dx S X X dx ωωρ ν μ − ++++ ν μ ρ ν μ ρ ν μ ρ ν μ + = m X (0)X (0) E J r4 4 ν μ ) 4 4 ⋅ ( ) ( ) 32 32 X (e )D X (e ) DX (e )D X (e ) X (e )D X (e ) DX (e )D X (e ) 4 4 44 4 4 44 44 4 4 44 4 4 μν μν ν μ ν μ⋅ − −− (44)

Because of Eq. (26), the right-hand-side is zero, hence

$$\begin{split} \left(\alpha\_{\nu}^{2} - \alpha\_{\mu}^{2}\right) \Big(\rho\_{1}\mathbb{S}\_{1} \int\_{0}^{\boldsymbol{\alpha}\_{1}} \mathbb{X}\_{1\nu} \mathcal{X}\_{1\mu} \, \mathrm{d}\mathbf{x} + \rho\_{2}\mathbb{S}\_{2} \int\_{0}^{\boldsymbol{\alpha}\_{1}} \mathbb{X}\_{2\nu} \mathcal{X}\_{2\mu} \, \mathrm{d}\mathbf{x} + \rho\_{3}\mathbb{S}\_{3} \int\_{0}^{\boldsymbol{\alpha}\_{1}} \mathbb{X}\_{3\nu} \mathcal{X}\_{3\mu} \, \mathrm{d}\mathbf{x} + \rho\_{4}\mathbb{S}\_{4} \int\_{0}^{\boldsymbol{\alpha}\_{4}} \mathbb{X}\_{4\nu} \mathcal{X}\_{4\mu} \, \mathrm{d}\mathbf{x} + \rho\_{5}\mathbb{S}\_{5} \int\_{0}^{\boldsymbol{\alpha}\_{1}} \mathbb{X}\_{4\nu} \mathcal{X}\_{4\mu} \, \mathrm{d}\mathbf{x} + \rho\_{6}\mathbb{S}\_{6} \int\_{0}^{\boldsymbol{\alpha}\_{1}} \mathbb{X}\_{4\nu} \mathcal{X}\_{4\mu} \, \mathrm{d}\mathbf{x} + \rho\_{7}\mathbb{S}\_{7} \int\_{0}^{\boldsymbol{\alpha}\_{1}} \mathbb{X}\_{4\nu} \mathcal{X}\_{4\mu} \, \mathrm{d}\mathbf{x} \right) = 0. \end{split} \tag{45}$$

Fig. 7. Geometry of set with nj −elements

The orthogonality condition, Eq. (45), may be generalized in a simple way. Let the system beam-actuators-masses be divided into nj −elements as depicted in Fig. 7. In this case one has

$$\mathbb{E}\left(\alpha\_{\nu}^{2} - \alpha\_{\mu}^{2}\right) \left[ \sum\_{\boldsymbol{\cdot}} \left( \rho\_{\boldsymbol{\cdot}} \mathbb{S}\_{\boldsymbol{\cdot}} \int\_{\boldsymbol{\cdot}} \mathcal{X}\_{\boldsymbol{\cdot}\boldsymbol{\cdot}\boldsymbol{\cdot}} \mathcal{X}\_{\boldsymbol{\cdot}\boldsymbol{\mu}} \, d\mathbf{x} + \mathbf{m}\_{\boldsymbol{\cdot}} \mathcal{X}\_{\boldsymbol{\cdot}\boldsymbol{\cdot}\boldsymbol{\cdot}}(\mathbf{0}) \mathcal{X}\_{\boldsymbol{\cdot}\boldsymbol{\mu}}(\mathbf{0}) \right) + \mathbf{m}\_{\boldsymbol{\cdot}\_{\boldsymbol{\cdot}} + 1} \mathcal{X}\_{\boldsymbol{\cdot}\_{\boldsymbol{\cdot}}\boldsymbol{\cdot}}(\mathbf{e}\_{\mathbf{n}\_{\boldsymbol{\cdot}}}) \mathcal{X}\_{\boldsymbol{\cdot}\_{\boldsymbol{\cdot}}\boldsymbol{\mu}}(\mathbf{e}\_{\mathbf{n}\_{\boldsymbol{\cdot}}}) \right] = 0 \tag{46}$$

Since the term 2 2 ω ω ν μ − is canceled for μ =ν, the general orthogonality condition is given by

$$\sum\_{\boldsymbol{\nu}} \left( \rho\_{\boldsymbol{\triangleright}} \mathbf{S}\_{\boldsymbol{\triangleright}} \Big| \, \mathbf{X}\_{\boldsymbol{\triangleright}\boldsymbol{\nu}} \mathbf{X}\_{\boldsymbol{\mu}\boldsymbol{\nu}} \mathbf{dx} + \mathbf{m}\_{\boldsymbol{\triangleright}} \mathbf{X}\_{\boldsymbol{\n}\boldsymbol{\nu}}(\mathbf{0}) \mathbf{X}\_{\boldsymbol{\mu}\boldsymbol{\nu}}(\mathbf{0}) \right) + \mathbf{m}\_{\mathbf{n}\_{\boldsymbol{\flat}} + \mathbf{1}} \mathbf{X}\_{\mathbf{n}\_{\boldsymbol{\flat}}\boldsymbol{\nu}}(\mathbf{e}\_{\mathbf{n}\_{\boldsymbol{\flat}}}) \mathbf{X}\_{\mathbf{n}\_{\boldsymbol{\flat}}\boldsymbol{\mu}}(\mathbf{e}\_{\mathbf{n}\_{\boldsymbol{\flat}}}) = \begin{cases} 0, & \boldsymbol{\nu} \neq \boldsymbol{\mu} \\ \boldsymbol{\mathcal{B}}\_{\nu}^{2}, & \boldsymbol{\nu} = \boldsymbol{\mu} \end{cases} \tag{47}$$

The Eq. (47) in particular case is used in deriving the solution to the forced vibration problem.

### **2.5 Forced vibrations with damping**

A point departure for further consideration is Eq. (7); for j −element one has

$$\mathbf{E}\_{\flat} \mathbf{J}\_{\flat} \mathbf{D}^{4} \mathbf{u}\_{\flat} + \mu\_{\flat} \mathbf{E}\_{\flat} \mathbf{J}\_{\flat} \mathbf{D}^{4} \left( \mathbf{D}\_{\flat} \mathbf{u}\_{\flat} \right) + \rho\_{\flat} \mathbf{S}\_{\flat} \mathbf{D}\_{\flat}^{2} \mathbf{u}\_{\flat} = -\mathbf{f}\_{\flat} \tag{48}$$

The solution to Eq. (48) is forced vibrations with damping. Let be the load force in the form

$$\mathbf{f}\_{\rangle}(\mathbf{x}, \mathbf{t}) = \mathbf{f}\_{\rangle}(\mathbf{x}) \exp(i\alpha\_{\uparrow}\mathbf{t})\tag{49}$$

where 1 2 i ( 1) = − , ωf – excited frequency. Applying separation of variables method, the solution to Eq. (48) is assumed as

$$\mathbf{u}\_{\rangle}(\mathbf{x}, \mathbf{t}) = \mathbf{X}\_{\neq}(\mathbf{x}) \exp(\mathbf{i}a\_{\parallel}\mathbf{t})\tag{50}$$

Substituting Eqs. (49) and (50) to Eq. (48) one obtains

An Optimal Distribution of Actuators in

Fig. 8. External line moments of the actuator

a

Fig. 9. External pair of forces of the actuator

of actuators and the glue on the beam.

Next, the separate forces are considered in Eq. (56).

**2.7 Beam vibration reduction through actuators** 

polarization.

down to the form

given by

j I ν

Active Beam Vibration – Some Aspects, Theoretical Considerations 409

where Ca – constant depending on geometry and mechanical properties of the actuator and plate, d31 − piezoelectric material strain constants, V – voltage in the direction of

The problem is to determine of the Ca , because it depends on the analysis method of the mutual interaction between beam-actuator (Hansen & Snyder 1997; Pietrzakowski, 2004). Let the static force coupling model is taken into account. If relatively thin actuator compared with beam thickness is assumed (so uniform normal stress distribution is accepted) and furthermore by ignoring the neutral axis displacement d , see Fig. 4, the constant Ca is come

( )

fa

<sup>+</sup> <sup>=</sup> + (59)

Mf 2 x aa = (60)

x

(61)

( ) bb aa b a

Eh Eh h h

2Eh Eh

Since the beam vibration equation is the forces equation then to consider the action of

fa

fa

For the problem presented in Fig. 5, the total load of the beam, described by Eq. (56), is

j 0j ( ) s 1s jss js 2s f (x) f f (x x ) 2f (x ) f (x x ) =− + − − + +

where the symbol ja f is replace by js f in order to express, in the future, the interaction sum

An expression in brackets is the sum of interacting forces actuator-beam. Hence, the integral

δδ

actuator with the beam, moments Mx are replaced with two couples of forces, Fig. 9,

bb aa

x

Mx Mx

x1 x <sup>a</sup> <sup>2</sup>

a

fa /2

δ

, Eq. (54), for j f (x) expressed by above equation is given by

a

C

$$\mathcal{X}\_{\parallel}(1+i\mu\rho\_{\hbar})\mathcal{D}^4\mathcal{X}\_{\neq}(\mathbf{x}) - a\rho^2\mathcal{X}\_{\neq}(\mathbf{x}) = -\frac{1}{\rho\_{\parallel}\mathcal{S}\_{\neq}}\mathbf{f}\_{\neq}(\mathbf{x})\tag{51}$$

The solution of the above equation is given by

$$\mathcal{X}\_{\restriction \mathbb{\!}}(\mathbf{x}) = \sum\_{\nu} \mathsf{C}\_{\restriction \nu} \mathsf{X}\_{\restriction \mathbb{\!}^{\nu}}(\mathbf{x}) \tag{52}$$

where Cjν − constants, X (x) <sup>j</sup>ν− Eq. (38).

After some calculation, the constants Cjνare expressed by

$$\mathbf{C}\_{\parallel^{\nu}} = \frac{1}{\rho\_{\parallel} \mathbf{S}\_{\parallel}} \frac{1}{\alpha\_{\parallel^{\nu}}^2} \frac{1}{\beta\_{\nu}^2} \mathbf{I}\_{\parallel^{\nu}} = \mathbf{C}\_{\parallel^{\nu}}^\* \mathbf{I}\_{\parallel^{\nu}} \tag{53}$$

where

$$\mathbf{C}'\_{\mid\nu} = \frac{1}{\rho\_{\rangle}\mathbf{S}\_{\mid\nu}} \frac{1}{\alpha\_{\mid\nu}^2} \frac{1}{\beta\_{\nu}^2}, \ \frac{1}{\alpha\_{\mid\nu}^2} = \frac{1}{(1 + \mathbf{i}\mu\_{\mid}\alpha\_{\mathbf{i}})\alpha\_{\nu}^2 - \alpha\_{\mathbf{i}}^2}, \ \mathbf{I}\_{\mid\nu} = -\int\_{\cdot} \mathbf{f}\_{\mid}(\mathbf{x}) \mathbf{X}\_{\mid\nu}(\mathbf{x}) d\mathbf{x} \tag{54}$$

In the end, the problem of the forced j −element beam vibration with damping, excited with the force j f (x) is solved; in the harmonic steady state it is given by

$$X\_{\mathbb{f}}(\mathbf{x}) = \sum\_{\mathbb{I}} X\_{\mathbb{f}}(\mathbf{x}) = \sum\_{\mathbb{I}^{\nu}} \mathbf{C}\_{\mathbb{I}^{\nu}} X\_{\mathbb{I}^{\nu}}(\mathbf{x}) \tag{55}$$

In current problem, two form of the forces have the practical meaning namely, the force with constant amplitude j0 0 f (x) f = and the force acting at discrete point ja i f (x ) . The former may be interpreted as the spread excitation forced, for example with plane acoustic wave, but the latter is the control force due to actuators, henceforth

$$\mathbf{f\_{i}(x) = f\_{0} + f\_{\mu}(x\_{i})} \tag{56}$$

### **2.6 Interaction between beam and actuators**

It is assumed that the actuator is perfectly bonded to the beam surface. Exciting actuator, the interaction between actuator and the beam is appeared. The interaction process is explained in (Hansen & Snyder, 1997; Fuller at al, 1997) in detail and references cited therein. Assuming the spatially uniform actuator, it provides boundary induction solely in terms of the external line moment distributed along its edges (Burke & Hubbard, 1991; Sullivan et al., 1996). So, the bending moment in y −direction is given by the formula (Hansen & Snyder 1997), Fig. 8,

$$\mathbf{M}\_{\mathbf{x}} = -\mathbf{M}\_0 \left( <\mathbf{x} - \mathbf{x}\_1 >^{-2} - <\mathbf{x} - \mathbf{x}\_2 >^{-2} \right) \tag{57}$$

where (.) . δ <sup>1</sup> =>< <sup>−</sup> and D. (.) <sup>2</sup> =>< δ<sup>−</sup> − doublet function, M0 – line moment amplitude

$$\mathbf{M}\_0 = \mathbf{C}\_\mathbf{a} \, \frac{\mathbf{d}\_{31}}{\mathbf{h}\_\mathbf{a}} \, \mathbf{V} \tag{58}$$

where Ca – constant depending on geometry and mechanical properties of the actuator and plate, d31 − piezoelectric material strain constants, V – voltage in the direction of polarization.

Fig. 8. External line moments of the actuator

408 Acoustic Waves – From Microdevices to Helioseismology

<sup>1</sup> (1 i )D X (x) X (x) f (x) <sup>S</sup>

X (x) C X (x) jf j j ν ν

ν

j j 2 2 j j jj j 1 11 <sup>C</sup> I CI

j j f f

In the end, the problem of the forced j −element beam vibration with damping, excited with

f jf j <sup>j</sup> j j X (x) X (x) C X (x)

In current problem, two form of the forces have the practical meaning namely, the force with constant amplitude j0 0 f (x) f = and the force acting at discrete point ja i f (x ) . The former may be interpreted as the spread excitation forced, for example with plane acoustic wave,

It is assumed that the actuator is perfectly bonded to the beam surface. Exciting actuator, the interaction between actuator and the beam is appeared. The interaction process is explained in (Hansen & Snyder, 1997; Fuller at al, 1997) in detail and references cited therein. Assuming the spatially uniform actuator, it provides boundary induction solely in terms of the external line moment distributed along its edges (Burke & Hubbard, 1991; Sullivan et al., 1996). So, the bending moment in y −direction is given by the formula (Hansen & Snyder

( ) 2 2 M M xx xx x0 1 <sup>2</sup>

0 a

<sup>d</sup> MC V

31

a

δ

are expressed by

ν

 ν μωω

 ω <sup>=</sup> + − , j jj <sup>j</sup> I f (x) X (x)dx ν

 νν

ν ν

ν

ω

j j

<sup>=</sup> (52)

<sup>∗</sup> = = (53)

= = (55)

<sup>j</sup> <sup>0</sup> ja i f (x) f f (x ) = + (56)

− − =− < − > − < − > (57)

<sup>h</sup> <sup>=</sup> (58)

<sup>−</sup> − doublet function, M0 – line moment amplitude

 ν= − (54)

ρ

+ −= − (51)

4 2 j j f jf f jf j

γ

ν

The solution of the above equation is given by

− constants, X (x) <sup>j</sup>

After some calculation, the constants Cj

j 2 2 jj j 1 11 <sup>C</sup> S

α ν βν

**2.6 Interaction between beam and actuators** 

where (.) . δ <sup>1</sup> =>< <sup>−</sup> and D. (.) <sup>2</sup> =><

ν ρ

where Cj

where

1997), Fig. 8,

ν

 μω

− Eq. (38).

ν

<sup>∗</sup> = , 2 2 <sup>2</sup>

α ν

the force j f (x) is solved; in the harmonic steady state it is given by

but the latter is the control force due to actuators, henceforth

ν

S

ρ αβ ν ν

1 1 (1 i ) The problem is to determine of the Ca , because it depends on the analysis method of the mutual interaction between beam-actuator (Hansen & Snyder 1997; Pietrzakowski, 2004). Let the static force coupling model is taken into account. If relatively thin actuator compared with beam thickness is assumed (so uniform normal stress distribution is accepted) and furthermore by ignoring the neutral axis displacement d , see Fig. 4, the constant Ca is come down to the form

$$\mathbf{C\_{a}} = \frac{\mathbf{E\_{b}h\_{b}}\mathbf{E\_{a}h\_{a}}(h\_{b} + h\_{a})}{2\left(\mathbf{E\_{b}h\_{b}} + \mathbf{E\_{a}h\_{a}}\right)}\tag{59}$$

Since the beam vibration equation is the forces equation then to consider the action of actuator with the beam, moments Mx are replaced with two couples of forces, Fig. 9,

$$\mathbf{M}\_{\mathbf{x}} = \mathbf{f}\_{\mathbf{a}} \ell\_{\mathbf{a}} / 2 \tag{60}$$

Fig. 9. External pair of forces of the actuator

Next, the separate forces are considered in Eq. (56).

### **2.7 Beam vibration reduction through actuators**

For the problem presented in Fig. 5, the total load of the beam, described by Eq. (56), is given by

$$\mathbf{f}\_{\rangle}(\mathbf{x}) = -\mathbf{f}\_{0} + \left(\mathbf{f}\_{\neq}\mathcal{S}(\mathbf{x} - \mathbf{x}\_{1s}) - 2\mathbf{f}\_{\neq}\mathcal{S}(\mathbf{x}\_{s}) + \mathbf{f}\_{\neq}\mathcal{S}(\mathbf{x} + \mathbf{x}\_{2s})\right) \tag{61}$$

where the symbol ja f is replace by js f in order to express, in the future, the interaction sum of actuators and the glue on the beam.

An expression in brackets is the sum of interacting forces actuator-beam. Hence, the integral j I ν, Eq. (54), for j f (x) expressed by above equation is given by

An Optimal Distribution of Actuators in

where

or

where

parameters:

ω

• <sup>s</sup> κ

•

The Eqs (66), (69) and (70) and may be written commonly

<sup>f</sup> − excited frequency, it is contained in C

• xs − distribution of the actuators on the beam,

**3. Optimal actuators distribution problem** 

**3.1 Reduction and effectiveness coefficients** 

<sup>R</sup> Ψ (x) are given together by Eq. (74), where

properties of the actuators,

• ns − number of actuators.

reduction should be defined.

Active Beam Vibration – Some Aspects, Theoretical Considerations 411

jf jf j jf j (x) (x) C A (x) ν

> νν

κ κ

j j j j (x) (x) (x)

jj j <sup>j</sup> (x) C A (x) CA (x) ν ν

0 ss s <sup>0</sup> <sup>s</sup> A C I C I f (x ) C I I

Φ = (x) { X(x),D (x), (x)} κ κ

It is appeared from Eq. (75) that the active vibration reduction depends on the following

(x ) − value of the beam curvature at the point of the actuators distribution, • sf − interacting forces between beam-actuators or more generally − mechanical

• <sup>s</sup> − actuators lengths or more generally – geometrical properties of the actuators,

important meaning and finding of the { x is the aim of the chapter. s}

Let be the difference between any quantities of the beam vibration

As mentioned above, the optimal actuators distribution described with { x has an s}

Before the optimization problem will be formulated, any coefficients of the vibration

where Ψ(x) , R Ψ (x) – quantities calculated without and with actuators respectively; Ψ(x) ,

A CI0

ν

<sup>∗</sup> ,

( ) ( ) <sup>2</sup>

κ

 ν

ν

Omitting for simplicity the index "f", for entire system beam-actuators one has

ν

jf jf jf jf Ψ = {X ,Q ,M }, j j jj {X ,D , }, Φ = ν

 ν ν

ν ν

Ψ= Ψ = Ψ (73)

Ψ= Φ = Φ (74)

∗ ∗ <sup>∗</sup> == + = + <sup>Σ</sup> (75)

<sup>R</sup> ΔΨ = Ψ − Ψ (x) (x) (x) (77)

<sup>∗</sup> = (78)

Ψ=Ψ = Φ (71)

C {1, E J , E J } j j =± ±j j j (72)

(76)

 ν

$$\begin{split} \mathbf{I}\_{|\boldsymbol{\nu}|} = -\int\_{\boldsymbol{\restriction}} \mathbf{f}\_{\boldsymbol{\restriction}}(\mathbf{x}) \mathbf{X}\_{|\boldsymbol{\nu}|}(\mathbf{x}) d\mathbf{x} = -\mathbf{f}\_{0} \Big[ \mathbf{X}\_{|\boldsymbol{\nu}|}(\mathbf{x}) d\mathbf{x} + \mathbf{f}\_{\boldsymbol{\restriction}} \Big[ \Big( \mathcal{S}(\mathbf{x} - \mathbf{x}\_{1:\boldsymbol{\nu}}) - 2\mathcal{S}(\mathbf{x}\_{\boldsymbol{\nu}}) + \mathcal{S}(\mathbf{x} + \mathbf{x}\_{2:\boldsymbol{\nu}}) \Big) \mathbf{X}\_{|\boldsymbol{\nu}|} d\mathbf{x} = \\ = -\mathbf{f}\_{0} \Big[ \mathbf{X}\_{|\boldsymbol{\nu}|}(\mathbf{x}) d\mathbf{x} + \mathbf{f}\_{\boldsymbol{\textnormal}} \Big[ \mathbf{X}\_{|\boldsymbol{\nu}|}(\mathbf{x}\_{1:\boldsymbol{\nu}}) - 2\mathbf{X}\_{|\boldsymbol{\nu}|}(\mathbf{x}\_{\boldsymbol{\nu}}) + \mathbf{X}\_{|\boldsymbol{\nu}|}(\mathbf{x}\_{2:\boldsymbol{\nu}}) \Big] \end{split} \tag{62}$$

The expression in square bracket constitutes the second-order central finite difference. Since the distance between nodes s is constant, then the difference can be transformed into

$$\frac{1}{\ell\_s^2} \left[ \mathbf{X}\_{\left| \nu \right>} (\mathbf{x}\_{1s}) - 2 \mathbf{X}\_{\left| \nu \right>} (\mathbf{x}\_s) + \mathbf{X}\_{\left| \nu \right>} (\mathbf{x}\_{2s}) \right] = \mathbf{D}^2 \mathbf{X}\_{\left| \nu \right>} (\mathbf{x}\_s) \tag{63}$$

where

$$\mathrm{D}^{2}\mathbb{X}\_{\left|\nu\right.}\left(\mathbf{x}\_{\ast}\right) = \pm\kappa\_{\left|\nu\right.}\left(\mathbf{x}\_{\ast}\right)\tag{64}$$

The j <sup>s</sup> (x ) κ ν is the curvature of the mode X (x) <sup>j</sup>ν at the point x x = s (Brański & Szela, 2007; Brański & Szela, 2008). The sign of the j <sup>s</sup> (x ) κ ν is contractual namely, if the bending of the beam is directed upwards, the sign is positive and vice versa. Substituting Eq. (64) into Eq. (62), one obtains

$$\mathbf{I}\_{\vert \nu} = -\mathbf{f}\_0 \int\_{\vert \cdot} \mathbf{X}\_{\vert \nu}(\mathbf{x}) \, \mathrm{d}\mathbf{x} + \mathbf{f}\_{\vert \ast} \ell\_s^2 \, \kappa\_{\vert \nu}(\mathbf{x}\_s) = \mathbf{I}\_{\vert \nu \vert 0} + \mathbf{I}\_{\vert \nu \ast} \tag{65}$$

Next, substituting Eq. (65) into Eq. (52) through Eq. (53), the reduction vibration is obtained

$$\mathbf{X}\_{\left[\nu\right]}\left(\mathbf{x}\right) = \sum\_{\nu} \mathbf{C}\_{\left]\nu\right]}^{\ast} \mathbf{I}\_{\left]\nu} \mathbf{X}\_{\left]\nu\right]}(\mathbf{x}) = \sum\_{\nu} \mathbf{C}\_{\left]\nu}^{\ast} \left(\mathbf{I}\_{\left]\nu 0} + \mathbf{I}\_{\left]\nu \mathbf{s}}\right) \mathbf{X}\_{\left]\nu\right]}(\mathbf{x}) = \sum\_{\nu} \mathbf{A}\_{\left]\nu} \mathbf{X}\_{\left]\nu}(\mathbf{x})\tag{66}$$

Note, that the amplitude Ajfν is the direct quantity which is liable to the reduction, in explicit form is

A CI C I I jfν νν ν ν ν jj j j ( ) <sup>0</sup> <sup>j</sup> <sup>s</sup> ∗ ∗ == + (67)

At the same time, together with the vibration reduction amplitude Ajfν , the curvature is subjected to the reduction and based on Eq. (66) is

$$\boldsymbol{\kappa}\_{\boldsymbol{\upmu}} = \pm \mathbf{D}^2 \mathbf{X}\_{\boldsymbol{\upmu}} = \sum\_{\boldsymbol{\upnu}} \boldsymbol{\kappa}\_{\boldsymbol{\upmu}\boldsymbol{\upnu}} = \pm \sum\_{\boldsymbol{\upnu}} \mathbf{A}\_{\boldsymbol{\upmu}\boldsymbol{\upnu}} \boldsymbol{\kappa}\_{\boldsymbol{\upnu}} \tag{68}$$

Furthermore, the reduction of the Ajfν leads to the reduction of the shear force Q (x) and jf bending moment M (x) (Bra jf ński et al., 2010; Kaliski, 1986; Kozień, 2006),

$$\mathbf{Q}\_{\not\!\!\!\!} (\mathbf{x}) = \pm \mathbf{E}\_{\not\!\!\!\!} \mathbf{J}\_{\not\!\!\!\!} \mathbf{D} \boldsymbol{\kappa}\_{\not\!\!\!\!\!\/} (\mathbf{x}) = \pm \mathbf{E}\_{\not\!\!\!\!\!\!\/\!\!\!\!} \sum\_{\not\!\!\!\!\/\!\!\!\/} \mathbf{D} \boldsymbol{\kappa}\_{\not\!\!\!\!\/} (\mathbf{x}) = \pm \mathbf{E}\_{\not\!\!\!\!\!\/\!\!\!\!\/} \sum\_{\not\!\!\!\!\/\!\!\!\/} \mathbf{A}\_{\not\!\!\!\!\!\!\/} \mathbf{D} \boldsymbol{\kappa}\_{\not\!\!\!\!\/} (\mathbf{x}) \tag{69}$$

$$\mathbf{M}\_{\boldsymbol{\upbeta}}(\mathbf{x}) = \pm \mathbf{E}\_{\boldsymbol{\upbeta}} \mathbf{J}\_{\boldsymbol{\upbeta}} \boldsymbol{\upkappa}\_{\boldsymbol{\upbeta}}(\mathbf{x}) = \pm \mathbf{E}\_{\boldsymbol{\upbeta}} \mathbf{J}\_{\boldsymbol{\upbeta}} \sum\_{\nu} \kappa\_{\boldsymbol{\upbeta}\nu}(\mathbf{x}) = \pm \mathbf{E}\_{\boldsymbol{\upbeta}} \mathbf{J}\_{\boldsymbol{\upbeta}} \sum\_{\nu} \mathbf{A}\_{\boldsymbol{\upbeta}\nu} \kappa\_{\boldsymbol{\upbeta}}(\mathbf{x}) \tag{70}$$

As can be seen, the vibration reduction undergo on the following amplitudes: of the beam vibration X (x) jf − Eq. (66), of the shear force Q (x) jf − Eq. (69) and of the bending moment M (x) jf − Eq. (70). Hereafter, the notion e.g. "shear force reduction" is used instead of "the reduction of the amplitude of the shear force", and so on.

The Eqs (66), (69) and (70) and may be written commonly

$$\Psi\_{\restriction \ell}(\mathbf{x}) = \sum\_{\nu} \, \_{\nu} \Psi\_{\restriction \nu}(\mathbf{x}) = \mathbf{C}\_{\restriction} \sum\_{\nu} \, \_{\n} \mathbf{A}\_{\restriction \nu} \, \_{\n} \Phi\_{\restriction \nu}(\mathbf{x}) \tag{71}$$

where

410 Acoustic Waves – From Microdevices to Helioseismology

 = − = − + −− ++ = δ

0 j js j 1s j s j 2s <sup>j</sup> f X (x)dx f X (x ) 2X (x ) X (x )

The expression in square bracket constitutes the second-order central finite difference. Since the distance between nodes s is constant, then the difference can be transformed into

2 j 1s j s j 2s j s

<sup>2</sup> D X (x ) (x ) <sup>j</sup>

ν

beam is directed upwards, the sign is positive and vice versa. Substituting Eq. (64) into

Next, substituting Eq. (65) into Eq. (52) through Eq. (53), the reduction vibration is obtained

X (x) C I X (x) C I I X (x) A X (x) jf jj j

A CI C I I jf

 ν ν

ν

ν

ν

Q (x) E J D (x) E J D (x) E J A D (x) jf j j jf j j jf

 =± =± κκ

ν

M (x) E J (x) E J (x) E J A (x) jf j j jf j j jfν

As can be seen, the vibration reduction undergo on the following amplitudes: of the beam vibration X (x) jf − Eq. (66), of the shear force Q (x) jf − Eq. (69) and of the bending moment M (x) jf − Eq. (70). Hereafter, the notion e.g. "shear force reduction" is used instead of "the

 =± =± κκ

ν

jj j j ( ) <sup>0</sup> <sup>j</sup> <sup>s</sup>

 νν

 ν

 j j0 js j νν

2 <sup>j</sup> <sup>0</sup> j js s <sup>j</sup> <sup>s</sup> <sup>j</sup> <sup>0</sup> <sup>j</sup> <sup>s</sup> <sup>j</sup> I f X (x)dx f (x ) I I

 ν + κ

> ν ν( ) jf j

Ajf

κκ

 ν

∗ ∗ = = += (66)

 ν

> ν νj

=± = = ± (68)

leads to the reduction of the shear force Q (x) and jf

 ν

 ν

 νj j jf j

 νj j jf j

 κ= ± (70)

= ± (69)

ν

 κ

ν

is the direct quantity which is liable to the reduction, in

∗ ∗ == + (67)

 = + ν

 ν <sup>s</sup> = ±κ

νν

<sup>1</sup> X (x ) 2X (x ) X (x ) D X (x )

νν= − + −+ (62)

2

−+ = (63)

 ν

<sup>j</sup> s (64)

at the point x x = s (Brański & Szela, 2007;

ν ν

ν

, the curvature is

ν

is contractual namely, if the bending of the

ν(65)

δδ

ν

j jj 0 j js 1 ( ) ss 2s j j jj I f (x) X (x)dx f X (x)dx f (x x ) 2 (x ) (x x ) X dx

 ν

ν

κ ν

ν

 ν

ν

ν

is the curvature of the mode X (x) <sup>j</sup>

s

Brański & Szela, 2008). The sign of the j <sup>s</sup> (x )

ν= −

ν

subjected to the reduction and based on Eq. (66) is

Furthermore, the reduction of the Ajf

κ

reduction of the amplitude of the shear force", and so on.

νν ν

ν

ν

At the same time, together with the vibration reduction amplitude Ajf

2 jf D Xjf jf

ν

bending moment M (x) (Bra jf ński et al., 2010; Kaliski, 1986; Kozień, 2006),

νν

where

The j <sup>s</sup> (x ) κ ν

Eq. (62), one obtains

explicit form is

Note, that the amplitude Ajf

$$\mathbf{^j\Psi\_{j\!\!\!\!=}} = \{ \mathbf{X}\_{\not\!\! \! / \prime} \mathbf{Q}\_{\not\!\! \! / \prime} \mathbf{M}\_{\not\!\! \! \! / \prime} \} \qquad \mathbf{\Phi}\_{\!\!\! \nu} = \{ \mathbf{X}\_{\!\! \! \nu} \mathbf{D} \, \mathbf{\tilde{\kappa}}\_{\!\! \! \nu} \, \mathbf{\tilde{\kappa}}\_{\!\! \! \nu} \} \qquad \mathbf{C}\_{\!\!\! \! \! \! \! \! \! \/ \prime} \mathbf{E} = \{ \mathbf{1} \, \! \! \! \pm \mathbf{E}\_{\!\! \! \! \! \! \! \! \! \! \/ \prime} \mathbf{E}\_{\!\! \! \! \! \! \! \! \! \! \! \! \! \! \! \/ \prime} \} \tag{72}$$

Omitting for simplicity the index "f", for entire system beam-actuators one has

$$\Psi(\mathbf{x}) = \sum\_{\parallel} \Psi\_{\parallel}(\mathbf{x}) = \sum\_{\parallel \nu} \Psi\_{\parallel \nu}(\mathbf{x}) \tag{73}$$

or

$$\mathbf{^\downarrow \Psi}(\mathbf{x}) = \sum\_{\restriction \nu} \mathbf{C}\_{\restriction} \mathbf{A}\_{\restriction \nu} \, \Phi\_{\restriction \nu}(\mathbf{x}) = \mathbf{C} \mathbf{A} \, \Phi(\mathbf{x}) \tag{74}$$

where

$$\mathbf{A} = \mathbf{C}^\* \mathbf{I} = \mathbf{C}^\* \left( \mathbf{I}\_0 + \sum\_{s} \mathbf{f}\_s \ell\_s^2 \, \kappa(\mathbf{x}\_s) \right) = \mathbf{C}^\* \left( \mathbf{I}\_0 + \mathbf{I}\_\varepsilon \right) \tag{75}$$

$$\Phi(\mathbf{x}) = \{ \mathbf{X}(\mathbf{x}), \mathbf{D}\kappa(\mathbf{x}), \kappa(\mathbf{x}) \}\tag{76}$$

It is appeared from Eq. (75) that the active vibration reduction depends on the following parameters:


As mentioned above, the optimal actuators distribution described with { x has an s} important meaning and finding of the { x is the aim of the chapter. s}

### **3. Optimal actuators distribution problem**

Before the optimization problem will be formulated, any coefficients of the vibration reduction should be defined.

### **3.1 Reduction and effectiveness coefficients**

Let be the difference between any quantities of the beam vibration

$$
\Delta\Psi(\mathbf{x}) = \Psi(\mathbf{x}) - \Psi\_{\mathbb{R}}(\mathbf{x}) \tag{77}
$$

where Ψ(x) , R Ψ (x) – quantities calculated without and with actuators respectively; Ψ(x) , <sup>R</sup> Ψ (x) are given together by Eq. (74), where

$$\mathbf{A} = \mathbf{C}^\* \mathbf{I}\_0 \tag{78}$$

An Optimal Distribution of Actuators in

Active Beam Vibration – Some Aspects, Theoretical Considerations 413

means p-reduction; such instance is considered in (Brański et al., 2010; Brański & Lipiński,

Let the energy provided to the actuators be constant and hence, the sf is always constant. Now, for clarity of the disquisition, rewrite the effectiveness coefficient in explicit form

R R (x) (x) CA (x) CA (x) R (x) (x) CA (x) <sup>Ψ</sup>

Working out on above assumption, the R (x) <sup>Ψ</sup> will be maximal, if R Ψ (x) is minimal. Hence, the optimal condition R (x) R (x) Ψ Ψ = ;max leads to the next condition R R;min Ψ ≡Ψ (x) (x) . Note, that the R Ψ (x) depends on the reduction amplitude AR . So, the R R;min Ψ ≡Ψ (x) (x) , if the

Note, that the amplitude AR comprises the factor C 0 <sup>∗</sup> ≠ , but it is constant and this is the factor 0 R III + = <sup>Σ</sup> which is changed. In practice, instead of the condition (87), the following

R 0 <sup>0</sup> <sup>s</sup> s s s min I I I I f (x ) I =+=+ = <sup>Σ</sup>

For future considerations the sign of RI is very important. The vibrations are reduced with actuators, if the I is positive, but must be fulfilled the following condition: R 0 II I0 =+ ≥ <sup>Σ</sup> ; RI 0 = assures the total reduction. If this condition is not fulfilled, the actuators excite vibrations and thereby they are not accomplished owns role. Note, that the sign of 0 I is always negative, see Eq. (65). Then, the sign of I must be positive and it depends on the

κ

positive, if the bending of the beam is directed upwards. Then, for many actuators one has 2 2 2 s 1 11 1 22 2 s s <sup>s</sup> <sup>s</sup> I f ( ) (x ) f ( ) (x ) ... f ( 1) (x ) 0

To obtain the positive sign of I , the signs of s f should alternates; this problem clearly

22 s 1 2 s 1 11 1 22 2 s s <sup>s</sup> <sup>s</sup> I ( )f ( ) (x ) ( )f ( ) (x ) ... ( 1) f ( 1) (x ) 0

negative, and they take their extremes. To fulfill this requirement, the actuators should be

they are bended at {x . Hence, the distribution has a grea <sup>s</sup>} t significance; this problem was solved in (Brański & Szela, 2007; Brański & Szela, 2008; Brański et al., 2010; Brański & Szela 2010; Szela, 2009). Interpreting Eq. (88) through Eq. (90) it is appear that the actuators ought

=+ + +− − + = − − > (90)

 κ

specially distributed on the beam. An idea of description of the sign of s

κ

2

κ

= + + − += − > (89)

Ψ −Ψ Φ − Φ = = Ψ Φ (86)

A A R R ;min = (87)

(x) is changed and as established above; it is

 κ+

+ +

(x ) and sf are the same, namely positive or

(x ) is determined by means of the distribution of the actuators;

κ

κ

(x ) is advance

(88)

2011) and it seems that it is possible only in for separate mode.

amplitude AR is minimal and instead of the above condition, it leads to

κ(x ) .

κκ

**3.3 Heuristic analysis of the optimization problem** 

condition of the reduction must be fulfilled

signs both forces sf and curvatures s

expressed in the following way

determined. The value of s

From physical point of view, the sign of

κ

κ

First at all, it is possible if the signs of s

and

$$\mathbf{A}\_{\mathbb{R}} = \mathbf{C}^\* \left( \mathbf{I}\_0 + \mathbf{I}\_{\mathbb{Z}} \right) \tag{79}$$

The difference ΔΨ(x) is interpreted as the quantity of the vibration reduction and it is the first measure of this reduction namely, the quantity reduction coefficient. The second measure of the vibration reduction is defined as

$$\mathbb{R}\_{\boldsymbol{\Psi}}(\mathbf{x}) = \frac{\Delta \boldsymbol{\Psi}(\mathbf{x})}{\boldsymbol{\Psi}(\mathbf{x})} = \frac{\boldsymbol{\Psi}(\mathbf{x}) - \boldsymbol{\Psi}\_{\boldsymbol{\mathcal{R}}}(\mathbf{x})}{\boldsymbol{\Psi}(\mathbf{x})} \tag{80}$$

It is called as the reduction coefficient and it may be expressed in per cent. Note, that if the reduction coefficient equals one, the vibration reduction is total, <sup>R</sup> Ψ = (x) 0 .

An effectiveness of the vibration reduction is defined as a quotient of some vibration reduction measure by an amount of the energy W provided to the system in order to excite actuators. Hence, thirst measure of the vibration reduction may be defined by so called the effectiveness coefficient

$$\mathcal{E}\_{\Psi}(\mathbf{x}) = \mathcal{R}\_{\Psi}(\mathbf{x})/\mathcal{W} \tag{81}$$

The energy W provided to the system is translated into couples of forces, Fig. 9. Therefore, the energy W may be replaced by forces R s <sup>s</sup> f 4f <sup>=</sup> , hence

$$\mathbf{E}\_{\Psi}(\mathbf{x}) = \mathbf{R}\_{\Psi}(\mathbf{x})/\mathbf{f}\_{\mathbb{R}} \tag{82}$$

The Eqs. (77) – (82) define the appropriate factors of the vibration reduction at the point x . In many cases, it is convenient to calculate mean values of these coefficients at whole beam domain or at the beam sub-domains. First of them is the mean quantity reduction coefficient and it is defined by the formula

$$
\Delta \Psi\_{\rm m} = \frac{1}{\mathbf{n}\_{\rm i}} \sum\_{\rm i} \left( \Psi(\mathbf{x}\_{\rm i}) - \Psi\_{\rm R}(\mathbf{x}\_{\rm i}) \right) \qquad \mathbf{i} = \mathbf{1}, \mathbf{2}, \dots, \mathbf{n}\_{\rm i} \tag{83}
$$

Consequently, the mean reduction coefficient and the mean effectiveness coefficient are defined respectively

$$\mathbf{R}\_{\Psi\mathbf{m}} = \Delta\Psi\_{\mathbf{m}} \Big/ \Psi\_{\mathbf{m}} \Big/ \tag{84} = \sum\_{i} \Psi(\mathbf{x}\_{i}) \Big/ \mathbf{n}\_{i} \tag{84}$$

$$\mathbf{E}\_{\Psi m} = \mathbf{R}\_{\Psi m} / \mathbf{f}\_{\mathbb{R}} \tag{85}$$

The coefficients defined above may constitute the base to formulate the optimization problem; hereafter the R (x) <sup>Ψ</sup> is chosen.

### **3.2 Formulation of the optimization problem**

In this chapter, one formulates the following problem: find the optimal actuators distribution { x which maximize of the reduction coefficient s} R (x) <sup>Ψ</sup> ; hence R (x) <sup>Ψ</sup> is assumed as an objective function. In this case the maximal value of R (x) <sup>Ψ</sup> equals one and it

The difference ΔΨ(x) is interpreted as the quantity of the vibration reduction and it is the

<sup>R</sup> (x) (x) (x) R (x) (x) (x)

It is called as the reduction coefficient and it may be expressed in per cent. Note, that if the

An effectiveness of the vibration reduction is defined as a quotient of some vibration reduction measure by an amount of the energy W provided to the system in order to excite actuators. Hence, thirst measure of the vibration reduction may be defined by so called the

The energy W provided to the system is translated into couples of forces, Fig. 9. Therefore,

The Eqs. (77) – (82) define the appropriate factors of the vibration reduction at the point x . In many cases, it is convenient to calculate mean values of these coefficients at whole beam domain or at the beam sub-domains. First of them is the mean quantity reduction coefficient

Consequently, the mean reduction coefficient and the mean effectiveness coefficient are

The coefficients defined above may constitute the base to formulate the optimization

In this chapter, one formulates the following problem: find the optimal actuators distribution { x which maximize of the reduction coefficient s} R (x) <sup>Ψ</sup> ; hence R (x) <sup>Ψ</sup> is assumed as an objective function. In this case the maximal value of R (x) <sup>Ψ</sup> equals one and it

m i ( ) <sup>R</sup> <sup>i</sup> <sup>i</sup>

<sup>1</sup> (x ) (x )

first measure of this reduction namely, the quantity reduction coefficient.

reduction coefficient equals one, the vibration reduction is total, <sup>R</sup> Ψ = (x) 0 .

The second measure of the vibration reduction is defined as

the energy W may be replaced by forces R s <sup>s</sup> f 4f <sup>=</sup> , hence

i

n

ψ

A CI I R 0 ( ) <sup>∗</sup> = + <sup>Σ</sup> (79)

ΔΨ Ψ −Ψ = = Ψ Ψ (80)

E (x) R (x) W Ψ Ψ = (81)

E (x) R (x) f Ψ Ψ = R (82)

ΔΨ = Ψ − Ψ i i 1,2,...,n <sup>=</sup> (83)

<sup>R</sup><sup>Ψ</sup>m mm = ΔΨ Ψ , m ii <sup>i</sup> Ψ= Ψ (x ) n (84)

E Rf Ψ Ψ m mR = (85)

and

effectiveness coefficient

and it is defined by the formula

problem; hereafter the R (x) <sup>Ψ</sup> is chosen.

**3.2 Formulation of the optimization problem** 

defined respectively

means p-reduction; such instance is considered in (Brański et al., 2010; Brański & Lipiński, 2011) and it seems that it is possible only in for separate mode.

Let the energy provided to the actuators be constant and hence, the sf is always constant. Now, for clarity of the disquisition, rewrite the effectiveness coefficient in explicit form

$$\mathcal{R}\_{\mathbf{v}}(\mathbf{x}) = \frac{\Psi(\mathbf{x}) - \Psi\_{\mathbf{x}}(\mathbf{x})}{\Psi(\mathbf{x})} = \frac{\mathbf{C} \mathbf{A} \,\Phi(\mathbf{x}) - \mathbf{C} \mathbf{A}\_{\mathbf{x}} \,\Phi(\mathbf{x})}{\mathbf{C} \mathbf{A} \,\Phi(\mathbf{x})} \tag{86}$$

Working out on above assumption, the R (x) <sup>Ψ</sup> will be maximal, if R Ψ (x) is minimal. Hence, the optimal condition R (x) R (x) Ψ Ψ = ;max leads to the next condition R R;min Ψ ≡Ψ (x) (x) . Note, that the R Ψ (x) depends on the reduction amplitude AR . So, the R R;min Ψ ≡Ψ (x) (x) , if the amplitude AR is minimal and instead of the above condition, it leads to

$$\mathbf{A}\_{\mathbb{R}} = \mathbf{A}\_{\mathbb{R}; \text{min}} \tag{87}$$

### **3.3 Heuristic analysis of the optimization problem**

Note, that the amplitude AR comprises the factor C 0 <sup>∗</sup> ≠ , but it is constant and this is the factor 0 R III + = <sup>Σ</sup> which is changed. In practice, instead of the condition (87), the following condition of the reduction must be fulfilled

$$\mathbf{I}\_{\mathbb{R}} = \mathbf{I}\_0 + \mathbf{I}\_{\mathbb{z}} = \mathbf{I}\_0 + \sum\_{\mathbf{s}} \mathbf{f}\_{\mathbf{s}} \ell\_{\mathbf{s}}^2 \mathbf{x}(\mathbf{x}\_{\mathbf{s}}) = \mathbf{I}\_{\text{min}} \tag{88}$$

For future considerations the sign of RI is very important. The vibrations are reduced with actuators, if the I is positive, but must be fulfilled the following condition: R 0 II I0 =+ ≥ <sup>Σ</sup> ; RI 0 = assures the total reduction. If this condition is not fulfilled, the actuators excite vibrations and thereby they are not accomplished owns role. Note, that the sign of 0 I is always negative, see Eq. (65). Then, the sign of I must be positive and it depends on the signs both forces sf and curvatures s κ(x ) .

From physical point of view, the sign of κ(x) is changed and as established above; it is positive, if the bending of the beam is directed upwards. Then, for many actuators one has

$$\mathbf{I}\_{\Sigma} = \mathbf{f}\_1 \,\ell\_1^2(+) \,\kappa(\mathbf{x}\_1) + \mathbf{f}\_2 \,\ell\_2^2(-) \,\kappa(\mathbf{x}\_2) + \dots = \sum\_{\ast} \mathbf{f}\_{\ast} \,\ell\_{\ast}^2(-1)^{\ast \ast 1} \,\kappa(\mathbf{x}\_{\ast}) > 0 \tag{89}$$

To obtain the positive sign of I , the signs of s f should alternates; this problem clearly expressed in the following way

$$\mathbf{I}\_{\Sigma} = \text{(+)}\mathbf{f}\_{1}\ell\_{1}^{2}(+)\,\kappa(\mathbf{x}\_{1}) + (-)\mathbf{f}\_{2}\ell\_{2}^{2}(-)\,\kappa(\mathbf{x}\_{2}) + \dots \\ = \sum\_{s} (-1)^{s+1} \mathbf{f}\_{s}\ell\_{s}^{2}(-1)^{s+1} \,\kappa(\mathbf{x}\_{s}) > 0 \tag{90}$$

First at all, it is possible if the signs of s κ(x ) and sf are the same, namely positive or negative, and they take their extremes. To fulfill this requirement, the actuators should be specially distributed on the beam. An idea of description of the sign of s κ(x ) is advance determined. The value of s κ(x ) is determined by means of the distribution of the actuators; they are bended at {x . Hence, the distribution has a grea <sup>s</sup>} t significance; this problem was solved in (Brański & Szela, 2007; Brański & Szela, 2008; Brański et al., 2010; Brański & Szela 2010; Szela, 2009). Interpreting Eq. (88) through Eq. (90) it is appear that the actuators ought

An Optimal Distribution of Actuators in

where DI (x ) DI (x x ) Rs R s = = and hence

and {x ,x max min} ones, at which

**4. Conclusion** 

Deriving the shape of

parameters were considered.

The problem of the signs of the s

κ

transmitted by the actuators to the structure".

following conclusion may be formulated.

methods give the same results.

on the shape of X(x) , and consistently on the shape of

Because of <sup>2</sup>

Active Beam Vibration – Some Aspects, Theoretical Considerations 415

11 1 12 2 f D (x ) f D (x ) ... 0

D (x ) 0 κ

From the condition (94), a set of stationary points {x is obtained. The sufficient condition s} for existing extreme demands, in order to the function be determined on either side of the point xs and D() κ *x* must change sign at this point (turning point); it is sufficient condition

> <sup>2</sup> D (x ) 0 κ

points {x are calculated and they are compared to s} the values calculated at the end points of the appropriate interval. In the future consideration, the ns points among stationary {x s}

Analytical analysis was applied for p-reduction and for the separate beam modes (Brański & Lipiński, 2011). As pointed out there, the analytical solution to the optimal actuators

As can be note, the actuators optimal distribution is attained assuming that the added energy to excite actuators is constant. It is translated into constant sf . Having the optimal distribution, the reduction coefficient may be improved by adding more energy or in order words, by increasing sf . This way, presented optimal method corresponds to that one presented in (Q. Wang & C. Wang, 2001), namely "maximization of the control forces

Based on theoretical considerations, and numerical ones presented in own papers, the

1. The optimization problem of the actuators distribution assuring the maximal active vibration reduction of the beam, measured with reduction coefficient, may be solved both heuristically and analytically. In analyzed problem, it turned out that both

 κ

2 2

κ

formulated in the first form. This condition is expressed in the other form

If this condition is not fulfilling, then this point should be omitted. One still needs to consider the biggest and the lowest values of

κ

κ

distribution problem confirms the results obtained with heuristic solution.

adaptation method must be applied. But after determining shape of

2. The following algorithm of analytical method may be worked out:

hypothetical points {x ,x max min} . In order to find them, the values of

s s f 0 ≠ then instead of Eq. (93) one has

DI (x ) 0 R s = (92)

+ += (93)

<sup>s</sup> = (94)

<sup>s</sup> ≠ (95)

(x) takes in turn its absolute values, are taken into account.

(x ) and sf is quite the same as in heuristic analysis.

(x) , the influence both masses and stiffness of the actuators and glue

κ

κ

(x) , were omitted; if not, an

κ

(x) , all these

κ

(x) ; they are in

(x) at the stationary

to be bonded on the beam sub-domains in which the curvatures reach their extremum and consequently the highest and lowest values respectively, see Fig. 10. This is so called quasioptimal actuators distribution and it is described with Q s x x ≡ points, their number is n n Q s ≡ .

Fig. 10. Optimal distribution of the actuators

As far as signs and values of { fs} are concerned, it was assumed that the added energy exciting actuators is constant. So, the values { fs} of the separate actuators are known and always are constant, while the sign sf springs from the physical interpretation of the interaction beam-actuator. As can be seen in Fig. 10, the forces s 2 f are placed at the point of local extreme, namely at the xs , with the opposite direction to the bending of the beam X(x) . At the same time, the forces sf on the actuator edges are in the direction of the beam bending and let assume that this sign of sf is positive. Another way, the vectors sf and the beam bending X(x) are in the same direction. In such sign convection, both sf and s κ(x ) in Eq. (90) have the same signs and all terms are positive. Furthermore, the actuators distribution described with Q s x x ≡ ensures the maximum of the reduction coefficient.

The heuristic analysis described above was substantiated numerically for the separate beam and triangular modes and the details may be found in own papers.

### **3.4 Analytical analysis of the optimization problem**

The aim of this section is to work out of the analytical method, which will describe such distribution of the actuators in order to assure the maximum of the reduction coefficient. It is expected that the analytical method will confirm the quasi-optimal distribution which has been found above with heuristic method. Therefore the assumptions are the same like in heuristic method, namely ns , sf and s are settled.

Let the distribution of actuators be marked with the set of unknown coordinates s {x } for the moment; that are exactly these coordinates s Q {x } {x } ≡ of which are looked for. One starts from Eq. (88), hence

$$\mathbf{I}\_{\mathbf{k}} = \mathbf{I}\_0 + \mathbf{I}\_{\mathbf{z}} = \mathbf{I}\_0 + \mathbf{f}\_1 \ell\_1^2 \,\mathbf{\kappa}(\mathbf{x}\_1) + \mathbf{f}\_1 \ell\_2^2 \,\mathbf{\kappa}(\mathbf{x}\_2) + \dots \tag{91}$$

Since the κ(x) is the function which changes the sign, it is appropriate to search the points xs which assure the extreme RI (x) , not minimum only. The function RI (x) can have the extreme only at points xs , at which DI (x) R is equal to zero or 2 2 11 1 12 2 f D (x ) f D (x ) ... 0 κ κ + += does not exist (Fichtenholtz, 1999). Because 0 I is constant than a necessary condition for existing extreme value is

$$\text{DI}\_{\mathbb{R}}(\mathbf{x}\_{\circ}) = \mathbf{0} \tag{92}$$

where DI (x ) DI (x x ) Rs R s = = and hence

414 Acoustic Waves – From Microdevices to Helioseismology

to be bonded on the beam sub-domains in which the curvatures reach their extremum and consequently the highest and lowest values respectively, see Fig. 10. This is so called quasioptimal actuators distribution and it is described with Q s x x ≡ points, their number is

fs

f s

xs

As far as signs and values of { fs} are concerned, it was assumed that the added energy exciting actuators is constant. So, the values { fs} of the separate actuators are known and always are constant, while the sign sf springs from the physical interpretation of the interaction beam-actuator. As can be seen in Fig. 10, the forces s 2 f are placed at the point of local extreme, namely at the xs , with the opposite direction to the bending of the beam X(x) . At the same time, the forces sf on the actuator edges are in the direction of the beam bending and let assume that this sign of sf is positive. Another way, the vectors sf and the beam bending X(x) are in the same direction. In such sign convection, both sf and s

Eq. (90) have the same signs and all terms are positive. Furthermore, the actuators distribution described with Q s x x ≡ ensures the maximum of the reduction coefficient. The heuristic analysis described above was substantiated numerically for the separate beam

The aim of this section is to work out of the analytical method, which will describe such distribution of the actuators in order to assure the maximum of the reduction coefficient. It is expected that the analytical method will confirm the quasi-optimal distribution which has been found above with heuristic method. Therefore the assumptions are the same like in

Let the distribution of actuators be marked with the set of unknown coordinates s {x } for the moment; that are exactly these coordinates s Q {x } {x } ≡ of which are looked for. One starts

> R 0 0 11 1 12 2 I I I I f (x ) f (x ) ... =+=+ + + <sup>Σ</sup> κ

xs which assure the extreme RI (x) , not minimum only. The function RI (x) can have the extreme only at points xs , at which DI (x) R is equal to zero or 2 2

2 2

(x) is the function which changes the sign, it is appropriate to search the points

+ += does not exist (Fichtenholtz, 1999). Because 0 I is constant

 κ

f s

and triangular modes and the details may be found in own papers.

**3.4 Analytical analysis of the optimization problem** 

heuristic method, namely ns , sf and s are settled.

than a necessary condition for existing extreme value is

from Eq. (88), hence

κ

11 1 12 2 f D (x ) f D (x ) ... 0

 κ

Since the

κ

Fig. 10. Optimal distribution of the actuators

f s

x

κ(x ) in

(91)

n n Q s ≡ .

$$\text{f.}\,\ell\_1^2 \,\text{D}\,\mathsf{x}(\mathbf{x}\_1) + \text{f.}\,\ell\_2^2 \,\mathsf{D}\,\mathsf{x}(\mathbf{x}\_2) + \dots = \mathbf{0} \tag{93}$$

Because of <sup>2</sup> s s f 0 ≠ then instead of Eq. (93) one has

$$\mathbf{D}\,\kappa(\mathbf{x}\_s) = \mathbf{0} \tag{94}$$

From the condition (94), a set of stationary points {x is obtained. The sufficient condition s} for existing extreme demands, in order to the function be determined on either side of the point xs and D() κ *x* must change sign at this point (turning point); it is sufficient condition formulated in the first form. This condition is expressed in the other form

$$\mathbf{D}^2 \mathbf{x}(\mathbf{x}\_s) \neq 0 \tag{95}$$

If this condition is not fulfilling, then this point should be omitted.

One still needs to consider the biggest and the lowest values of κ(x) ; they are in hypothetical points {x ,x max min} . In order to find them, the values of κ(x) at the stationary points {x are calculated and they are compared to s} the values calculated at the end points of the appropriate interval. In the future consideration, the ns points among stationary {x s} and {x ,x max min} ones, at which κ(x) takes in turn its absolute values, are taken into account. The problem of the signs of the s κ(x ) and sf is quite the same as in heuristic analysis.

Analytical analysis was applied for p-reduction and for the separate beam modes (Brański & Lipiński, 2011). As pointed out there, the analytical solution to the optimal actuators distribution problem confirms the results obtained with heuristic solution.
