A.1 A few reminders from RDM

### A.1.1 Calculation of the deflection of a bridge embedded at its 2 ends

Note: We take the case of pure bending, the shear force T is such that With M the bending moment applied to the piezoelectric bridge. The Casimir force in the z axis is applied in lp/2 at the center of the bridge (Figure 65).

Recall that the neutral axis G (x), is the place where the normal bending stress is zero. G is a point of this neutral axis which in the case of a symmetrical bridge is also on the median line of the center of gravity of the bridge. Let us name:


We will assume that the strains of the bridge subjected to the weak Casimir force are themselves small, so the induced strain angles d (ρ) are also small. Thus as it is stipulated in all the works of R.D.M. we can assimilate the radius of curvature ρ with Eq. (21) (Figure 66)

$$\tan\left(d\alpha\right) = \frac{d\mathbf{x}}{\rho} \approx \alpha \Rightarrow \frac{\mathbf{1}}{\rho} = \frac{d\alpha}{d\mathbf{x}} = -\frac{M\_Z(\mathbf{x})}{E\_P I\_P(\mathbf{x})}$$

Since the bridge is parallelepiped in shape, the bending moment of inertia along the z axis of the section of this bridge is Eq. (22) (Figure 67)

Figure 65.

General appearance a/ b/ of the device studied, forces and applied moments, c/ of the deformed bridge.

Figure 66. Piezoelectric Bridge Nomenclature.

Perspective Chapter: Device, Electronic,Technology for a M.E.M.S. Which Allow… DOI: http://dx.doi.org/10.5772/intechopen.105197

Figure 67.

(a) Forces, shear forces and moments applied on the bridge. (b) Variation of bending moment. (c) Shape and arrow of the bridge embedded at both ends. With: δ0 = inflection points, zmax = arrow of the bridge.

$$I\_{GZ} = \frac{b\_P a\_p^3}{12} = \text{Cte}$$

$$I\ln\text{x} = \frac{l\_P}{2} \Rightarrow \frac{d\mathbf{z}}{d\mathbf{x}\_{\mathbf{x} = \frac{l^p}{2}}} = \mathbf{0} \Rightarrow MAZ(\mathbf{x} = lp/2) = -\frac{F\_{CA}}{8}l\mathbf{P} \Rightarrow MAZ\mathbf{x} = \mathbf{0}\right) = -\frac{FCA}{8}l\mathbf{P}$$

The maximum deflection is in x = lp/2 which gives an arrow: z max ¼ FCA:lp 3 =ð Þ 192 EP:IP in x ¼ lP=2 (Eq. (23)). See RDM nomenclature

#### A.1.2 Calculation of the resonant frequency of the piezoelectric bridge

It is demonstrated (see for example: Vibrations of continuous media Jean-Louis Guyader (Hermes)) that the amplitude z (x, t) of the transverse displacement of a cross section of the beam is given by the partial differential equation: <sup>δ</sup>4<sup>z</sup> dx<sup>4</sup> <sup>þ</sup> <sup>ρ</sup><sup>S</sup> EPIP d2 z dt<sup>2</sup> ¼ 0 if one neglects the internal damping!

The linear mass m(x) being equal to ρ<sup>S</sup> (Kg/m), with ρ the density (Kg/m<sup>3</sup> ) and S the section (m2), hence the differential equation of free vibrations deduced is: (Eq. (28)) <sup>δ</sup>4<sup>z</sup> dx<sup>4</sup> ¼ � <sup>ρ</sup><sup>S</sup> EPIP d2 z dt<sup>2</sup> With k <sup>¼</sup> <sup>ρ</sup>Sω<sup>2</sup> =EpIp � �<sup>1</sup>=<sup>4</sup> , the solution of this differential equation is.

written in the general form: Z xð Þ¼ A1 exp k x ð Þþ A2 exp ð Þþ �k x A3 exp i k x ð Þþ A4 exp i k x ð Þ and in the more convenient form:

$$\mathbf{Z}(\mathbf{x}) = \mathbf{a} \cdot \sin\left(\mathbf{k} \cdot \mathbf{x}\right) + \mathbf{b} \cdot \cos\left(\mathbf{k} \cdot \mathbf{x}\right) + \mathbf{c} \cdot \sin(\mathbf{k} \cdot \mathbf{x}) + \mathbf{d} \cdot \epsilon \mathbf{h}(\mathbf{k} \cdot \mathbf{x})\tag{21}$$

#### A.1.2.1 Eigen modes and frequencies

In this part, we will assume that the moving part of the structure (Figure 28) vibrates at its resonant frequency. The only series of discrete pulsations wi (proper pulsations of vibration) will be authorized, these pulsations being obtained in the general form: with: wpi = resonance pulsation and therefore fpi = frequency resonant

$$
\omega\_{PR} = \left(\frac{a\_I}{lP}\right)^2 \sqrt{\left(\frac{E\_P I P}{\rho \mathcal{S}}\right)} \\
\Rightarrow f\_{PI} = \frac{1}{2\pi} \left(\frac{a\_I}{lP}\right)^2 \sqrt{\left(\frac{E\_P I P}{\rho \mathcal{S}}\right)}
$$

#### A.1.2.2 Boundary conditions

In our situation we have a recessed-recessed bridge. Z xð Þ¼ a � sin kð Þþ � x b � cos kð Þþ � x c � sh kð Þþ � x d � ch kð Þ � x (30).

In this case for x = 0 and for x = lp, we have z(x) = 0 and dz./dx = 0 (zero elongations and slopes). Let:

<sup>F</sup>ο<sup>r</sup> : <sup>x</sup> <sup>¼</sup> <sup>0</sup> : <sup>1</sup> : <sup>a</sup> <sup>þ</sup> <sup>c</sup> <sup>¼</sup> 0a ¼ �c2 : <sup>b</sup> <sup>þ</sup> <sup>d</sup> <sup>¼</sup> 0b <sup>¼</sup> ‐<sup>d</sup> Fοr x ¼ lp : 3 : a sin k lp ð Þþ b cos k lp ð Þþ c sh k lp ð Þþ d ch k lp ð Þ¼ 0; <sup>4</sup> : a cos k lp ð Þ‐b sin k lp ð Þþ c ch k lp ð Þþ d sh k lp ð Þ¼ <sup>0</sup>

We deduce from the preceding Eqs. (1–4) that: c. (sh (k lp) � sin (k lp)) + d (ch (k lp) � cos (k lp)) = 0; and c. (ch (k lp) � cos (k lp)) + d (sh (k lp) + sin (k lp)) = 0.

These last 2 equations lead to a transcendent equation in k lp: so, the determinant of this system is

zero! sh kð Þ :lp <sup>2</sup> ‐sin kð Þ :lp <sup>2</sup> h i–½ � ch kð Þ� :lp cos kð Þ :lp <sup>2</sup> <sup>¼</sup> 0 cos kð Þ¼ :lp <sup>1</sup>=ch kð Þ :lp .

The solutions of this equation can be solved graphically or numerically.

The numerical resolution (for example by the dichotomy method in Figure 68 of this equation gives for the first 5 solutions: a1 = 4.7300; a2 = 7.8532; a3 = 10.9956; a4 = 14.1317; a5 = 17.2787.

So, the first resonant frequency of the piezoelectric bridge is.

$$\rho\_{P1} = \left(\textbf{4.73}\right)^2 \sqrt{\frac{\textbf{E} \rho I\_P}{M\_{\text{S}l\_P^\circ}}} \Rightarrow f\_{P1} = \frac{1}{2\pi} \left(\textbf{4.73}\right)^2 \sqrt{\frac{\textbf{E} \rho I\_P}{M\_{\text{S}l\_P^\circ}}} \text{ (31)}.$$

For example, for a bridge embedded at both ends with the following characteristics,

For geometries:

$$\begin{aligned} \text{lp} &= 50 \,\upmu\text{m}, \text{bp} = 150 \,\upmu\text{m}, \text{ap} = 10 \,\upmu\text{m}; \text{Sp} = 1,510^{-9} \,\text{m}^2, \\ \text{lp} &= \text{bp} \cdot \text{ap} \text{3}/12 = 1.2510^{-20} \,\text{m}^4 \\ \text{li} &= \text{10 } \upmu\text{m}; \text{bi} = \text{150 } \upmu\text{m}; \text{ai} = \text{10 } \upmu\text{m}; \text{ls} = \text{1000 } \upmu\text{m}; \text{bs} = \text{150 } \upmu\text{m}; \text{as} = \text{10 } \upmu\text{m} \end{aligned}$$

For the material: PZT: density

<sup>r</sup> <sup>¼</sup> 7600 kg=m3 � �,Young<sup>0</sup> s modulus Ep <sup>¼</sup> <sup>6</sup><sup>∗</sup> <sup>10</sup><sup>10</sup>ð Þ Pa Kg m s�<sup>2</sup> � �

For the section inertia: Ip <sup>¼</sup> bp<sup>∗</sup> ap<sup>3</sup>=<sup>12</sup> <sup>¼</sup> <sup>1</sup>:25 10�<sup>20</sup> <sup>m</sup><sup>4</sup> ð Þ.

Then the calculated first resonance frequency is for the PZT material: fp1 = 1.1553 \* 10<sup>7</sup> hertz. For this embedded bridge, an ANSYS simulation (Figure 69) gives a resonant frequency of f1 = 1.02. 107 Hz which is close to that calculated in the draft calculation presented in this report and validates the orders of magnitude obtained with the equations for these preliminary calculations. If one carries out the calculation of the resonant frequency of the structure of Figure 5 which comprises a free sole of Casimir SS2 parallel to a fixed surface SS3 and transmitting by a mechanical link finger the force of Casimir, one finds that the resonant frequency have the same form but with Ms. a fixed mass applied in the middle of the bridge. Ms = the total mass of the structure! Ms ¼ r Sp <sup>∗</sup> ap <sup>þ</sup> Si <sup>∗</sup> ai <sup>þ</sup> Ss <sup>∗</sup> as � � <sup>¼</sup> r lp <sup>∗</sup> bp <sup>∗</sup> ap <sup>þ</sup> li <sup>∗</sup> bi <sup>∗</sup> ai <sup>þ</sup> ls <sup>∗</sup> bs ∗ as � �.

Figure 68. Numerical solution of Eq. (18).

Perspective Chapter: Device, Electronic,Technology for a M.E.M.S. Which Allow… DOI: http://dx.doi.org/10.5772/intechopen.105197

Figure 69. ANSYS simulation of the resonant frequency of the piezoelectric.

With r = the medium density of the piezoelectric material, of the connecting finger, of the Casimir electrode sole and Sp,Si,Ss the longitudinal surfaces of this bridge. Indeed, the presence of the Casimir sole connected by the Casimir force transmission finger in the middle of the piezoelectric bridge, modifies the resonant frequencies of this bridge.

The calculated resonance frequency then becomes for the same geometries and materials. fs1 = 2.509 10<sup>6</sup> hertz. With these characteristics, an ANSYS simulation of this structure gives a close resonance frequency: fs1 = 2.62 106 Hertz. This approach greatly simplifies these preliminary calculations because otherwise the curvature of the piezoelectric bridge makes the Casimir force strongly depend on the longitudinal and transverse positions x and z of the facing surfaces!
