Structural Scheme of an Engine for Nanomedicine and Nanotechnology

Structural Model of an Engine

The electro magneto elasticity expression [1, 2, 3, 4, 5, 6, 7, 8] is used

$$S_i = s_{ij}^{E,H} T_j + d_{mi}^{H} E_m + d_{mi}^{E} H_m$$

where $S_i$ is the relative deformation on axis $i$, $E_m$ is the electric field strength on axis $m$, $H_m$ is the magnetic field strength on axis $m$, $s_{ij}^{E,H}$ is the elastic compliance for $E = \text{const}$, $H = \text{const}$, $T_j$ is the mechanical stress on the axis $j$, $d_{mi}^{H}$ is the piezo module, $d_{mi}^{E}$ is the magnetostriction coefficient.

The expression of the reverse piezo effect [1, 2, 3, 4, 5, 6, 7, 8] has the form

$$S_i = d_{mi} E_m + s_{ij}^{E} T_j$$

where $S_i$, $d_{mi}$, $E_m$, $s_{ij}^{E}$, $T_j$ are the relative displacement, piezo module, strength electric field, elastic compliance, strength mechanical field.

The expression of the longitudinal inverse piezo effect [1, 2, 3, 4, 5, 6, 7, 8] has the form

$$S_3 = d_{33} E_3 + s_{33}^{E} T_3$$

The differential equation of an engine is used [4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56]

$$\frac{d^2 \Xi(x,s)}{dx^2} - \gamma^2 \Xi(x,s) = 0$$

where $\Xi(x,s)$, $s$, $x$, $\gamma$ are the transformation Laplace for displacement, the parameter, the coordinate, the coefficient of propagation. For the longitudinal piezo engine we have at $x = 0$ the deformation $\Xi(0,s) = \Xi_1(s)$ and at $x = \delta \Xi(\delta,s) = \Xi_2(s)$. The decision is calculated.

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$$\Xi(x, s) = \left\{ \Xi_1(s) \text{ sh} \left[ (\delta - x) \gamma \right] + \Xi_2(s) \text{ sh} \left(x \gamma \right) \right\} / \text{sh} \left(\delta \gamma \right)$$

The set of equations for boundary conditions is determined [4, 5, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32]

$$T_3(0, s) = \frac{1}{s_{33}^E} \frac{d\Xi(x, s)}{dx} - \frac{d_{33}}{s_{33}^E} E_3(s)$$

$$T_3(\delta, s) = \frac{1}{s_{33}^E} \frac{d\Xi(x, s)}{dx} - \frac{d_{33}}{s_{33}^E} E_3(s)$$

Its structural model is written in the form

$$\Xi_1(s) = \left( M_1 s^2 \right)^{-1} \left\{ -F_1(s) + \left( \chi_{33}^E \right)^{-1} \left[ d_{33} E_3(s) - \left[ \gamma / \text{sh} \left(\delta \gamma \right) \right] \right] \times \left[ \text{ch} \left(\delta \gamma \right) \Xi_1(s) - \Xi_2(s) \right] \right\}$$

$$\Xi_2(s) = \left( M_2 s^2 \right)^{-1} \left\{ -F_2(s) + \left( \chi_{33}^E \right)^{-1} \left[ d_{33} E_3(s) - \left[ \gamma / \text{sh} \left(\delta \gamma \right) \right] \right] \times \left[ \text{ch} \left(\delta \gamma \right) \Xi_2(s) - \Xi_1(s) \right] \right\}$$

$$\chi_{33}^E = s_{33}^E / S_0$$

where $\Xi_1(s), \Xi_2(s)$ are the transforms of the displacements, $S_0$ is the area.

The expression of the longitudinal magnetostriction [1, 2, 3, 4, 5, 6, 7, 8] has the form

$$S_3 = d_{33} H_3 + s_{33}^H T_3$$

The structural model of the longitudinal magnetostriction engine is transformed in the form

$$\Xi_1(s) = \left( M_1 s^2 \right)^{-1} \left\{ -F_1(s) + \left( \chi_{33}^H \right)^{-1} \left[ d_{33} H_3(s) - \left[ \gamma / \text{sh} \left(\delta \gamma \right) \right] \right] \times \left[ \text{ch} \left(\delta \gamma \right) \Xi_1(s) - \Xi_2(s) \right] \right\}$$

$$\Xi_2(s) = \left( M_2 s^2 \right)^{-1} \left\{ -F_2(s) + \left( \chi_{33}^H \right)^{-1} \left[ d_{33} H_3(s) - \left[ \gamma / \text{sh} \left(\delta \gamma \right) \right] \right] \times \left[ \text{ch} \left(\delta \gamma \right) \Xi_2(s) - \Xi_1(s) \right] \right\}$$

$$\chi_{33}^H = s_{33}^H / S_0$$

The expression of the transverse inverse piezo effect [1, 2, 3, 4, 5, 6, 7, 8] has the form

$$S_1 = d_{31} E_3 + s_{11}^E T_1$$

The decision of the differential equation is written

$$\Xi(x, s) = \left\{ \Xi_1(s) \text{ sh} \left[ (h - x) \gamma \right] + \Xi_2(s) \text{ sh} \left(x \gamma \right) \right\} / \text{sh} \left(h \gamma \right)$$

The set of equations is obtained

$$T_1(0, s) = \frac{1}{s_{11}^E} \frac{d\Xi(x, s)}{dx} - \frac{d_{31}}{s_{11}^E} E_3(s)$$

$$T_1(h, s) = \frac{1}{s_{11}^E} \frac{d\Xi(x, s)}{dx} - \frac{d_{31}}{s_{11}^E} E_3(s)$$

The structural model of the transverse piezo engine is written

$$\Xi_1(s) = \left( M_1 s^2 \right)^{-1} \left\{ -F_1(s) + \left( \chi_{11}^E \right)^{-1} \left[ d_{31} E_3(s) - \left[ \gamma / \text{sh} \left(h \gamma \right) \right] \right] \times \left[ \text{ch} \left(h \gamma \right) \Xi_1(s) - \Xi_2(s) \right] \right\}$$

$$\Xi_2(s) = \left( M_2 s^2 \right)^{-1} \left\{ -F_2(s) + \left( \chi_{11}^E \right)^{-1} \left[ d_{31} E_3(s) - \left[ \gamma / \text{sh} \left(h \gamma \right) \right] \right] \times \left[ \text{ch} \left(h \gamma \right) \Xi_2(s) - \Xi_1(s) \right] \right\}$$

$$\chi_{11}^E = s_{11}^E / S_0$$

The expression of the transverse magnetostriction [1, 2, 3, 4, 5, 6, 7, 8] has the form

$$S_1 = d_{31} H_3 + s_{11}^H T_1$$

The structural model is determined in the form

$$\Xi_1(s) = \left( M_1 s^2 \right)^{-1} \left\{ -F_1(s) + \left( \chi_{11}^H \right)^{-1} \left[ d_{31} H_3(s) - \left[ \gamma / \text{sh} \left(h \gamma \right) \right] \right] \times \left[ \text{ch} \left(h \gamma \right) \Xi_1(s) - \Xi_2(s) \right] \right\}$$

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$$\Xi_2(s) = \left( M_2 s^2 \right)^{-1} \begin{cases} -F_2(s) + \left( \chi_{11}^H \right)^{-1} \\ \times \left[ d_{31} H_3(s) - \left[ \gamma/\text{sh}(h\gamma) \right] \\ \times \left[ \text{ch}(h\gamma) \Xi_2(s) - \Xi_1(s) \right] \end{cases}$$

$$\chi_{11}^H = s_{11}^H / S_0$$

The expression of the shift inverse piezo effect [1, 2, 3, 4, 5, 6, 7, 8] has the form

$$S_5 = d_{15} E_1 + s_{55}^E T_5$$

The decision of the differential equation is calculated

$$\Xi(x,s) = \left\{ \Xi_1(s) \text{sh} \left[ (b-x)\gamma \right] + \Xi_2(s) \text{sh}(x\gamma) \right\} / \text{sh}(b\gamma)$$

The set of equations for the shift piezo engine is written

$$T_3(0,s) = \frac{1}{s_{55}^E} \frac{d\Xi(x,s)}{dx} \Bigg|_{x=0} - \frac{d_{15} E_1(s)}{s_{55}^E} E_1(s)$$

$$T_5(b,s) = \frac{1}{s_{55}^E} \frac{d\Xi(x,s)}{dx} \Bigg|_{x=b} - \frac{d_{15} E_1(s)}{s_{55}^E} E_1(s)$$

Its structural model has the form

$$\Xi_1(s) = \left( M_1 s^2 \right)^{-1} \begin{cases} -F_1(s) + \left( \chi_{55}^E \right)^{-1} \\ \times \left[ d_{15} E_1(s) - \left[ \gamma/\text{sh}(b\gamma) \right] \\ \times \left[ \text{ch}(b\gamma) \Xi_1(s) - \Xi_2(s) \right] \end{cases}$$

$$\Xi_2(s) = \left( M_2 s^2 \right)^{-1} \begin{cases} -F_2(s) + \left( \chi_{55}^E \right)^{-1} \\ \times \left[ d_{15} E_1(s) - \left[ \gamma/\text{sh}(b\gamma) \right] \\ \times \left[ \text{ch}(b\gamma) \Xi_2(s) - \Xi_1(s) \right] \end{cases}$$

$$\chi_{55}^E = s_{55}^E / S_0$$

The expression of the shift magnetostriction [1, 2, 3, 4, 5, 6, 7, 8] has the form

$$S_5 = d_{15} H_1 + s_{55}^H T_5$$

The structural model of the shift magnetostrictive engine is transformed

$$v_{mi} = \begin{cases} d_{33}, d_{31}, d_{15} \\ g_{33}, g_{31}, g_{15} \\ d_{33}, d_{31}, d_{15} \end{cases}$$

At $l = \{\delta, h, b\}$ the decision of the differential equation of an engine in general has the form

$$\Xi(x,s) = \left\{ \Xi_1(s) \text{sh} \left[ (l-x)\gamma \right] + \Xi_2(s) \text{sh}(x\gamma) \right\} / \text{sh}(l\gamma)$$

The set of equations is determined

$$T_j(0,s) = \frac{1}{s_{ij}^E} \frac{d\Xi(x,s)}{dx} \Bigg|_{x=0} - \frac{v_{mi}}{s_{ij}^E} \Psi_m(s)$$

$$T_j(l,s) = \frac{1}{s_{ij}^E} \frac{d\Xi(x,s)}{dx} \Bigg|_{x=l} - \frac{v_{mi}}{s_{ij}^E} \Psi_m(s)$$

The structural model in general of an engine on Figure 1 is calculated

$$\Xi_1(s) = \left( M_1 s^2 \right)^{-1} \begin{cases} -F_1(s) + \left( \chi_{ij}^E \right)^{-1} \\ \times \left[ v_{mi} \Psi_m(s) - \left[ \gamma/\text{sh}(l\gamma) \right] \\ \times \left[ \text{ch}(l\gamma) \Xi_1(s) - \Xi_2(s) \right] \end{cases}$$

$$\Xi_2(s) = \left( M_2 s^2 \right)^{-1} \begin{cases} -F_2(s) + \left( \chi_{ij}^E \right)^{-1} \\ \times \left[ v_{mi} \Psi_m(s) - \left[ \gamma/\text{sh}(l\gamma) \right] \\ \times \left[ \text{ch}(l\gamma) \Xi_2(s) - \Xi_1(s) \right] \end{cases}$$

$$\chi_{ij}^E = s_{ij}^E / S_0$$

here

$$v_{mi} = \begin{cases} d_{33}, d_{31}, d_{15} \\ g_{33}, g_{31}, g_{15} \\ d_{33}, d_{31}, d_{15} \end{cases}$$

, , , E E D D H H

$$ \Psi_ {m} = \left\{ \begin{array}{l} I \\ I \\ I \end{array} \right. $$

3 1

3 1 m

3 1 $$ \begin{array}{l} s _ {i j} ^ {\Psi} = \left\{ \begin{array}{l l} s _ {3 3} ^ {E}, s _ {1 1} ^ {E}, s _ {5 5} ^ {E} \\ s _ {3 3} ^ {D}, s _ {1 1} ^ {D}, s _ {5 5} ^ {D} \\ s _ {3 3} ^ {H}, s _ {1 1} ^ {H}, s _ {5 5} ^ {H} \end{array} \right. \\ \gamma = \left\{\gamma^ {E}, \gamma^ {D}, \gamma^ {H} \right. \\ \end{array} $$ E E E , , , , , , s s s s s s s s s s

33 11 55 D D D ij H H H

33 11 55

33 11 55 $$ c ^ {\Psi} = \left\{c ^ {E}, c ^ {D}, c ^ {H} \right. $$

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The matrix of deformations is calculated ( ) ( ) ( )

m s s W s W s W s F s s W s W s W s F s

$$ \left( \begin{array}{l} \Xi_ {1} (s) \\ \Xi_ {2} (s) \end{array} \right) = \left( \begin{array}{l l l} W _ {1 1} (s) & W _ {1 2} (s) & W _ {1 3} (s) \\ W _ {2 1} (s) & W _ {2 2} (s) & W _ {2 3} (s) \end{array} \right) \left( \begin{array}{l} \Psi_ {m} (s) \\ F _ {1} (s) \\ F _ {2} (s) \end{array} \right) $$ ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

1 11 12 13 1 2 21 22 23 2

$$ W _ {1 1} (s) = \Xi_ {1} (s) / \Psi_ {m} (s) = \nu_ {m i} \left[ M _ {2} \chi_ {i j} ^ {\Psi} s ^ {2} + \gamma \operatorname {t h} \left(l \gamma / 2\right) \right] / A _ {i j} $$ $$ \begin{array}{l} A _ {i j} = M _ {1} M _ {2} \left(\chi_ {i j} ^ {\Psi}\right) ^ {2} s ^ {4} + \left\{\left(M _ {1} + M _ {2}\right) \chi_ {i j} ^ {\Psi} / \left[ c ^ {\Psi} \operatorname {t h} \left(l \gamma\right) \right] \right\} s ^ {3} + \left[ \left(M _ {1} + M _ {2}\right) \chi_ {i j} ^ {\Psi} \alpha / \operatorname {t h} \left(l \gamma\right) + 1 / \left(c ^ {\Psi}\right) ^ {2} \right] s ^ {2} + 2 \alpha s / c ^ {\Psi} + \alpha^ {2} \\ W _ {2 1} (s) = \Xi_ {2} (s) / \Psi_ {m} (s) = \nu_ {m i} \left[ M _ {1} \chi_ {i j} ^ {\Psi} s ^ {2} + \gamma \operatorname {t h} \left(l \gamma / 2\right) \right] / A _ {i j} \\ \end{array} $$ $$ W _ {1 2} (s) = \Xi_ {1} (s) / F _ {1} (s) = - \chi_ {i j} ^ {\Psi} \left[ M _ {2} \chi_ {i j} ^ {\Psi} s ^ {2} + \gamma / \operatorname {t h} (l \gamma) \right] / A _ {i j} $$ ( ) ( ) ( ) ( ) ( ) ( ) ( ) 13 1 2 22 2 1 sh ij ij W s s F s W s s F s l A χ γ γ Ψ   = Ξ = = Ξ =   ( ) ( ) ( ) ( ) 2 23 2 2 1 th ij ij ij W s s F s M s l A χ χ γ γ Ψ Ψ   = Ξ = − +   The static longitudinal deformations are obtained $$ \xi_ {1} = d _ {3 3} U M _ {2} / \left(M _ {1} + M _ {2}\right) $$ $$ \xi_ {2} = d _ {3 3} U M _ {1} / \left(M _ {1} + M _ {2}\right) $$ For 33 d = 4·10-10 m/V, U= 25 V, M1= 1 kg, M2= 4 kg we have $$ \xi_ {1} = 8 \mathrm {n m}, \xi_ {2} = 2 \mathrm {n m} \mathrm {a n d} \xi_ {1} + \xi_ {2} $$

= 10 nm at error 10%.

The expression of the direct piezo effect [8, 9, 10, 11, 12, 13, 14, 15, 16] is used $$ D _ {m} = d _ {m i} T _ {i} + \varepsilon_ {m k} ^ {E} E _ {k} $$ Where m D , E mk ε are the electric induction and the permittivity, The direct coefficient at voltage control is written $$ k _ {d} = \frac {d _ {m i} S _ {0}}{\delta s _ {i j} ^ {E}} $$

0 mi d E ij The expression of the transform voltage negative feedback at voltage control on Figure 2 is used d S R U s s k R s s δ $$ ) = \frac {d _ {m i} S _ {0} R}{\delta s _ {n} ^ {E}} \dot {\Xi} _ {n} (s) = k _ {d} R \dot {\Xi} _ {n} (s), n = 1, 2 $$ ( ) ( ) ( ) 0 mi d d n n E ij The maximum force at voltage control is determined $$ F _ {\max } = E _ {m} d _ {m i} S _ {0} / s _ {i j} ^ {E} $$ $$ T _ {j \max } = E _ {m} d _ {m i} / s _ {i j} ^ {E} $$

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The maximum force at current control has the form $$ F _ {\max } = \frac {U}{\delta} d _ {m i} \frac {S _ {0}}{s _ {i j} ^ {E}} + \frac {F _ {\max }}{S _ {0}} d _ {m i} S _ {c} \frac {1}{\varepsilon_ {m k} ^ {T} S _ {c} / \delta} \frac {1}{\delta} d _ {m i} \frac {S _ {0}}{s _ {i j} ^ {E}} $$

1 1

0 max 0 max 0 mi mi c mi E T E ij mk c ij

$$ + \left(1 - \frac {d _ {m i} ^ {2}}{\varepsilon_ {m k} ^ {T} s _ {i j} ^ {E}}\right) s _ {i j} ^ {E} = 1 $$

2 max F d s E d S s ε

0 1 E mi ij m mi T E mk ij

$$ \left| T _ {j \max } \left(1 - k _ {m i} ^ {2}\right) s _ {i j} ^ {E} = E _ {m} d _ {m i} \right| $$

here mi k is the coefficient of electromechanical coupling.

The expressions at current control are used $$ F _ {\max } = E _ {m} d _ {m i} S _ {0} / s _ {i j} ^ {D} $$ $$ T _ {j \max } = E _ {m} d _ {m i} / s _ {i j} ^ {D} $$ $$ s _ {i j} ^ {D} = \left(1 - k _ {m i} ^ {2}\right) s _ {i j} ^ {E} $$ The mechanical characteristic [11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26] is determined $$ \left. S _ {i} \left(T _ {j}\right) \right| _ {\Psi = c o n s t} = \nu_ {m i} \Psi_ {m} \bigg | _ {\Psi = c o n s t} + s _ {i j} ^ {\Psi} T _ {j} $$ The adjustment characteristic [11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26] is obtained ( ) i m mi m ij j T const T const S s T ν Ψ $$ \left(\Psi_ {m}\right) \bigg | _ {T = c o n s t} = \nu_ {m i} \Psi_ {m} + s _ {i j} ^ {\Psi} T _ {j} \bigg | _ {T = c c} $$ Therefore, the mechanical characteristics of the engine has the form $$ \Delta l = \Delta l _ {\max } \left(1 - F / F _ {\max }\right) $$ $$ \Delta l _ {\max } = \nu_ {m i} \Psi_ {m} l $$ $$ F _ {\max } = T _ {j \max } S _ {0} = \nu_ {m i} \Psi_ {m} S _ {0} / s _ {i j} ^ {\Psi} $$ The expression for the transverse piezo engine is calculated $$ \Delta h = \Delta h _ {\max } \left(1 - F / F _ {\max }\right) $$ $$ \Delta h _ {\max } = d _ {3 1} E _ {3} h $$ $$ F _ {\max } = d _ {3 1} E _ {3} S _ {0} / s _ {1 1} ^ {E} $$ At 31 d

= 2∙10-10 m/V,

$$ \Delta h _ {\max } = 1 2 5 \mathrm {n m}, F _ {\max } $$

= 5 N with error 10% The deformation of an engine at elastic load has the form Ψ ∆= Ψ − , e F C l = ∆ ij mi m s l F l S ν

0 The adjustment characteristic is determined l l C C ν $$ \Delta l = \frac {\nu_ {m i} l \Psi_ {m}}{1 + C _ {e} / C _ {i j} ^ {\Psi}} $$ mi m

1 e ij The coefficients in general are calculated $$ k _ {d} = k _ {r} = \frac {d _ {m i} S _ {0}}{\delta s _ {i j}} $$

0 mi d r ij $$ s _ {i j} = k _ {s} s _ {i j} ^ {E} $$

$$ \left| \left(1 - k _ {m i} ^ {2}\right) \leq k _ {s} \leq 1 \right| $$

At one fixed face and elastic-inertial load Figure 2 is transformed to Figure 3.

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Therefore, the function of piezo engine for 0 = R has the form ( ) ( ) ( )

s k W s U s T s T s ξ $$ ) = \frac {\Xi (s)}{U (s)} = \frac {k _ {3 1} ^ {U}}{T _ {t} ^ {2} s ^ {2} + 2 T _ {t} \xi_ {t} s + 1} $$ U

31 2 2 2 1 t t t

$$ k _ {3 1} ^ {U} = d _ {3 1} \left(h / \delta\right) / \left(1 + C _ {l} / C _ {1 1} ^ {E}\right) $$ $$ T _ {t} = \sqrt {M / \left(C _ {l} + C _ {1 1} ^ {E}\right)}, \omega_ {t} = 1 / T _ {t} $$ At M= 1 kg, l C = 0.1·107, 11 E C = 1.5·107, 31 d = 2·10-10 m/V, h δ = 20 the parameters of piezo engine are calculated tT = 0.25·10-3 s, t ω = 4·103 s-1, 31 U k = 3.75 nm/V with error 10%.