1 Beam with connecting circuit
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The constitutive law is
\[\left[\begin{array}{l} \mathbf{D} \\ \mathbf{S} \end{array}\right]=\left[\begin{array}{ll} \mathbf{e}^T & \mathbf{d} \\ \mathbf{d}_t & \mathbf{s}^E \end{array}\right]\left[\begin{array}{l} \mathbf{E} \\ \mathbf{T} \end{array}\right]\]Assume the field in the piezo patch is uniform, we have the relation after Laplace transform
\[\mathbf{V}(s)=\mathbf{L} \cdot \mathbf{E}(s), \quad \mathbf{I}(s)=s \mathbf{A} \cdot \mathbf{D}(s)\]Then the constitutive law can be transformed as
\[\left[\begin{array}{l} \mathbf{I} \\ \mathbf{S} \end{array}\right]=\left[\begin{array}{cc} s \mathbf{A} \varepsilon^T \mathbf{L}^{-1} & s\mathrm{\boldsymbol{Ad}} \\ \mathbf{d}_t \mathbf{L}^{-1} & \mathbf{s}^{\mathrm{E}} \end{array}\right]\left[\begin{array}{l} \mathbf{V} \\ \mathbf{T} \end{array}\right]\]We define the capacity of piezo patch as
\[A_i \varepsilon_i^T / L_i=C_{p i}^T\]Then Eq. (3) can be rewritten as
\[\left[\begin{array}{c} \mathbf{I} \\ \mathbf{S} \end{array}\right]=\left[\begin{array}{cc} s \mathbf{C}_p^T & s \mathbf{A d} \\ \mathbf{d}_t \mathbf{L}^{-1} & \mathbf{s}^E \end{array}\right]\left[\begin{array}{l} \mathbf{V} \\ \mathbf{T} \end{array}\right]=\left[\begin{array}{cc} \mathbf{Y}^D(s) & s \mathbf{A d} \\ \mathbf{d}_t \mathbf{L}^{-1} & \mathbf{s}^E \end{array}\right]\left[\begin{array}{l} \mathbf{V} \\ \mathbf{T} \end{array}\right],\]where $\mathbf{Y}^D(s)$ is the open circuit admittance of the piezoelectric patch.
For shunted piezoelectric applications,
\[\left[\begin{array}{l} \mathbf{I} \\ \mathbf{S} \end{array}\right]=\left[\begin{array}{cc} \mathbf{Y}^{E L} & s \mathbf{A d} \\ \mathbf{d}_{\mathbf{t}} \mathbf{L}^{-1} & \mathbf{S}^E \end{array}\right]\left[\begin{array}{l} \mathbf{V} \\ \mathbf{T} \end{array}\right] \text { where } \quad \mathbf{Y}^{E L}=\mathbf{Y}^D+\mathbf{Y}^{S U}\]The first equation gives
\[\mathbf{V}=\left(\mathbf{Z}^{E L}\right) \mathbf{I}-\left(\mathbf{Z}^{E L} s \mathbf{A d}\right) \mathbf{T}\]and inserting into second equation gives
\[\mathbf{S}=\left[\mathbf{s}^E-\mathbf{d}_t \mathbf{L}^{-1} \mathbf{Z}^{E L} s \mathbf{A d}\right] \mathbf{T}+\left[\mathbf{d}_t \mathbf{L}^{-1} \mathbf{Z}^{E L}\right] \mathbf{I} .\]The mechanical compliance is
\[\mathbf{s}^{S U}=\left[\mathbf{s}^E-\mathbf{d}_t \mathbf{L}^{-1} \mathbf{Z}^{E L} \boldsymbol{s A d}\right] .\]Upon noting that with constant stress
\[\begin{gathered} \mathbf{Z}^E(\mathrm{~s})=\mathbf{0}=\text { short circuit electrical impedance, } \\ \mathbf{Z}^D(s)=\left(\mathbf{C}_p^T s\right)^{-1}=\text { open circuit electrical impedance, } \end{gathered}\]and that
\[s \mathbf{L}^{-1} \boldsymbol{\varepsilon}^T \mathbf{A}=\mathbf{C}_p^T s,\]equation (16) can be put in the form
\[\mathbf{s}^{S U}=\left[\mathbf{s}^E-\mathbf{d}_{\mathbf{t}} \overline{\mathbf{Z}}^{E L}\left(\boldsymbol{\varepsilon}^T\right)^{-1} \mathbf{d}\right],\]where the matrix of non-dimensional electrical impedances is defined as
\[\overline{\mathbf{Z}}^{E L}=\mathbf{Z}^{E L}\left(\mathbf{Z}^D\right)^{-1}=\left(s \mathbf{C}_p^T+\mathbf{Y}^{S U}\right)^{-1} s \mathbf{C}_p^T\]Here, we introduce the electromechanical coupling coefficients
\[\begin{gathered} \text { shear, } k_{15}=d_{15} / \sqrt{s_{55}^E \varepsilon_1^T}=k_{24}, \quad \text { transverse, } k_{31}=d_{31} / \sqrt{s_{11}^E \varepsilon_3^T}=k_{32}, \\ \text { longitudinal, } k_{33}=d_{33} / \sqrt{s_{33}^E \varepsilon_3^T}, \end{gathered}\]or, in the notation used, for force in the $j$ th direction and field in the $i$ th direction
\[k_{i j}=d_{i j} / \sqrt{s_{j j}^E \varepsilon_i^T} \text {. }\]Substituting equation (25) into (23) gives
\[s_{j i}^{S U}=s_{j j}^E\left[1-k_{i j}^2 \bar{Z}_i^{E L}\right] .\]1.1 Shunted with negative capacity
For our problem, the electric field is along $z$ direction and stress is along $x$ direction. So we have
\[s_{33}^{S U} = s_{33}^E\left[1-k_{31}^2 \bar{Z}_1^{E L}\right] = s_{33}^E\left[1- \frac{ s C_p^T k_{31}^2}{s C_p^T+Y^{S U}}\right] = s_{33}^E \left[\frac{ s C_p^T (1-k_{31}^2)+Y^{SU}}{s C_p^T+Y^{S U}}\right]\]Then we have
\[E_{\mathrm{p}}^{\mathrm{SU}}(\omega)=E_{\mathrm{p}}^E \frac{\mathrm{i} \omega C_{\mathrm{p}}^T+Y^{\mathrm{SU}}}{\mathrm{i} \omega C_{\mathrm{p}}^T\left(1-k_{31}^2\right)+Y^{\mathrm{SU}}}\]where $s=i\omega$.
If shunted with a negative capacity, we have
\[Y^{SU} = -i\omega C'=-i\omega H C_0\]So the shunting Young’s modulus is (see reference [1])
\[E_{\mathrm{p}}^{\mathrm{SU}}(\omega)=E_{\mathrm{p}}^E \frac{C' - C_{\mathrm{p}}^T}{C' - C_{\mathrm{p}}^T\left(1-k_{31}^2\right)}\]or
\[E_{\mathrm{p}}^{\mathrm{SU}}(\omega)=E_{\mathrm{p}}^E \frac{C' - C_{\mathrm{p}}^T}{C' - C_{\mathrm{p}}^S}\]where $C_p^S = (1-k_{31}^2)C_p^T, \ C_{p}^T = A \varepsilon_{33}^T / d$, which is the result from reference [3,4].
Finally,
\[\begin{aligned} &E_{e q}=\frac{E_b I_b+2 E_p^{S U} I_p}{I_b+2 I_p}\\ &I_b=\frac{b H^3}{12}, \quad I_p=\frac{b h_p^3}{12}+b h_p\left(\frac{H}{2}+\frac{h_p}{2}\right)^2 \end{aligned}\]1.2 Shunting circuit
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| Circuit diagram |
The impedance is derived in the following picture.
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| Derivation of impedance |
Now we have
\[Z_2 = \frac{1}{j\omega C + 1/R_0}, \quad Z_3 = R_1, \quad Z_4 = R_2\]Then we have impedance
\[Z_{\text{in}} =- \frac{R_1}{R_2\left(j\omega C + 1/R_0\right)} = \frac{1}{j\omega C_N}\]where $R_0$ is very big and $1/R_0$ can be neglected. The negative capacity is
\[C_N = -\frac{R_2}{R_1} C\]2 Reference
[1] Hagood, N. W., & Von Flotow, A. (1991). Damping of structural vibrations with piezoelectric materials and passive electrical networks. Journal of sound and vibration, 146(2), 243-268.
[2] Beck, B. S., Cunefare, K. A., & Collet, M. (2013). The power output and efficiency of a negative capacitance shunt for vibration control of a flexural system. Smart Materials and Structures, 22(6), 065009.
[3] Trainiti, G., Xia, Y., Marconi, J., Cazzulani, G., Erturk, A., & Ruzzene, M. (2019). Time-periodic stiffness modulation in elastic metamaterials for selective wave filtering: Theory and experiment. Physical review letters, 122(12), 124301.