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513.160 Microelectronics and Micromechanics | |
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Beam vibrationsThe Euler–Lagrange equation for a beam vibrating in the $x$-$z$ plane is, \[ \begin{equation} \frac{\partial^2}{\partial x^2}\left(EI\frac{\partial^2w}{\partial x^2}\right) = -\mu\frac{\partial^2 w}{\partial t^2}+q(x). \end{equation} \]Here $x$ is the position, $t$ is the time, $w$ is the displacement of the beam in the $z$ direction, $E$ is the elastic modulus, $\mu$ is the mass per unit length, $q$ is the load on the beam acting in the $z$-direction, and $I$ is the second moment of area of the beam's cross-section, \[ \begin{equation} I=\int\int z^2 dydz. \end{equation} \]Here $(y=0,z=0)$ is the centroid of the cross section of the beam. For a beam with a rectangular cross section with a width $L_y$ in the $y$-direction and a width $L_z$ in the $z$-direction, $I=L_yL_z^3/12$. If the beam is made of homogeneous material and the cross section is constant, then the product $EI$ (known as the flexural rigidity) is a constant and the Euler–Lagrange equation can be written, \[ \begin{equation} EI\frac{\partial^4w}{\partial x^4} = -\mu\frac{\partial^2 w}{\partial t^2}+q(x). \end{equation} \]The general solution for an unloaded beam ($q=0$) has the form, \[ \begin{equation} w(x,t) = \left(A_1\sin(kx)+A_2\cos(kx)+A_3\sinh(kx)+A_4\cosh(kx)\right)e^{-i\omega t}, \end{equation} \]where the constants $A_i$ are determined by the boundary conditions. The relationship between the frequency $\omega$ and the wavenumber $k$ is, \[ \begin{equation} \omega = \sqrt{\frac{EI}{\mu}}k^2. \end{equation} \]This can be checked by substituting the general solution into the Euler–Lagrange equation. |