Lateral vibration of a simply supported beam

(a) Euler-Bernoulli beam equation

Consider an element $dx$ of a beam with bending stiffness $EI$, mass per unit length $\rho A$, deflection $y(x,t)$, shear force $V$ and bending moment $M$.

     V+dV
 M ->|====|<- M+dM
     V

Vertical equilibrium: $\;V - (V + dV) = \rho A\,dx\,\dfrac{\partial^2 y}{\partial t^2}$, that is $-\dfrac{\partial V}{\partial x} = \rho A\dfrac{\partial^2 y}{\partial t^2}$.

Moments (neglecting rotary inertia and shear deformation): $V = \dfrac{\partial M}{\partial x}$ and $M = EI\dfrac{\partial^2 y}{\partial x^2}$. Hence

$$EI\frac{\partial^4 y}{\partial x^4} + \rho A\frac{\partial^2 y}{\partial t^2} = 0$$

Put $y = Y(x)\sin\omega t$ and $\beta^4 = \dfrac{\rho A\omega^2}{EI}$:

$$\frac{d^4Y}{dx^4} - \beta^4Y = 0 \;\Rightarrow\; Y = C_1\cos\beta x + C_2\sin\beta x + C_3\cosh\beta x + C_4\sinh\beta x$$

Simply supported ends: $Y(0) = Y''(0) = 0 \Rightarrow C_1 = C_3 = 0$. At $x = l$, $Y(l) = Y''(l) = 0$ gives $C_4 = 0$ and

$$\sin\beta l = 0 \;\Rightarrow\; \beta_nl = n\pi$$ $$\omega_n = (n\pi)^2\sqrt{\frac{EI}{\rho Al^4}} = \frac{n^2\pi^2}{l^2}\sqrt{\frac{EI}{\rho A}}, \qquad Y_n = \sin\frac{n\pi x}{l}$$

(b) Numerical

$$I = \frac{bh^3}{12} = \frac{0.05 \times 0.1^3}{12} = 4.1667\times10^{-6}\ \text{m}^4, \qquad A = bh = 0.005\ \text{m}^2$$ $$\sqrt{\frac{EI}{\rho A}} = \sqrt{\frac{200\times10^9 \times 4.1667\times10^{-6}}{7850 \times 0.005}} = 145.710\ \text{m}^2/\text{s}$$ $$\omega_1 = \frac{\pi^2}{3^2}\times 145.710 = 159.79\ \text{rad/s}, \qquad f_1 = 25.43\ \text{Hz}$$ $$f_2 = 4f_1 = 101.72\ \text{Hz}, \qquad f_3 = 9f_1 = 228.88\ \text{Hz}$$

Answer: $f_1 = 25.43$ Hz, $f_2 = 101.72$ Hz, $f_3 = 228.88$ Hz.