Rod longitudinal and shaft torsional frequencies

(a) Longitudinal vibration of a rod

For a rod of area $A$, modulus $E$ and density $\rho$, let $u(x,t)$ be the axial displacement. The axial force is $P = EA\dfrac{\partial u}{\partial x}$. For an element $dx$:

$$\left(P + \frac{\partial P}{\partial x}dx\right) - P = \rho A\,dx\,\frac{\partial^2 u}{\partial t^2} \;\Rightarrow\; \frac{\partial^2 u}{\partial t^2} = c^2\frac{\partial^2 u}{\partial x^2}, \qquad c = \sqrt{\frac{E}{\rho}}$$

With $u = U(x)\sin\omega t$, $\;U = A_1\cos\dfrac{\omega x}{c} + B_1\sin\dfrac{\omega x}{c}$.

Fixed at $x = 0$, free at $x = l$:

• Fixed end: $U(0) = 0 \Rightarrow A_1 = 0$.
• Free end (no stress): $\dfrac{dU}{dx}(l) = 0 \Rightarrow \cos\dfrac{\omega l}{c} = 0$.

$$\frac{\omega_nl}{c} = \frac{(2n-1)\pi}{2} \;\Rightarrow\; f_n = \frac{(2n-1)c}{4l}, \quad n = 1, 2, 3, \ldots$$

Mode shape: $U_n = B\sin\dfrac{(2n-1)\pi x}{2l}$.

(For fixed-fixed or free-free ends, $f_n = nc/(2l)$.)

(b) Numerical values

Longitudinal:

$$c = \sqrt{\frac{200\times10^9}{7850}} = 5047.5\ \text{m/s}$$ $$f_1 = \frac{c}{4l} = \frac{5047.5}{4 \times 2} = 630.9\ \text{Hz}, \quad f_2 = 3f_1 = 1892.8\ \text{Hz}, \quad f_3 = 5f_1 = 3154.7\ \text{Hz}$$

Torsional shaft: The equation is the same with $E \to G$ and $u \to \theta$, so the wave speed is $c_t = \sqrt{G/\rho}$. Taking $\rho = 7850$ kg/m³ for steel:

$$c_t = \sqrt{\frac{80\times10^9}{7850}} = 3192.3\ \text{m/s}$$ $$f_1 = \frac{c_t}{4l} = \frac{3192.3}{4 \times 1.2} = 665.1\ \text{Hz}, \qquad f_2 = 3f_1 = 1995.2\ \text{Hz}$$

The torsional frequencies do not depend on the shaft diameter, because $I_p$ cancels out of the equation.

Answer: Longitudinal: $630.9$, $1892.8$, $3154.7$ Hz. Torsional: $665.1$ and $1995.2$ Hz.