Natural Frequency of Shaft Vibration

This section explains how to determine the natural frequency of free transverse vibrations for a simply supported shaft carrying a point load. The calculation is closely linked to the static deflection caused by this load.

Understanding Natural Frequency

In mechanics, the natural frequency is the frequency at which a system tends to oscillate in the absence of any driving or damping force. For a simple system involving mass and stiffness, the natural frequency ($f_n$) in Hertz (Hz) is typically calculated using the formula:

$$ f_n = \frac{1}{2\pi} \sqrt{\frac{k}{m}} $$

Here, $k$ represents the stiffness of the system (like a spring constant) and $m$ is the mass involved in the vibration.

Relating Load, Deflection, and Stiffness

For a simply supported shaft with a point load ($W$), the static deflection ($\delta_s$) at the point of load application is related to the load and the shaft's stiffness ($k$). The relationship is:

$$ \delta_s = \frac{W}{k} $$

This implies that the stiffness of the shaft under the specific loading condition can be expressed as:

$$ k = \frac{W}{\delta_s} $$

Deriving the Frequency Formula with Static Deflection

To find the natural frequency in terms of static deflection, we make a key assumption: the point load ($W$) is the weight of the mass ($m$) that is oscillating. Thus, we can write $W = m \times g$, where $g$ is the acceleration due to gravity (approximately $9.81 \, m/s^2$).

Substituting $W = mg$ into the stiffness equation gives:

$$ k = \frac{m \times g}{\delta_s} $$

Now, substitute this expression for $k$ back into the natural frequency formula:

$$ f_n = \frac{1}{2\pi} \sqrt{\frac{k}{m}} = \frac{1}{2\pi} \sqrt{\frac{(m \times g) / \delta_s}{m}} $$

Simplifying the term inside the square root:

$$ f_n = \frac{1}{2\pi} \sqrt{\frac{g}{\delta_s}} $$

This formula can be rearranged to show the direct relationship between frequency and static deflection:

$$ f_n = \left( \frac{\sqrt{g}}{2\pi} \right) \frac{1}{\sqrt{\delta_s}} $$

Calculating the Numerical Value

The constant part of the formula, $\frac{\sqrt{g}}{2\pi}$, needs to be calculated. Using $g \approx 9.81 \, m/s^2$:

$$ \frac{\sqrt{g}}{2\pi} \approx \frac{\sqrt{9.81}}{2 \times \pi} \approx \frac{3.13209}{6.283185} \approx 0.4985 $$

Therefore, the natural frequency ($f_n$) is approximately:

$$ f_n \approx \frac{0.4985}{\sqrt{\delta_s}} $$

This result indicates that the natural frequency is inversely proportional to the square root of the static deflection, assuming the deflection $\delta_s$ is measured in consistent units (e.g., meters).