Fundamental frequency of a shaft: Rayleigh and Dunkerley

Data: $L = 1.5$ m, $d = 0.05$ m, $E = 200\times10^9$ Pa, masses 20, 30, 15 kg at $x = 0.4, 0.8, 1.2$ m.

$$I = \frac{\pi d^4}{64} = \frac{\pi (0.05)^4}{64} = 3.068\times10^{-7}\ \text{m}^4, \qquad EI = 61359\ \text{N·m}^2$$

Influence coefficients

For a unit load at distance $a$ from the left end (with $b = L - a$), the deflection at $x \le a$ is

$$\alpha(x,a) = \frac{b\,x\,(L^2 - b^2 - x^2)}{6EIL}, \qquad \alpha_{ij} = \alpha_{ji}\ (\text{Maxwell})$$

The influence-coefficient matrix (m/N):

$$[\alpha] = 10^{-6}\begin{bmatrix} 0.701 & 0.811 & 0.435 \\ 0.811 & 1.136 & 0.661 \\ 0.435 & 0.661 & 0.469 \end{bmatrix}$$

(i) Rayleigh's method

Static deflections under the weights $W_i = m_ig$: $y_i = \sum_j\alpha_{ij}W_j$.

$$\omega_n^2 = \frac{g\sum m_iy_i}{\sum m_iy_i^2} = \frac{9.81 \times 31.7554\times10^{-3}}{16.16626\times10^{-6}} = 19270\ \text{rad}^2/\text{s}^2$$ $$\omega_n = 138.82\ \text{rad/s}, \qquad f_n = 22.09\ \text{Hz}$$

(ii) Dunkerley's method

The frequency with each mass acting alone is $\omega_i = \dfrac{1}{\sqrt{\alpha_{ii}m_i}}$:

$$\frac{1}{\omega_n^2} = \sum\frac{1}{\omega_i^2} = 5.514\times10^{-5} \Rightarrow \omega_n = 134.67\ \text{rad/s}, \qquad f_n = 21.43\ \text{Hz}$$

Comparison

Rayleigh's method gives an upper bound and Dunkerley's method a lower bound on the true fundamental frequency. Solving the eigenvalue problem exactly gives $\omega_n = 138.78$ rad/s ($f = 22.087$ Hz), which lies between them and only just below the Rayleigh value.

Answer: Rayleigh: $f_n \approx 22.09$ Hz; Dunkerley: $f_n \approx 21.43$ Hz.