---
title: "Fundamental frequency of a shaft: Rayleigh and Dunkerley"
subject: "Theory of Mechanism and Machine II"
chapter: 9
key: "semester_7/theory_of_mechanism_and_machine_ii:shaft-rayleigh-dunkerley-numerical"
source: https://www.bctengineer.com/q/semester_7/theory_of_mechanism_and_machine_ii/shaft-rayleigh-dunkerley-numerical
site: BCT Engineering
---

# Fundamental frequency of a shaft: Rayleigh and Dunkerley

Theory of Mechanism and Machine II · Chapter 9: Vibration of Multi Degree of Freedom Systems

- Asked in: year not given, 10 marks
- Video: [Rayleigh's method to find the natural frequency of transverse vibrations](https://www.youtube.com/watch?v=eo7319v7xYg) — Mechanical Engineering Online Classes
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/theory_of_mechanism_and_machine_ii/shaft-rayleigh-dunkerley-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_7/theory_of_mechanism_and_machine_ii/chapter-9

## Question

A shaft of diameter 50 mm and length 1.5 m is simply supported at its ends and carries three discs of mass 20 kg, 30 kg and 15 kg at distances of 0.4 m, 0.8 m and 1.2 m from one end. Neglecting the mass of the shaft and taking $E = 200$ GPa, estimate the fundamental frequency of transverse vibration by (i) Rayleigh's method and (ii) Dunkerley's method.

## Short answer

Rayleigh’s method (upper bound) uses static deflections under weight to compute ω_n from g Σ(my)/Σ(my²), yielding f_n ≈ 22.09 Hz. Dunkerley’s method (lower bound) uses 1/ω_n² = Σ 1/ω_i² with ω_i = 1/√(α_ii m_i), yielding f_n ≈ 21.43 Hz.

## Answer

**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 $W_i = m_ig$:  $y_i = \sum_j\alpha_{ij}W_j$.

| Disc | $m$ (kg) | $x$ (m) | $\alpha_{ii}\times10^6$ | $y$ (mm) | $my\ (\times10^{-3})$ | $my^2\ (\times10^{-6})$ |
|---|---|---|---|---|---|---|
| 1 | 20 | 0.4 | 0.701 | 0.4403 | 8.8054 | 3.87673 |
| 2 | 30 | 0.8 | 1.136 | 0.5906 | 17.7188 | 10.46515 |
| 3 | 15 | 1.2 | 0.469 | 0.3487 | 5.2312 | 1.82437 |

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

$$
\omega_n = 138.82\ \text{rad/s}, \qquad f_n = 22.09\ \text{Hz}
$$

#### (ii) Dunkerley's method
The frequency with each mass acting alone is $\omega_i = \dfrac{1}{\sqrt{\alpha_{ii}m_i}}$:

| Disc | $\alpha_{ii}\times10^6$ | $\omega_i$ (rad/s) | $1/\omega_i^2$ (s²) |
|---|---|---|---|
| 1 | 0.701 | 267.04 | $1.402\times10^{-5}$ |
| 2 | 1.136 | 171.32 | $3.407\times10^{-5}$ |
| 3 | 0.469 | 376.88 | $0.704\times10^{-5}$ |

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

#### 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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Rayleigh's method gives the fundamental frequency from static deflections y_i as:
   A. w^2 = g sum(m y) / sum(m y^2)
   B. w^2 = g sum(m y^2) / sum(m y)
   C. w^2 = sum(m y) / g
   D. w^2 = g / sum(y)
   Answer: A. Equating maximum potential energy (1/2 g sum m y) and kinetic energy (1/2 w^2 sum m y^2). (Key point: Rayleigh formula)

2. In Rayleigh's method the deflections y_i are obtained from:
   A. the exact mode shape
   B. the weights W_j = m_j g with influence coefficients: y_i = sum alpha_ij W_j
   C. random trial values
   D. the shaft diameter only
   Answer: B. The static deflection curve under the weights is used as the assumed mode shape. (Key point: Static deflection source)

3. Dunkerley's method gives the fundamental frequency from:
   A. w_n = sum of w_i
   B. w_n^2 = product of w_i^2
   C. 1/w_n^2 = sum of 1/w_i^2, with w_i = 1/sqrt(alpha_ii m_i)
   D. w_n = largest w_i
   Answer: C. Each load acting alone gives w_i = 1/sqrt(alpha_ii m_i); their reciprocal squares add. (Key point: Dunkerley's formula)

4. Compared with the true fundamental frequency:
   A. Rayleigh gives a lower bound, Dunkerley an upper bound
   B. both give lower bounds
   C. Rayleigh gives an upper bound, Dunkerley a lower bound
   D. both are exact
   Answer: C. Rayleigh (22.09 Hz) overestimates and Dunkerley (21.43 Hz) underestimates, bracketing the true value. (Key point: Bounds)

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