Rayleigh's quotient in vibrations analysis

The Rayleigh's quotient represents a quick method to estimate the natural frequency of a multi-degree-of-freedom vibration system, in which the mass and the stiffness matrices are known.

The eigenvalue problem for a general system of the form [math]\displaystyle{ M\,\ddot{\textbf{q}}(t) + C\,\dot{\textbf{q}}(t) + K\,\textbf{q}(t) = \textbf{Q}(t) }[/math] in absence of damping and external forces reduces to [math]\displaystyle{ M\,\ddot{\textbf{q}}(t) + K\,\textbf{q}(t) = 0 }[/math]

The previous equation can be written also as the following: [math]\displaystyle{ K\,\textbf{u} = \lambda\,M\,\textbf{u} }[/math] where [math]\displaystyle{ \lambda=\omega^2 }[/math], in which [math]\displaystyle{ \omega }[/math] represents the natural frequency, M and K are the real positive symmetric mass and stiffness matrices respectively.

For an n-degree-of-freedom system the equation has n solutions [math]\displaystyle{ \lambda_m }[/math], [math]\displaystyle{ \textbf{u}_m }[/math] that satisfy the equation [math]\displaystyle{ K\,\textbf{u}_m = \lambda_m\,M\,\textbf{u}_m }[/math]

By multiplying both sides of the equation by [math]\displaystyle{ \textbf{u}_{m}^{T} }[/math] and dividing by the scalar [math]\displaystyle{ \textbf{u}_m^T\,M\,\textbf{u}_m }[/math], it is possible to express the eigenvalue problem as follow: [math]\displaystyle{ \lambda_m = \omega_m^2 = \frac{\textbf{u}_m^T\,K\,\textbf{u}_m}{\textbf{u}_m^T\,M\,\textbf{u}_m} }[/math] for m = 1, 2, 3, ..., n.

In the previous equation it is also possible to observe that the numerator is proportional to the potential energy while the denominator depicts a measure of the kinetic energy. Moreover, the equation allow us to calculate the natural frequency only if the eigenvector (as well as any other displacement vector) [math]\displaystyle{ \textbf{u}_{m} }[/math] is known. For academic interests, if the modal vectors are not known, we can repeat the foregoing process but with [math]\displaystyle{ \lambda = \omega^2 }[/math] and [math]\displaystyle{ \textbf{u} }[/math] taking the place of [math]\displaystyle{ \lambda_{m} = \omega_{m}^2 }[/math] and [math]\displaystyle{ \textbf{u}_{m} }[/math], respectively. By doing so we obtain the scalar [math]\displaystyle{ R(\textbf{u}) }[/math], also known as Rayleigh's quotient: [math]\displaystyle{ R(\textbf{u}) = \lambda = \omega^2 = \frac{\textbf{u}^T\,K\,\textbf{u}}{\textbf{u}^T\,M\,\textbf{u}} }[/math]

Therefore, the Rayleigh's quotient is a scalar whose value depends on the vector [math]\displaystyle{ \textbf{u} }[/math] and it can be calculated with good approximation for any arbitrary vector [math]\displaystyle{ \textbf{u} }[/math] as long as it lays reasonably far from the modal vectors [math]\displaystyle{ \textbf{u}_{i} }[/math], i = 1,2,3,...,n.

Since, it is possible to state that the vector [math]\displaystyle{ \textbf{u} }[/math] differs from the modal vector [math]\displaystyle{ \textbf{u}_m }[/math] by a small quantity of first order, the correct result of the Rayleigh's quotient will differ not sensitively from the estimated one and that's what makes this method very useful. A good way to estimate the lowest modal vector [math]\displaystyle{ (u_1) }[/math], that generally works well for most structures (even though is not guaranteed), is to assume [math]\displaystyle{ (u_1) }[/math] equal to the static displacement from an applied force that has the same relative distribution of the diagonal mass matrix terms. The latter can be elucidated by the following 3-DOF example.

Example – 3DOF

As an example, we can consider a 3-degree-of-freedom system in which the mass and the stiffness matrices of them are known as follows: [math]\displaystyle{ M = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{bmatrix} \; , \quad K = \begin{bmatrix} 3 & -1 & 0 \\ -1 & 3 & -2 \\ 0 & -2 & 2 \end{bmatrix} }[/math]

To get an estimation of the lowest natural frequency we choose a trial vector of static displacement obtained by loading the system with a force proportional to the masses: [math]\displaystyle{ \textbf{F} = k\begin{bmatrix} m_1 \\ m_2 \\ m_3 \end{bmatrix} = 1 \begin{bmatrix} 1 \\ 1 \\ 3 \end{bmatrix} }[/math]

Thus, the trial vector will become [math]\displaystyle{ \textbf{u} = K^{-1}\textbf{F} = \begin{bmatrix} 2.5 \\ 6.5 \\ 8 \end{bmatrix} }[/math] that allow us to calculate the Rayleigh's quotient: [math]\displaystyle{ R = \frac{\textbf{u}^T\,K\,\textbf{u}}{\textbf{u}^T\,M\,\textbf{u}} = \cdots = 0.137214 }[/math]

Thus, the lowest natural frequency, calculated by means of Rayleigh's quotient is: [math]\displaystyle{ w_\text{Ray} = 0.370424 }[/math]

Using a calculation tool is pretty fast to verify how much it differs from the "real" one. In this case, using MATLAB, it has been calculated that the lowest natural frequency is: [math]\displaystyle{ w_\text{real} = 0.369308 }[/math] that has led to an error of [math]\displaystyle{ 0.302315 \% }[/math] using the Rayleigh's approximation, that is a remarkable result.

The example shows how the Rayleigh's quotient is capable of getting an accurate estimation of the lowest natural frequency. The practice of using the static displacement vector as a trial vector is valid as the static displacement vector tends to resemble the lowest vibration mode.