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Rayleigh-Ritz and finite difference methods

Short answer

Rayleigh-Ritz approximates mode shape as a series of admissible functions and makes Rayleigh’s quotient stationary, giving an eigenvalue problem for upper-bound frequencies. Finite difference replaces spatial derivatives with difference formulas at grid points, forming a discrete eigenvalue problem whose accuracy increases with finer discretization.

Write short notes on the Rayleigh-Ritz method and the finite difference method for finding natural frequencies.

Rayleigh-Ritz method

It improves Rayleigh's quotient by using a trial function with several adjustable constants.

  1. Assume the mode shape as a series of admissible functions (satisfying the geometric boundary conditions):
y(x)=c1ϕ1(x)+c2ϕ2(x)+⋯+cnϕn(x)y(x) = c_1\phi_1(x) + c_2\phi_2(x) + \cdots + c_n\phi_n(x)
  1. Write the Rayleigh quotient ω2=UmaxTmax∗\omega^2 = \dfrac{U_{max}}{T^*_{max}}, where UmaxU_{max} is the maximum strain energy and Tmax∗T^*_{max} is the kinetic energy expression with ω2\omega^2 taken out.
  2. Make ω2\omega^2 stationary with respect to every constant:
∂∂ci(Umax−ω2Tmax∗)=0,i=1,2,…,n\frac{\partial}{\partial c_i}\left(U_{max} - \omega^2T^*_{max}\right) = 0, \qquad i = 1,2,\ldots,n

This gives ([K]−ω2[M]){c}=0\left([K] - \omega^2[M]\right)\{c\} = 0, an eigenvalue problem of order nn. 4. The nn roots approximate the lowest nn natural frequencies, always as upper bounds. More terms give better accuracy, and the higher modes are less accurate.

Finite difference method

It replaces the derivatives in the differential equation by differences between displacements at equally spaced stations.

For a beam divided into stations a distance hh apart:

d2ydx2∣i≈yi+1−2yi+yi−1h2,d4ydx4∣i≈yi+2−4yi+1+6yi−4yi−1+yi−2h4\left.\frac{d^2y}{dx^2}\right|_i \approx \frac{y_{i+1} - 2y_i + y_{i-1}}{h^2}, \qquad \left.\frac{d^4y}{dx^4}\right|_i \approx \frac{y_{i+2} - 4y_{i+1} + 6y_i - 4y_{i-1} + y_{i-2}}{h^4}

Procedure:

  1. Divide the system into nn segments and write the governing equation, e.g. EId4ydx4=ω2ρAyEI\dfrac{d^4y}{dx^4} = \omega^2\rho Ay, at every interior station.
  2. Apply the boundary conditions (e.g. y=0y = 0 and y′′=0y'' = 0 at a simple support) using extra fictitious stations if necessary.
  3. The result is [A]{y}=ω2[B]{y}[A]\{y\} = \omega^2[B]\{y\}, solved as an eigenvalue problem.

Accuracy improves as hh decreases, but the order of the matrix grows. For a uniform beam, 4 to 6 segments already give the first frequency within a few percent.

PointRayleigh-RitzFinite difference
Based onEnergy (variational)Differential equation
UnknownsConstants of trial functionsDisplacements at stations
Result boundUpper boundNo fixed bound

✓ Answer checked by Claude Opus 5.5 (high effort)

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