# 7.4: Fourier Series Expansion and the Ritz Method

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Consider a symetrically loaded simply supported plate by the point force at the center. The total potential energy of the system is

$\prod = \int_{0}^{l} \frac{EI}{2} (w^{\prime\prime})^2 dx − P w$

The objective is to find the amplitude and shape of the deflection function that is in equilibrium with the prescribed load $$P$$. In other words we will be looking the deflection and shape that will make the total potential energy stationary.

Assume the solution as a Fourier expansion function

$w(x) = \sum^{N}_{n=1} a_n \phi_n(x) \label{8.24}$

where $$\phi_n(x)$$ is a complete system of coordinate function satisfying kinematic boundary conditions. In the rectangular coordinate system this system consists of hormonic functions, in the polar coordinate system these are Bessel function, and in the spherical coordinate system this role is taken by the Legender functions. In our case

$\phi_n(x) = \sin \frac{ n \pi x}{l} \label{8.25}$

The kinematic boundary conditions $$\phi_n(x = 0) = \phi_n(x = l)$$ are identically satisfied. Furthermore, because of the symmetry of the problem, only the symmetric function will contribute to the solution. Figure $$\PageIndex{1}$$: Asymmetric modes do not satisfy boundary condition $$w^{\prime} (x = \frac{l}{2}) = 0$$ at the center of the beam.

The solution is then represented as

$w(x) = a_1 \sin \frac{\pi x}{l} + a_3 \sin \frac{3 \pi x}{l} + \dots \label{8.26}$

Consider first one-term approximation

$w(x) = a_1 \sin \frac{\pi x}{l}$

$w^{\prime}(x) = a_1 \left( \frac{\pi}{l} \right) \cos \frac{\pi x}{l}$

$w^{\prime\prime}(x) = −a_1 \left( \frac{\pi}{l} \right)^2 \sin \frac{\pi x}{l}$

The expression for the total potential energy is

$\prod = \frac{1}{2} EI \left( \frac{\pi}{l} \right)^4 a^2_1 \int_{0}^{l} \sin \frac{\pi x}{l}- P a_1$

where the integral is simply $$l/2$$. Equation \ref{8.26} reduces then to

$\prod = \frac{1}{4} EI \frac{\pi^4}{l^3} a^2_1 − P a_1$

For equilibrium $$\frac{d\prod}{da_1} = 0$$, which yields

$\frac{1}{2} EI \frac{\pi^4}{l^3} a_1 − P = 0$

The load-displacement relation is finally given by

$(a_1)_{\text{opt}} = \frac{Pl^3}{\frac{\pi^4}{2}EI} = \frac{Pl^3}{48.7EI}$

The numerical coefficient in the exact solution of this problem is 48. The error of the approximate solution is 1.4%. Such a good accuracy of just one-term approximation can be explained by making the Taylor series expansion of the sign function

$\sin \frac{\pi x}{l} = \frac{\pi x}{l} − \frac{1}{6} \left( \frac{\pi x}{l} \right)^3 + \dots$

The two term expansion has a linear and cubic terms in $$x$$, the same as the exact solution.

Let’s examine next the stationary property of the functional $$\prod$$. Defining the normalized total potential energy as $$\bar{\prod} = \frac{\prod}{P w_o}$$, one gets

$\bar{\prod} = \frac{1}{2} \frac{a_1}{w_o} \left(\frac{a_1}{w_o} − 2 \right)$

where $$a_1$$ is the amplitude of the trial function Equation \ref{8.25} and $$w_o$$ is the exact amplitude. The plot of the function $$\bar{\prod} (a_1)$$ is shown in Figure ($$\PageIndex{2}$$). Figure $$\PageIndex{2}$$: By varying the amplitude around $$\frac{a_1}{w_o} = 1$$, $$\bar{\prod}$$ does not change.

The function $$\bar{\prod}(a_1)$$ is a parabola with a stationary point at $$a_1 = w_o$$. The stationary point is at the same time the minimum. The negative value of the minimum (actual) value of the total potential energy comes from the choice of the reference level of the potential energy. In mechanics, the reference level is the position of the undeformed axis of the beam. Upon loading, the beam is loosing the potential energy and the second term in Equation (7.3.3) is negative and larger than the first term of the stored elastic energy.

Even though the accuracy of one term approximation in the Fourier series expansion, Equation (7.3.4) gave a very good approximation (1.4% error), the solution can be further improved by considering the next term in the expansion, according to Equation \ref{8.24}. In this case the total potential energy is the function of two unknown amplitudes, $$\prod = \prod(a_1, a_3)$$ and the solution is obtained from two algebraic equations

$\frac{\partial \prod}{\partial a_1} = 0, \quad \frac{\partial \prod}{\partial a_3} = 0$

This page titled 7.4: Fourier Series Expansion and the Ritz Method is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Tomasz Wierzbicki (MIT OpenCourseWare) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.