# 5.4: Galerkin Method of Solving Non-linear Differential Equation

Beris Galerkin, a Russian scientist, mathematician and engineer was active in the first forty ears of the 20th century. He is an example of a university professor who applied methods of structural mechanics to solve engineering problems. At that time (World War I), the unsolved problem was moderately large deflections of plates. In 1915, he developed an approximate method of solving the above problem and by doing it made an important and everlasting contribution to mechanics.

The theoretical foundation of the Galerkin method goes back to the Principle of Virtual Work. We will illustrate his idea on the example of the moderately large theory of beams. If we go back to Chapter 2 and follow the derivation of the equations of equilibrium from the variational principle, the so called “weak” form of the equilibrium is given by Equation (2.5.10). Adding the non-linear term representing the contribution of finite rotations, this equation can be re-written as

$\int_{0}^{l} (M^{\prime\prime} + N^{\prime\prime} + q) \delta w dx + \int_{0}^{l} N^{\prime} \delta u dx + \text{ Boundary terms}$

where

$M = −EIw^{\prime\prime}$

$N = EA[u^{\prime} + \frac{1}{2}(w^{\prime})^2]$

From the weak (global) equilibrium one can derive the strong (local) equilibrium by considering an infinite class of variations. But, what happens if, instead of a “class”, we consider only one specific variation (shape) that satisfies kinematic boundary conditions? The equilibrium will be violated locally, but can be satisfied globally in average

$\int_{0}^{l} \left[ −EIw^{\mathrm{IV}} + EA \left( u^{\prime} + \frac{1}{2}(w^{\prime})^2 w^{\prime\prime} + q \right) \right] \delta w dx = 0 \label{6.45}$

Consider the example of a simply supported beam, restrained from axial motion. The exact solution of this problem fro the sinusoidal distribution of load was given in the previous section. Assume now that the same beam is loaded by a uniform line load $$q(x) = q$$. No exact solution of this problem exists.

Let’s solve this problem approximately by means of the Galerkin method. As a trial approximate deflected shape, we take the same shape that was found as a particular solution of the full equation

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

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

With the condition of ends fixity in the axial direction, $$u = u^{\prime} = 0$$, and Equation \ref{6.45} yields

$\delta C \int_{0}^{l} \left[ −EIw^{\mathrm{IV}} + \frac{EA}{2}(w^{\prime})^2 w^{\prime\prime} + q \right] \sin \frac{\pi x}{l} dx = 0$

Evaluating the derivatives and integrating, the following expression is obtained

$\frac{l}{2} C + \frac{l}{8}\frac{C^3A}{2I} - \frac{q_1}{EI(\frac{\pi}{l})^4\pi}\frac{2l}{\pi} = 0$

After re-arranging, the dimensionless deflection amplitude $$\frac{C}{h} = \frac{w_o}{h}$$ is related to the remaining

$\frac{w_o}{h} + \frac{3}{2}\left( \frac{w_o}{h}\right)^3 = \left(\frac{q_1}{EI}\right) \frac{48}{\pi^5}\left(\frac{l}{h}\right)^4 \label{6.49}$

The above cubic equation has a simple solution.

Let’s discuss the two limiting cases. Without the non-linear term, Equation \ref{6.49} predicts the following deflection of the beam under pure bending action for the square section

$\frac{w_o}{h} = \left(\frac{q_1}{Eh}\right)\frac{48}{\pi^5} \left(\frac{l}{h}\right)^4$

In the exact solution of the same problem, the numerical coefficient is $$\frac{60}{384} = \frac{1}{6.4}$$, which is only 1.5% smaller than the present approximate solution $$\frac{48}{\pi^5} = \frac{1}{6.3}$$. If on the other hand the flexural resistance is small, $$EI \rightarrow 0$$, the first term in Equation \ref{6.49} vanishes giving a cubic load-deflection relation

$\left( \frac{w_o}{h}\right)^3 = \frac{32}{\pi^5}\left(\frac{q_1}{EI}\right) \left(\frac{l}{h}\right)^4$

There is no closed-form solution for the pure membrane response of the beam under uniform pressure. However, the present prediction compares favorably with the Equation (5.3.28) for the moderately large deflection, if the total load under the uniform and sinusoidal pressure is the same

$P = q_1l = q_o \int_{0}^{l} \sin \frac{\pi x}{l} dx = q_o \frac{2 l}{\pi}$

Replacing $$q_1$$ by $$\frac{2}{\pi}q_o$$, the pure membrane solution takes the final form

$\left( \frac{w_o}{h}\right)^3 = \frac{6.4}{\pi^4}\left(\frac{q_o}{Eh}\right) \left(\frac{l}{h}\right)^4 \label{6.53}$

One can see that not only the dimensionless form of the exact and approximate solutions are identical, but also the coefficient 6.4 in Equation \ref{6.53} is of the same order as the coefficient 4 in Equation (5.3.29).