Tutorial : Finite Element Method Fundamentals

hello again,

In the previous post I left you guys with a simple code which gives a one dimensional finite element approximation to longitudinal deformation of a bar. Let us learn about the problem in more details.

Problem Definition and its Differential Equation

The problem is explained with the help of the following diagram

Fig. 1

As shown in Fig. 1, a bar fixed at one end is subjected to uniform axial force $f_{o}$ and an end load $P$. At the fixed end the deformation $u$ is zero. The objective is to obtain an expression for the deformation $u$ as a function of $x$. To elaborate the implementation of FEM on this simple problem, let us consider its differential equation.

This equation in terms of the unknown function $u(x)$ is valid in the domain $0 < x< l$ .

Boundary Conditions Application

The boundary conditions given at the two extremities of $x$ are:

Since the bar is fixed at $x=0$, it is very easy to see that there is no deformation and consequently $u(x)$ is zero at this point. This is called as the "Dirichlet" type of boundary condition. At the other end, no deformation is imposed and consequently the bar is free to deform under the applied load. It can be observed from Fig. 1 that there is an end load present. This end load which is a point force can be expressed in terms of the first derivative of the unknown deformation function $u(x)$. This expression is demonstrated by the second boundary condition and it also goes by the name "Neumann" type of boundary condtion.

Obtaining the Weak Form

The next step is to find the weak form of the differential equation. Let $r(x)$ be the residual of the differential equation given by:

$$r(x)=-\frac{d}{dx} \left( EA(x)\frac{du}{dx} \right ) - f(x)$$

Let us now introduce a test function $v(x)$ which is an admissible function and behaves more or less like the original trial function $u(x)$. Note that we are intentionally skipping the part where the admissible function $v(x)$ is termed as weight function $w(x)$ and multiplied by the residue $r(x)$. This omitted step has the name "weighted residual". However, we are going to follow almost identical steps as with weighted residual without having to use it explicitly. Since, the residual is zero we may write;

$$\int_{0}^{l}\left \{ -\frac{d}{dx} \left ( EA (x)\frac{du}{dx} \right ) - f(x) \right \} v(x) dx = 0$$

The most important step in FEM formulation is the application of integration by parts. Let us apply this technique first and then see why was it necessary.

$$\int_{0}^{l}EA(x)\frac{du}{dx} \frac{dv}{dx} dx = \int_{0}^{l}f(x)v(x) + \left. EA(x)\frac{du}{dx} v(x) \right |_{0}^{l}$$

The second term on the right hand side is the boudary term. At $x=0$, $u=0$ and therefore $v$ is also zero. This is one of the assumptions for the test function to be admissible as mentioned earlier. At $x=l$:

$$\left. EA(x)\frac{du}{dx} v(x) \right |_{x=l}=\left. Pv(x)\right |_{x=l}$$

Simplifying and substituting the boundary conditions, the weak form is finally given as:

$$\int_{0}^{l}EA(x)\frac{du}{dx} \frac{dv}{dx} dx = \int_{0}^{l}f(x)v(x) +\left. Pv(x)\right |_{x=l}$$

This topic is continued until the next post. Till then take care.