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We shall start our dive into the finite element method by analyzing differential equations in 1D and get familiar with the concepts, methods and various school of thoughts in a easy setting. The 1D problems are trivial to solve using conventional methods taught in calculus, and we will compare the approximation solutions from the finite element method to the analytical solutions. 1D problem often have analytical solutions compared to 2D or 3D problems for which there might not exist analytical solutions. In the 1D setting we can focus on the central ideas without the distractions of the more complicated multivariate math or the more detailed physics that stems from a 2D geometry.
In what follows we shall introduce two simple problems, one simple mechanical and the other a heat problem. The problems will be used as a model problem, since both are described by the same second order differential equation. There are of course many more problems in 1D that can be analyzed and solved using the FEM, but we shall start with the simplest, non-trivial problems.
Figure 1 shows a model of an elastic rod with a constant cross sectional area, \(A\text{ [mm}^{2}\text{]}\), exposed to a boundary force \(F\text{ [N]}\) at the far end of the rod at the length \(L\text{ [mm]}\), as well as a body load \(f(x)\text{[N/mm]}\) acting along the entirety of the rod. The body load can be seen as a gravitational force.
Problem definition
The problem is to derive an expression for the displacement, \(u\), at every point \(x\) with respect to the load case, which in this case consists of the two boundary values: \(u(0)=0\) as well as the load \(F\) at \(x=L\).
Solution
We shall begin by examining an infinitesimal element, a cut of size \(\Delta x\), of the rod at an arbitrary location \(x\), see the figure below
Force equilibrium at \(x\) along the rod yields:
\[\downarrow:\ N(x+\Delta x)-N(x)+\int_{x}^{x+\Delta x}f(\xi)d\xi=0\]where \(N\) denotes the normal force and the third term is the body load summed up over the length of the infinitesimal element. Note that at the boundary \(x=L\) we have \(N(L)=F\). If we now let \(\Delta x \rightarrow 0\) then naturally \(\int_{x}^{x+\Delta x}f(\xi)d\xi\rightarrow f(x)\) and using the definition of the derivative we get
\[\lim_{\Delta x \rightarrow 0}\dfrac{N(x+\Delta x)-N(x)}{\Delta x}=f(x)\]which leads to the point-wise equilibrium equation
\[\boxed{-\dfrac{dN}{dx}=f}\quad{\color{#6dbbff}\text{Equilibrium relation}}\]The relation between forces and stresses assumes a constant cross-sectional area:
\[\boxed{N=\sigma A}\]where \(\sigma\) denotes the stress with a unit of MPa or \(\text{N/mm}^{2}\).
The relation between the stress and the strain is assumed to be linear, hence Hooke's law is used to describe it
\[\boxed{\sigma=E\varepsilon}\ {\color{#6dbbff}\text{Constitutive relation}}\]The strain is assumed to be small such that it can be defined as
\[\varepsilon:=\dfrac{\text{elongation}}{\text{original length}}\](this is also know as the small strain assumption) and with the displacements in our infinitesimal element this becomes
\[\varepsilon:=\lim_{\Delta x\rightarrow0}\dfrac{u(x+\Delta x)-u(x)}{\Delta x}\rightarrow\boxed{\varepsilon=\dfrac{du(x)}{dx}}\ {\color{#6dbbff}\text{Geometric relation}}\]Now we insert (7) into (5) and the result into (4) and (3) and we get the point-wise equilibrium ordinary differential equation
\[\boxed{\text{(BVP)}\begin{cases} -\dfrac{d}{dx}\left(EA\dfrac{du}{dx}\right)=f\quad\text{for }x\in[0,L] & \text{Differential Equation}\\ u(0)=0 & \text{Essential boundary condition}\\ EA\dfrac{du}{dx}(L)=F & \text{Natural (or Force) boundary condition} \end{cases}}\]A 1D heat equation can be derived similarly to the 1D elastic rod differential equation in the previous section. Interestingly, both physical phenomena are described with the same second order differential equation. It is a common engineering task to analyze both structural problems and heat problems and many commercial Finite Element Analysis tools can handle both fields.
Problem definition
Consider a homogeneous heat conductive rod of length \(L\text{ [mm]}\) that is heated by some heat source \(f\text{ [W]}\). The rod can be heated by e.g., an electric current, such that the rod is uniformly being heated, in which \(f\) is constant, or by e.g., a gas torch such that \(f\) varies along the length.
We wish to find the temperature \(u(x)\) at the steady state, i.e., after long enough time such that the temperature field no longer changes.
Solution
From thermodynamics we know that heat energy \(Q\) per unit time, \(\dot{Q}\) is a balance between heat flow in, heat flow out and internal heat. Let's consider our rod being heated by an internal heat source, \(f(x)\), the heat flow in is denoted by \(q(0)\) and \(q(L)\) denotes the heat flow out. The total heat movement must be conserved:
\[\begin{aligned} \dot{Q} & =0\Rightarrow\int_{0}^{L}f(x)dx-q(L)+q(0)=0\\ \Rightarrow & \int_{0}^{L}f(x)dx=q(L)-q(0)\overset{\text{2nd theorem of calculus}}{=}\int_{0}^{L}\dfrac{dq}{dx}dx\end{aligned}\]or
\[\boxed{\int_{0}^{L}f(x)dx=\int_{0}^{L}\dfrac{dq}{dx}dx}\]which is known as the first law of thermodynamics or conservation of energy.
Next we have the second law of thermodynamics: heat flows from high to low temp. Fourier's law of heat conduction states:
\[\boxed{q=-k\dfrac{du}{dx}}\]meaning that heat flows towards decreasing temperature (hence the negative derivative) with some rate, \(k\) (heat conductivity), that is dependent on the material.
Insert (11) into (10) and we get
\[-\int_{0}^{L}\frac{d}{dx}\left(k\dfrac{du}{dx}\right)dx=\int_{0}^{L}f(x)dx\]note that
\[-\int_{x_{i}}^{x_{i+1}}\dfrac{dq}{dx}dx=\int_{x_{i}}^{x_{i+1}}f(x)dx\]must hold for any \(x_{i} < x_{i+1}\), which means that \(-\frac{dq}{dx}=f\) must hold everywhere in \([0,L]\)!
Thus the strong form (or point-wise) heat equation in 1D is
\[\boxed{-\frac{d}{dx}\left(k\dfrac{du}{dx}\right)=f(x)}\]This problem contains both natural boundary conditions i.e., the derivative of \(u\) is known, as well as essential boundary conditions, i.e., \(u\) itself is know somewhere on the boundary.
