Introduction to the Finite Element Method

Modern bridges, car bodies, aircraft wings and even hip implants are analysed with the finite element method (FEM) before anything is built. Behind the colourful stress plots in software such as ANSYS or Abaqus lies a simple idea: break a complicated object into many small, simple pieces, solve each piece with algebra, and stitch the results together. In this introduction we build that idea from scratch using axial bars, so you can see every matrix and every number.

The Core Idea

Real structures have complicated shapes and loads, and their governing differential equations rarely have neat closed-form solutions. FEM replaces the continuous problem with a discrete one:

1. Discretise: divide the body into small regions called elements, joined at points called nodes.
2. Approximate: within each element, assume a simple variation of the unknown (for example, displacement varying linearly).
3. Element equations: derive a small matrix equation for each element.
4. Assemble: combine element equations into one global system \( [K]\{u\} = \{F\} \).
5. Apply boundary conditions: fix known displacements.
6. Solve for the nodal unknowns, then compute strains and stresses.

As elements are made smaller (mesh refinement), the approximate answer converges toward the exact one.

The Simplest Element: An Axial Bar

Consider a straight bar element of length \( L \), cross-sectional area \( A \) and Young's modulus \( E \), with a node at each end. Each node can move only along the bar axis, with displacements \( u_1 \) and \( u_2 \).

The bar behaves like a spring of stiffness

$$ k = \frac{EA}{L} $$

Equilibrium of the two nodal forces gives the element stiffness matrix:

$$ \begin{bmatrix} f_1 \\ f_2 \end{bmatrix} = \frac{EA}{L}\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}\begin{bmatrix} u_1 \\ u_2 \end{bmatrix} $$

Notice the matrix is symmetric, and its rows sum to zero: moving both nodes equally (a rigid-body shift) produces no force. On its own the matrix is singular, which is why we must apply supports before solving.

Once displacements are known, the stress in the element is

$$ \sigma = E\varepsilon = E\,\frac{u_2 - u_1}{L} $$

Assembly: Building the Global Matrix

When elements share a node, their stiffnesses add at that node. For two elements in a line with nodes 1, 2 and 3, stiffnesses \( k_1 \) and \( k_2 \):

$$ [K] = \begin{bmatrix} k_1 & -k_1 & 0 \\ -k_1 & k_1 + k_2 & -k_2 \\ 0 & -k_2 & k_2 \end{bmatrix} $$

Node 2 belongs to both elements, so its diagonal term is \( k_1 + k_2 \). This "direct stiffness" assembly is exactly what FEM software does, only for millions of nodes.

Solved Problem 1: A Stepped Steel Bar

A stepped steel bar is fixed at its left end (node 1). Element 1 has \( A_1 = 200 \) mm² and \( L_1 = 500 \) mm; element 2 has \( A_2 = 100 \) mm² and \( L_2 = 500 \) mm. \( E = 200 \) GPa \( = 200\,000 \) N/mm². A tensile load \( P = 20 \) kN acts at the free end (node 3). Find the nodal displacements, element stresses and support reaction.

Step 1: Element stiffnesses. Working in N and mm:

$$ k_1 = \frac{(200\,000)(200)}{500} = 80\,000\ \text{N/mm}, \qquad k_2 = \frac{(200\,000)(100)}{500} = 40\,000\ \text{N/mm} $$

Step 2: Assemble. In units of kN/mm:

$$ [K] = \begin{bmatrix} 80 & -80 & 0 \\ -80 & 120 & -40 \\ 0 & -40 & 40 \end{bmatrix} $$

Step 3: Apply the boundary condition. Since \( u_1 = 0 \), delete row 1 and column 1. The loads are 0 at node 2 and 20 kN at node 3:

$$ \begin{bmatrix} 120 & -40 \\ -40 & 40 \end{bmatrix}\begin{bmatrix} u_2 \\ u_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 20 \end{bmatrix} $$

Step 4: Solve. From the first equation, \( 120u_2 = 40u_3 \), so \( u_3 = 3u_2 \). Substituting into the second:

$$ -40u_2 + 40(3u_2) = 20 \Rightarrow 80u_2 = 20 \Rightarrow u_2 = 0.25\ \text{mm} $$

$$ u_3 = 0.75\ \text{mm} $$

Step 5: Stresses.

$$ \sigma_1 = 200\,000\,\frac{0.25 - 0}{500} = 100\ \text{N/mm}^2 = 100\ \text{MPa} $$

$$ \sigma_2 = 200\,000\,\frac{0.75 - 0.25}{500} = 200\ \text{MPa} $$

Step 6: Reaction. Row 1 of the full system gives \( R_1 = 80(0) - 80(0.25) + 0 = -20 \) kN, that is 20 kN acting to the left.

Sanity check. Statics alone says each element carries 20 kN, so \( \sigma_1 = 20\,000/200 = 100 \) MPa and \( \sigma_2 = 20\,000/100 = 200 \) MPa, exactly matching. The reaction balances the applied load, and both stresses are below the yield strength of mild steel (about 250 MPa). For this simple bar FEM is exact, because the true displacement really is linear within each uniform element.

Solved Problem 2: A Tapered Aluminium Bar and Convergence

Now try a bar whose area varies linearly from 400 mm² at the fixed end to 200 mm² at the free end, with length 1000 mm, \( E = 70\,000 \) N/mm² and an end load of 10 kN. Here FEM is only approximate, so we can watch it converge.

Step 1: Exact solution. Integrating \( du/dx = P/(EA(x)) \) over a linear taper gives

$$ \delta_{\text{exact}} = \frac{PL}{E(A_1 - A_2)}\ln\frac{A_1}{A_2} = \frac{(10\,000)(1000)}{(70\,000)(200)}\ln 2 = 0.7143 \times 0.6931 = 0.4951\ \text{mm} $$

Step 2: One element. Use the average area of 300 mm²:

$$ k = \frac{(70\,000)(300)}{1000} = 21\,000\ \text{N/mm}, \qquad \delta = \frac{10\,000}{21\,000} = 0.4762\ \text{mm} $$

The error is \( (0.4951 - 0.4762)/0.4951 = 3.8\% \).

Step 3: Two elements. Each is 500 mm long, with average areas 350 mm² and 250 mm²:

$$ k_1 = \frac{(70\,000)(350)}{500} = 49\,000\ \text{N/mm}, \qquad k_2 = \frac{(70\,000)(250)}{500} = 35\,000\ \text{N/mm} $$

Because the elements are in series and each carries the full 10 kN,

$$ \delta = \frac{10\,000}{49\,000} + \frac{10\,000}{35\,000} = 0.2041 + 0.2857 = 0.4898\ \text{mm} $$

The error drops to about 1.1%.

Sanity check. Refining the mesh moved the answer toward the exact value, and it approaches from below, meaning the FE model is slightly too stiff. That is a known property of displacement-based finite elements, and it reassures us the method is behaving correctly.

Where FEM Goes Next

• 2D and 3D elements (triangles, quadrilaterals, tetrahedra) for plates, shells and solids.
• Beam elements with rotations as well as displacements.
• Thermal, fluid and electromagnetic problems, which use the same assembly logic with different element matrices.
• Nonlinear and dynamic analysis for crashes, plasticity and vibration.

Did You Know?

• Engineers at Boeing, including M. J. Turner and colleagues, published the stiffness approach for aircraft structures in 1956, and Ray Clough coined the name "finite element method" around 1960.
• A modern car crash simulation can contain several million elements and still finish overnight on a computing cluster.
• Mathematician Richard Courant used a very similar idea with triangles in 1943, before computers made it practical.

Key Takeaways

• FEM divides a structure into elements, writes simple equations for each and assembles them into \( [K]\{u\} = \{F\} \).
• The bar element stiffness is \( EA/L \) times the matrix with rows \( (1, -1) \) and \( (-1, 1) \).
• Boundary conditions must be applied before solving, or the stiffness matrix is singular.
• Refining the mesh converges toward the exact solution; always check results against hand calculations.

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