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Finite element basics

Advanced and Design Structural Analysis Reviewed 24 Oct 2026 2 min read

The finite element method breaks a structure into small pieces, the elements, joined at nodes. Within each element the displacement is assumed to follow a simple shape, and from that assumption the element's stiffness matrix is derived. Assembling all the element matrices gives a global stiffness matrix K linking nodal forces F to nodal displacements u, so the whole analysis reduces to solving K u = F.

The simplest element is a bar carrying axial force, with stiffness A E over L and two degrees of freedom. Beam elements add rotations and bending stiffness; plate and shell elements model slabs, walls and tanks; solid elements model foundations, dams and soil. Boundary conditions are applied by fixing the restrained degrees of freedom, and loads by placing forces on nodes, with distributed loads converted to equivalent nodal loads.

Diagram: Indiacon original, not to scale.

Accuracy depends on the element type and the mesh. A coarse mesh with simple elements is usually too stiff, giving deflections that are too small. Refining the mesh, or using higher-order elements, makes results converge towards the true solution; a mesh convergence check, re-running with a finer mesh until key results stop changing, is basic good practice. Stress concentrations at re-entrant corners and point supports never fully converge and must be interpreted, not read blindly.

Software makes finite element analysis routine, which makes checking more important, not less. Equilibrium of reactions against applied loads, a hand estimate of the main moment or deflection, and inspection of the deformed shape catch most modelling errors. Units, local axes, release conditions and load directions are the commonest sources of mistakes.

Tip

Before trusting any stress plot, check that reactions balance the applied loads.

Key formulas
  • Bar element: k = (A E / L) [[1, −1], [−1, 1]]
  • Global system: K u = F
  • Element force from displacements: N = (A E / L)(u_j − u_i)
Common mistake

Reading peak stresses at point supports or sharp corners as real design values.

Solved example

Problem. A stepped bar is fixed at its left end. Element 1: A = 200 mm², L = 1 000 mm. Element 2: A = 100 mm², L = 500 mm. E = 200 GPa. A 10 kN pull acts at the free right end. Find the nodal displacements by assembling the stiffness matrix.

  1. k₁ = 200 × 200 000 / 1 000 = 40 000 N/mm; k₂ = 100 × 200 000 / 500 = 40 000 N/mm.
  2. After fixing node 0, the free nodes 1 and 2 give K = [[k₁ + k₂, −k₂], [−k₂, k₂]] = [[80 000, −40 000], [−40 000, 40 000]] and F = [0, 10 000].
  3. Solving: u₁ = 0.25 mm and u₂ = 0.50 mm.
  4. Hand check: Σ P L / (A E) = 10 000 × 1 000 / (200 × 200 000) + 10 000 × 500 / (100 × 200 000) = 0.25 + 0.25 = 0.50 mm. It matches.

Answer. u₁ = 0.25 mm, u₂ = 0.50 mm

Quick quiz

1Is a coarse mesh of simple elements usually too stiff or too flexible?

Too stiff; deflections come out too small.

2What is a mesh convergence check?

Re-running with a finer mesh until key results stop changing.

3Name one quick check of an FE model.

Sum of reactions equals applied load, or a hand estimate of moment or deflection.

References and further reading

Standard texts, listed in no order of merit. We do not copy text, figures, tables or solved problems from any book; our explanations and examples are original.

  • R. C. Hibbeler, Structural Analysis, Pearson, latest edition.
  • C. S. Reddy, Basic Structural Analysis, McGraw Hill Education (India), latest edition.
  • Devdas Menon, Structural Analysis, Narosa, latest edition.
  • A. K. Chopra, Dynamics of Structures, Pearson, latest edition.

Free and official sources

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