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Matrix stiffness method

Intermediate Structural Analysis Reviewed 24 Oct 2026 2 min read

The matrix stiffness method is how almost all structural analysis software works. It treats a structure as an assembly of members (elements) connected at joints (nodes). The unknowns are the displacements and rotations at the nodes. For each element, a stiffness matrix relates the forces at its ends to the displacements at its ends. These are transformed from local member axes to common global axes and added into one structure stiffness matrix K. Then K d = F, where F holds the loads, is solved for the displacements d, and member forces follow.

A two-dimensional truss member has two degrees of freedom per node, giving a 4 × 4 matrix based on A E / L. A plane frame member has three per node (two translations and one rotation), giving a 6 × 6 matrix with axial terms A E / L and bending terms 12 E I / L³, 6 E I / L² and 4 E I / L or 2 E I / L, the same coefficients seen in the slope-deflection method. Supports are applied by removing restrained degrees of freedom from the system. Loads between nodes are converted to equivalent nodal loads using fixed-end forces.

Diagram: Indiacon original, not to scale.

The method is systematic, which makes it perfect for computers, but it can also produce nonsense quickly if the model is wrong. Engineers check that reactions balance the loads, that deflected shapes look sensible and that a few key results agree with hand methods. The same framework extends to finite elements, second-order effects and dynamics.

Key formulas
  • K d = F
  • Truss member (local): k = (AE/L) [[1, −1], [−1, 1]]
  • Beam member terms: 12EI/L³, 6EI/L², 4EI/L, 2EI/L
  • Global: K = Σ Tᵀ k T
Solved example

Problem. Two springs (bars) in series: bar 1 from a fixed support to node 1 with stiffness 2 000 kN/m, bar 2 from node 1 to node 2 with 1 000 kN/m. A 30 kN load acts at node 2. Find node displacements.

  1. Structure stiffness: K = [[3 000, −1 000], [−1 000, 1 000]] kN/m.
  2. Equations: 3 000 d₁ − 1 000 d₂ = 0; −1 000 d₁ + 1 000 d₂ = 30.
  3. Adding: 2 000 d₁ = 30, so d₁ = 0.015 m; d₂ = 0.045 m.
  4. Bar forces: bar 1 = 2 000 × 0.015 = 30 kN; bar 2 = 1 000 × 0.030 = 30 kN. Equilibrium checks.

Answer. d₁ = 15 mm, d₂ = 45 mm, both bars carry 30 kN

Quick quiz

1What are the unknowns in the stiffness method?

Nodal displacements and rotations.

2How many degrees of freedom per node in a plane frame?

Three.

3Name one check on software results.

Reactions should balance the applied loads.

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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