Finite Element Methods · Chapter 2 of 10 · Intermediate
The Direct Stiffness Method: Spring Elements
The whole machinery of FEM appears in its simplest setting: springs. Assemble their stiffness matrices, fix a support, solve for displacements, and recover the forces. Every later element follows this exact recipe.
Readiness check
This chapter is the direct stiffness method in miniature. Tick only what you can do closed-notes.
- Recall a spring force F = kx.
- Solve a 2-by-2 linear system.
- Add matrices entry by entry.
- Apply a fixed boundary condition.
- Check that forces are in equilibrium.
The core idea
Each spring contributes a 2-by-2 stiffness matrix to the nodes it connects. Overlapping these at shared nodes assembles the global K; fixing a support and solving Ku = F gives the displacements.
k(e) = [k −k; −k k]K assembled by adding at shared DOFsolve reduced Ku = F, then recover R and fA spring between two nodes relates its two nodal forces to its two displacements through the element matrix [k, −k; −k, k]. Assembly is superposition: each element matrix is added into the global stiffness at the rows and columns of the nodes it touches, so a shared node accumulates contributions from every element meeting there. The global system Ku = F is singular until a support is applied, because the structure could float freely; fixing a degree of freedom removes that rigid-body motion and lets the reduced system be solved. From the displacements, the reaction at the support and the force in each element follow directly, and they must balance the applied load, the built-in check.
The skills, taught in order
Five skills cover the element matrix, assembly, boundary conditions, solving, and recovery.
2.1 The spring element matrix
For a spring of stiffness k connecting nodes i and j, the element stiffness is [k, −k; −k, k], relating the forces at i and j to their displacements. The negative off-diagonals say a displacement at one node pulls on the other.
2.2 Assembly by superposition
Each element matrix is added into the global K at the locations of the nodes it connects. A node shared by two springs accumulates both diagonal contributions, so its global diagonal becomes the sum of the connected stiffnesses. This direct addition is why the method is called direct stiffness.
2.3 Boundary conditions
Without supports the global K is singular: the assembly allows rigid-body motion. Fixing a degree of freedom (setting its displacement to zero) removes the corresponding row and column, leaving a reduced, solvable system. At least enough supports must be applied to prevent free motion.
2.4 Solving the reduced system
The reduced Ku = F is solved for the unknown displacements, by hand for small systems and by elimination or iteration for large ones. The displacements are the primary unknowns from which everything else is recovered.
2.5 Recovering reactions and element forces
Substituting the displacements back into the full system gives the reactions at the supports, and each element force is k times the relative displacement of its nodes. These must satisfy equilibrium with the applied load, the verification step built into every analysis.
Engineering connection: this assemble-constrain-solve-recover recipe is identical for bars, beams, and two-dimensional elements; only the element matrix grows.
Worked example 1: assembling and solving two springs
Two springs in series connect three nodes: k₁ = 1000 N/mm (nodes 1 to 2) and k₂ = 2000 N/mm (nodes 2 to 3). Node 1 is fixed and a 3000 N force acts at node 3. Find the displacements of nodes 2 and 3.
- ProblemFind u₂ and u₃ for the spring system in Figure 1.
- Given / findk₁ = 1000, k₂ = 2000 N/mm, node 1 fixed, F₃ = 3000 N. Find u₂, u₃.
- AssumptionsLinear springs; one-dimensional motion along the axis.
- ModelAssemble the global K, delete the fixed row and column, and solve the reduced system.
- EquationsK = [1000 −1000 0; −1000 3000 −2000; 0 −2000 2000]reduced: [3000 −2000; −2000 2000][u₂; u₃] = [0; 3000]
- SolveFrom the reduced system, u₂ = 3 mm and u₃ = 4.5 mm.
- CheckIn series the equivalent stiffness is 1/(1/1000 + 1/2000) = 666.7 N/mm, so the end displacement is 3000/666.7 = 4.5 mm, matching u₃. The shared node 2 carries the assembled stiffness 3000.
- ConclusionAssembly, a fixed support, and a 2-by-2 solve produce the displacements. This is the full direct stiffness method in its smallest form.
Worked example 2: reactions and element forces
Using the displacements from Worked Example 1 (u₂ = 3 mm, u₃ = 4.5 mm), find the reaction at the fixed node 1 and the internal force in each spring, and confirm equilibrium.
- ProblemFind R₁ and the spring forces in Figure 2 and confirm equilibrium.
- Given / findu₁ = 0, u₂ = 3 mm, u₃ = 4.5 mm, k₁ = 1000, k₂ = 2000 N/mm, F₃ = 3000 N. Find R₁, f₁, f₂.
- AssumptionsLinear springs; reactions recovered from the full stiffness rows.
- ModelThe reaction is the first row of Ku; each element force is k times the relative displacement of its nodes.
- EquationsR₁ = k₁(u₁ − u₂) = −1000(3)f₁ = k₁(u₂ − u₁), f₂ = k₂(u₃ − u₂)
- SolveR₁ = −1000 × 3 = −3000 N (the support pushes back with 3000 N). f₁ = 1000 × 3 = 3000 N; f₂ = 2000 × (4.5 − 3) = 2000 × 1.5 = 3000 N.
- CheckBoth springs carry 3000 N, as series springs must, and the reaction magnitude equals the applied 3000 N, so equilibrium holds. The solution is validated.
- ConclusionRecovering reactions and element forces and checking equilibrium is the final, non-negotiable step. A solution that fails this check is wrong somewhere upstream.
Misconceptions and diagnostics
| Mistake | Symptom | Diagnostic question | Correction |
|---|---|---|---|
| No support applied | K is singular, no solution | "Can the structure float freely?" | Fix enough DOF to remove rigid-body motion. |
| Wrong assembly location | Stiffness placed at the wrong DOF | "Which nodes does this element connect?" | Add the element matrix at the rows and columns of its nodes. |
| Forgetting to share the node | Diagonal too small at a junction | "Do both elements meet at this node?" | The shared diagonal is the sum of connected stiffnesses. |
| Skipping the equilibrium check | An error goes undetected | "Do reactions balance the load?" | Recover reactions and confirm they sum to the applied force. |
Practice ladder
A single spring k = 500 N/mm has one end fixed and a 2000 N force on the other. Find the free-end displacement.
Show answer
u = F/k = 2000/500 = 4 mm.
For the Worked Example 1 system, what is the displacement of node 3 if the force is doubled to 6000 N?
Show answer
Linear, so displacements double: u₃ = 9 mm, u₂ = 6 mm. The equivalent stiffness is unchanged.
Two springs k₁ = 800 and k₂ = 800 N/mm in series, node 1 fixed, 1600 N at node 3. Find u₂ and u₃.
Show answer
Equal springs share the force: f = 1600 N each. u₂ = 1600/800 = 2 mm; u₃ = u₂ + 1600/800 = 4 mm.
Model a real series load path (a bolt and clamped plates, stacked mounts) as springs, assemble, and predict the displacement, then state the equilibrium check.
What good work looks like
A spring model with assembled stiffness, a displacement from Ku = F, and a reaction that balances the applied load.
Working with AI, and proving it yourself
Use AI as an examiner, not a solver
Portfolio task
Assemble and solve a small spring system end to end: build the global K, apply a support, solve for displacements, and recover reactions and element forces with an equilibrium check.
Retrieval and spaced review
Closed notes. Answer out loud, then reveal.
1. Write the spring element stiffness matrix.
[k, −k; −k, k] for the two connected nodes.
2. How is the global K assembled?
By adding each element matrix at the rows and columns of its nodes; shared nodes accumulate contributions.
3. Why is K singular without supports?
The structure can move as a rigid body; a support removes that motion.
4. How is an element force found?
f = k times the relative displacement of its two nodes.
5. What is the final check?
Reactions must balance the applied loads (equilibrium).
Textbook mapping
| Item | Mapping |
|---|---|
| Primary source | Logan, A First Course in the Finite Element Method, Chapter 2 (Introduction to the Stiffness Method) |
| Cross-reference | Hutton, Ch. 2 · Mechanics of Materials, Ch. 2 |
| Core topics | 2.1 Spring element matrix · 2.2 Assembly · 2.3 Boundary conditions · 2.4 Solving · 2.5 Reactions and element forces |
| Engineering connection | The same recipe scales to bars, beams, and 2D elements. |
| Read next | Chapter 3: Bar and Truss Elements. |