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Finite Element Methods · Chapter 4 of 10 · Advanced

Beam Elements

A beam node carries a deflection and a rotation, so the element grows to a 4-by-4 stiffness matrix. Built from the cubic deflection of bending, a single element reproduces the cantilever exactly.

01

Readiness check

This chapter adds bending degrees of freedom. Tick only what you can do closed-notes.

  • Recall beam bending and EI.
  • Recall the cantilever deflection PL³/3EI.
  • Multiply a 2-by-2 matrix by a vector.
  • Distinguish a deflection from a rotation (slope).
  • Apply a fixed (clamped) boundary condition.
0 or 1 weak itemsContinue with this chapter.
2 weak itemsRevisit beam deflection in Mechanics of Materials, Chapter 9.
3 or more weak itemsReview bending stress in Mechanics of Materials, Chapter 4.
02

The core idea

Each beam node has two degrees of freedom, a transverse deflection and a rotation, so a two-node element has four. Its 4-by-4 stiffness, built from the cubic bending shape, exactly captures end loads.

DOF: v₁, θ₁, v₂, θ₂k entries: 12EI/L³, 6EI/L², 4EI/L, 2EI/Lcantilever: δ = PL³/3EI, θ = PL²/2EI

Bending links a transverse load not only to deflection but to slope, so each beam node carries two degrees of freedom: a transverse displacement v and a rotation θ. A two-node beam element therefore has a 4-by-4 stiffness matrix, whose entries 12EI/L³, 6EI/L², 4EI/L, and 2EI/L come from integrating the cubic deflection that Euler-Bernoulli bending predicts. Because that cubic is the exact deformed shape for end loads and moments, a single beam element reproduces the classic cantilever result δ = PL³/3EI exactly, with no mesh refinement needed. Assembling beam elements and recovering moments and shear gives the full bending response of a frame.

The skill works when: you track both deflection and rotation at each node and use the 4-by-4 element.
The skill breaks down when: rotations are dropped, collapsing the beam to a bar that cannot bend.
The concept. A beam element bends into a cubic shape. Each node has a transverse deflection v and a rotation θ, giving four degrees of freedom and a 4-by-4 element stiffness.
03

The skills, taught in order

Five skills set the beam degrees of freedom, the stiffness matrix, assembly, the cantilever solution, and recovery.

4.1 Beam degrees of freedom

Bending couples force to both deflection and slope, so each node has two degrees of freedom: a transverse displacement v and a rotation θ. The element's four DOF are v₁, θ₁, v₂, θ₂, and its stiffness matrix is 4-by-4.

4.2 The beam stiffness matrix

Integrating the cubic Hermite shape functions over the element gives the stiffness entries: 12EI/L³ couples the two deflections, 6EI/L² couples deflection to rotation, and 4EI/L and 2EI/L are the rotational stiffnesses. Together they form the standard symmetric 4-by-4 matrix.

EntryExpressionCouples
Translational12EI/L³deflection to deflection
Cross6EI/L²deflection to rotation
Rotational (direct)4EI/La node's own rotation
Rotational (coupling)2EI/Lrotation to the far node

4.3 Assembling beams

Beam elements assemble like bars, but each shared node now contributes two rows and columns. Continuous beams and frames build up by adding these 4-by-4 matrices at the shared deflection and rotation degrees of freedom.

4.4 The cantilever solution

Clamping one node sets both its deflection and rotation to zero. Solving the reduced 2-by-2 system for the free node returns the exact cantilever results δ = PL³/3EI and θ = PL²/2EI, because the element's cubic shape matches the true deflection for an end load.

4.5 Recovering moments and shear

From the nodal deflections and rotations, the bending moment and shear within an element follow from the second and third derivatives of the deflection. These let bending stress σ = Mc/I be recovered and checked against beam theory.

Engineering connection: combining bar and beam degrees of freedom gives the frame element, and the same Hermite functions appear in the shape-function chapter next.

04

Worked example 1: the beam stiffness entries

A steel beam element has E = 200 GPa, second moment of area I = 4×10⁶ mm⁴, and length L = 1 m. Compute the four characteristic entries of its stiffness matrix.

Figure 1. The four entries scale with EI and inverse powers of L, so a longer or softer beam is far less stiff in bending; the translational term falls with L cubed.
  1. ProblemFind the four stiffness entries for the beam element in Figure 1.
  2. Given / findE = 200 000 MPa, I = 4×10⁶ mm⁴, L = 1000 mm. Find 12EI/L³, 6EI/L², 4EI/L, 2EI/L.
  3. AssumptionsEuler-Bernoulli beam; consistent units (N, mm).
  4. ModelCompute EI, then evaluate each entry.
  5. EquationsEI = 200 000 × 4×10⁶ = 8×10¹¹ N·mm²entries: 12EI/L³, 6EI/L², 4EI/L, 2EI/L
  6. Solve12EI/L³ = 12(8×10¹¹)/10⁹ = 9600 N/mm. 6EI/L² = 4.8×10⁶. 4EI/L = 3.2×10⁹. 2EI/L = 1.6×10⁹ (N·mm units for the rotational terms).
  7. CheckThe translational stiffness 9600 N/mm is far softer than the bar's AE/L would be, as bending should be; the rotational entries are much larger numerically because they carry moment per radian. The 4:2 ratio of the rotational terms is the standard beam signature.
  8. ConclusionThese four numbers populate the symmetric 4-by-4 element matrix. Their dependence on L (cubed, squared, linear) is why beam stiffness is so length-sensitive.
Result. 12EI/L³ = 9600 N/mm; 6EI/L² = 4.8×10⁶; 4EI/L = 3.2×10⁹; 2EI/L = 1.6×10⁹.
05

Worked example 2: a cantilever from one element

The same beam (EI = 8×10¹¹ N·mm², L = 1 m) is clamped at one end and loaded by 1000 N at the free end. Using a single beam element, find the tip deflection and rotation, and compare to beam theory.

Figure 2. A single beam element gives the exact cantilever deflection and rotation, because its cubic shape function matches the true deformed shape for an end load.
  1. ProblemFind the tip deflection and rotation of the cantilever in Figure 2.
  2. Given / findEI = 8×10¹¹ N·mm², L = 1000 mm, P = 1000 N. Find δ and θ.
  3. AssumptionsClamped end (v₁ = θ₁ = 0); single beam element; end load only.
  4. ModelReduce the 4-by-4 to the free node's 2-by-2 and solve, or use the closed-form results the element reproduces.
  5. Equationsδ = PL³/3EIθ = PL²/2EI
  6. Solveδ = 1000(1000)³/(3 × 8×10¹¹) = 10¹²/2.4×10¹² = 0.417 mm. θ = 1000(1000)²/(2 × 8×10¹¹) = 10⁹/1.6×10¹² = 6.25×10⁻⁴ rad.
  7. CheckThese equal the mechanics-of-materials cantilever formulas exactly, so the single-element FEM and beam theory agree to the digit. No mesh refinement improves them, because the element is already exact for this load.
  8. ConclusionFor end loads, one beam element is exact, the ideal validation case. Distributed loads need the work-equivalent nodal loads and benefit from more elements.
Result. δ = 0.417 mm, θ = 6.25×10⁻⁴ rad, matching beam theory exactly.
06

Misconceptions and diagnostics

MistakeSymptomDiagnostic questionCorrection
Dropping rotationsBeam cannot bend correctly"Does each node have a rotation DOF?"Keep both v and θ at every beam node.
Clamping only the deflectionCantilever rotates at the wall"Is the slope fixed too?"A clamped end fixes both v and θ.
Lumping a distributed load at a nodeWrong deflection for UDL"Did I use work-equivalent nodal loads?"Convert distributed loads to consistent nodal forces and moments.
Mixing unit systemsEntries off by powers of ten"Are E, I, L all in N and mm?"Keep EI in N·mm² and L in mm.
07

Practice ladder

Level 1 · Direct skill

For EI = 8×10¹¹ N·mm² and L = 2000 mm, find the translational stiffness 12EI/L³.

Show answer

12EI/L³ = 12(8×10¹¹)/(2000)³ = 9.6×10¹²/8×10⁹ = 1200 N/mm, one eighth of the L = 1000 mm value.

Level 2 · Mixed concept

The Worked Example 2 cantilever is loaded with 2500 N. What is the tip deflection?

Show answer

δ scales with load: δ = 0.417 × (2500/1000) = 1.04 mm. Linear response.

Level 3 · Independent problem

A cantilever has EI = 5×10¹¹ N·mm², L = 800 mm, P = 600 N. Find the tip deflection.

Show answer

δ = PL³/3EI = 600(800)³/(3 × 5×10¹¹) = 600 × 5.12×10⁸/1.5×10¹² = 3.072×10¹¹/1.5×10¹² = 0.205 mm.

Level 4 · Transfer to real engineering

Take a real beam or bracket, model it with beam elements, predict the tip deflection, and state the hand-calc you would validate it against.

What good work looks like

A beam model with deflection and rotation DOF, a predicted deflection, and a closed-form benchmark (PL³/3EI or 5wL⁴/384EI) for validation.

08

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"Check that I kept both deflection and rotation DOF at each node."
"Give me three beams; I will compute their 12EI/L³ stiffness."
"Solve this cantilever." Reducing and solving the element yourself is the skill.
"Is the deflection right?" Comparing to PL³/3EI is the point.

Portfolio task

Model a beam with one or more beam elements: form the stiffness, clamp the support, solve for tip deflection and rotation, and validate against the closed-form beam formula.

Must include: a 4-by-4 element stiffness, a solved tip deflection, and a beam-theory comparison.
09

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. How many DOF does a beam node have?

Two: a transverse deflection v and a rotation θ.

2. Name the four stiffness entries.

12EI/L³, 6EI/L², 4EI/L, and 2EI/L.

3. What does a clamped end fix?

Both the deflection and the rotation at that node.

4. Why is one element exact for a cantilever end load?

The cubic shape function matches the true deflected shape for end loads.

5. How are moments recovered?

From derivatives of the deflection, giving M and then σ = Mc/I.

TodayFinish this quiz and Levels 1 and 2 of the ladder.
+1 dayRe-derive the stiffness entries and cantilever result from a blank page.
+3 daysSolve two new beam cases.
+7 daysCarry the shape functions into Chapter 5, interpolation.
+30 daysReuse the cantilever as a validation benchmark for beam FEA.
10

Textbook mapping

ItemMapping
Primary sourceLogan, A First Course in the Finite Element Method, Chapter 4 (Development of Beam Equations)
Cross-referenceHutton, Ch. 4 · Mechanics of Materials, Ch. 9
Core topics4.1 Beam DOF · 4.2 Stiffness matrix · 4.3 Assembly · 4.4 Cantilever solution · 4.5 Moments and shear
Engineering connectionBar plus beam DOF gives the frame element; the Hermite functions return next.
Read nextChapter 5: Shape Functions and Interpolation.