Finite Element Methods · Chapter 9 of 10 · Advanced
Dynamics and Modal Analysis
Add mass and the structure can vibrate. A new mass matrix joins the stiffness, and an eigenvalue problem returns the natural frequencies and mode shapes that decide whether a design will resonate.
Readiness check
This chapter adds inertia. Tick only what you can do closed-notes.
- Recall the natural frequency √(k/m).
- Set up an eigenvalue problem.
- Compute a 2-by-2 determinant.
- Recall what a mode shape is.
- Convert rad/s to Hz.
The core idea
Dynamics adds a mass matrix to the stiffness: Mü + Ku = F. Free vibration turns this into an eigenvalue problem whose roots are the squared natural frequencies and whose vectors are the mode shapes.
Mü + Ku = F(t)(K − ω²M)φ = 0det(K − ω²M) = 0 gives the frequenciesA static analysis ignores inertia, but a vibrating structure cannot. The finite element method builds a mass matrix M alongside the stiffness K, and the equation of motion becomes Mü + Ku = F. With no forcing, the structure vibrates freely, and assuming harmonic motion turns the equation into the generalised eigenvalue problem (K − ω²M)φ = 0. Its eigenvalues ω² are the squared natural frequencies, the speeds at which the structure wants to oscillate, and its eigenvectors φ are the mode shapes, the patterns of motion at each frequency. For a single degree of freedom this reduces to the familiar √(k/m); for many degrees of freedom it gives a frequency and shape for each. Knowing these tells a designer which excitation frequencies to avoid.
The skills, taught in order
Five skills set the dynamic equation, the mass matrix, the eigenvalue problem, the frequencies, and the mode shapes.
9.1 The dynamic equation
Including inertia gives the equation of motion Mü + Ku = F, where M is the mass matrix, K the stiffness, and ü the nodal accelerations. The static problem is the special case with no acceleration. Damping, when included, adds a Cu̇ term.
9.2 The mass matrix
The mass matrix can be lumped, placing each element's mass at its nodes to give a diagonal matrix, or consistent, derived from the same shape functions as the stiffness. Lumped masses are simpler and often adequate; consistent masses are more accurate for higher modes.
9.3 The eigenvalue problem
Free, undamped vibration assumes harmonic motion u = φ sin ωt, which substituted into Mü + Ku = 0 gives (K − ω²M)φ = 0. A nontrivial solution requires det(K − ω²M) = 0, the characteristic equation for the natural frequencies.
9.4 Natural frequencies
The roots ω² of the characteristic equation are the squared natural frequencies; there are as many as degrees of freedom. The lowest, the fundamental frequency, usually matters most, since it is the easiest to excite. Frequencies convert to hertz by f = ω/2π.
9.5 Mode shapes
Each natural frequency has a mode shape φ, the relative pattern of nodal motion when the structure vibrates at that frequency. Mode shapes show where motion concentrates, guiding where to stiffen or add mass to shift a frequency away from an excitation.
Engineering connection: matching these natural frequencies against operating speeds is how rotating machinery and structures are kept clear of resonance.
Worked example 1: a single natural frequency
A single-degree-of-freedom model has stiffness k = 1×10⁵ N/m and mass m = 10 kg. Find its natural frequency in rad/s and in hertz from the eigenvalue problem.
- ProblemFind the natural frequency of the system in Figure 1.
- Given / findk = 1×10⁵ N/m, m = 10 kg. Find ωn and f.
- AssumptionsSingle degree of freedom; undamped free vibration.
- ModelThe eigenvalue problem (k − ω²m)φ = 0 gives ω² = k/m.
- Equationsωn = √(k/m)f = ωn/2π
- Solveωn = √(1×10⁵/10) = √(10⁴) = 100 rad/s. f = 100/2π = 15.9 Hz.
- CheckThe units of √(N/m / kg) = √(1/s²) = 1/s = rad/s, as required. A stiffer or lighter system would vibrate faster.
- ConclusionFor one degree of freedom the eigenvalue problem is just √(k/m). The same logic, in matrix form, gives every natural frequency of a large model.
Worked example 2: a two-degree-of-freedom system
Two equal masses m = 1 kg are connected by three springs of stiffness k = 100 N/m (wall, mass, mass, wall), giving K = [200, −100; −100, 200] and M = the identity. Find the two natural frequencies and describe the mode shapes.
- ProblemFind the two natural frequencies and mode shapes for the system in Figure 2.
- Given / findm = 1 kg, k = 100 N/m, K = [200, −100; −100, 200], M = I. Find ω₁, ω₂ and the shapes.
- AssumptionsUndamped free vibration; equal masses and springs.
- ModelSolve det(K − ω²M) = 0 for ω², then find each eigenvector.
- Equationsdet([200 − ω², −100; −100, 200 − ω²]) = 0(200 − ω²)² = 100²
- Solve200 − ω² = ±100, so ω² = 100 or 300. ω₁ = √100 = 10 rad/s (φ₁ = φ₂, masses in phase); ω₂ = √300 = 17.3 rad/s (φ₁ = −φ₂, masses out of phase).
- CheckThe lower mode does not stretch the middle spring (both masses move together), so it is softer; the higher mode stretches it most, so it is stiffer, consistent with ω₂ > ω₁. Two degrees of freedom give two modes.
- ConclusionThe eigenvalue problem returns a frequency and a shape for each degree of freedom. The mode shapes reveal which springs are working hardest at each frequency.
Misconceptions and diagnostics
| Mistake | Symptom | Diagnostic question | Correction |
|---|---|---|---|
| Omitting the mass matrix | No frequencies predicted | "Did I build M as well as K?" | Dynamics needs both M and K. |
| Confusing ω and f | Frequency off by 2π | "Is this rad/s or Hz?" | f = ω/2π. |
| Expecting one frequency | Higher modes ignored | "How many degrees of freedom?" | There is a natural frequency per DOF. |
| Ignoring mode shapes | Stiffening the wrong location | "Where does this mode move most?" | Read the mode shape before modifying the design. |
Practice ladder
Find the natural frequency of a system with k = 4×10⁴ N/m and m = 4 kg.
Show answer
ωn = √(k/m) = √(10⁴) = 100 rad/s, or 15.9 Hz.
For the Worked Example 1 system, what mass would halve the natural frequency to 50 rad/s?
Show answer
ω² = k/m, so halving ω quarters ω², needing four times the mass: m = 40 kg.
For the Worked Example 2 system, what is the fundamental frequency in hertz?
Show answer
f₁ = ω₁/2π = 10/2π = 1.59 Hz, the lowest and most easily excited frequency.
For a real component, explain how you would use its natural frequencies and mode shapes to avoid resonance with a known excitation.
What good work looks like
Comparing the fundamental frequency to the excitation, and using the mode shape to decide where to add stiffness or mass to shift the frequency clear.
Working with AI, and proving it yourself
Use AI as an examiner, not a solver
Portfolio task
Set up a dynamic FEM for a small system: assemble M and K, solve the eigenvalue problem for the natural frequencies and mode shapes, and identify the fundamental mode.
Retrieval and spaced review
Closed notes. Answer out loud, then reveal.
1. Write the dynamic equation.
Mü + Ku = F, adding the mass matrix to the stiffness.
2. What is the free-vibration eigenvalue problem?
(K − ω²M)φ = 0, from assuming harmonic motion.
3. How are natural frequencies found?
From det(K − ω²M) = 0; there is one per degree of freedom.
4. What is a mode shape?
The eigenvector φ, the relative motion pattern at a natural frequency.
5. Lumped versus consistent mass?
Lumped is diagonal and simple; consistent comes from the shape functions and is more accurate for higher modes.
Textbook mapping
| Item | Mapping |
|---|---|
| Primary source | Logan, A First Course in the Finite Element Method, Chapter 16 (Structural Dynamics) |
| Cross-reference | Hutton, Ch. 10 · Dynamics, Ch. 10 |
| Core topics | 9.1 Dynamic equation · 9.2 Mass matrix · 9.3 Eigenvalue problem · 9.4 Natural frequencies · 9.5 Mode shapes |
| Engineering connection | Natural frequencies and modes keep machinery clear of resonance. |
| Read next | Chapter 10: Convergence, Errors, and Modeling. |