Module 9 · Lesson 9.2
The eigenvalue problem, solved by hand
Is there a deformed shape that, once set moving, keeps its shape? For two degrees of freedom the question is a quadratic, and the answer is two shapes.
Why this matters
A coupled system looks intractable: floor 1 pushes floor 2, which pushes back, and neither can be solved without the other.
The way out is to stop asking what each floor does and ask a different question: are there any special deformed SHAPES in which the whole structure vibrates as a unit, with every point reaching its maximum at the same instant and passing through zero at the same instant?
For a two-degree-of-freedom system there are exactly two such shapes, and finding them turns one coupled problem into two independent ones we already know how to solve.
By the end of this lesson you should be able to
- Set up the free-vibration eigenvalue problem
- Solve the resulting quadratic for the two natural frequencies
- Find the mode shapes and explain their arbitrary scaling
- Verify orthogonality and say what it makes possible
From first principles
The two-degree-of-freedom eigenvalue problem
We want to show: Find the frequencies and shapes in which a 2DOF system can vibrate freely without changing shape, by assuming such a motion exists and seeing what it requires.
Suppose the building vibrates so that both floors move sinusoidally at the same frequency and in step — reaching their extremes together and passing through zero together. Then the SHAPE of the deformation never changes; only its amplitude scales up and down. If we assume such a motion and substitute it into the equations, we will find that it is only possible at certain particular frequencies — and each of those comes with its own shape.
Worked example
Solving a two-storey building completely, by hand
Given
- m₁ = 320 000 kg, m₂ = 240 000 kg
- k₁ = k₂ = 50.3 MN/m (from the previous lesson)
Find
Both natural frequencies, both mode shapes, and a check on each.
Assumptions
- Undamped, for the eigenvalue problem
Decomposing an arbitrary motion
The modes are useful because ANY motion of the structure can be written as a combination of them:
u(t) = φ₁q₁(t) + φ₂q₂(t)
The q's are the modal coordinates — how much of each mode is present at time t.
Given an initial displacement u₀, how much of each mode does it contain? Orthogonality answers it immediately. Pre-multiply by φₙᵀM:
φₙᵀMu₀ = φₙᵀM(φ₁q₁ + φ₂q₂) = qₙ(φₙᵀMφₙ)
because every cross term vanishes. So:
qₙ(0) = φₙᵀMu₀ / (φₙᵀMφₙ)
That is what orthogonality is FOR. It lets each mode's contribution be extracted without solving anything — the modes act as an independent set of measuring instruments, each blind to the others.
And that is why modal analysis works
Substituting the modal expansion into the equations of motion and pre-multiplying by φₙᵀ decouples them completely. Each modal coordinate then satisfies
q̈ₙ + ωₙ²qₙ = (something known)
which is an SDOF equation — a problem solved in Module 3. The coupled two-degree-of-freedom problem has become two independent single-degree-of-freedom problems.
Module 12 does this properly, with damping and for any number of degrees of freedom. The mechanism is exactly the one above.
Predict first
A two-storey building is displaced so that its initial shape is EXACTLY proportional to mode 1, then released. What happens?
Practice
A 2DOF system has m₁ = m₂ = 1 000 kg and k₁ = k₂ = 100 kN/m. What is the first natural frequency, in rad/s?
Practice
For that system, what is the mode-1 shape ratio φ₂/φ₁?
Practice
A 2DOF system has mode shapes φ₁ = {1, 1.6} and φ₂ = {1, −0.9}, with m₁ = 4 000 kg and m₂ = 2 500 kg. Compute φ₁ᵀMφ₂ to check orthogonality. The answer should be near zero — give it in kg.
Check yourself
Why can only the RATIO of a mode shape's components be determined, and not their absolute values?
Check yourself
In the second mode of a uniform two-storey shear frame, how do the two floors move?
Worked example
Two floors, two frequencies
Given
- A two-storey frame, 100 tonnes at each floor
- Storey stiffness 40 MN/m at each level
Find
The two natural frequencies
Worked example
Beating, and why it is not resonance
Given
- A machine forcing a structure at 10.6 rad/s
- The structure's natural frequency is 10.0 rad/s
- Damping is very light
Find
What the response looks like, and its period
Summary
- Assume synchronous motion u = φ sin(ωt + θ) and the differential equation becomes algebraic
- (K − ω²M)φ = 0 requires det(K − ω²M) = 0 for a non-trivial answer
- 2DOF gives a quadratic in ω², hence two frequencies and two shapes
- Only the shape RATIO is determined — the magnitude is arbitrary
- Mode 1 has the floors moving together; mode 2 has them opposed
- φ₁ᵀMφ₂ = 0 is orthogonality, and it is a free check on any computed shapes
- Orthogonality lets each mode's contribution be extracted without solving anything
This is educational material. It uses simplified examples to teach principles, and must not be relied on for real design or safety-critical decisions. Module overview and checkpoint