🌀Eigenvalue Problems

Vibration Modes and Generalised Eigenproblems

Kφ = ω²Mφ, mass-orthogonal modes, modal superposition, shear buildings and footbridges

London, 10 June 2000

The Millennium Bridge, a slender steel suspension footbridge across the Thames between St Paul's Cathedral and Tate Modern, opens to the public. Its designers have called it a 'blade of light'. Between 80,000 and 100,000 people cross it that day. Within minutes of the crowds arriving, the deck begins to sway sideways, slowly, about once a second — enough that pedestrians stop walking and grab the handrails. The bridge is closed on 12 June. The engineers' investigation found the cause in a mode they had not been worried about: a lateral mode of the span at around 1 Hz. A walking person pushes slightly sideways with each step, alternately left and right — a sideways force at half the pace frequency, about 1 Hz. When the deck began to sway a little, pedestrians unconsciously adjusted their gait to keep their balance, falling into step with the motion and pushing it harder. The more people, the larger the sway. The fix cost about £5 million: dozens of viscous dampers and tuned mass dampers to absorb energy in the troublesome modes. The bridge reopened in February 2002 and has been still ever since. Finding those modes before construction is an eigenvalue problem — but not quite the one of the last three lessons.

For an undamped structure with mass matrix

M

and stiffness matrix

K

, free vibration obeys

M\ddot{\mathbf x} + K\mathbf x = \mathbf 0

. Try a solution in which every point moves in the same shape at one frequency,

\mathbf x(t) = \boldsymbol\phi\cos\omega t

K\boldsymbol\phi = \omega^2 M\boldsymbol\phi

The generalised eigenvalue problem. Eigenvalues ω² give natural frequencies f = ω/2π; eigenvectors φ give mode shapes. For a structure, K and M are both symmetric and M is positive definite.

💡Solving Kφ = ω²Mφ

Never form

M^{-1}K

— it is not symmetric, and symmetric algorithms are faster and more accurate. Instead use the Cholesky factor

M = LL^{\mathsf T}

(lesson 3-3): with

\boldsymbol\psi = L^{\mathsf T}\boldsymbol\phi

, the problem becomes

(L^{-1}KL^{-\mathsf T})\,\boldsymbol\psi = \omega^2\boldsymbol\psi,

an ordinary symmetric eigenproblem. This is what scipy.linalg.eigh(K, M) does. For large sparse models, scipy.sparse.linalg.eigsh(K, k, M, sigma=0) applies shift-and-invert Lanczos (lesson 6-2) to the generalised problem directly, never touching most of the spectrum.

💡Mass-orthogonality and modal coordinates

The mode shapes are orthogonal with respect to the mass matrix, and can be scaled so that

\Phi^{\mathsf T}M\Phi = I, \qquad \Phi^{\mathsf T}K\Phi = \Omega^2 = \operatorname{diag}(\omega_1^2, \dots, \omega_n^2).

Substitute

\mathbf x = \Phi\mathbf q

M\ddot{\mathbf x} + K\mathbf x = \mathbf f(t)

and multiply by

\Phi^{\mathsf T}

: the coupled system falls apart into

n

independent single-degree-of-freedom oscillators,

\ddot q_i + \omega_i^2 q_i = \boldsymbol\phi_i^{\mathsf T}\mathbf f(t).

This is modal superposition, the backbone of structural dynamics: solve each oscillator (often analytically), and add up. Usually a few modes suffice, because loads like earthquakes and wind excite mainly the lowest ones.

Derivation

Why the modes are mass-orthogonal

Take two modes,

K\boldsymbol\phi_i = \omega_i^2M\boldsymbol\phi_i

and

K\boldsymbol\phi_j = \omega_j^2M\boldsymbol\phi_j

. Multiply the first on the left by

\boldsymbol\phi_j^{\mathsf T}

and the second by

\boldsymbol\phi_i^{\mathsf T}

\boldsymbol\phi_j^{\mathsf T}K\boldsymbol\phi_i = \omega_i^2\,\boldsymbol\phi_j^{\mathsf T}M\boldsymbol\phi_i, \qquad \boldsymbol\phi_i^{\mathsf T}K\boldsymbol\phi_j = \omega_j^2\,\boldsymbol\phi_i^{\mathsf T}M\boldsymbol\phi_j.

K

and

M

are symmetric, so the two left-hand sides are the same number, and so are the two

M

products. Subtract:

0 = (\omega_i^2 - \omega_j^2)\,\boldsymbol\phi_j^{\mathsf T}M\boldsymbol\phi_i.

For two different frequencies the bracket is not zero, so

\boldsymbol\phi_j^{\mathsf T}M\boldsymbol\phi_i = 0

— and then the first equation gives

\boldsymbol\phi_j^{\mathsf T}K\boldsymbol\phi_i = 0

as well. Scaling each mode so that

\boldsymbol\phi_i^{\mathsf T}M\boldsymbol\phi_i = 1

turns the first equation, with

j = i

, into

\boldsymbol\phi_i^{\mathsf T}K\boldsymbol\phi_i = \omega_i^2

. That is exactly

\Phi^{\mathsf T}M\Phi = I

and

\Phi^{\mathsf T}K\Phi = \Omega^2

: in modal coordinates both matrices are diagonal, which is what lets the equations of motion fall apart.

📐A three-storey frame

Floor masses 400, 350 and 300 t; storey stiffnesses 300, 260 and 200 MN/m. eigh(K, M) gives:

mode 1: 2.042 Hz ( $T = 0.490$ s), shape $(0.40,\ 0.75,\ 1)$ — all floors one way;
mode 2: 5.117 Hz, shape $(-0.98,\ -0.55,\ 1)$ — one node;
mode 3: 7.435 Hz, shape $(1.89,\ -2.27,\ 1)$ — two nodes.

\Phi^{\mathsf T}M\Phi = I

to rounding, and mode 1 carries 88.5% of the total mass as effective modal mass: it alone captures most of the response to ground shaking. Released from a pushed-over shape, the roof displacement 1.5 s later is 44.94 mm from mode 1 alone, 44.35 mm with two modes and 44.34 mm with three — exactly the reference solution, computed with the matrix exponential that lesson 9-6 will explain.

Derivation

Modal starting values and effective modal mass

Starting values. A displacement

\mathbf x_0

is a combination of the modes,

\mathbf x_0 = \Phi\mathbf q_0

. Multiply by

\Phi^{\mathsf T}M

and use

\Phi^{\mathsf T}M\Phi = I

\mathbf q_0 = \Phi^{\mathsf T}M\,\mathbf x_0.

Released from rest, each modal oscillator then moves as

q_i(t) = q_{0i}\cos\omega_it

, and

\mathbf x(t) = \sum_i\boldsymbol\phi_i\,q_{0i}\cos\omega_it

— the lab's sum over 1, 2 and 3 modes.

Ground shaking. If the ground accelerates by

a_g(t)

, every floor feels the inertial force of its own mass,

-M\mathbf 1\,a_g

, where

\mathbf 1

is a vector of ones. In modal coordinates each oscillator is driven by

-\Gamma_i a_g

, with the participation factor

\Gamma_i = \boldsymbol\phi_i^{\mathsf T}M\mathbf 1

. The base force a mode produces is proportional to

\Gamma_i^2

, which is called its effective modal mass.

Why the effective masses add up to the total mass. Write the rigid shape

\mathbf 1

itself in modes:

\mathbf 1 = \Phi\mathbf c

with

\mathbf c = \Phi^{\mathsf T}M\mathbf 1

, so

c_i = \Gamma_i

. Then

\mathbf 1^{\mathsf T}M\mathbf 1 = \mathbf c^{\mathsf T}\Phi^{\mathsf T}M\Phi\,\mathbf c = \sum_i\Gamma_i^2.

The left side is the total mass. Each mode takes a share, and 88.5% for mode 1 means one oscillator carries most of the building's response to an earthquake.

Philosophy is written in this grand book, which stands continually open before our eyes (I say the 'Universe'), but can not be understood without first learning to comprehend the language and know the characters as it is written. It is written in mathematical language

– Galileo Galilei, Il Saggiatore (1623)

📐A footbridge in the walking band

Model a 144 m footbridge span as a simply supported beam in lateral bending, with illustrative values

EI = 1.6 \times 10^{11}

N m² and

\mu = 2000

kg/m. Forty beam finite elements with consistent mass matrices (their stiffness and mass matrices are supplied in the lab; lesson 10-5 derives them) give lateral frequencies 0.678, 2.710, 6.098 and 10.841 Hz — identical to four digits with beam theory,

f_n = \frac{n^2\pi}{2L^2}\sqrt{EI/\mu}

. The first mode lies inside the 0.5–1.2 Hz band where pedestrians' sideways forces act. Modern footbridge guidance requires exactly this check, and either a stiffer design or added damping when a lateral mode falls in the band.

🕰Why 1 Hz matters

People walk at about 2 steps per second, so vertical forces come at ~2 Hz and lateral forces — left foot, right foot — at ~1 Hz. Many footbridges had swayed before London's (the Solférino bridge in Paris in 1999, the Auckland Harbour Bridge during a march in 1975), but the Millennium Bridge was the case that made engineers study 'synchronous lateral excitation' systematically. Arup's research after the closure showed the effect switches on above a critical number of pedestrians — a stability threshold, which is itself an eigenvalue problem for the coupled bridge–crowd system.

💡Where you will use this

Structures and Earthquake Engineering: response-spectrum analysis combines modal responses; codes require enough modes to capture 90% of the mass. Aircraft and Spacecraft: launch-vehicle coupled loads analysis and flutter clearance start from the modes of the structure; spacecraft are designed so their lowest modes stay above the launcher's limits. Rotating machinery and Aircraft Engines: Campbell diagrams track blade and shaft modes against engine speed. Attitude Control: flexible solar arrays and antennas enter the control model as a few modal oscillators.

📚Further reading

Anil Chopra, Dynamics of Structures (5th ed., Pearson, 2017), chapters 10–13 — free vibration, modal analysis and earthquake response of multi-storey buildings, exactly as in this lesson.
Ray Clough and Joseph Penzien, Dynamics of Structures (2nd ed., McGraw-Hill, 1993) — the classic text on the generalised eigenproblem in structural dynamics.
Pat Dallard and co-authors, 'The London Millennium Footbridge', The Structural Engineer 79(22), 17–33 (2001) — Arup's account of the sway and its cure.

↓But there is a problem

Six units in, we can solve, fit and decompose: linear systems of any size, nonlinear equations, data that do not agree, matrices and their hidden directions. Everything so far has worked with numbers we were given. Real data are sparse. A precise GNSS ephemeris gives each satellite's position only every 15 minutes; a receiver needs it at the exact nanosecond a signal left. A rocket engineer has gas properties tabulated every 100 K; the chamber is at 3,412 K. A wind tunnel gives lift at a dozen angles of attack; the flight simulator needs every angle in between. Knowing a function between the points you measured sounds easy. In 1901 Carl Runge showed that the obvious way to do it can go spectacularly wrong.