Damped free vibration

Tidewater has designed a new fabricated baseplate for the TP-80, made of formed channel with welded cross members, to replace the cast iron base that customers found heavy and expensive. Before it is released, Lucie wants its first structural mode measured. Owen bolts a pump and motor to a prototype on the test floor, attaches an accelerometer at the pump-end feet and taps the base with an instrumented hammer. The trace on the analyzer rings at a little over 33 Hz and dies away in a fraction of a second. From that decay he must extract two numbers: the natural frequency and the damping. This lesson explains how, and why the damping value, small as it is, controls how violently the base will respond if anything ever drives it at that frequency.

The viscously damped oscillator

Adding a viscous damper with coefficient c to the mass-spring system gives

Trying \(x = Ce^{st}\) gives the characteristic equation \(ms^2 + cs + k = 0\), with roots

The damping ratio ζ compares the actual damping with the critical damping c_cr, the smallest value for which the system returns to rest without oscillating. Three cases follow. If ζ > 1, the system is overdamped: both roots are real and negative, and the mass creeps back to equilibrium. If ζ = 1, it is critically damped and returns fastest without overshoot; door closers and some instrument movements are designed near this value. If ζ < 1, the system is underdamped and oscillates within a decaying envelope:

The damped natural frequency ω_d is slightly lower than ω_n. For ζ = 0.05, the difference is only 0.13 percent, so for machinery structures the distinction rarely matters for frequency, but it matters a great deal for amplitude, as Lesson 4 shows. Almost every machine and structure Owen will meet is underdamped, with ζ somewhere between 0.005 and 0.10.

Measuring damping: the logarithmic decrement

Because the envelope decays exponentially, the ratio of successive peaks is constant. Its natural logarithm is the logarithmic decrement:

Taking the ratio over n cycles rather than one averages out reading errors and noise. Solving for the damping ratio, \(\zeta = \delta/\sqrt{4\pi^2 + \delta^2}\). The relation holds for displacement, velocity or acceleration peaks, because each is the same exponential envelope multiplied by a constant.

# Damping from a bump test on the TP-80 baseplate, and friction (Coulomb) decay of a pipe support
import math
fs = 2000.0                        # sample rate of the recorded decay, Hz
# Peak accelerations (m/s^2) read from successive cycles of the ring-down after a hammer tap
peaks = [9.80, 8.23, 6.92, 5.81, 4.88, 4.10, 3.45]
t_peaks = [0.0, 0.0302, 0.0604, 0.0906, 0.1208, 0.1510, 0.1812]   # s
n = len(peaks) - 1
delta = math.log(peaks[0] / peaks[-1]) / n
zeta = delta / math.sqrt(4 * math.pi**2 + delta**2)
Td = (t_peaks[-1] - t_peaks[0]) / n
fd = 1 / Td
fn = fd / math.sqrt(1 - zeta**2)
print(f"Log decrement over {n} cycles: delta = {delta:.4f}")
print(f"Damping ratio zeta = {zeta:.4f} ({zeta*100:.2f} % of critical)")
print(f"Damped frequency {fd:.2f} Hz, natural frequency {fn:.2f} Hz")
print(f"Cycles to halve the amplitude: {math.log(2)/delta:.1f}")

# Equivalent mass and stiffness of the baseplate mode from a static push test
k = 4.9e6                          # N/m, measured with a jack and dial gauge at the pump feet
m = k / (2 * math.pi * fn)**2
c_cr = 2 * math.sqrt(k * m); c = zeta * c_cr
print(f"\nModal mass {m:.0f} kg, critical damping {c_cr/1e3:.1f} kN s/m, actual c = {c:.0f} N s/m")

# Free response of the damped model, x(t) = X e^(-zeta wn t) cos(wd t)
wn = 2 * math.pi * fn; wd = wn * math.sqrt(1 - zeta**2)
for t in (0.0, 0.05, 0.10, 0.20, 0.40):
    print(f"  t = {t:.2f} s: envelope {math.exp(-zeta*wn*t):.3f} of the first peak")

# Coulomb friction: a sliding pipe support with friction force F loses 4F/k of amplitude per cycle
F = 25.0                           # friction force at the support, N
k_pipe = 1.2e6                     # lateral stiffness of the pipe span, N/m
X0 = 0.50e-3                       # initial amplitude after a valve slam, m
loss = 4 * F / k_pipe
cycles = 0; X = X0
while X > F / k_pipe:              # motion stops inside the dead band +/- F/k
    X -= loss; cycles += 1
print(f"\nCoulomb decay: {loss*1e6:.0f} um lost per cycle, motion stops after about {cycles} cycles")
print(f"  stopping band +/- {F/k_pipe*1e6:.0f} um; decay is linear, not exponential")

Log decrement over 6 cycles: delta = 0.1740
Damping ratio zeta = 0.0277 (2.77 % of critical)
Damped frequency 33.11 Hz, natural frequency 33.13 Hz
Cycles to halve the amplitude: 4.0

Modal mass 113 kg, critical damping 47.1 kN s/m, actual c = 1303 N s/m
  t = 0.00 s: envelope 1.000 of the first peak
  t = 0.05 s: envelope 0.750 of the first peak
  t = 0.10 s: envelope 0.562 of the first peak
  t = 0.20 s: envelope 0.316 of the first peak
  t = 0.40 s: envelope 0.100 of the first peak

Coulomb decay: 83 um lost per cycle, motion stops after about 6 cycles
  stopping band +/- 21 um; decay is linear, not exponential

Owen reads seven successive positive peaks from the trace, from 9.80 down to 3.45 m/s² over six cycles. The program gives δ = 0.174 and ζ = 0.0277, just under 3 percent of critical. Six cycles take 0.181 s, so the damped frequency is 33.11 Hz and the undamped natural frequency 33.13 Hz, the difference being invisible in practice. The amplitude halves every four cycles and is down to 10 percent after 0.4 s. A separate static test, jacking the pump feet upward and reading a dial gauge, gave a stiffness of 4.9 MN/m at the same point; dividing by ω_n² gives a modal mass of 113 kg, less than half the weight of the pump and motor because only part of the mass moves fully in this bending mode. The viscous coefficient that reproduces the decay is about 1,300 N·s/m, under 3 percent of the critical 47 kN·s/m.

Energy dissipated per cycle

A viscous damper dissipates energy in each cycle of harmonic motion of amplitude X at frequency ω:

Comparing this with the maximum strain energy \(U = \tfrac12 kX^2\) gives a measure called the loss factor, \(\eta = \Delta E/(2\pi U)\). At resonance, \(\eta = 2\zeta\). Materials data and rubber-mount catalogs often quote a loss factor rather than a damping ratio: natural rubber has η of about 0.05 to 0.1, highly damped elastomers 0.2 or more, and steel well under 0.001. The low material damping of steel is why a welded steel baseplate rings so clearly, and why most damping in a real structure comes from bolted joints, grout and the attached piping rather than from the metal.

Hysteretic and structural damping

Measurements on many structures show that the energy lost per cycle hardly depends on frequency, which is inconsistent with a purely viscous model. The usual fix is hysteretic, or structural, damping: a complex stiffness \(k(1 + i\eta)\) used in harmonic analysis. Over a narrow frequency band around a single resonance, the viscous and hysteretic models give almost identical results if \(\zeta = \eta/2\), so Owen can use whichever is convenient. For time-domain simulation of transients (Lesson 7) the viscous model is simpler, and finite element programs usually let the user specify modal damping ratios directly.

Coulomb (dry friction) damping

Friction at a sliding support or a loose joint produces a force of constant magnitude F opposing the velocity. The equation of motion is \(m\ddot x + F\,\mathrm{sgn}(\dot x) + kx = 0\), which is nonlinear but can be solved half-cycle by half-cycle. Each half-cycle is a harmonic motion about a shifted equilibrium at ±F/k, and the amplitude falls by 2F/k per half-cycle, or 4F/k per full cycle. The decay is linear rather than exponential, and the motion stops when the amplitude falls inside the band ±F/k, where the spring force can no longer overcome friction. The frequency stays at ω_n.

The program shows the effect on a pipe span with a sliding support. With a friction force of 25 N and a lateral stiffness of 1.2 MN/m, each cycle loses 83 µm, and a 0.5 mm disturbance from a valve slam dies in about six cycles, leaving the pipe stuck somewhere inside a band of ±21 µm. A plot of peak amplitude against cycle number distinguishes the two mechanisms at a glance: a straight line on linear axes means friction; a straight line on a logarithmic amplitude scale means viscous damping. Real structures often show both: friction at large amplitudes and viscous-like behavior at small ones.

Damping and design

Lucie's guidance to the design group is that damping is a safety margin, not a design tool. It is uncertain, it changes with temperature, bolt preload and age, and it cannot be relied on to make a resonance acceptable. The primary defense is frequency placement: keep natural frequencies away from forcing frequencies. Damping becomes the main tool only when the forcing covers a broad band, as with variable-speed drives that must run through a resonance, or with random excitation. The 2.8 percent of the new baseplate is typical of bolted, ungrouted steel; once grouted on site the damping will probably rise and the frequency will rise too, which Lesson 4 exploits.

Owen's test report records f_n = 33.1 Hz and ζ = 0.028 at the pump end, with a note that the value will change once the base is grouted. Farah, reviewing it, immediately asks about the variable-speed option: Tidewater offers the TP-80 on a drive up to 2,000 rpm, which is 33.3 Hz. Lesson 4 answers her question.