Exercises — Random vibration — PSD, RMS acceleration

It says: in a 1 Hz-wide band centred on 250 Hz, the mean-square acceleration is $0.05{\text{g}}^2$. WHAT we did: read the vertical-axis value and remembered its units are "power per hertz." WHY: PSD is a density — it only becomes a physical ${\text{g}}^2$ once you multiply by a bandwidth (here $1\text{Hz}$). It is not itself an acceleration; you cannot say "the box sees $0.05\text{g}$."

(c) $a_{\text{RMS}}$. The chain of units is: $G_{xx}$ is ${\text{g}}^2/\text{Hz}$ → integrate over $\text{Hz}$ → $\overline{a^2}$ is ${\text{g}}^2$ → square-root → $a_{\text{RMS}}$ is $\text{g}$. WHY the square root exists: we want a number in the same units as the original acceleration signal so it is physically interpretable.

Bigger. Area $=0.02\times (600-100)=0.02\times 500=10{\text{g}}^2$. WHAT it looks like: a rectangle of height $0.02$ and width $\Delta f=500$ — a wide thin strip whose area is easily above $1$. This trains the "PSD = rectangle area" reflex before any hard problem.

Step 1 — mean-square (area): $\overline{a^2}=0.03\times (2000-20)=0.03\times 1980=59.4{\text{g}}^2$. WHY $G\times \Delta f$: the PSD is constant, so its integral is just base times height of one rectangle. Step 2 — RMS: $a_{\text{RMS}}=\sqrt{59.4}\approx 7.71\text{g}$. WHAT it means: the part feels roughly like a steady $7.71g$ push, but delivered as noise across the whole band.

Step 1 — three rectangles, three areas (in the figure, the two short cyan blocks and the tall amber block):

• Block 1: $0.008\times (150-20)=0.008\times 130=1.04{\text{g}}^2$
• Block 2: $0.05\times (800-150)=0.05\times 650=32.5{\text{g}}^2$
• Block 3: $0.008\times (2000-800)=0.008\times 1200=9.6{\text{g}}^2$ WHY separate: the PSD is piecewise constant, so no single rectangle covers it — add the areas. Step 2 — total: $\overline{a^2}=1.04+32.5+9.6=43.14{\text{g}}^2$. Step 3 — RMS: $a_{\text{RMS}}=\sqrt{43.14}\approx 6.57\text{g}$. Insight: the middle "bump" (the amber block) carries $32.5$ of $43.14{\text{g}}^2$ — about $75\%$ of the energy sits in one band.

Fraction $=\frac{32.5}{43.14}=0.753$, i.e. about $75.3\%$. WHY this matters: energy fractions tell you where to spend design effort. If a Modal Analysis shows a resonance inside $150$–$800\text{Hz}$, that block will dominate the response — see L4.

Step 0 — decode the language. An octave is a doubling of frequency. "$+3\text{dB/octave}$" means: every time $f$ doubles, the PSD rises by $3\text{dB}$, and $3\text{dB}$ in power is a factor of $10^{3/10}=1.995\approx 2$. So the PSD doubles per octave. On a log–log plot that is a straight line of slope $m=1$ because $G\propto f^m$ with $2=2^m\Rightarrow m=1$.

Step 1 — value at 80 Hz. From $20$ to $80\text{Hz}$ is two octaves ($20\to 40\to 80$), so the PSD doubles twice: $G(80)=0.01\times 2\times 2=0.04{\text{g}}^2/\text{Hz}$. WHY this works: each octave multiplies the PSD by the "$+3\text{dB}$" factor of $2$ established in Step 0, and two octaves stack two doublings ($\times 4$). Check with the power law: $G(f)=0.01(f/20{)}^1$, so $G(80)=0.01\times 4=0.04$. ✓

Step 2 — area of a sloped segment (use the power-law rule from the top of the page). Write $G(f)=0.01(f/20{)}^1=5\times 10^{-4}f$, so here $C=5\times 10^{-4}$ and $m=1$. Applying $\int C{f}^{m}df=\frac{C}{m+1}(f_2^{m+1}-f_1^{m+1})$ with $m+1=2$: ${\overline{a^2}}_{\text{ramp}}=\frac{5\times 10^{-4}}{2}(80^2-20^2).$ WHY the power rule and not a rectangle: the height changes with $f$, so no single rectangle fits — but $G$ is a pure power of $f$, and the antiderivative of a power is the one rule we stated up front. Evaluate: $80^2-20^2=6400-400=6000$, so ${\overline{a^2}}_{\text{ramp}}=\frac{5\times 10^{-4}}{2}\times 6000=1.5{\text{g}}^2$.

WHAT it looks like: the amber region under the sloped line in the figure — a curved sliver whose exact area we just computed.

Step 1 — plateau area: $0.04\times (2000-80)=0.04\times 1920=76.8{\text{g}}^2$. WHY $G\times \Delta f$ here: this segment is flat, so it is a genuine rectangle — height $0.04$ times width $\Delta f=1920$ — and the general integral collapses to base times height (it is the $m=0$ case of the power rule, where the area is just $C\Delta f$). Step 2 — total mean-square: ramp $+$ plateau $=1.5+76.8=78.3{\text{g}}^2$. WHY add: the two segments occupy disjoint frequency bands, so their areas simply sum to the total area under the whole curve. Step 3 — RMS: $a_{\text{RMS}}=\sqrt{78.3}\approx 8.85\text{g}$. Insight: the ramp contributes only $1.5$ of $78.3{\text{g}}^2$ — low-frequency ramps carry little energy because they are both low in height and narrow in bandwidth.

Step 0 — meaning. Transmissibility $|T(f){|}^2$ (full formula in the blue box near the top of this page) is the factor by which the base-input PSD $G_{\ddot{x}}$ is multiplied to get the mass's absolute-acceleration response PSD $G_{\ddot{y}}$. $\zeta$ is the damping ratio: $\zeta =0$ is undamped, $\zeta =1$ is critically damped. Small $\zeta$ = sharp, tall resonance.

(a) Quality factor: $Q=\frac{1}{2\zeta }=\frac{1}{2\times 0.04}=12.5$. WHY $Q$: it is the shorthand "how many times bigger is resonance than the flat response," and it appears everywhere in Structural Damping.

(b) Peak power: start from the transmissibility formula quoted at the top of the page, $|T(f){|}^2=\frac{1+{\left[2\zeta \frac{f}{f_n}\right]}^2}{{\left[1-{\left(\frac{f}{f_n}\right)}^2\right]}^2+{\left[2\zeta \frac{f}{f_n}\right]}^2}.$ At $f=f_n$ the term $1-(f/f_n{)}^2=0$ vanishes and $f/f_n=1$, leaving $|T(f_n){|}^2=\frac{1+(2\zeta {)}^2}{(2\zeta {)}^2}=1+\frac{1}{4{\zeta }^2}=1+\frac{1}{4(0.04{)}^2}=1+\frac{1}{0.0064}=\mathrm{157.25.}$ So the response PSD at resonance is $\approx 157$ times the input PSD. Note $\sqrt{157.25}\approx 12.54\approx Q$ — the amplitude gain is $Q$, the power gain is $Q^2$ (plus the small $+1$).

Step 0 — why we need a shortcut. The exact response RMS is $\sqrt{\int |T(f){|}^2{G}_{\ddot{x}}(f)df}$. That integral is nasty because $|T{|}^2$ has a sharp peak. For a flat input and a lightly damped SDOF, the integral has a beautiful closed form — Miles' equation — because the area under the resonance peak depends only on its height and width.

Step 1 — plug in: $a_{\text{RMS}}=\sqrt{\frac{\pi }{2}\times 300\times 12.5\times 0.02}$. Step 2 — inside the root: $\frac{\pi }{2}\times 300=471.24$; $\times 12.5=5890.5$; $\times 0.02=117.81{\text{g}}^2$. Step 3 — root: $a_{\text{RMS}}=\sqrt{117.81}\approx 10.85\text{g}$. WHAT it means: although the raw input RMS across a wide band might be modest, the resonance concentrates energy and the box itself sees about $10.9g$ RMS — the number that feeds a Fatigue Analysis.

Step 1 — invert Miles. Require $a_{\text{RMS}}\le 8\text{g}$, so $a_{\text{RMS}}^2\le 64{\text{g}}^2$. WHY square both sides first: Miles gives $a_{\text{RMS}}$ under a square root, and $Q$ lives inside that root; squaring removes the root so $Q$ becomes a plain linear factor we can isolate by simple division. From $a_{\text{RMS}}^2=\frac{\pi }{2}{f}_{n}QG$, $Q\le \frac{64}{\frac{\pi }{2}{f}_{n}G}=\frac{64}{\frac{\pi }{2}\times 300\times 0.02}=\frac{64}{9.4248}=\mathrm{6.79.}$ Step 2 — convert to damping. Use $\zeta =\frac{1}{2Q}$: $\zeta \ge \frac{1}{2\times 6.79}=\mathrm{0.0736.}$ WHY the inequality flips direction: $\zeta =1/(2Q)$ is a decreasing function of $Q$ — a bigger allowed $Q$ means a smaller $\zeta$, so an upper bound on $Q$ ($Q\le 6.79$) turns into a lower bound on $\zeta$. ANSWER: the largest allowable quality factor is $Q\approx 6.79$, requiring a minimum damping ratio $\zeta \approx 0.074$ (about $7.4\%$). WHAT it means: you must add Structural Damping until $\zeta \gtrsim 0.074$, or the box will exceed the $8\text{g}$ RMS limit.

Step 1 — Miles ratio. With $Q$ and $G$ identical, $a_{\text{RMS}}\propto \sqrt{f_n}$. WHY only $\sqrt{f_n}$ survives: in $\sqrt{\frac{\pi }{2}{f}_{n}QG}$ every factor except $f_n$ is the same for both boxes, so it cancels in the ratio, leaving just the square root of the frequency ratio. So $\frac{a_B}{a_A}=\sqrt{\frac{600}{150}}=\sqrt{4}=2.$ Box B sees exactly twice the response RMS. WHY physically: a higher natural frequency spreads the same peak height over a wider effective bandwidth ($\propto f_n$), so it integrates more energy. Design lesson: stiff, high-$f_n$ hardware is not automatically safer under random vibration.

(a) Input RMS — three rectangles:

• A: $0.03\times (200-20)=0.03\times 180=5.4{\text{g}}^2$
• B: $0.005\times (400-200)=0.005\times 200=1.0{\text{g}}^2$
• C: $0.03\times (2000-400)=0.03\times 1600=48.0{\text{g}}^2$ WHY add: the three regions are disjoint bands, so their areas sum. Total $=5.4+1.0+48.0=54.4{\text{g}}^2$; $a_{\text{RMS,in}}=\sqrt{54.4}\approx 7.38\text{g}$.

(b) Response via Miles. Miles uses the input PSD at the resonance frequency, because a sharp resonance only "sees" the input in the narrow band around $f_n$. Here $f_n=300\text{Hz}$ sits inside the notch, so the local input is $G_{\ddot{x}}(f_n)=0.005{\text{g}}^2/\text{Hz}$, not the broadband $0.03$: $a_{\text{RMS}}=\sqrt{\frac{\pi }{2}\times 300\times 10\times 0.005}=\sqrt{4712.4\times 0.005}=\sqrt{23.562}\approx 4.85\text{g}.$ WHY the notch helps: by lowering the input exactly where the box resonates, the damper cuts the response from what would have been $\sqrt{\frac{\pi }{2}\times 300\times 10\times 0.03}=\sqrt{141.37}\approx 11.9\text{g}$ down to $4.85\text{g}$ — a factor of $\sqrt{0.03/0.005}=\sqrt{6}\approx 2.45$ reduction. This is the whole point of tuned damping in a launch spectrum.

Step 0 — the statistical picture. For a stationary random signal the instantaneous acceleration follows a bell-shaped (Gaussian) distribution whose standard deviation $\sigma$ equals the RMS. So "RMS" and "$1\sigma$" are the same number, $\sigma =4.85\text{g}$.

Step 1 — 3-sigma peak: $a_{3\sigma }=3\times 4.85=14.55\text{g}$. WHY $3\sigma$: a Gaussian exceeds $3\sigma$ only about $0.27\%$ of the time — rare, but it will occur many times over a launch. Designing to $3\sigma$ covers essentially all the peaks the structure actually feels. WHY not RMS alone: the RMS is a typical value; the material fails or fatigues on the large excursions, so the equivalent static load must be a high percentile, conventionally $3\sigma$.

Step 1 — translate the requirement into an RMS limit. $3\sigma =3a_{\text{RMS}}\le 30\text{g}\Rightarrow a_{\text{RMS}}\le 10\text{g}\Rightarrow a_{\text{RMS}}^2\le 100{\text{g}}^2$. WHY square: as in 4.2, $\zeta$ hides inside a square root through $Q$; squaring exposes it as a linear factor.

Step 2 — Miles, solved for $Q$: $a_{\text{RMS}}^2=\frac{\pi }{2}{f}_{n}QG\Rightarrow Q\le \frac{100}{\frac{\pi }{2}\times 250\times 0.04}=\frac{100}{15.708}=\mathrm{6.366.}$

Step 3 — damping: $\zeta =\frac{1}{2Q}\ge \frac{1}{2\times 6.366}=0.0785$. WHY the flip again: $\zeta =1/(2Q)$ decreases with $Q$, so the upper bound on $Q$ becomes a lower bound on $\zeta$. Minimum $\zeta \approx 0.0785$ (about $7.9\%$). WHAT we synthesised: PSD area thinking (L2) + Miles' equation (L4) + the $3\sigma$ statistical rule (L5) into one design number that a real spacecraft engineer would report. Below this damping the box would fail its qualification against Sine Vibration Testing and random-vibration requirements alike.