Exercises — Random vibration — PSD, RMS acceleration

Ye kehta hai: 250 Hz par centred ek 1 Hz-wide band mein, mean-square acceleration $0.05{\text{g}}^2$ hai. WHAT we did: vertical-axis value padhi aur yaad kiya ki uske units "power per hertz" hain. WHY: PSD ek density hai — ye tabhi physical ${\text{g}}^2$ banta hai jab aap ise ek bandwidth se multiply karo (yahan $1\text{Hz}$). Ye khud ek acceleration nahi hai; aap nahi keh sakte "box ko $0.05\text{g}$ feel hota hai."

(c) $a_{\text{RMS}}$. Units ki chain hai: $G_{xx}$ ${\text{g}}^2/\text{Hz}$ hai → $\text{Hz}$ par integrate karo → $\overline{a^2}$ ${\text{g}}^2$ hai → square-root → $a_{\text{RMS}}$ $\text{g}$ hai. WHY square root exist karta hai: hume ek number chahiye jo original acceleration signal ke same units mein ho taaki wo physically interpretable ho.

Bada. Area $=0.02\times (600-100)=0.02\times 500=10{\text{g}}^2$. WHAT it looks like: height $0.02$ aur width $\Delta f=500$ ka ek rectangle — ek wide thin strip jiska area aasaani se $1$ se upar hai. Ye kisi bhi mushkil problem se pehle "PSD = rectangle area" reflex train karta hai.

Step 1 — mean-square (area): $\overline{a^2}=0.03\times (2000-20)=0.03\times 1980=59.4{\text{g}}^2$. WHY $G\times \Delta f$: PSD constant hai, isliye uska integral bas ek rectangle ka base times height hai. Step 2 — RMS: $a_{\text{RMS}}=\sqrt{59.4}\approx 7.71\text{g}$. WHAT it means: part roughly ek steady $7.71g$ push jaisa feel karta hai, lekin poore band mein noise ke roop mein deliver hota hai.

Step 1 — teen rectangles, teen areas (figure mein, do short cyan blocks aur 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 alag-alag: PSD piecewise constant hai, isliye koi akela rectangle usse cover nahi karta — areas add karo. 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: beech wala "bump" (amber block) $43.14{\text{g}}^2$ mein se $32.5$ carry karta hai — lagbhag $75\%$ energy ek hi band mein baiṭhi hai.

Fraction $=\frac{32.5}{43.14}=0.753$, yani lagbhag $75.3\%$. WHY ye matter karta hai: energy fractions batate hain kahan design effort lagani chahiye. Agar ek Modal Analysis dikhata hai ki resonance $150$–$800\text{Hz}$ ke andar hai, toh wo block response dominate karega — L4 dekhein.

Step 0 — bhasha decode karo. Ek octave frequency ki doubling hai. "$+3\text{dB/octave}$" ka matlab: jab bhi $f$ double hoti hai, PSD $3\text{dB}$ se badhti hai, aur power mein $3\text{dB}$ ka matlab $10^{3/10}=1.995\approx 2$ ka factor hai. Toh PSD har octave mein double hoti hai. Log–log plot par ye slope $m=1$ ki straight line hai kyunki $G\propto f^m$ aur $2=2^m\Rightarrow m=1$.

Step 1 — 80 Hz par value. $20$ se $80\text{Hz}$ tak do octaves hain ($20\to 40\to 80$), isliye PSD do baar double hoti hai: $G(80)=0.01\times 2\times 2=0.04{\text{g}}^2/\text{Hz}$. WHY ye kaam karta hai: har octave PSD ko Step 0 mein established "$+3\text{dB}$" factor $2$ se multiply karta hai, aur do octaves do doublings stack karte hain ($\times 4$). Power law se check karo: $G(f)=0.01(f/20{)}^1$, toh $G(80)=0.01\times 4=0.04$. ✓

Step 2 — sloped segment ka area (page ke upar se power-law rule use karo). $G(f)=0.01(f/20{)}^1=5\times 10^{-4}f$ likho, toh yahan $C=5\times 10^{-4}$ aur $m=1$ hai. $m+1=2$ ke saath $\int C{f}^{m}df=\frac{C}{m+1}(f_2^{m+1}-f_1^{m+1})$ apply karo: ${\overline{a^2}}_{\text{ramp}}=\frac{5\times 10^{-4}}{2}(80^2-20^2).$ WHY power rule aur rectangle nahi: height $f$ ke saath change hoti hai, isliye koi akela rectangle fit nahi karta — lekin $G$, $f$ ki pure power hai, aur power ka antiderivative wo akela rule hai jo humne upar state kiya tha. Evaluate karo: $80^2-20^2=6400-400=6000$, toh ${\overline{a^2}}_{\text{ramp}}=\frac{5\times 10^{-4}}{2}\times 6000=1.5{\text{g}}^2$.

WHAT it looks like: figure mein sloped line ke neeche amber region — ek curved sliver jiska exact area humne abhi compute kiya.

Step 1 — plateau area: $0.04\times (2000-80)=0.04\times 1920=76.8{\text{g}}^2$. WHY yahan $G\times \Delta f$: ye segment flat hai, isliye ye ek genuine rectangle hai — height $0.04$ times width $\Delta f=1920$ — aur general integral base times height mein collapse ho jaata hai (ye power rule ka $m=0$ case hai, jahan area bas $C\Delta f$ hai). Step 2 — total mean-square: ramp $+$ plateau $=1.5+76.8=78.3{\text{g}}^2$. WHY add karo: do segments disjoint frequency bands occupy karte hain, isliye unke areas simply poore curve ke neeche total area mein sum ho jaate hain. Step 3 — RMS: $a_{\text{RMS}}=\sqrt{78.3}\approx 8.85\text{g}$. Insight: ramp sirf $78.3{\text{g}}^2$ mein se $1.5$ contribute karta hai — low-frequency ramps thodi energy carry karte hain kyunki wo height mein bhi kam hain aur bandwidth mein bhi narrow.

Step 0 — meaning. Transmissibility $|T(f){|}^2$ (is page ke upar blue box mein full formula) wo factor hai jisse base-input PSD $G_{\ddot{x}}$ ko multiply karte hain mass ke absolute-acceleration response PSD $G_{\ddot{y}}$ paane ke liye. $\zeta$ damping ratio hai: $\zeta =0$ undamped hai, $\zeta =1$ critically damped hai. Chhota $\zeta$ = sharp, tall resonance.

(a) Quality factor: $Q=\frac{1}{2\zeta }=\frac{1}{2\times 0.04}=12.5$. WHY $Q$: ye shorthand hai "resonance flat response se kitni baar bada hai," aur ye Structural Damping mein har jagah aata hai.

(b) Peak power: page ke upar quote ki gayi transmissibility formula se shuru karo, $|T(f){|}^2=\frac{1+{\left[2\zeta \frac{f}{f_n}\right]}^2}{{\left[1-{\left(\frac{f}{f_n}\right)}^2\right]}^2+{\left[2\zeta \frac{f}{f_n}\right]}^2}.$ $f=f_n$ par term $1-(f/f_n{)}^2=0$ vanish ho jaata hai aur $f/f_n=1$, isse ye bachta hai: $|T(f_n){|}^2=\frac{1+(2\zeta {)}^2}{(2\zeta {)}^2}=1+\frac{1}{4{\zeta }^2}=1+\frac{1}{4(0.04{)}^2}=1+\frac{1}{0.0064}=\mathrm{157.25.}$ Toh resonance par response PSD input PSD ka $\approx 157$ times hai. Note karo $\sqrt{157.25}\approx 12.54\approx Q$ — amplitude gain $Q$ hai, power gain $Q^2$ hai (plus small $+1$).

Step 0 — shortcut kyun chahiye. Exact response RMS hai $\sqrt{\int |T(f){|}^2{G}_{\ddot{x}}(f)df}$. Wo integral nasty hai kyunki $|T{|}^2$ mein ek sharp peak hai. Flat input aur lightly damped SDOF ke liye, integral ka ek sundar closed form hai — Miles' equation — kyunki resonance peak ke neeche area sirf uski height aur width par depend karta hai.

Step 1 — plug in: $a_{\text{RMS}}=\sqrt{\frac{\pi }{2}\times 300\times 12.5\times 0.02}$. Step 2 — root ke andar: $\frac{\pi }{2}\times 300=471.24$; $\times 12.5=5890.5$; $\times 0.02=117.81{\text{g}}^2$. Step 3 — root: $a_{\text{RMS}}=\sqrt{117.81}\approx 10.85\text{g}$. WHAT it means: halanki ek wide band mein raw input RMS modest ho sakta hai, resonance energy concentrate karta hai aur box khud lagbhag $10.9g$ RMS dekhta hai — woh number jo Fatigue Analysis ko feed karta hai.

Step 1 — Miles ko invert karo. Require karo $a_{\text{RMS}}\le 8\text{g}$, toh $a_{\text{RMS}}^2\le 64{\text{g}}^2$. WHY pehle dono sides square karo: Miles $a_{\text{RMS}}$ ek square root ke neeche deta hai, aur $Q$ us root ke andar rehta hai; square karne se root hat jaata hai toh $Q$ ek plain linear factor ban jaata hai jise hum simple division se isolate kar sakte hain. $a_{\text{RMS}}^2=\frac{\pi }{2}{f}_{n}QG$ se, $Q\le \frac{64}{\frac{\pi }{2}{f}_{n}G}=\frac{64}{\frac{\pi }{2}\times 300\times 0.02}=\frac{64}{9.4248}=\mathrm{6.79.}$ Step 2 — damping mein convert karo. $\zeta =\frac{1}{2Q}$ use karo: $\zeta \ge \frac{1}{2\times 6.79}=\mathrm{0.0736.}$ WHY inequality direction flip hoti hai: $\zeta =1/(2Q)$ $Q$ ki decreasing function hai — zyada allowed $Q$ matlab chhota $\zeta$, isliye $Q$ par ek upper bound ($Q\le 6.79$) $\zeta$ par lower bound ban jaata hai. ANSWER: sabse bada allowable quality factor $Q\approx 6.79$ hai, jisme minimum damping ratio $\zeta \approx 0.074$ (lagbhag $7.4\%$) chahiye. WHAT it means: aapko Structural Damping add karna hoga jab tak $\zeta \gtrsim 0.074$ na ho, nahi toh box $8\text{g}$ RMS limit exceed karega.

Step 1 — Miles ratio. $Q$ aur $G$ identical hone par, $a_{\text{RMS}}\propto \sqrt{f_n}$. WHY sirf $\sqrt{f_n}$ bachta hai: $\sqrt{\frac{\pi }{2}{f}_{n}QG}$ mein $f_n$ ke alawa har factor dono boxes ke liye same hai, toh ratio mein cancel ho jaata hai, sirf frequency ratio ka square root bachta hai. Toh $\frac{a_B}{a_A}=\sqrt{\frac{600}{150}}=\sqrt{4}=2.$ Box B exactly do baar response RMS dekhta hai. WHY physically: zyada natural frequency same peak height ko ek wider effective bandwidth ($\propto f_n$) par spread karta hai, isliye zyada energy integrate hoti hai. Design lesson: stiff, high-$f_n$ hardware random vibration mein automatically safer nahi hota.

(a) Input RMS — teen 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 karo: teen regions disjoint bands hain, isliye unke areas sum hote hain. 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) Miles ke through Response. Miles resonance frequency par input PSD use karta hai, kyunki ek sharp resonance sirf $f_n$ ke aas-paas narrow band mein input "dekhti" hai. Yahan $f_n=300\text{Hz}$ notch ke andar hai, isliye local input $G_{\ddot{x}}(f_n)=0.005{\text{g}}^2/\text{Hz}$ hai, broadband $0.03$ nahi: $a_{\text{RMS}}=\sqrt{\frac{\pi }{2}\times 300\times 10\times 0.005}=\sqrt{4712.4\times 0.005}=\sqrt{23.562}\approx 4.85\text{g}.$ WHY notch help karta hai: input ko exactly wahan kam karke jahan box resonate karta hai, damper response ko us se kaat deta hai jo $\sqrt{\frac{\pi }{2}\times 300\times 10\times 0.03}=\sqrt{141.37}\approx 11.9\text{g}$ hota — $\sqrt{0.03/0.005}=\sqrt{6}\approx 2.45$ ka reduction factor. Ye hi launch spectrum mein tuned damping ka pura point hai.

Step 0 — statistical picture. Ek stationary random signal ke liye instantaneous acceleration ek bell-shaped (Gaussian) distribution follow karta hai jiska standard deviation $\sigma$ RMS ke barabar hota hai. Toh "RMS" aur "$1\sigma$" same number hain, $\sigma =4.85\text{g}$.

Step 1 — 3-sigma peak: $a_{3\sigma }=3\times 4.85=14.55\text{g}$. WHY $3\sigma$: ek Gaussian $3\sigma$ sirf lagbhag $0.27\%$ time exceed karta hai — rare, lekin ek launch mein ye kaafi baar hoga. $3\sigma$ ke liye design karna essentially un sabhi peaks ko cover karta hai jo structure actually feel karta hai. WHY sirf RMS nahi: RMS ek typical value hai; material bade excursions par fail ya fatigue hota hai, isliye equivalent static load ek high percentile hona chahiye, conventionally $3\sigma$.

Step 1 — requirement ko RMS limit mein translate karo. $3\sigma =3a_{\text{RMS}}\le 30\text{g}\Rightarrow a_{\text{RMS}}\le 10\text{g}\Rightarrow a_{\text{RMS}}^2\le 100{\text{g}}^2$. WHY square karo: jaise 4.2 mein, $\zeta$ $Q$ ke through ek square root ke andar chhupa hai; squaring usse linear factor ki tarah expose karta hai.

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

Step 3 — damping: $\zeta =\frac{1}{2Q}\ge \frac{1}{2\times 6.366}=0.0785$. WHY phir flip: $\zeta =1/(2Q)$ $Q$ ke saath decrease karta hai, isliye $Q$ par upper bound $\zeta$ par lower bound ban jaata hai. Minimum $\zeta \approx 0.0785$ (lagbhag $7.9\%$). WHAT we synthesised: PSD area thinking (L2) + Miles' equation (L4) + $3\sigma$ statistical rule (L5) ek design number mein jo ek real spacecraft engineer report karta — is damping se neeche box apni qualification Sine Vibration Testing aur random-vibration requirements dono ke against fail kar deta.