Modal analysis — natural frequencies, mode shapes

3.6.10 · Physics › Spacecraft Structures & Systems Engineering

Overview

Modal analysis wo process hai jisme ek structure ki natural frequencies aur mode shapes determine ki jaati hain—yaani wo frequencies jisme wo naturally vibrate karta hai aur un frequencies par motion ke characteristic patterns. Spacecraft ke liye yeh critical hai: launch vibrations, thruster firings, aur attitude maneuvers in natural modes ko excite kar sakte hain, jisse structural failure, instrument misalignment, ya mission-critical damage ho sakta hai.

Yeh kyun matter karta hai: Agar aap ek spacecraft ko uski natural frequency par excite karo, toh resonance hoti hai—amplitudes catastrophically badhti hain. Modal properties jaanne se hum structures design kar sakte hain jo natural frequencies ko forcing frequencies se door rakhein, strategically damping add karein, aur dynamic response predict karein.

Connections

Free vibration fundamentals
Multi-degree-of-freedom systems
Structural damping mechanisms
Finite element analysis (FEA)
Launch vehicle load environments
Orthogonality of mode shapes
Frequency response functions

Core Concepts

Intuition Physical Intuition

Socho tum ek guitar string pluck kar rahe ho. Woh specific pitches (frequencies) par vibrate karta hai, har ek ka ek characteristic shape hota hai—first mode ek single hump hai, second mode mein do humps hain jo ek still point (node) se alag hote hain. String in patterns mein vibrate "karna chahti" hai.

Ek spacecraft structure ek 3D "guitar string" hai. Iske multiple natural frequencies hain (first bending, first torsion, panel modes, etc.) aur corresponding mode shapes. Launch vibrations aise hain jaise ise pluck kiya—structure apne natural modes ke combinations mein vibrate karke respond karta hai.

Danger yeh hai: Agar launch vibrations mein kisi natural frequency par energy ho, toh woh mode "strongly pluck" ho jaata hai, aur structure violently shake karta hai (resonance). Modal analysis in vulnerable frequencies ko dhundhti hai taaki hum ya toh structure ko stiffen karein (frequencies ko forcing range se upar le jaayein) ya dampers lagaayein.

Derivation from First Principles

Starting Point: Multi-DOF Free Vibration

Maano ek structure $n$ degrees of freedom mein discretize ki gayi hai (springs se connected masses). Damping ya forcing ke bina equation of motion:

[M] {\overset{x}{¨}} + [K] {x} = {0}

jahan:

$[M]$ : mass matrix (symmetric, positive definite)
$[K]$ : stiffness matrix (symmetric, positive semi-definite)
${x}$ : displacement vector

Yeh form kyun? Har row ek mass ke liye Newton's second law hai: $m_{i} \overset{x}{¨}_{i} = - k_{i, i} x_{i} - k_{i, j} x_{j} + \dots$ (neighbors se spring forces).

Assuming Harmonic Solution

Free vibration ke liye, maano saare points same frequency $ω$ par fixed relative amplitudes ke saath oscillate karte hain:

{x (t)} = {ϕ} sin (ω t + θ)

Harmonic kyun? Bina damping ke linear systems sinusoidally oscillate karte hain. ${ϕ}$ mode shape hai (spatial pattern), $ω$ natural frequency hai (time pattern).

Equation of motion mein substitute karo:

\overset{x}{¨} = - ω^{2} {ϕ} sin (ω t + θ)

[M] (- ω^{2} {ϕ}) sin (ω t + θ) + [K] {ϕ} sin (ω t + θ) = {0}

$sin (ω t + θ)$ cancel karo (nonzero hai):

([K] - ω^{2} [M]) {ϕ} = {0}

Eigenvalue problem kyun? Humein nontrivial solutions chahiye ( ${ϕ} \neq = {0}$ ). Yeh tabhi hota hai jab determinant zero ho—characteristic polynomial jiske roots $ω_{i}^{2}$ hain.

Orthogonality of Mode Shapes

Do alag modes $i$ aur $j$ ke liye:

[K] {ϕ_{i}} = ω_{i}^{2} [M] {ϕ_{i}}

[K] {ϕ_{j}} = ω_{j}^{2} [M] {ϕ_{j}}

Pehle ko ${ϕ_{j}}^{T}$ se pre-multiply karo, doosre ko ${ϕ_{i}}^{T}$ se:

{ϕ_{j}}^{T} [K] {ϕ_{i}} = ω_{i}^{2} {ϕ_{j}}^{T} [M] {ϕ_{i}}

{ϕ_{i}}^{T} [K] {ϕ_{j}} = ω_{j}^{2} {ϕ_{i}}^{T} [M] {ϕ_{j}}

Kyunki $[K]$ aur $[M]$ symmetric hain, ${ϕ_{j}}^{T} [K] {ϕ_{i}} = {ϕ_{i}}^{T} [K] {ϕ_{j}}$ . Equations subtract karo:

0 = (ω_{i}^{2} - ω_{j}^{2}) {ϕ_{j}}^{T} [M] {ϕ_{i}}

Agar $ω_{i} \neq = ω_{j}$ , toh:

{ϕ_{j}}^{T} [M] {ϕ_{i}} = 0 (mass orthogonality)

Similarly:

{ϕ_{j}}^{T} [K] {ϕ_{i}} = 0 (stiffness orthogonality)

Orthogonality kyun matter karti hai? Mode shapes independent hain. Hum coupled equations ko $n$ independent single-DOF oscillators (modal coordinates) mein transform kar sakte hain.

Modal matrix $[Φ] = [{ϕ_{1}}, {ϕ_{2}}, \dots, {ϕ_{n}}]$ define karo (columns mode shapes hain).

Physical coordinates ko modal coordinates ${q}$ mein transform karo:

{x} = [Φ] {q}

Physical meaning: $q_{i} (t)$ yeh hai ki mode $i$ total response mein kitna participate karta hai.

Equation of motion mein substitute karo:

[M] [Φ] {\overset{q}{¨}} + [K] [Φ] {q} = {F (t)}

$[Φ]^{T}$ se pre-multiply karo:

[Φ]^{T} [M] [Φ] {\overset{q}{¨}} + [Φ]^{T} [K] [Φ] {q} = [Φ]^{T} {F (t)}

Orthogonality se, $[Φ]^{T} [M] [Φ]$ aur $[Φ]^{T} [K] [Φ]$ diagonal matrices hain:

[Φ]^{T} [M] [Φ] = [M^{*}] = diag (M_{1}^{*}, M_{2}^{*}, \dots) (generalized mass)

[Φ]^{T} [K] [Φ] = [K^{*}] = diag (K_{1}^{*}, K_{2}^{*}, \dots) (generalized stiffness)

Har modal equation ban jaata hai:

M_{i}^{*} \overset{q}{¨}_{i} + K_{i}^{*} q_{i} = Q_{i} (t)

jahan $Q_{i} = {ϕ_{i}}^{T} {F}$ modal force hai.

Yeh $n$ independent single-DOF oscillators hain! Har mode $ω_{i} = K_{i}^{*} / M_{i}^{*}$ par independently vibrate karta hai.

Worked Examples

Worked example Example 1: Two-DOF Spring–Mass System

Given: Do equal masses $m$ springs se connected hain. Pehla mass wall se spring $k$ se attached hai, masses spring $k$ se connected hain, doosra mass free hai.

|--[k]--[m1]--[k[m2]

Find: Natural frequencies aur mode shapes.

Solution: Step 1: Equations of motion likho.

Mass 1: $m \overset{x}{¨}_{1} = - k x_{1} - k (x_{1} - x_{2}) = - 2 k x_{1} + k x_{2}$
Mass 2: $m \overset{x}{¨}_{2} = - k (x_{2} - x_{1}) = k x_{1} - k x_{2}$

Matrix form:

[M] = [m 0 0 m], [K] = [2 k - k - k k]

Yeh matrices kyun? $M$ diagonal hai kyunki masses directly ek doosre ko affect nahi karte. $K$ mein off-diagonal terms hain kyunki springs motion ko couple karte hain.

Step 2: Eigenvalue problem solve karo $det ([K] - ω^{2} [M]) = 0$ .

det [2 k - m ω^{2} - k - k k - m ω^{2}] = 0

(2 k - m ω^{2}) (k - m ω^{2}) - k^{2} = 0

2 k^{2} - 2 k m ω^{2} - k m ω^{2} + m^{2} ω^{4} - k^{2} = 0

m^{2} ω^{4} - 3 k m ω^{2} + k^{2} = 0

Expand kyun kiya? $ω^{2}$ mein characteristic polynomial ke liye.

Quadratic formula: $ω^{2} = \frac{3 k \pm 9 k ^{2} - 4 k ^{2}}{2 m} = \frac{3 k \pm 5 k}{2 m}$

ω_{1} = \frac{( 3 - 5 ) k}{2 m} \approx 0.618 k / m

ω_{2} = \frac{( 3 + 5 ) k}{2 m} \approx 1.618 k / m

Step 3: Mode shapes dhundo. $ω_{1}^{2} = (3 - 5) k / (2 m)$ ke liye:

[K] {ϕ_{1}} = ω_{1}^{2} [M] {ϕ_{1}}

[2 k - k - k k] [ϕ_{11} ϕ_{21}] = \frac{( 3 - 5 ) k}{2} [ϕ_{11} ϕ_{21}]

First row: $2 k ϕ_{11} - k ϕ_{21} = \frac{( 3 - 5 ) k}{2} ϕ_{11}$

ϕ_{21} = (2 - \frac{3 - 5}{2}) ϕ_{11} = \frac{1 + 5}{2} ϕ_{11} \approx 1.618 ϕ_{11}

Normalize karo (set $ϕ_{11} = 1$ ):

{ϕ_{1}} = [1 1.618] (dono masses in-phase move karte hain, mass 2 ki amplitude zyada)

$ω_{2}^{2}$ ke liye, similarly:

{ϕ_{2}} = [1 - 0.618] (masses out-of-phase move karte hain)

Physical interpretation:

Mode 1 (lower frequency): dono masses saath move karte hain, "breathing" mode
Mode 2 (higher frequency): masses opposite direction mein move karte hain, "scissor" mode, center spring zyada stretch/compress hoti hai → higher stiffness → higher frequency

Worked example Example 2: Cantilever Beam Modal Analysis

Given: Uniform cantilever beam, length $L$ , mass per unit length $μ$ , bending stiffness $E I$ .

Find: Pehli natural frequency aur mode shape (continuous system).

Solution:

Step 1: Beam vibration PDE (Euler-Bernoulli):

E I \frac{\partial ^{4} w}{\partial x ^{4}} + μ \frac{\partial ^{2} w}{\partial t ^{2}} = 0

Yeh PDE kyun? Bending moment $M = E I \frac{\partial ^{2} w}{\partial x ^{2}}$ , shear force balance se fourth derivative aata hai.

Step 2: Separation of variables maano $w (x, t) = ϕ (x) sin (ω t)$ .

E I \frac{d ^{4} ϕ}{d x ^{4}} - μ ω^{2} ϕ = 0

β^{4} = \frac{μ ω ^{2}}{E I}

\frac{d ^{4} ϕ}{d x ^{4}} - β^{4} ϕ = 0

General solution: $ϕ (x) = A cos (β x) + B sin (β x) + C cosh (β x) + D sinh (β x)$

Step 3: Boundary conditions apply karo (cantilever: $x = 0$ par fixed, $x = L$ par free).

$ϕ (0) = 0$ (koi displacement nahi)
$ϕ^{'} (0) = 0$ (koi rotation nahi)
$ϕ^{''} (L) = 0$ (koi moment nahi)
$ϕ^{'''} (L) = 0$ (koi shear nahi)

Algebra ke baad (standard beam problem), characteristic equation:

cos (β L) cosh (β L) = - 1

First root: β_{1} L \approx 1.875

ω_{1} = β_{1}^{2} \frac{E I}{μ} = 3.516 \frac{E I}{μ L ^{4}} [rad/s]

Mode shape (normalized):

ϕ_{1} (x) = cosh (β_{1} x) - cos (β_{1} x) - σ_{1} (sinh (β_{1} x) - sin (β_{1} x))

jahan $σ_{1} = \frac{c o s h ( β _{1} L ) + c o s ( β _{1} L )}{s i n h ( β _{1} L ) + s i n ( β _{1} L )} \approx 0.7341$

Physical interpretation: Pehla mode ek single smooth curve hai, free end par maximum deflection, characteristic "cantilever bend."

Common Pitfalls

Common mistake Mistake 1: Natural Frequency aur Forcing Frequency Ko Confuse Karna

Galat soch: "Spacecraft launch ke dauran engine ki frequency par vibrate karta hai."

Yeh sahi kyun lagta hai: External forces apni frequency structure par impose karti hain.

Sach yeh hai: Natural frequencies intrinsic properties hain (jaise bell ki pitch)—yeh exist karti hain chahe structure forced ho ya na ho. Forcing frequency external hai. Resonance tab hoti hai jab forcing frequency natural frequency se match kare.

Steel-man: Aap sahi hain ki structure forcing frequency par respond karta hai. Lekin resonance par, $ω_{n}$ par ek chhoti si force bhi badi amplitudes cause karti hai kyunki mode "in tune" hai. $ω_{n}$ se door, response chhota hota hai.

Fix: Natural frequency = sirf structure ki property. Forcing frequency = external input. Dono match karein → disaster.

Common mistake Mistake 2: Mode Shape Normalization Ko Ignore Karna

Galat soch: "Mode shapes ke definite magnitudes hote hain— $ϕ_{1} = 2$ ka matlab 2 meters displacement hai."

Yeh sahi kyun lagta hai: Mode shapes displacement patterns hain, toh units length honi chahiye.

Sach yeh hai: Mode shapes relative patterns hain, absolute displacements nahi. Hum kisi mode shape ko kisi bhi constant se multiply kar sakte hain aur woh still ek valid solution hai (eigenvectors scaling tak hi unique hote hain). Typically normalize kiya jaata hai taaki ${ϕ}^{T} [M] {ϕ} = 1$ (unit modal mass).

Actual displacement modal coordinates se aata hai: $x (t) = ϕ_{i} q_{i} (t)$ , jahan $q_{i} (t)$ actual amplitude hai.

Fix: Mode shapes sirf shape dikhate hain. Amplitude modal coordinates se aati hai, jo initial conditions ya forcing se determine hoti hai.

Active Recall Practice

Recall Explain Like I'm Twelve (Feynman Technique)

Socho tumhare haath mein ek slinky latki hai. Agar tum dheerey se bottom ko tap karo, woh ek specific wiggly pattern mein upar-neeche bounce karta hai—woh uska mode shape hai. Aur woh ek certain speed par bounce karta hai (kitne bounces per second)—woh uski natural frequency hai.

Ab socho slinky sirf ek spring nahi, balki springs aur masses se bani ek poori jungle gym hai jo saath connected hai. Agar tum ise tap karo, alag-alag parts complicated patterns mein wigle karte hain. Lekin yahan cool baat yeh hai: kuch special patterns hote hain jahan har part ek dance routine ki tarah saath move karta hai, sab ek hi rhythm mein. Yahi natural modes hain.

Ek spacecraft us jungle gym jaisi hai. Jab rocket use launch ke dauran shake karta hai, toh woh jungle gym ko tap karne jaisa hai. Spacecraft apne favorite patterns (modes) mein shake karna chahta hai. Agar rocket bilkul sahi rhythm par shake kare (natural frequency), toh spacecraft zyada se zyada shake karta hai jab tak cheezein toot na jaayein—yahi resonance hai.

Engineers modal analysis karte hain taaki saare favorite patterns aur rhythms pata karein, aur yeh ensure karein ki rocket kabhi bhi un exact rhythms par na shake kare, spacecraft ko safe rakhte hue.

Flashcards

#flashcards/physics

Natural frequency kya hoti hai? :: Ek intrinsic frequency jisme ek structure disturb kiye jaane ke baad freely vibrate karta hai, bina kisi external forcing ke. Yeh sirf mass aur stiffness properties par depend karti hai.

Mode shape kya hoti hai?

Ek structure mein saari jagah relative displacements ka spatial pattern jab woh kisi particular natural frequency par vibrate karta hai. Yeh dimensionless aur normalized hoti hai.

Mode shapes ek orthogonal set kyun banate hain?

Kyunki mass aur stiffness matrices symmetric aur positive definite hain. Mathematically,

i \neq = j

ke liye:

{ϕ_{j}}^{T} [M] {ϕ_{i}} = 0

aur

{ϕ_{j}}^{T} [K] {ϕ_{i}} = 0

Structural dynamics mein resonance kya hai?

Jab forcing frequency natural frequency se match kare (

ω_{forcing} = ω_{n}

), amplitude bina bound ke badhne lagti hai (undamped) ya bahut badi values tak (lightly damped), jo structural failure ka kaaran ban sakti hai.

Modal analysis mein equations of motion ko decouple kaise karte hain?

Physical coordinates

{x}

ko modal matrix use karke modal coordinates

{q}

mein transform karo:

{x} = [Φ] {q}

. Yeh system ko

n

independent single-DOF oscillators mein diagonalize karta hai.

Mode

i

ke liye generalized mass kya hai?

Structure ka effective mass jab woh mode

i

mein vibrate kar raha ho,

M_{i}^{*} = {ϕ_{i}}^{T} [M] {ϕ_{i}}

se calculate kiya jaata hai. Natural frequency compute karne ke liye use hota hai:

ω_{i} = K_{i}^{*} / M_{i}^{*}

Spacecraft dynamics mein higher modes less important kyun hote hain?

Kyunki typical forcing functions (launch vibrations, thrusters) ki zyaatar energy low frequencies par hoti hai, aur higher modes ke smaller modal participation factors hote hain. Pehle 10-50 modes typically >95% response capture karte hain.

Characteristic equation

det ([K] - ω^{2} [M]) = 0

kya represent karti hai?

Eigenvalue problem ke nontrivial solutions ki condition. Iske roots

ω_{i}^{2}

hain, yaani squared natural frequencies.

n

-DOF system ke liye, yeh

n

th order polynomial hai.

Stiffness natural frequency ko kaise affect karti hai?

Natural frequency stiffness ke square root ke saath badhti hai:

ω = k / m

. Stiffness double karne se frequency

2 \approx 1.41 \times

badhti hai, use low-frequency launch loads se door le jaati hai.

Modal truncation kya hai?

Modal analysis mein sirf pehle

m

modes rakhna (jahan

m < n

), higher modes jo response mein negligibly contribute karte hain unhe discard karna. Typically modes 2× highest forcing frequency tak retain kiye jaate hain.