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ME-502 · Mechanical Vibrations/Quick Revision Short Notes

Mechanical Vibrations (ME-502) - Unit 1 Short Notes

UNIT 1: FUNDAMENTALS OF VIBRATION & SINGLE-DEGREE-OF-FREEDOM SYSTEMS


1.0 INTRODUCTION & BASIC CONCEPTS

1.1 Importance and Study of Vibrations

  • Causes: Unbalanced forces, friction, elastic deformation, wind, seismic activity.

  • Effects: Fatigue failure, noise, discomfort, malfunction, structural failure.

  • Applications: Vibratory feeders, compactors, ultrasonic cleaners, musical instruments, earthquake-resistant design.

  • Goal: Understand, predict, control (reduce or utilize) vibrations.

1.2 Basic Elements of a Vibratory System

Element Symbol Function Idealization
Mass/Inertia $m$ or $J$ Stores kinetic energy Lumped mass or rotary inertia
Spring $k$ (linear) or $$\displaystyle k_t $$ (torsional) Stores potential energy, provides restoring force Linear elastic, stiffness constant
Damper $c$ (viscous) Dissipates energy (heat) Force $\propto$ velocity: $$\displaystyle F_d = c\dot{x} $$

1.3 Degree of Freedom (DOF)

  • Definition: Minimum number of independent coordinates required to define the system's configuration at any instant.

  • Types:

    • Translational DOF: Linear motion (e.g., $x$).

    • Rotational DOF: Angular motion (e.g., $\theta$).

  • Examples:

    • 1-DOF: Simple spring-mass-damper, pendulum (small angle).

    • 2-DOF: Two masses on three springs, car suspension (heave & pitch).

    • Multi-DOF: Building frames, multi-mass shafts.

[!TIP] Exam Focus: Be able to identify DOF from diagrams. A rigid body in plane has 3 DOF (2 trans + 1 rotation).

1.4 Modeling: Lumped vs. Distributed Parameters

  • Lumped Parameter: Mass, stiffness, damping concentrated at discrete points. Used for SDOF & MDOF. Governing ODEs.

  • Distributed Parameter: Properties vary continuously (e.g., beam, rod). Governing PDEs. Approximated by lumped models.

1.5 Classification of Vibrations

Basis Types Key Characteristics
Energy Input Free Initial energy only; response decays (damped) or persists (undamped).
Forced Continuous external force; steady-state + transient response.
Damping Undamped No energy loss; perpetual oscillation.
Damped Energy dissipation; amplitude decays.
Linearity Linear Superposition holds; $$\displaystyle m\ddot{x}+c\dot{x}+kx=F(t) $$.
Non-linear Superposition fails; e.g., Coulomb friction, large deflection.
Force Nature Deterministic Precisely known (harmonic, periodic).
Random Probabilistic description (e.g., road roughness, wind).

1.6 Vibration Analysis Approach

  1. Modeling: Idealize physical system → mathematical model (lumped parameters).

  2. Equation of Motion (EOM): Derive using Newton's laws or energy methods (Lagrange).

  3. Solution: Solve ODE/PDE (analytical/numerical) for displacement $x(t)$.

  4. Interpretation: Extract natural frequency, damping ratio, mode shapes, response amplitude.


2.0 UNDAMPED SINGLE-DEGREE-OF-FREEDOM (SDOF) SYSTEMS

2.1 EOM for Undamped Spring-Mass System

System: Mass $m$, spring stiffness $k$, no damping, force $F(t)$.

DiagramCANVAS: Simple horizontal spring-mass system with coordinate x from equilibrium

Derivation (Newton's 2nd Law):

$$\sum F = m\ddot{x} \quad \Rightarrow \quad -kx + F(t) = m\ddot{x}$$

Standard Form:

$$m\ddot{x} + kx = F(t)$$

For free vibration ($$\displaystyle F=0 $$):

$$\boxed{m\ddot{x} + kx = 0}$$

2.2 Simple Harmonic Motion (SHM)

  • Mathematical Definition: Solution to $$\displaystyle \ddot{x} + \omega_n^2 x = 0 $$ is $$\displaystyle x(t) = C_1 \sin(\omega_n t) + C_2 \cos(\omega_n t) $$.

  • Natural Frequency ($$\displaystyle \omega_n $$):

$$\omega_n = \sqrt{\frac{k}{m}} \quad \text{(rad/s)}$$

  • Time Period ($$\displaystyle T_n $$):

$$T_n = \frac{2\pi}{\omega_n} = 2\pi\sqrt{\frac{m}{k}} \quad \text{(s)}$$

  • Vector Representation:

    DiagramCANVAS: Phasor diagram showing rotating vector for displacement x, velocity (90° lead), acceleration (180° lead)
    • Displacement: $$\displaystyle x = X \sin(\omega_n t) $$ or $$\displaystyle X \cos(\omega_n t) $$.

    • Velocity: $$\displaystyle \dot{x} = \omega_n X \cos(\omega_n t) $$ (leads $x$ by $$\displaystyle 90^\circ $$).

    • Acceleration: $$\displaystyle \ddot{x} = -\omega_n^2 X \sin(\omega_n t) $$ (lags $x$ by $$\displaystyle 180^\circ $$).

2.3 Energy Method (Rayleigh's Method)

  • Principle: For undamped, conservative systems, max KE = max PE.

$$\left( \frac{1}{2}m\dot{x}_{max}^2 \right)_{t=0} = \left( \frac{1}{2}kx_{max}^2 \right)_{t=\frac{T_n}{4}}$$

  • Natural Frequency Estimate:

$$\omega_n^2 \approx \frac{k_{eq}}{m} \quad \text{where} \quad k_{eq} \text{ is stiffness "seen" by mass.}$$

> [!CAUTION] Rayleigh's method gives **upper bound** for $$\displaystyle \omega_n $$ if the assumed mode shape is exact; otherwise an approximation.

2.4 Natural Frequency with Multiple Springs

  • Equivalent Stiffness:

    • Parallel: $$\displaystyle k_{eq} = k_1 + k_2 + ... $$

    • Series: $$\displaystyle \frac{1}{k_{eq}} = \frac{1}{k_1} + \frac{1}{k_2} + ... $$

  • Complex Systems: Reduce stepwise to single $$\displaystyle k_{eq} $$.

2.5 Natural Frequency: Rotational & Pendulum Systems

  • Torsional System: $$\displaystyle J\ddot{\theta} + k_t\theta = 0 $$

$$\omega_n = \sqrt{\frac{k_t}{J}}$$

> 
DiagramCANVAS: Torsional system with disk of inertia J, torsional spring kt
  • Simple Pendulum (small $\theta$):

$$\omega_n = \sqrt{\frac{g}{L}}$$

  • Compound Pendulum:

$$\omega_n = \sqrt{\frac{mgh}{J_O}} \quad \text{where } h \text{ is distance from pivot to CG.}$$

2.6 Natural Frequency: Rigid Body Motion

  • Example: Mass on inclined plane with spring.

    DiagramCANVAS: Mass on frictionless incline angle α, spring parallel to incline
    • Static equilibrium: $$\displaystyle mg\sin\alpha = k\delta_{st} $$.

    • EOM about equilibrium: $$\displaystyle m\ddot{x} + kx = 0 $$ (same as horizontal!).

    • $$\displaystyle \omega_n = \sqrt{k/m} $$ independent of gravity/incline angle.


3.0 DAMPED SINGLE-DEGREE-OF-FREEDOM SYSTEMS

3.1 Types of Damping

Type Force Law Characteristics Common in
Viscous $$\displaystyle F_d = c\dot{x} $$ Linear, frequency-independent, easy math. Hydraulic dampers, lubricated parts. HIGHLY FREQUENT
Coulomb (Dry Friction) $$\displaystyle F_d = \mu N \cdot \text{sgn}(\dot{x}) $$ Constant magnitude, non-linear, stick-slip. Brakes, unlubricated joints.
Structural/Hysteresis $$\displaystyle F_d = \beta \dot{x} $$ (complex $k$) Energy loss per cycle $\propto$ displacement^2. Materials (metal fatigue, rubber).
Solid Damping $$\displaystyle F_d = b\dot{x} + d\dot{x}^3 $$ Combination of viscous & non-linear. Some polymers at large amplitudes.

3.2 EOM for Viscously Damped System

$$m\ddot{x} + c\dot{x} + kx = F(t)$$

Free vibration ($$\displaystyle F=0 $$):

$$m\ddot{x} + c\dot{x} + kx = 0 \quad \text{or} \quad \ddot{x} + 2\zeta\omega_n\dot{x} + \omega_n^2 x = 0$$

where:

  • $$\displaystyle \omega_n = \sqrt{k/m} $$ (undamped natural frequency)

  • Critical Damping Constant: $$\displaystyle c_c = 2\sqrt{km} = 2m\omega_n $$

  • Damping Ratio: $$\displaystyle \zeta = \frac{c}{c_c} $$ (dimensionless, key parameter)

3.3 Solution for Free Vibration (Viscous)

Assume solution $$\displaystyle x = e^{st} $$. Substituting gives characteristic equation:

$$ms^2 + cs + k = 0 \quad \Rightarrow \quad s^2 + 2\zeta\omega_n s + \omega_n^2 = 0$$

Roots: $$\displaystyle s_{1,2} = -\zeta\omega_n \pm \omega_n\sqrt{\zeta^2 - 1} $$

Damping Case Condition Roots $s$ Response $x(t)$ Behavior
Underdamped $$\displaystyle \zeta < 1 $$ $$\displaystyle -\zeta\omega_n \pm i\omega_d $$ $$\displaystyle e^{-\zeta\omega_n t}(C_1\sin\omega_d t + C_2\cos\omega_d t) $$ Oscillatory, amplitude decays.
Critically Damped $$\displaystyle \zeta = 1 $$ $$\displaystyle -\omega_n $$ (repeated) $$\displaystyle (C_1 + C_2 t)e^{-\omega_n t} $$ Fastest return to equilibrium without oscillation.
Overdamped $$\displaystyle \zeta > 1 $$ Two real, negative $$\displaystyle C_1e^{s_1 t} + C_2e^{s_2 t} $$ Slow return, no oscillation.

Damped Natural Frequency ($$\displaystyle \zeta < 1 $$):

$$\omega_d = \omega_n\sqrt{1 - \zeta^2}$$

3.4 Logarithmic Decrement ($\delta$)

  • Definition: Natural log of ratio of successive amplitudes $n$ cycles apart.

$$\delta = \ln\left(\frac{x(t)}{x(t+nT_d)}\right) \approx \frac{2\pi\zeta}{\sqrt{1-\zeta^2}} \quad \text{(for $$\displaystyle \zeta < 0.2 $$)}$$

> 
DiagramCANVAS: Decaying envelope of underdamped response showing peaks at $$\displaystyle x_1, x_2, ... $$
  • Exact Formula:

$$\boxed{\delta = \frac{2\pi\zeta}{\sqrt{1-\zeta^2}}}$$

  • For small $\zeta$ ($$\displaystyle <0.2 $$): $\delta \approx 2\pi\zeta$.

  • Measurement: $$\displaystyle \zeta = \frac{\delta}{\sqrt{4\pi^2 + \delta^2}} $$.

[!TIP] Exam Focus: Interconversion between $c$, $$\displaystyle c_c $$, $\zeta$, $\delta$ is VERY FREQUENT. Remember: $$\displaystyle \zeta = c/c_c $$, $\delta \approx 2\pi\zeta$ for light damping.

3.5 Measurement of Damping

  • Logarithmic Decrement Method: Measure $\delta$ from free decay response → compute $\zeta$.

  • Half-Power Bandwidth Method (Forced Vibration): From MF curve, measure bandwidth $\Delta\omega$ at $$\displaystyle MF = 1/\sqrt{2} $$ of peak. $$\displaystyle \zeta \approx \frac{\Delta\omega}{2\omega_n} $$.

3.6 Comparison: Underdamped vs. Coulomb Damping

Feature Viscous Damping Coulomb Damping
Force Law $$\displaystyle F_d = c\dot{x} $$ $$\displaystyle F_d = \mu N \cdot \text{sgn}(\dot{x}) $$
Amplitude Decay Exponential: $$\displaystyle x \propto e^{-\zeta\omega_n t} $$ Linear: $$\displaystyle x = x_0 - \frac{4\mu N}{m\omega_n}t $$ per half-cycle
Frequency Slightly less than $$\displaystyle \omega_n $$: $$\displaystyle \omega_d = \omega_n\sqrt{1-\zeta^2} $$ Independent of amplitude: $$\displaystyle \omega_n' \approx \omega_n\sqrt{1 - \frac{4\mu N}{\pi m\omega_n x_0}} $$
Math Linear, solvable analytically. Non-linear; requires approximate or numerical methods.

3.7 Special Cases & Interconversion

Given any two of: $c$, $$\displaystyle c_c $$, $\zeta$, $\delta$ → find others.

  • $$\displaystyle \zeta = c / c_c = c / (2\sqrt{km}) $$

  • $$\displaystyle \delta = \frac{2\pi\zeta}{\sqrt{1-\zeta^2}} $$

  • $$\displaystyle c = \zeta \cdot 2\sqrt{km} $$


4.0 FORCED VIBRATION OF SDOF SYSTEMS

4.1 Forcing Functions

  • Harmonic: $$\displaystyle F(t) = F_0 \sin \omega t $$ (most common).

  • Periodic: Can be expressed as Fourier series (sum of harmonics).

  • Arbitrary: Solved via convolution/Laplace transform.

4.2 EOM

$$m\ddot{x} + c\dot{x} + kx = F_0 \sin \omega t$$

4.3 Steady-State Solution (Particular Solution)

Method: Complex Notation (Phasors).

Assume complex force: $$\displaystyle \tilde{F} = F_0 e^{i\omega t} $$ and complex response: $$\displaystyle \tilde{x} = X e^{i\omega t} $$.

Substitute into EOM (with $$\displaystyle d/dt \rightarrow i\omega $$):

$$(-\omega^2 m + i\omega c + k)\tilde{x} = \tilde{F}$$

$$\tilde{x} = \frac{F_0}{k - m\omega^2 + i c\omega} = \frac{F_0/k}{1 - r^2 + i(2\zeta r)}$$

where $$\displaystyle r = \omega / \omega_n $$ (frequency ratio).

Magnitude (Magnification Factor, MF):

$$X = \frac{F_0/k}{\sqrt{(1-r^2)^2 + (2\zeta r)^2}}$$

$$\boxed{M = \frac{X}{X_{st}} = \frac{1}{\sqrt{(1-r^2)^2 + (2\zeta r)^2}}}$$

where $$\displaystyle X_{st} = F_0/k $$ (static deflection).

Phase Angle ($\phi$): Phase of displacement lag behind force.

$$\tan\phi = \frac{2\zeta r}{1 - r^2}$$

DiagramCANVAS: MF vs r curve for different ζ, showing resonance peak and phase transition from 0° to 180°

4.4 Analysis of MF Curve

  • Effect of Damping ($\zeta$):

    • Decreases peak magnification $$\displaystyle M_{max} $$.

    • Widens bandwidth (frequency range where $$\displaystyle M > 1 $$).

    • Eliminates infinite resonance at $$\displaystyle r=1 $$ for $$\displaystyle \zeta > 0 $$.

  • Resonance: Occurs near $$\displaystyle r=1 $$. For light damping ($$\displaystyle \zeta < 0.2 $$), resonant frequency:

$$r_{max} \approx \sqrt{1 - 2\zeta^2} \quad \Rightarrow \quad \omega_{max} \approx \omega_n\sqrt{1 - 2\zeta^2}$$

$$\displaystyle M_{max} \approx \frac{1}{2\zeta} $$.
  • Regions:

    • $r \ll 1$ (Rigid Body): $M \approx 1$, $$\displaystyle \phi \approx 0^\circ $$.

    • $r \gg 1$ (High Freq/Isolation): $$\displaystyle M \approx 1/r^2 $$, $$\displaystyle \phi \approx 180^\circ $$.

4.5 Total Response

$$x(t) = x_c(t) + x_p(t)$$

  • $$\displaystyle x_c(t) $$: Complementary (Transient) solution of homogeneous equation (free vibration). Decays with time if $$\displaystyle \zeta > 0 $$.

  • $$\displaystyle x_p(t) $$: Particular (Steady-State) solution: $$\displaystyle x_p(t) = X \sin(\omega t - \phi) $$.

  • Given ICs: $$\displaystyle x(0)=x_0 $$, $$\displaystyle \dot{x}(0)=v_0 $$, determine $$\displaystyle C_1, C_2 $$ in $$\displaystyle x_c(t) $$.

[!TIP] Exam Focus: "Find total response" problems are FREQUENT. Compute $$\displaystyle \omega_n $$, $\zeta$, $X$, $\phi$, then apply ICs to $$\displaystyle x_c + x_p $$ at $$\displaystyle t=0 $$.

4.6 Forced Vibration with Base Excitation

System: Base motion $y(t)$, mass relative displacement $$\displaystyle z = x - y $$.

DiagramCANVAS: Mass-spring-damper on moving base y(t), relative coordinate z

EOM (relative to base):

$$m\ddot{z} + c\dot{z} + kz = -m\ddot{y}$$

If $$\displaystyle y = Y \sin \omega t $$, then RHS is $$\displaystyle -m\omega^2 Y \sin \omega t $$.

  • Displacement Transmissibility ($$\displaystyle T_d $$): Ratio of transmitted base amplitude to machine amplitude.

$$T_d = \frac{X}{Y} = \frac{r^2}{\sqrt{(1-r^2)^2 + (2\zeta r)^2}}$$

  • Force Transmissibility ($$\displaystyle T_f $$): Ratio of transmitted force to static force.

$$T_f = \frac{\sqrt{(kX)^2 + (c\omega X)^2}}{kY} = \sqrt{1 + (2\zeta r)^2} \cdot T_d$$

Vibration Isolation:

  • Goal: Reduce transmitted force/acceleration to foundation ($$\displaystyle T_f < 1 $$).

  • Condition: Requires $$\displaystyle r > \sqrt{2} $$ and some damping ($$\displaystyle \zeta > 0 $$).

  • Design: Use soft springs (low $$\displaystyle \omega_n $$) to achieve $$\displaystyle r = \omega/\omega_n > \sqrt{2} $$ for operating frequency $\omega$.


5.0 APPLICATIONS OF SDOF FORCED VIBRATION

5.1 Vibration Isolation Design

  • Given: Machine weight $$\displaystyle W = mg $$, operating speed $\omega$, desired transmissibility $$\displaystyle T_f $$.

  • Steps:

    1. Find required $$\displaystyle r = \omega/\omega_n $$ from $$\displaystyle T_f $$ formula (often iterative).

    2. Compute $$\displaystyle \omega_n = \omega / r $$.

    3. Stiffness: $$\displaystyle k = m\omega_n^2 $$.

    4. Damping: Choose $\zeta$ (typically 0.05–0.2) → $$\displaystyle c = \zeta \cdot 2\sqrt{km} $$.

5.2 Rotating Unbalance

  • Model: Mass $$\displaystyle m_u $$ rotating at radius $e$ at speed $\omega$ → harmonic force $$\displaystyle F_0 = m_u e \omega^2 $$.

  • Steady-State Amplitude:

$$X = \frac{m_u e r^2}{\sqrt{(1-r^2)^2 + (2\zeta r)^2}}$$

where $$\displaystyle r = \omega / \omega_n $$, $$\displaystyle \omega_n = \sqrt{k/(m+m_u)} $$.
  • Critical Speed: Speed at which $$\displaystyle r=1 $$ → amplitude large unless damped.

5.3 Speed-Dependent Excitation (Vehicle on Sinusoidal Road)

  • Model: Base excitation $$\displaystyle y = Y \sin(\frac{2\pi}{\lambda} v t) = Y \sin \omega t $$, where $$\displaystyle \omega = 2\pi v / \lambda $$.

  • Amplitude of Vibration (relative to base):

$$Z = Y \cdot \frac{r^2}{\sqrt{(1-r^2)^2 + (2\zeta r)^2}}$$

  • Critical Speed: $$\displaystyle v_{cr} = \frac{\lambda \omega_n}{2\pi} $$ (when $$\displaystyle r=1 $$).

5.4 Recoil Mechanisms (Gun Recoil)

  • Goal: Limit recoil distance $$\displaystyle x_{max} $$ and time.

  • Model: Gun mass $M$, recoil spring $k$, damper $c$. Initial recoil velocity $$\displaystyle v_0 $$.

  • Design for Critical Damping ($$\displaystyle \zeta=1 $$):

    • $$\displaystyle c_c = 2\sqrt{kM} $$.

    • Maximum recoil: $$\displaystyle x_{max} = v_0 / \omega_n $$ (since critically damped, no overshoot).

    • Given $$\displaystyle x_{max} $$ and $$\displaystyle v_0 $$ → $$\displaystyle \omega_n = v_0 / x_{max} $$ → $$\displaystyle k = M \omega_n^2 $$.

5.5 Whirling of Shafts

  • Definition: Lateral vibration of a rotating shaft due to unbalance, causing the shaft axis to whirl (rotate in a curved path).

  • Critical Speed ($$\displaystyle N_c $$): Rotational speed at which $\omega$ (rotational) = $$\displaystyle \omega_n $$ (natural frequency of transverse vibration).

  • Proof: For a simple rotor on massless shaft:

    • Unbalance force: $$\displaystyle F = m e \omega^2 $$.

    • EOM: $$\displaystyle m\ddot{x} + kx = m e \omega^2 \sin \omega t $$.

    • Steady-state amplitude: $$\displaystyle X = \frac{m e \omega^2 / k}{1 - (\omega/\omega_n)^2} $$.

    • As $$\displaystyle \omega \to \omega_n $$, $X \to \infty$ (theoretically). This speed is critical.

  • Factors Affecting $$\displaystyle N_c $$: Shaft length, diameter, material ($E$), support stiffness, rotor mass & eccentricity.


6.0 MULTI-DEGREE-OF-FREEDOM (MDOF) SYSTEMS - INTRODUCTION

6.1 Modeling & Coordinates

  • DOF: Number of independent coordinates needed.

  • Coordinates: Cartesian ($$\displaystyle x_1, x_2 $$), angular ($$\displaystyle \theta_1, \theta_2 $$), or modal (principal coordinates).

6.2 Formulation of EOM

  • Newton's 2nd Law (FBD): Draw free body diagram for each mass → write $$\displaystyle \sum F = m_i\ddot{x}_i $$. FREQUENT IN EXAMS.

    DiagramCANVAS: 2-DOF spring-mass system with masses m1, m2, springs k1, k2, k3, coordinates x1, x2
  • Rayleigh's Energy Method: For approximate $$\displaystyle \omega_n $$ (usually fundamental). Assume mode shape → equate max KE & PE.

6.3 Matrix Form

$$\boxed{[M]\ddot{\mathbf{x}} + [C]\dot{\mathbf{x}} + [K]\mathbf{x} = \mathbf{F}(t)}$$

where:

  • $[M]$: Mass matrix (symmetric, positive definite).

  • $[K]$: Stiffness matrix (symmetric, positive semi-definite).

  • $[C]$: Damping matrix (often proportional: $$\displaystyle [C] = \alpha[M] + \beta[K] $$).

  • $\mathbf{x}, \mathbf{F}$: Displacement & force vectors.

6.4 Free Vibration of Undamped MDOF

EOM: $$\displaystyle [M]\ddot{\mathbf{x}} + [K]\mathbf{x} = \mathbf{0} $$.

Assume harmonic solution: $$\displaystyle \mathbf{x}(t) = \boldsymbol{\phi} \sin(\omega t + \phi) $$.

Substitute → Eigenvalue Problem:

$$\left([K] - \omega^2[M]\right)\boldsymbol{\phi} = \mathbf{0}$$

  • Natural Frequencies ($$\displaystyle \omega_i $$): Roots of characteristic equation $$\displaystyle \det([K] - \omega^2[M]) = 0 $$. For $n$-DOF, $n$ eigenvalues $$\displaystyle \omega_1^2 \le \omega_2^2 \le ... \le \omega_n^2 $$.

  • Mode Shapes ($$\displaystyle \boldsymbol{\phi}^{(i)} $$): Corresponding eigenvectors (shape of vibration at $$\displaystyle \omega_i $$). Defined up to a scale factor. Often mass-normalized: $$\displaystyle {\boldsymbol{\phi}^{(i)}}^T[M]\boldsymbol{\phi}^{(i)} = 1 $$.

  • Nodes: Points in a mode shape with zero displacement.

  • Orthogonality Properties:

    1. Mass Orthogonality: $$\displaystyle {\boldsymbol{\phi}^{(i)}}^T[M]\boldsymbol{\phi}^{(j)} = 0 $$ for $i \neq j$.

    2. Stiffness Orthogonality: $$\displaystyle {\boldsymbol{\phi}^{(i)}}^T[K]\boldsymbol{\phi}^{(j)} = 0 $$ for $i \neq j$.

    3. Generalized Orthogonality: $$\displaystyle {\boldsymbol{\phi}^{(i)}}^T([K] - \omega_j^2[M])\boldsymbol{\phi}^{(j)} = 0 $$ for $i \neq j$.

Solution Methods:

  • Determinant (Characteristic Equation): For small $n$ (2-3 DOF). Solve $$\displaystyle \det([K] - \omega^2[M]) = 0 $$ for $$\displaystyle \omega_i $$, then find $$\displaystyle \boldsymbol{\phi}^{(i)} $$ from $$\displaystyle ([K] - \omega_i^2[M])\boldsymbol{\phi} = \mathbf{0} $$.

  • Matrix Iteration (Power Method): Finds fundamental ($$\displaystyle \omega_1 $$) mode shape. Start with guess vector $$\displaystyle \mathbf{x}_0 $$, compute $$\displaystyle \mathbf{x}_1 = [K]^{-1}[M]\mathbf{x}_0 $$, normalize, repeat → converges to $$\displaystyle \boldsymbol{\phi}^{(1)} $$.

  • Jacobi Method: For all eigenvalues/eigenvectors (conceptual).

6.5 Free Vibration Response via Mode Superposition

  • Assumption: Total response is linear combination of normal modes:

$$\mathbf{x}(t) = \sum_{i=1}^{n} q_i(t) \boldsymbol{\phi}^{(i)}$$

where $$\displaystyle q_i(t) $$ are **generalized (modal) coordinates**.
  • **Substitute into EOM:**利用 orthogonality → uncoupled equations:

$$\ddot{q}_i + \omega_i^2 q_i = 0 \quad \text{(undamped)}$$

Solution: $$\displaystyle q_i(t) = A_i \sin(\omega_i t) + B_i \cos(\omega_i t) $$.
  • Determine $$\displaystyle A_i, B_i $$ from ICs: Expand initial conditions $$\displaystyle \mathbf{x}_0, \dot{\mathbf{x}}_0 $$ in terms of mode shapes:

$$\mathbf{x}_0 = \sum_{i=1}^{n} q_i(0) \boldsymbol{\phi}^{(i)} = [\boldsymbol{\Phi}]\mathbf{q}_0$$

where $$\displaystyle [\boldsymbol{\Phi}] = [\boldsymbol{\phi}^{(1)} \boldsymbol{\phi}^{(2)} ... \boldsymbol{\phi}^{(n)}] $$.

Multiply by $$\displaystyle {\boldsymbol{\phi}^{(j)}}^T[M] $$ → $$\displaystyle q_i(0) = \frac{{\boldsymbol{\phi}^{(i)}}^T[M]\mathbf{x}_0}{{\boldsymbol{\phi}^{(i)}}^T[M]\boldsymbol{\phi}^{(i)}} $$ (if mass-normalized, denominator=1).

6.6 Principal (Modal) Coordinates

  • Definition: Coordinates $$\displaystyle q_i(t) $$ that decouple the equations of motion. They represent the amplitude of each normal mode.

  • Determination: $$\displaystyle \mathbf{q} = [\boldsymbol{\Phi}]^{-1}\mathbf{x} $$ if $[\boldsymbol{\Phi}]$ is invertible (non-defective systems). For orthogonal modes, $$\displaystyle [\boldsymbol{\Phi}]^T[M][\boldsymbol{\Phi}] = [I] $$ (mass-normalized) → $$\displaystyle \mathbf{q} = [\boldsymbol{\Phi}]^T[M]\mathbf{x} $$.

  • Significance: In principal coordinates, each DOF vibrates independently at its own natural frequency.


7.0 SPECIAL TOPICS & NOISE FUNDAMENTALS

7.1 Fourier Series Expansion

  • Purpose: Represent periodic, non-sinusoidal forcing function $F(t)$ as sum of sinusoids.

  • Fourier Series (General):

$$F(t) = a_0 + \sum_{n=1}^{\infty} \left( a_n \cos n\omega_0 t + b_n \sin n\omega_0 t \right)$$

where $$\displaystyle \omega_0 = 2\pi/T $$ (fundamental frequency).
  • For Impact Force (e.g., forging hammer): Often represented by half-range sine/cosine series depending on symmetry.

    DiagramCANVAS: Periodic impulse force F(t) with period T, showing its Fourier series approximation
  • Response: System responds at each harmonic $$\displaystyle n\omega_0 $$ with amplitude $$\displaystyle X_n = \frac{F_n / k}{\sqrt{(1-r_n^2)^2 + (2\zeta r_n)^2}} $$, where $$\displaystyle r_n = n\omega_0 / \omega_n $$.

  • Total Steady-State Response: $$\displaystyle x(t) = \sum_{n=1}^{\infty} X_n \sin(n\omega_0 t - \phi_n) $$.

7.2 Torsional Vibration Systems

  • Elements: Rotational inertia $J$ (kg·m²), torsional stiffness $$\displaystyle k_t $$ (N·m/rad).

  • EOM for 2-DOF Torsional System:

    DiagramCANVAS: Two disks with inertias J1, J2 connected by shaft with stiffness kt1, kt2, fixed at one end

$$J_1\ddot{\theta}_1 + k_{t1}(\theta_1 - \theta_2) = 0$$

$$J_2\ddot{\theta}_2 + k_{t1}(\theta_2 - \theta_1) + k_{t2}\theta_2 = 0$$

  • Matrix Form: $$\displaystyle [J]\ddot{\boldsymbol{\theta}} + [K_t]\boldsymbol{\theta} = \mathbf{0} $$.

  • Natural Frequencies: Solve $$\displaystyle \det([K_t] - \omega^2[J]) = 0 $$.

  • Mode Shapes: Displacement ratios $$\displaystyle \theta_1/\theta_2 $$ at each $$\displaystyle \omega_i $$.

7.3 Noise Fundamentals (Applied Vibration)

Sound Pressure Level (SPL)

  • Definition: Logarithmic measure of sound pressure relative to reference.

$$L_p = 10 \log_{10}\left(\frac{p^2}{p_{ref}^2}\right) = 20 \log_{10}\left(\frac{p}{p_{ref}}\right) \quad \text{(dB)}$$

where $$\displaystyle p_{ref} = 20\ \mu\text{Pa} $$ (threshold of human hearing at 1 kHz).
  • Key Points:

    • 0 dB SPL does not mean no sound; it means sound pressure equals $$\displaystyle p_{ref} $$.

    • 6 dB increase ≈ doubling of sound pressure.

    • 10 dB increase ≈ perceived doubling of loudness.

Sound Power Level (SWL)

  • Definition: Logarithmic measure of total acoustic power radiated.

$$L_W = 10 \log_{10}\left(\frac{W}{W_{ref}}\right) \quad \text{(dB)}$$

where $$\displaystyle W_{ref} = 10^{-12}\ \text{W} $$.
  • Difference from SPL: SPL is location-dependent (distance from source), SWL is source property (total power output). SWL is fixed for a machine; SPL varies with distance.

Inverse Square Law

  • For a point source in free field, sound intensity $$\displaystyle I \propto 1/r^2 $$, sound pressure $p \propto 1/r$.

  • Doubling distance: SPL decreases by 6 dB.

$$\Delta L_p = 20 \log_{10}\left(\frac{r_2}{r_1}\right)$$

If $$\displaystyle r_2 = 2r_1 $$, $$\displaystyle \Delta L_p = 20 \log_{10}(2) \approx 6\ \text{dB} $$.

Octave Band Analysis

  • Purpose: Break complex noise into frequency bands to identify dominant frequencies (source diagnosis).

  • Octave Band: Frequency range where upper limit = 2 × lower limit. Center frequencies: 31.5, 63, 125, 250, 500, 1000, 2000, 4000, 8000 Hz.

  • 1/3 Octave Bands: More precise; each octave split into 3 bands (e.g., 1000 Hz octave → 800, 1000, 1250 Hz bands).

  • Output: SPL in each band → noise spectrum (plot of dB vs. band center freq).

Human Response & Hearing Conservation

  • Audible Range: 20 Hz – 20 kHz (decreases with age).

  • Damage Risk Criteria (Permissible Exposure Limits - PELs):

    • Based on A-weighting (dBA) to mimic human ear sensitivity.

    • OSHA/NIOSH standards: e.g., 90 dBA/8 hr (OSHA), 85 dBA/8 hr (NIOSH) with 3 dB exchange rate (NIOSH) or 5 dB (OSHA).

    • Exposure Time: Halved for every 3–5 dB increase above threshold.

  • Precautions & Remedies:

    1. Source Control: Balance rotors, use quieter processes, isolate vibrations.

    2. Path Control: Enclosures, barriers, acoustic lining, vibration isolation mounts.

    3. Receiver Protection: Earplugs, earmuffs, administrative controls (shift rotation).

[!TIP] Exam Focus: Know SPL formula, reference pressure, 6 dB/doubling distance rule, octave band purpose, and basic hearing conservation (dBA, exposure time trade-off).

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