ME 321: Dynamics of Machines and Mechanical Vibrations

Homeyra Pourmohammadali

Estimated study time: 1 hr 55 min

Table of contents

Sources and References

Primary textbooks — W.L. Cleghorn & N. Dechev, Mechanics of Machines, 2nd ed., Oxford University Press, 2015; Daniel J. Inman, Engineering Vibration, 5th ed., Pearson, 2022.

Online resources — MIT OCW 2.003SC Engineering Dynamics (Vandiver et al.); MIT OCW 2.14 Analysis and Design of Feedback Control Systems; S.S. Rao, Mechanical Vibrations, 6th ed., Pearson, 2017; W.T. Thomson & M.D. Dahleh, Theory of Vibration with Applications, 5th ed., Prentice Hall, 1998; B. Tongue, Principles of Vibration, 2nd ed., Oxford University Press, 2002; Cambridge Engineering Part IB Mechanics notes; NPTEL Lecture Series on Theory of Machines (IIT Kharagpur).


Chapter 1: Mechanisms — Kinematics and Degrees of Freedom

1.1 Machines, Mechanisms, and Kinematic Chains

A machine is a device that transforms energy and transmits or modifies motion to do useful work. A mechanism is the kinematic skeleton of a machine — the arrangement of rigid bodies (links) connected by joints that constrains and guides relative motion. Understanding the distinction is the first step toward systematic analysis.

Every mechanism is built from two primitive concepts: links and kinematic pairs. A link is a rigid body forming one member of a mechanism. A kinematic pair (joint) is a constraint between two links that permits certain relative motions while preventing others. When links and joints are assembled into a closed network they form a kinematic chain; when one link of the chain is fixed to a frame of reference the result is a mechanism.

Links are classified by the number of joints they carry:

  • Binary link: two joints
  • Ternary link: three joints
  • Quaternary link: four joints

The fixed link (ground) is called the frame or fixed link. The link that receives input motion is the crank or driver. The link that delivers output is the follower. The link connecting driver to follower (when neither is grounded) is the coupler or connecting rod.

1.1.2 Classification of Kinematic Pairs

Kinematic pairs are classified by the number of degrees of freedom they permit and by the type of surface contact.

Lower kinematic pairs have surface (area) contact between mating links. The six standard lower pairs in three-dimensional space are:

  • Revolute joint (R) — one rotational DOF; a pin or hinge. Symbol: circle at the pivot.
  • Prismatic joint (P) — one translational DOF; a slider in a guide.
  • Helical joint (H) — one DOF coupling rotation and translation (screw).
  • Cylindric joint (C) — two DOF: rotation and translation along a common axis.
  • Spherical joint (S) — three rotational DOF; a ball-and-socket.
  • Planar joint (E) — three DOF: two translations and one rotation in a plane.

Higher kinematic pairs have point or line contact (cam-follower, gear teeth). They generally impose one constraint and leave five DOF in 3D.

For the planar mechanisms studied in this course every revolute joint removes two DOF (prevents two translations in the plane, leaving one rotation) and every prismatic joint also removes two DOF (prevents one translation perpendicular to the slide direction and one rotation, leaving one translation). This accounting leads directly to Grübler’s equation.

1.2 Degrees of Freedom and Grübler’s Equation

Before any kinematic or kinetic analysis can be performed, one must know how many independent inputs are needed to uniquely specify the configuration of the mechanism. This count is the mobility or degrees of freedom of the mechanism.

Degrees of Freedom (DOF) of a Mechanism

The degrees of freedom (mobility) \( M \) of a mechanism is the number of independent generalized coordinates required to completely specify the positions of all links relative to the fixed frame. Equivalently, it is the number of independent driver inputs needed to produce definite, constrained motion.

1.2.1 Constraint Counting in the Plane

A single rigid body moving freely in the plane possesses 3 DOF (two translations and one rotation). An assembly of \( n \) links (excluding the fixed frame, which contributes no DOF) therefore has \( 3n \) DOF before any joints are introduced.

Each lower kinematic pair in the plane removes 2 DOF (it imposes 2 scalar constraints). Each higher pair removes 1 DOF. Let:

  • \( L \) = total number of links including the fixed frame
  • \( n = L - 1 \) = number of moving links
  • \( J_1 \) = number of lower pairs (full joints: revolute or prismatic)
  • \( J_2 \) = number of higher pairs (half joints: rolling contact, gear teeth)

Grübler’s Equation (Kutzbach criterion for planar mechanisms)

\[ M = 3(L - 1) - 2J_1 - J_2 \]

where \( L \) is the total link count (including the fixed frame), \( J_1 \) is the number of full joints, and \( J_2 \) is the number of half joints. A mechanism with \( M = 1 \) has a single input; \( M = 0 \) is a structure; \( M < 0 \) is overconstrained (statically indeterminate frame); \( M > 1 \) requires multiple simultaneous inputs for definite motion.

Derivation of Grübler’s Equation.

Start with \( L \) links. Fix one (the frame), leaving \( n = L - 1 \) moving links, each with 3 DOF in the plane. Total unconstrained DOF \( = 3(L-1) \). Each full joint (lower pair) connecting two links removes 2 DOF from the system. Each half joint (higher pair) removes 1 DOF. Therefore:

\[ M = 3(L - 1) - 2J_1 - J_2. \]

The equation follows directly from linear constraint counting and is exact provided (i) all constraints are independent, (ii) there are no special geometric conditions (idle DOF or paradoxes such as the Bennett linkage), and (iii) the mechanism is planar.

Example 1.1 — Four-Bar Linkage.

A four-bar linkage consists of a fixed frame (link 1), a crank (link 2), a coupler (link 3), and a rocker (link 4), connected at four revolute joints.

  • \( L = 4 \)
  • \( J_1 = 4 \) revolute joints, \( J_2 = 0 \)
\[ M = 3(4 - 1) - 2(4) - 0 = 9 - 8 = 1 \]

The four-bar linkage has one DOF — a single input (e.g., crank angle) fully determines the positions of all other links. This is the most important planar mechanism.

Example 1.2 — Slider-Crank Mechanism.

The slider-crank has: fixed frame (1), crank (2), connecting rod (3), slider/piston (4). There are three revolute joints (at the crank pin, crank–coupler, coupler–slider) and one prismatic joint (slider in guide).

  • \( L = 4 \), \( J_1 = 4 \) (3 revolute + 1 prismatic), \( J_2 = 0 \)
\[ M = 3(3) - 2(4) = 9 - 8 = 1 \]

One DOF: the crank angle drives piston translation. The slider-crank converts rotary to reciprocating motion and is the basis of virtually all internal combustion engines and piston compressors.

1.3 Mechanism Inversion

Mechanism Inversion

An inversion of a kinematic chain is obtained by fixing a different link as the frame while keeping all relative motions between links the same. An \( n \)-link chain has \( n \) distinct inversions, though some may be geometrically equivalent.

Inversions of the four-bar chain yield the four-bar linkage, the crank-rocker, the double-crank, and the double-rocker, depending on the relative link lengths. Inversions of the slider-crank chain yield the slider-crank (engine form), the Scotch yoke mechanism, oscillating-cylinder engines, and the fixed-block crank mechanism. Recognizing inversions allows the analyst to apply one set of kinematic equations to many superficially different devices.

1.4 Grashof’s Criterion

Not every four-bar linkage allows continuous rotation of all links. Grashof’s criterion determines which link can make a full rotation.

Grashof’s Criterion

For a four-bar linkage with link lengths \( s \) (shortest), \( l \) (longest), \( p \), and \( q \), define the Grashof condition:

\[ s + l \leq p + q \]
  • If the condition is satisfied, at least one link (specifically the shortest link or the link adjacent to the shortest) can make a complete revolution. The mechanism is called a Grashof mechanism.
  • If the condition is not satisfied (\( s + l > p + q \)), no link can make a full revolution — all links only oscillate. This is a non-Grashof mechanism.
  • If \( s + l = p + q \), the mechanism is a change-point (Grashof-special) mechanism capable of toggling between configurations.

Depending on which link is grounded, a Grashof chain yields:

  • Crank-rocker: ground the link adjacent to the shortest — the shortest link rotates fully (crank), and the opposite link rocks.
  • Double-crank (drag-link): ground the shortest link — both adjacent links rotate fully.
  • Double-rocker: ground the link opposite to the shortest — both cranks only oscillate. The coupler, however, can make a full revolution.

This classification is essential for mechanism synthesis: an engine needs a crank-rocker so the crank makes full revolutions continuously; a windshield wiper mechanism uses a double-rocker so both arms oscillate.

Example 1.3 — Grashof Classification.

A four-bar linkage has link lengths: \( l_1 = 40 \) mm (frame), \( l_2 = 15 \) mm, \( l_3 = 45 \) mm, \( l_4 = 30 \) mm.

Identify \( s = 15 \) mm, \( l = 45 \) mm, \( p = 40 \) mm, \( q = 30 \) mm.

Check Grashof: \( s + l = 15 + 45 = 60 \) mm; \( p + q = 40 + 30 = 70 \) mm.

Since \( 60 < 70 \), the condition is satisfied — Grashof mechanism. The shortest link is link 2 (the crank). Grounding link 1 (frame) gives a crank-rocker: link 2 (adjacent to shortest) fully rotates, link 4 rocks.


Chapter 2: Analytical Kinematic Analysis of Planar Mechanisms

2.1 Vector Loop Equations

The most systematic approach to kinematic analysis of planar mechanisms is the vector loop method. Each link is represented as a vector; the closed-loop condition (the vector sum around any closed loop must be zero) provides algebraic equations relating link positions, which can then be differentiated to obtain velocity and acceleration equations.

2.1.1 Position Analysis — The Loop Closure Equation

For a planar mechanism, represent link \( i \) of length \( r_i \) directed at angle \( \theta_i \) from the positive \( x \)-axis as the complex vector \( \mathbf{R}_i = r_i e^{i\theta_i} = r_i(\cos\theta_i + i\sin\theta_i) \). For a single closed loop:

\[ \mathbf{R}_2 + \mathbf{R}_3 - \mathbf{R}_4 - \mathbf{R}_1 = 0 \]

Separating real and imaginary parts gives two scalar equations:

\[ r_2 \cos\theta_2 + r_3 \cos\theta_3 - r_4 \cos\theta_4 - r_1 \cos\theta_1 = 0 \]\[ r_2 \sin\theta_2 + r_3 \sin\theta_3 - r_4 \sin\theta_4 - r_1 \sin\theta_1 = 0 \]

With \( \theta_1 = 0 \) (frame along \( x \)-axis) and \( \theta_2 \) as the known input, these two nonlinear equations in \( \theta_3 \) and \( \theta_4 \) can be solved analytically (Freudenstein equation) or numerically.

2.2 Velocity Analysis by Differentiation

Derivation of the Velocity Loop Equation.

Starting from the position loop equation

\[ r_2 e^{i\theta_2} + r_3 e^{i\theta_3} - r_4 e^{i\theta_4} - r_1 = 0, \]

differentiate with respect to time \( t \). The link lengths \( r_i \) are constant (rigid links). Using \( \frac{d}{dt}e^{i\theta_k} = i\dot{\theta}_k e^{i\theta_k} \):

\[ i r_2 \dot{\theta}_2 e^{i\theta_2} + i r_3 \dot{\theta}_3 e^{i\theta_3} - i r_4 \dot{\theta}_4 e^{i\theta_4} = 0. \]

Separating real and imaginary parts:

\[ -r_2 \dot{\theta}_2 \sin\theta_2 - r_3 \dot{\theta}_3 \sin\theta_3 + r_4 \dot{\theta}_4 \sin\theta_4 = 0 \]\[ r_2 \dot{\theta}_2 \cos\theta_2 + r_3 \dot{\theta}_3 \cos\theta_3 - r_4 \dot{\theta}_4 \cos\theta_4 = 0 \]

This is a linear system in the two unknowns \( \dot{\theta}_3 \) and \( \dot{\theta}_4 \) (with \( \dot{\theta}_2 = \omega_2 \) given). In matrix form:

\[ \begin{bmatrix} -r_3 \sin\theta_3 & r_4 \sin\theta_4 \\ r_3 \cos\theta_3 & -r_4 \cos\theta_4 \end{bmatrix} \begin{bmatrix} \dot{\theta}_3 \\ \dot{\theta}_4 \end{bmatrix} = \begin{bmatrix} r_2 \dot{\theta}_2 \sin\theta_2 \\ -r_2 \dot{\theta}_2 \cos\theta_2 \end{bmatrix}. \]

The system is solved by Cramer’s rule or matrix inversion, provided the determinant \( D = r_3 r_4 \sin(\theta_4 - \theta_3) \neq 0 \) (i.e., links 3 and 4 are not parallel — a singular configuration).

The angular velocities \( \dot{\theta}_3 = \omega_3 \) and \( \dot{\theta}_4 = \omega_4 \) found here completely describe how fast each link rotates. The velocity of any point on the coupler can then be found by vector addition.

2.3 Acceleration Analysis by Differentiation

Derivation of the Acceleration Loop Equation.

Differentiate the velocity loop equation with respect to time again. Starting from

\[ ir_2\omega_2 e^{i\theta_2} + ir_3\omega_3 e^{i\theta_3} - ir_4\omega_4 e^{i\theta_4} = 0, \]

differentiate, using \( \frac{d}{dt}(\omega_k e^{i\theta_k}) = \dot{\omega}_k e^{i\theta_k} + i\omega_k^2 e^{i\theta_k} \):

\[ i r_2 \alpha_2 e^{i\theta_2} - r_2\omega_2^2 e^{i\theta_2} + i r_3 \alpha_3 e^{i\theta_3} - r_3\omega_3^2 e^{i\theta_3} - ir_4\alpha_4 e^{i\theta_4} + r_4\omega_4^2 e^{i\theta_4} = 0, \]

where \( \alpha_k = \ddot{\theta}_k \). With \( \alpha_2 = 0 \) (constant-speed input), separating real and imaginary parts:

\[ -r_3\alpha_3\sin\theta_3 + r_4\alpha_4\sin\theta_4 = r_2\omega_2^2\cos\theta_2 + r_3\omega_3^2\cos\theta_3 - r_4\omega_4^2\cos\theta_4 \]\[ r_3\alpha_3\cos\theta_3 - r_4\alpha_4\cos\theta_4 = r_2\omega_2^2\sin\theta_2 + r_3\omega_3^2\sin\theta_3 - r_4\omega_4^2\sin\theta_4 \]

The right-hand sides are entirely known from the position and velocity analyses. Again a \( 2\times2 \) linear system in \( \alpha_3, \alpha_4 \) with the same coefficient matrix as the velocity system.

The acceleration equations always produce centripetal (centrifugal) terms \( r_k\omega_k^2 \) on the right-hand side — these are the normal (radial) acceleration components of each link’s coupler point. The tangential terms \( r_k\alpha_k \) appear on the left as unknowns. This structure is universal for single-loop planar mechanisms.

2.4 Kinematic Analysis of the Slider-Crank Mechanism

The slider-crank is analytically tractable in closed form. Let the crank (link 2) have length \( r \) and angle \( \theta \), the connecting rod (link 3) have length \( l \) and angle \( \phi \), and let \( x \) be the piston displacement from the crank center. The offset (eccentricity) is zero for a centered slider-crank.

2.4.1 Position

The closure equation in the \( x \)-direction gives:

\[ x = r\cos\theta + l\cos\phi \]

The closure in \( y \):

\[ 0 = r\sin\theta + l\sin\phi \implies \sin\phi = -\frac{r}{l}\sin\theta \]

Defining the crank ratio \( \lambda = r/l \):

\[ \cos\phi = \sqrt{1 - \lambda^2\sin^2\theta} \]

2.4.2 Velocity

Differentiating with respect to time:

\[ \dot{x} = -r\omega\sin\theta - l\dot{\phi}\sin\phi, \qquad 0 = r\omega\cos\theta + l\dot{\phi}\cos\phi \]

From the second equation: \( \dot{\phi} = -\frac{r\omega\cos\theta}{l\cos\phi} \). Substituting:

\[ \dot{x} = -r\omega\sin\theta + \frac{r^2\omega\cos\theta\sin\phi}{\cos\phi \cdot l} \]

For most engine analyses the approximation \( \lambda \ll 1 \) (long connecting rod) is used, giving:

\[ \dot{x} \approx -r\omega\left(\sin\theta + \frac{\lambda}{2}\sin 2\theta\right) \]

2.4.3 Acceleration

\[ \ddot{x} \approx -r\omega^2\left(\cos\theta + \lambda\cos 2\theta\right) \]

The slider-crank piston acceleration contains a fundamental harmonic \( \cos\theta \) and a second harmonic \( \lambda\cos 2\theta \). These harmonics are the physical source of engine vibration. In multi-cylinder engines, cranks are phased to cancel selected harmonics — the mechanical engineering basis of inline-6 and V8 balance strategies.

Example 2.1 — Velocity and Acceleration in a Four-Bar Linkage.

Given: \( r_1 = 100 \) mm, \( r_2 = 35 \) mm, \( r_3 = 85 \) mm, \( r_4 = 70 \) mm. Crank angle \( \theta_2 = 60° \), angular velocity \( \omega_2 = 10 \) rad/s (CCW). From position analysis (not shown here), \( \theta_3 = 320.5° \), \( \theta_4 = 278.3° \).

Setting up the velocity system:

\[ \begin{bmatrix} -85\sin(320.5°) & 70\sin(278.3°) \\ 85\cos(320.5°) & -70\cos(278.3°) \end{bmatrix} \begin{bmatrix} \omega_3 \\ \omega_4 \end{bmatrix} = \begin{bmatrix} 35(10)\sin(60°) \\ -35(10)\cos(60°) \end{bmatrix} \]

Evaluating: \( \sin(320.5°) \approx -0.648 \), \( \cos(320.5°) \approx 0.762 \), \( \sin(278.3°) \approx -0.990 \), \( \cos(278.3°) \approx 0.143 \).

\[ \begin{bmatrix} 55.1 & -69.3 \\ 64.8 & -10.0 \end{bmatrix} \begin{bmatrix} \omega_3 \\ \omega_4 \end{bmatrix} = \begin{bmatrix} 303.1 \\ -175.0 \end{bmatrix} \]

Solving by Cramer’s rule: determinant \( D = (55.1)(-10.0) - (-69.3)(64.8) = -551 + 4490 = 3939 \).

\( \omega_3 = \frac{(303.1)(-10.0) - (-69.3)(-175.0)}{3939} = \frac{-3031 - 12128}{3939} = \frac{-15159}{3939} \approx -3.85 \) rad/s (CW).

\( \omega_4 = \frac{(55.1)(-175.0) - (64.8)(303.1)}{3939} = \frac{-9643 - 19641}{3939} \approx -7.43 \) rad/s (CW).


Chapter 3: Kinetic Analysis, Force Transmission, and Balancing

3.1 Force Analysis of Linkages

Kinetic (dynamic) analysis determines the forces and moments acting at each joint and the inertial loads on each moving link. This information is essential for component sizing, bearing selection, and power calculation.

The analysis proceeds in two stages. First, kinematics is solved (Chapters 1–2) to find positions, velocities, and accelerations. Then Newton’s second law is applied to each moving link treated as a free body.

For a planar rigid link \( k \) with mass \( m_k \) and moment of inertia \( I_{G_k} \) about its own centre of mass \( G_k \):

\[ \sum \mathbf{F}_k = m_k \mathbf{a}_{G_k} \]\[ \sum M_{G_k} = I_{G_k} \alpha_k \]

These are six scalar equations per moving link (three in 3D, three in planar form: two force balances + one moment balance). With \( n \) moving links there are \( 3n \) equations and \( 3n \) unknown joint forces (each full joint has two force components). For a mechanism with one DOF and one unknown (input torque/force), the system is determinate.

3.2 Force Analysis of the Four-Bar Mechanism

Example 3.1 — Static Force Analysis of a Four-Bar Linkage.

Consider a four-bar linkage in static equilibrium with a known external torque \( T_4 = 50 \) N·m on the rocker (link 4). Find the required input torque \( T_2 \) on the crank.

Using the principle of virtual work for a system in equilibrium: the total virtual work done by all applied forces and torques for any virtual displacement consistent with constraints is zero:

\[ \delta W = T_2 \delta\theta_2 + T_4 \delta\theta_4 = 0 \]

The velocity (kinematic) relationship gives \( \delta\theta_4 / \delta\theta_2 = \omega_4/\omega_2 \). Denoting this ratio as the mechanical advantage ratio \( \mu = \omega_4/\omega_2 \):

\[ T_2 = -T_4 \mu = -T_4 \frac{\omega_4}{\omega_2} \]

If from kinematics \( \omega_4 = 3.2 \) rad/s and \( \omega_2 = 10 \) rad/s:

\[ T_2 = 50 \times \frac{3.2}{10} = 16 \text{ N·m} \]

The negative sign convention indicates the input torque opposes the positive velocity direction. This result shows how mechanical advantage is built into the gear ratio \( \omega_4/\omega_2 \).

3.3 Dynamic Force Analysis

Dynamic force analysis accounts for inertia effects (\( m\mathbf{a} \) and \( I\alpha \) terms). For a crank-slider mechanism, the piston’s inertia force \( F_i = m_p \ddot{x} \) alternates in direction with each stroke, creating unbalanced forces transmitted to the engine frame.

3.3.1 Inertia Forces and D’Alembert’s Principle

D’Alembert’s principle reformulates dynamics as a statics problem by introducing an inertia force \( -m_k \mathbf{a}_{G_k} \) and an inertia torque \( -I_{G_k}\alpha_k \) as fictitious applied loads. The link is then in “dynamic equilibrium” under real plus inertia loads, and all static analysis methods apply.

3.4 Balancing of Rotating Machinery

3.4.1 Static Balancing

A rotating shaft carrying multiple masses is statically balanced if the resultant centrifugal force is zero — equivalently, the centre of mass lies on the axis of rotation.

Static Balance Condition

For \( n \) masses \( m_i \) at radii \( r_i \) from the rotation axis, at angular positions \( \theta_i \), rotating at angular speed \( \omega \):

\[ \sum_{i=1}^{n} m_i r_i e^{i\theta_i} = 0 \]

Separating into \( x \) and \( y \) components:

\[ \sum m_i r_i \cos\theta_i = 0, \qquad \sum m_i r_i \sin\theta_i = 0 \]

If these are satisfied, no net centrifugal force acts on the bearings, regardless of speed.

3.4.2 Dynamic Balancing

Static balance is necessary but not sufficient to eliminate bearing forces. Even when the net force is zero, a couple may exist if the masses are in different planes along the shaft. Dynamic balance requires both force and moment balance.

Dynamic Balance Conditions

For masses \( m_i \) at radii \( r_i \), angular positions \( \theta_i \), and axial positions \( z_i \) along the shaft:

\[ \sum_{i} m_i r_i e^{i\theta_i} = 0 \quad \text{(force balance)} \]\[ \sum_{i} m_i r_i z_i e^{i\theta_i} = 0 \quad \text{(moment balance)} \]

Dynamic balance implies static balance, but not vice versa. A dynamically balanced rotor produces no net forces or couples on its bearings at any speed (ignoring shaft flexibility and resonance effects).

Example 3.2 — Two-Plane Balancing.

A shaft carries three unbalanced masses: \( m_1 r_1 = 0.2 \) kg·m at \( \theta_1 = 0° \), \( z_1 = 0 \); \( m_2 r_2 = 0.15 \) kg·m at \( \theta_2 = 120° \), \( z_2 = 0.3 \) m; \( m_3 r_3 = 0.25 \) kg·m at \( \theta_3 = 240° \), \( z_3 = 0.6 \) m.

Two correction masses are to be placed in planes A (\( z_A = 0 \)) and B (\( z_B = 0.9 \) m).

Step 1 — Moment balance about plane A to find correction mass in plane B:

\[ m_B r_B e^{i\theta_B} = -\frac{1}{z_B}\sum_{i} m_i r_i z_i e^{i\theta_i} \]

Computing the moment sum (real and imaginary parts):

Real: \( 0.2(0)(1) + 0.15(0.3)\cos120° + 0.25(0.6)\cos240° = 0 - 0.0225 - 0.075 = -0.0975 \) kg·m²

Imaginary: \( 0 + 0.15(0.3)\sin120° + 0.25(0.6)\sin240° = 0.03897 - 0.12990 = -0.09093 \) kg·m²

\( m_B r_B = \frac{\sqrt{0.0975^2 + 0.09093^2}}{0.9} = \frac{0.1334}{0.9} \approx 0.148 \) kg·m at \( \theta_B = \arctan(-0.09093/-0.0975) + 180° \approx 223° \).

Step 2 — Force balance to find correction mass in plane A. (Left as an exercise.)

3.4.3 Partial Balancing of the Slider-Crank

For a single-cylinder engine the reciprocating piston cannot be fully balanced by a rotating counterweight. Adding a counterweight of mass \( m_b \) at radius \( r_b \) on the crank, oriented to oppose the piston, can cancel the primary (first-harmonic) force. However, this introduces a new lateral (transverse) force of the same magnitude. The engineer must choose the fraction of primary balance: typically 50% is used for single-cylinder engines as a compromise.


Chapter 4: Gears and Gear Trains

4.1 Fundamentals of Gear Geometry

Gears transmit motion and torque between shafts by the meshing of teeth. Unlike friction drives (which slip under overload), toothed gears enforce a positive velocity ratio provided tooth geometry obeys the fundamental law of gearing.

4.1.1 The Fundamental Law of Toothed Gearing

Fundamental Law of Toothed Gearing

For two mating gears to transmit a constant angular velocity ratio, the common normal to the tooth profiles at any point of contact must pass through a fixed point on the line of centers. This fixed point is called the pitch point.

Equivalently, the velocity ratio \( \omega_2/\omega_1 \) must be constant throughout the mesh — not just at average. Only conjugate tooth profiles (primarily the involute) satisfy this law perfectly.

Proof via the Theorem of Three Centres.

Let gear 1 rotate at \( \omega_1 \) and gear 2 at \( \omega_2 \), with centers \( O_1 \) and \( O_2 \). At the contact point \( K \), the components of the velocity of material points of each gear along the common normal must be equal (no separation or penetration). Denoting the common normal direction \( \hat{n} \):

\[ (\boldsymbol{\omega}_1 \times \mathbf{r}_{O_1 K}) \cdot \hat{n} = (\boldsymbol{\omega}_2 \times \mathbf{r}_{O_2 K}) \cdot \hat{n} \]

This geometrically requires the foot of the common normal to divide the center distance \( O_1 O_2 \) in the ratio \( O_1 P : O_2 P = \omega_2 : \omega_1 \), where \( P \) is a fixed point. For \( \omega_1/\omega_2 = \text{const} \), \( P \) must be fixed — the pitch point.

4.1.2 Involute Tooth Geometry

The involute of a circle is the curve traced by the end of a taut string unwinding from the circle. Involute tooth profiles automatically satisfy the fundamental law because the common normal at any contact point is the common tangent to the two base circles, which passes through a fixed pitch point.

Key involute gear parameters:

  • Module \( m = d/N \): pitch diameter divided by number of teeth (SI). All meshing gears must have the same module.
  • Pitch circle diameter \( d = mN \)
  • Pressure angle \( \phi \): angle between the common normal and the pitch circle tangent, standardized at \( 20° \) (formerly \( 14.5° \))
  • Base circle diameter \( d_b = d\cos\phi \)
  • Addendum \( a = m \) (tooth height above pitch circle)
  • Dedendum \( b = 1.25m \) (tooth depth below pitch circle)
  • Center distance for meshing gears: \( C = \frac{m(N_1 + N_2)}{2} \)
  • Gear ratio: \( i = \omega_1/\omega_2 = N_2/N_1 = d_2/d_1 \)

4.2 Types of Gears

4.2.1 Spur Gears

Spur gears have teeth parallel to the rotation axis. They transmit motion between parallel shafts. Spur gears are the simplest and most common. They produce no axial thrust but are noisier than helical gears due to abrupt engagement of the full tooth width.

4.2.2 Helical Gears

Helical gears have teeth inclined at a helix angle \( \psi \) to the rotation axis. Engagement begins at one end of the tooth and progresses smoothly across the face — resulting in quieter, smoother operation. However, helical gears generate an axial (thrust) force component \( F_a = F_t \tan\psi \) that must be absorbed by thrust bearings. Crossed helical gears can connect non-parallel, non-intersecting shafts.

4.2.3 Bevel Gears

Bevel gears transmit motion between intersecting shafts (usually at \( 90° \)). Teeth are formed on the surface of a cone. Straight bevel gears are analogous to spur gears; spiral bevel gears provide smoother engagement analogous to helical gears. The pitch surfaces are cones meeting at the apex.

4.2.4 Worm Gears

A worm gear pair consists of a worm (essentially a screw) meshing with a worm wheel. It transmits motion between non-parallel, non-intersecting (crossed) shafts, typically at \( 90° \). Very large speed reductions (\( 10:1 \) to \( 500:1 \)) are achievable in a single stage. The lead angle of the worm determines whether the drive is reversible (backdrivable) or self-locking. Self-locking occurs when the lead angle is less than the friction angle.

4.3 Gear Trains

A gear train is a series of meshing gears. It amplifies torque, reduces speed (or vice versa), and can reverse the direction of rotation.

4.3.1 Simple Gear Trains

In a simple gear train each shaft carries exactly one gear. The overall velocity ratio is the product of successive gear ratios:

\[ \frac{\omega_{\text{out}}}{\omega_{\text{in}}} = \pm\frac{N_1}{N_2} \cdot \frac{N_3}{N_4} \cdots \]

Intermediate gears between driver and driven gear are called idler gears; they change rotation direction but do not affect the speed ratio magnitude.

4.3.2 Compound Gear Trains

In a compound gear train, at least one shaft carries more than one gear. Multiple reductions can be achieved in a compact package. The velocity ratio:

\[ \frac{\omega_{\text{out}}}{\omega_{\text{in}}} = \pm \prod_{k} \frac{N_{\text{driver},k}}{N_{\text{driven},k}} \]

Example 4.1 — Compound Gear Train.

A compound train: Gear A (20 teeth) drives Gear B (80 teeth) on shaft 2. On shaft 2, Gear C (25 teeth) drives Gear D (100 teeth). The input is shaft 1 (gear A), output is shaft 3 (gear D).

\[ \frac{\omega_{\text{out}}}{\omega_{\text{in}}} = \frac{N_A}{N_B} \times \frac{N_C}{N_D} = \frac{20}{80} \times \frac{25}{100} = \frac{1}{4} \times \frac{1}{4} = \frac{1}{16} \]

The output rotates at 1/16th the input speed — a 16:1 reduction. If input torque is \( T_{\text{in}} = 5 \) N·m, the output torque (assuming 100% efficiency) is \( T_{\text{out}} = 5 \times 16 = 80 \) N·m.

4.4 Epicyclic (Planetary) Gear Trains

An epicyclic or planetary gear train differs from ordinary (simple or compound) trains in that at least one gear’s axis is not fixed to the frame — it revolves around another gear’s axis. This gives planetary trains exceptional versatility: very high ratios in compact packages, multiple operating modes by holding different members, and the ability to add or subtract speeds (differentials).

4.4.1 Components of a Planetary Train

A basic planetary set consists of:

  • Sun gear (S): central gear, axis fixed to frame
  • Planet gears (P): mesh with sun and ring, carried on planet carrier
  • Ring gear (R, annular gear): internal gear, axis fixed to frame
  • Planet carrier (C, arm): the link carrying the planet gear axes

Planetary Gear Train Relationship (Tabular/Relative-Motion Method)

Fix the carrier and calculate gear ratios for the sun-planet-ring sub-train; then add the carrier’s rotation back to all members. If the gear ratio of the gear train from sun to ring with the arm fixed is \( e \), then for any combination of two fixed inputs (or one fixed + one input):

\[ e = \frac{\omega_S - \omega_C}{\omega_R - \omega_C} = -\frac{N_R}{N_S} \]

(negative because the ring is internal and the sun rotates opposite to the ring when the arm is fixed).

Derivation via the Relative-Motion (Tabulation) Method.

The planetary gear set is analyzed by superimposing two states:

Step 1: Lock the arm (carrier, \( \omega_C = 0 \)) and give the sun angular velocity \( \omega_S' \). The planet and ring rotate by ordinary gear relationships. The ring rotates at \( \omega_R' = -(N_S/N_R)\omega_S' \) (negative: internal mesh).

Step 2: Rotate the entire locked assembly (sun, planet, ring, arm together) at \( +\omega_C \). All members gain \( +\omega_C \).

Superposing:

\[ \omega_S = \omega_S' + \omega_C, \quad \omega_R = \omega_R' + \omega_C = -\frac{N_S}{N_R}\omega_S' + \omega_C \]

From the first equation \( \omega_S' = \omega_S - \omega_C \). Substituting:

\[ \omega_R - \omega_C = -\frac{N_S}{N_R}(\omega_S - \omega_C) \]

Hence:

\[ \frac{\omega_S - \omega_C}{\omega_R - \omega_C} = -\frac{N_R}{N_S} \equiv e \]

This is the fundamental planetary equation. Given any two of \( \omega_S, \omega_R, \omega_C \), the third is determined.

Example 4.2 — Planetary Gear Ratio.

A planetary set: \( N_S = 30 \) teeth, \( N_P = 20 \) teeth, \( N_R = 70 \) teeth. The sun is the input at \( \omega_S = 1000 \) rpm. The ring is held fixed (\( \omega_R = 0 \)). Find the arm (carrier) output speed.

\[ e = -\frac{N_R}{N_S} = -\frac{70}{30} = -\frac{7}{3} \]\[ \frac{\omega_S - \omega_C}{\omega_R - \omega_C} = -\frac{7}{3} \]\[ \frac{1000 - \omega_C}{0 - \omega_C} = -\frac{7}{3} \]\[ 3(1000 - \omega_C) = 7\omega_C \implies 3000 = 10\omega_C \implies \omega_C = 300 \text{ rpm} \]

Speed ratio (sun to arm) \( = 1000/300 = 10/3 \). The arm output is 300 rpm in the same direction as the sun — a 3.33:1 reduction in a very compact package.

Automatic transmissions in vehicles use planetary gear sets with clutches and brakes that selectively hold or connect the sun, ring, or carrier to achieve multiple forward gear ratios and reverse — all within a single planetary unit. The Ravigneaux compound planetary set and the Simpson gear set are standard examples. Understanding which member is held or driven in each gear is a direct application of the planetary equation above.


Chapter 5: Free Vibration of Single-Degree-of-Freedom Systems

5.1 Motivation and Overview

Mechanical vibration is the oscillatory motion of a system about an equilibrium position. Vibrations arise whenever stored energy alternates between potential and kinetic forms. While controlled vibration is useful (musical instruments, vibrating conveyors, ultrasonic machining), uncontrolled vibration causes fatigue, noise, structural failure, and reduced precision. This chapter establishes the mathematical framework for the simplest vibrating system: a single-DOF mass-spring-damper.

5.2 Equation of Motion

5.2.1 Newton’s Law Approach

Consider a mass \( m \) on a spring of stiffness \( k \) with a viscous dashpot of coefficient \( c \). Let \( x(t) \) be displacement from the static equilibrium position. The equation of motion (Newton’s 2nd law, with static deflection terms cancelling):

\[ m\ddot{x} + c\dot{x} + kx = 0 \]

5.2.2 Energy Method Derivation

Derivation of Natural Frequency via the Energy Method.

For an undamped system, total mechanical energy \( E = T + V = \text{const} \):

\[ T = \tfrac{1}{2}m\dot{x}^2, \qquad V = \tfrac{1}{2}kx^2 \]\[ \frac{dE}{dt} = m\dot{x}\ddot{x} + kx\dot{x} = \dot{x}(m\ddot{x} + kx) = 0 \]

Since \( \dot{x} \neq 0 \) in general: \( m\ddot{x} + kx = 0 \). Assuming simple harmonic motion \( x = A\cos(\omega_n t + \phi) \):

\[ -m\omega_n^2 A\cos(\omega_n t + \phi) + kA\cos(\omega_n t + \phi) = 0 \]\[ (k - m\omega_n^2)A = 0 \implies \omega_n = \sqrt{\frac{k}{m}} \]

The energy method identifies \( \omega_n \) directly from the maximum kinetic and potential energies (Rayleigh’s method): at maximum displacement \( x_{\max} = A \), all energy is potential: \( V_{\max} = \tfrac{1}{2}kA^2 \). At \( x = 0 \) all energy is kinetic: \( T_{\max} = \tfrac{1}{2}m\omega_n^2 A^2 \). Setting \( T_{\max} = V_{\max} \) yields \( \omega_n = \sqrt{k/m} \).

Natural Frequency and Period

For an undamped SDOF system with mass \( m \) and stiffness \( k \):

  • Natural angular frequency: \( \omega_n = \sqrt{k/m} \) (rad/s)
  • Natural frequency: \( f_n = \omega_n / (2\pi) \) (Hz)
  • Period: \( \tau = 1/f_n = 2\pi/\omega_n \) (s)

The natural frequency depends only on the system parameters \( k \) and \( m \) — not on initial conditions or amplitude (for linear systems). The general undamped free response is:

\[ x(t) = A\cos\omega_n t + B\sin\omega_n t = C\cos(\omega_n t - \phi) \]

where \( C = \sqrt{A^2 + B^2} \), \( \phi = \arctan(B/A) \), with \( A = x(0) \), \( B = \dot{x}(0)/\omega_n \).

5.3 Damped Free Vibration

Dividing the equation of motion \( m\ddot{x} + c\dot{x} + kx = 0 \) by \( m \) and introducing standard parameters:

Damping Ratio

The critical damping coefficient is \( c_c = 2\sqrt{km} = 2m\omega_n \). The dimensionless damping ratio is:

\[ \zeta = \frac{c}{c_c} = \frac{c}{2m\omega_n} \]

The equation of motion becomes:

\[ \ddot{x} + 2\zeta\omega_n\dot{x} + \omega_n^2 x = 0 \]

The character of motion depends entirely on \( \zeta \):

  • Underdamped (\( \zeta < 1 \)): oscillatory decay
  • Critically damped (\( \zeta = 1 \)): fastest return to equilibrium without oscillation
  • Overdamped (\( \zeta > 1 \)): exponential decay, no oscillation

5.3.1 Underdamped Response (\( \zeta < 1 \))

The characteristic equation \( s^2 + 2\zeta\omega_n s + \omega_n^2 = 0 \) has roots:

\[ s_{1,2} = -\zeta\omega_n \pm i\omega_n\sqrt{1-\zeta^2} = -\sigma \pm i\omega_d \]

where \( \sigma = \zeta\omega_n \) is the decay rate and \( \omega_d = \omega_n\sqrt{1-\zeta^2} \) is the damped natural frequency. The general solution:

\[ x(t) = e^{-\zeta\omega_n t}\left(A\cos\omega_d t + B\sin\omega_d t\right) = Ce^{-\zeta\omega_n t}\cos(\omega_d t - \phi) \]

with initial conditions \( x(0) = x_0 \) and \( \dot{x}(0) = v_0 \):

\[ A = x_0, \qquad B = \frac{v_0 + \zeta\omega_n x_0}{\omega_d} \]

The amplitude decays as an exponential envelope \( Ce^{-\zeta\omega_n t} \) while the response oscillates at frequency \( \omega_d \).

5.3.2 Critically Damped Response (\( \zeta = 1 \))

Repeated roots \( s_{1,2} = -\omega_n \) yield:

\[ x(t) = (A + Bt)e^{-\omega_n t} \]

with \( A = x_0 \), \( B = v_0 + \omega_n x_0 \). This is the minimum damping for which no oscillation occurs; it corresponds to the fastest return to rest without overshoot and is the design target for instrument needles and door closers.

5.3.3 Overdamped Response (\( \zeta > 1 \))

Two distinct real roots \( s_{1,2} = \omega_n\left(-\zeta \pm \sqrt{\zeta^2 - 1}\right) \):

\[ x(t) = Ae^{s_1 t} + Be^{s_2 t} \]

The system returns to equilibrium without oscillating, but more slowly than the critically damped case.

5.4 Logarithmic Decrement

The logarithmic decrement \( \delta \) is the standard experimental method for measuring damping ratio from a free-vibration decay record.

Logarithmic Decrement

For an underdamped system, successive peaks \( x_1, x_2, \ldots \) separated by the damped period \( \tau_d = 2\pi/\omega_d \) decay in a constant ratio:

\[ \frac{x_n}{x_{n+1}} = e^{\zeta\omega_n \tau_d} = e^{2\pi\zeta/\sqrt{1-\zeta^2}} \]

The logarithmic decrement is:

\[ \delta = \ln\frac{x_n}{x_{n+1}} = \frac{2\pi\zeta}{\sqrt{1-\zeta^2}} \]

For measuring over \( m \) cycles (to improve accuracy with noise):

\[ \delta = \frac{1}{m}\ln\frac{x_n}{x_{n+m}} \]

Inverting for the damping ratio:

\[ \zeta = \frac{\delta}{\sqrt{4\pi^2 + \delta^2}} \]

For small damping (\( \zeta \ll 1 \)): \( \delta \approx 2\pi\zeta \), so \( \zeta \approx \delta/(2\pi) \).

Example 5.1 — Identifying Damping from Decay Data.

An underdamped SDOF system’s free-vibration record shows peaks: \( x_1 = 12.5 \) mm, \( x_5 = 4.2 \) mm. Find \( \zeta \) and \( \omega_d / \omega_n \).

Over \( m = 4 \) cycles:

\[ \delta = \frac{1}{4}\ln\frac{12.5}{4.2} = \frac{\ln(2.976)}{4} = \frac{1.091}{4} = 0.273 \]\[ \zeta = \frac{0.273}{\sqrt{4\pi^2 + 0.273^2}} = \frac{0.273}{\sqrt{39.48 + 0.075}} = \frac{0.273}{6.286} \approx 0.0434 \]\[ \frac{\omega_d}{\omega_n} = \sqrt{1 - \zeta^2} = \sqrt{1 - 0.00188} \approx 0.999 \]

The damped frequency is essentially equal to the natural frequency (4.3% damping barely shifts the oscillation frequency). This is typical of lightly damped structural systems.

Example 5.2 — Spring-Mass System Natural Frequency.

A machine of mass \( m = 80 \) kg is mounted on a rubber isolator pad of stiffness \( k = 320 \) kN/m. Find the natural frequency and the static deflection.

\[ \omega_n = \sqrt{\frac{k}{m}} = \sqrt{\frac{320{,}000}{80}} = \sqrt{4000} = 63.25 \text{ rad/s} \]\[ f_n = \frac{63.25}{2\pi} = 10.07 \text{ Hz} \]

Static deflection: \( \delta_{st} = mg/k = (80)(9.81)/320{,}000 = 2.45 \times 10^{-3} \) m = 2.45 mm.

Note the relationship: \( \omega_n = \sqrt{g/\delta_{st}} \). If static deflection is known, \( f_n = \frac{1}{2\pi}\sqrt{9.81/\delta_{st}}\) (Hz, with \( \delta_{st} \) in metres). This is useful in practice because static deflection is easily measured.


Chapter 6: Forced Harmonic Response and Vibration Isolation

6.1 Undamped Forced Vibration

Apply a harmonic force \( F(t) = F_0\cos\omega t \) to an undamped SDOF system:

\[ m\ddot{x} + kx = F_0\cos\omega t \]

The particular solution (steady-state response):

\[ x_p(t) = \frac{F_0/k}{1 - (\omega/\omega_n)^2}\cos\omega t = X\cos\omega t \]

The amplitude \( X = \frac{F_0/k}{1 - r^2} \) where \( r = \omega/\omega_n \) is the frequency ratio. The static deflection \( X_{st} = F_0/k \).

At \( r \to 1 \) (\( \omega \to \omega_n \)), the denominator vanishes and the theoretical amplitude grows without bound — this is resonance. In physical undamped systems (theoretical idealization) the amplitude actually grows linearly with time: \( x(t) = \frac{F_0}{2m\omega_n}t\sin\omega_n t \). In real (damped) systems the amplitude is finite but very large at resonance, limited only by damping. Resonance is destructive because even a small harmonic force can excite enormous oscillations if applied at the natural frequency — bridges, aircraft wings, and rotating machinery can all fail at resonance if proper isolation and/or detuning is not provided.

6.2 Damped Forced Vibration — Frequency Response

Derivation of the Frequency Response Function for Damped Harmonic Excitation.

The equation of motion for harmonic excitation:

\[ m\ddot{x} + c\dot{x} + kx = F_0 e^{i\omega t} \]

(taking the real part gives the response to \( F_0\cos\omega t \)). Assume a steady-state solution \( x_p = X e^{i\omega t} \):

\[ (-m\omega^2 + ic\omega + k)Xe^{i\omega t} = F_0 e^{i\omega t} \]\[ X = \frac{F_0}{k - m\omega^2 + ic\omega} = \frac{F_0/k}{1 - r^2 + 2i\zeta r} \]

where \( r = \omega/\omega_n \). The complex amplitude \( X \) can be written \( X = |X|e^{-i\phi} \) where the magnitude is the dynamic magnification factor (DMF):

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

and the phase lag:

\[ \phi = \arctan\frac{2\zeta r}{1 - r^2} \]

The steady-state response is:

\[ x_p(t) = \frac{F_0/k}{\sqrt{(1-r^2)^2 + (2\zeta r)^2}}\cos(\omega t - \phi) \]

The frequency response function (FRF) or receptance is the complex ratio:

\[ H(\omega) = \frac{X}{F_0} = \frac{1/k}{1 - r^2 + 2i\zeta r} \]

Dynamic Magnification Factor and Phase Angle

The dynamic magnification factor (also called amplitude ratio or magnification factor) is:

\[ M = \frac{|X|}{X_{st}} = \frac{1}{\sqrt{(1 - r^2)^2 + (2\zeta r)^2}} \]

where \( X_{st} = F_0/k \) and \( r = \omega/\omega_n \). The phase angle \( \phi \) is the lag of response behind excitation:

\[ \phi = \arctan\frac{2\zeta r}{1 - r^2} \]

Key observations:

  • At \( r = 0 \): \( M = 1 \), \( \phi = 0° \) — quasi-static response, in phase with force.
  • At \( r = 1 \): \( M = 1/(2\zeta) \), \( \phi = 90° \) for all \( \zeta \) — the phase jump through 90° at resonance is a universal feature.
  • At \( r \gg 1 \): \( M \to 0 \), \( \phi \to 180° \) — the system cannot follow the rapidly oscillating force; response is out of phase.
  • Peak of \( M \) occurs at \( r = \sqrt{1 - 2\zeta^2} \) (for \( \zeta < 1/\sqrt{2} \approx 0.707 \)).

Example 6.1 — Steady-State Response to Harmonic Force.

A system: \( m = 10 \) kg, \( k = 40\,000 \) N/m, \( c = 200 \) N·s/m. Excitation: \( F(t) = 100\cos(120t) \) N.

\[ \omega_n = \sqrt{40{,}000/10} = 63.25 \text{ rad/s}, \quad c_c = 2m\omega_n = 2(10)(63.25) = 1265 \text{ N·s/m} \]\[ \zeta = 200/1265 = 0.158, \quad r = 120/63.25 = 1.897 \]\[ M = \frac{1}{\sqrt{(1-1.897^2)^2 + (2\times0.158\times1.897)^2}} = \frac{1}{\sqrt{(1-3.599)^2 + (0.600)^2}} = \frac{1}{\sqrt{6.752 + 0.360}} = \frac{1}{\sqrt{7.112}} = 0.375 \]\[ X_{st} = 100/40{,}000 = 0.0025 \text{ m}, \qquad X = M \cdot X_{st} = 0.375 \times 0.0025 = 0.938 \text{ mm} \]\[ \phi = \arctan\frac{0.600}{1 - 3.599} = \arctan\frac{0.600}{-2.599} = 180° - 13.0° = 167.0° \]

The response amplitude is 0.938 mm at a phase lag of 167° (nearly opposite to the force). Operating above resonance (\( r > 1 \)) gives low amplitude but large phase shift.

6.3 Rotating Unbalance Excitation

Many vibration problems arise from rotating machinery with mass imbalance — eccentrically mounted rotors, unbalanced fans, crank mechanisms. A mass \( m_e \) (unbalanced mass) rotating at eccentricity \( e \) generates a centrifugal force:

\[ F(t) = m_e e \omega^2 \cos\omega t \]

For a machine of total mass \( M \) (including \( m_e \)):

\[ M\ddot{x} + c\dot{x} + kx = m_e e \omega^2 \cos\omega t \]

The steady-state amplitude:

\[ MX = \frac{m_e e r^2}{\sqrt{(1-r^2)^2 + (2\zeta r)^2}} \]

or in dimensionless form with \( r = \omega/\omega_n \):

\[ \frac{MX}{m_e e} = \frac{r^2}{\sqrt{(1-r^2)^2 + (2\zeta r)^2}} \]

Unlike the direct force case (where \( M \to 1 \) as \( r \to 0 \)), the rotating unbalance response goes to zero as \( r \to 0 \) (the force itself goes to zero at low speed) and approaches unity as \( r \to \infty \). At high speeds above resonance, the unbalanced machine runs smoothly because it self-centers — the centre of mass of the total system approaches a fixed point. This is the physical basis of self-balancing washing machines in spin mode.

6.4 Base Excitation

When the support (base) of an SDOF system moves harmonically — \( y(t) = Y\sin\omega t \) — the relative displacement \( z = x - y \) and the absolute displacement \( x \) of the mass are both of interest.

The equation of motion for the mass:

\[ m\ddot{x} + c(\dot{x} - \dot{y}) + k(x - y) = 0 \]

Or equivalently, in terms of absolute displacement \( x \):

\[ m\ddot{x} + c\dot{x} + kx = kY\sin\omega t + c\omega Y\cos\omega t \]

The steady-state amplitude of the mass:

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

This ratio \( X/Y \) is the displacement transmissibility.

6.5 Transmissibility and Vibration Isolation

Force Transmissibility

The force transmissibility \( T_r \) is the ratio of the force transmitted to the base (through the spring and damper) to the applied harmonic force amplitude. For direct force excitation:

\[ T_r = \frac{F_T}{F_0} = \sqrt{\frac{1 + (2\zeta r)^2}{(1-r^2)^2 + (2\zeta r)^2}} \]

Note this is identical in form to the displacement transmissibility for base excitation. For vibration isolation (reducing forces or displacements transmitted), the goal is \( T_r < 1 \).

Key isolation design principles:

  1. Transmissibility \( T_r < 1 \) requires \( r > \sqrt{2} \) (for any \( \zeta \)).
  2. In the isolation region (\( r > \sqrt{2} \)), less damping gives better isolation — high damping actually degrades isolation performance above \( \sqrt{2}\omega_n \).
  3. However, during startup (passing through \( r = 1 \)), low damping allows large resonant amplitude. Practical isolators balance these competing demands.
  4. Natural rubber, elastomeric mounts, and air springs are common isolator materials providing both stiffness and damping.

Example 6.2 — Vibration Isolator Design.

A reciprocating compressor of mass 150 kg operates at 1200 rpm. Specify an isolator to achieve 85% isolation (transmissibility \( T_r = 0.15 \)) at operating speed. Assume low damping (\( \zeta \approx 0.05 \)).

Operating frequency: \( \omega = 1200 \times 2\pi/60 = 125.7 \) rad/s.

From the transmissibility formula (low damping approximation, \( r \gg \sqrt{2} \)):

\[ T_r \approx \frac{1}{r^2 - 1} = 0.15 \implies r^2 - 1 = 6.67 \implies r = 2.76 \]

Required natural frequency: \( \omega_n = \omega/r = 125.7/2.76 = 45.5 \) rad/s, so \( f_n = 7.24 \) Hz.

Required stiffness: \( k = m\omega_n^2 = 150 \times (45.5)^2 = 310{,}000 \) N/m = 310 kN/m.

If four isolators are used in parallel, each has stiffness \( k_i = 310/4 = 77.5 \) kN/m.

Static deflection under weight: \( \delta_{st} = mg/k = 150(9.81)/310{,}000 = 4.75 \) mm. The isolators must support this static deflection plus dynamic amplitude.


Chapter 7: Multi-Degree-of-Freedom Systems and Modal Analysis

7.1 Equations of Motion for Two-DOF Systems

Real structures have infinitely many DOF, but reduced-order models with two (or a few) DOF capture essential dynamics and introduce the key concepts of natural frequencies, mode shapes, and modal analysis that extend to full-scale finite element models.

7.1.1 Derivation of the 2-DOF Equations

Consider two masses \( m_1 \) and \( m_2 \) connected in series by springs \( k_1 \), \( k_2 \), \( k_3 \) (with \( k_1 \) to ground, \( k_2 \) between masses, \( k_3 \) from \( m_2 \) to ground or absent). Applying Newton’s second law to each mass:

\[ m_1\ddot{x}_1 + (k_1 + k_2)x_1 - k_2 x_2 = F_1(t) \]\[ m_2\ddot{x}_2 - k_2 x_1 + (k_2 + k_3)x_2 = F_2(t) \]

7.1.2 Matrix Form of the Equations of Motion

Matrix Equation of Motion for an N-DOF System

The equations of motion for an \( N \)-DOF undamped system are written compactly as:

\[ \left[ M \right]\ddot{\mathbf{x}} + \left[ K \right]\mathbf{x} = \mathbf{F}(t) \]

where:

  • \( \left[ M \right] \in \mathbb{R}^{N\times N} \) is the mass matrix (symmetric, positive definite)
  • \( \left[ K \right] \in \mathbb{R}^{N\times N} \) is the stiffness matrix (symmetric, positive semi-definite)
  • \( \mathbf{x}(t) \in \mathbb{R}^N \) is the displacement vector
  • \( \mathbf{F}(t) \in \mathbb{R}^N \) is the force vector

For the two-mass, three-spring system with no damping:

\[ \begin{bmatrix} m_1 & 0 \\ 0 & m_2 \end{bmatrix} \begin{bmatrix} \ddot{x}_1 \\ \ddot{x}_2 \end{bmatrix} + \begin{bmatrix} k_1 + k_2 & -k_2 \\ -k_2 & k_2 + k_3 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} F_1 \\ F_2 \end{bmatrix} \]

The mass matrix for this lumped-parameter model is diagonal (no inertial coupling between coordinates). The stiffness matrix is tridiagonal and symmetric. Coupling enters through the off-diagonal stiffness terms — spring \( k_2 \) connects the two masses and creates the \( -k_2 \) coupling terms. Modal analysis decouples these equations by transforming to a new coordinate system aligned with the system’s natural oscillation patterns.

7.2 The Eigenvalue Problem — Natural Frequencies and Mode Shapes

7.2.1 Setting Up the Eigenvalue Problem

For free vibration (\( \mathbf{F} = \mathbf{0} \)), assume harmonic motion \( \mathbf{x}(t) = \boldsymbol{\Phi} e^{i\omega t} \):

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

This has a non-trivial solution only when:

\[ \det\left(\left[ K \right] - \omega^2\left[ M \right]\right) = 0 \]

This is the characteristic equation. Its \( N \) roots \( \omega_1^2 \leq \omega_2^2 \leq \cdots \leq \omega_N^2 \) are the squared natural frequencies.

Natural Frequencies and Mode Shapes

The natural frequencies \( \omega_r \) (\( r = 1, 2, \ldots, N \)) are the \( N \) positive square roots of the eigenvalues of the generalized eigenvalue problem:

\[ \left[ K \right]\boldsymbol{\phi}^{(r)} = \omega_r^2 \left[ M \right]\boldsymbol{\phi}^{(r)} \]

The corresponding vectors \( \boldsymbol{\phi}^{(r)} \) are the mode shapes (eigenvectors). Each mode shape describes the relative displacement pattern of all DOF when the system oscillates at the corresponding natural frequency. Mode shapes are determined only up to an arbitrary scalar multiple; they are normalized by convention (e.g., unit modal mass or unit maximum entry).

The lowest natural frequency \( \omega_1 \) is the fundamental frequency; the corresponding mode is the first mode.

Example 7.1 — Natural Frequencies and Mode Shapes of a 2-DOF System.

A two-mass system: \( m_1 = m_2 = m \), \( k_1 = k_3 = 0 \) (no ground springs), \( k_2 = k \) (one connecting spring). The matrices are:

\[ \left[ M \right] = \begin{bmatrix} m & 0 \\ 0 & m \end{bmatrix}, \quad \left[ K \right] = \begin{bmatrix} k & -k \\ -k & k \end{bmatrix} \]

Characteristic equation:

\[ \det\left(\begin{bmatrix} k-m\omega^2 & -k \\ -k & k-m\omega^2 \end{bmatrix}\right) = (k-m\omega^2)^2 - k^2 = 0 \]\[ (k-m\omega^2-k)(k-m\omega^2+k) = 0 \implies (-m\omega^2)(2k-m\omega^2) = 0 \]

Roots: \( \omega_1^2 = 0 \) and \( \omega_2^2 = 2k/m \).

Mode 1 (\( \omega_1 = 0 \)): \( (k - 0)\phi_1^{(1)} - k\phi_2^{(1)} = 0 \implies \phi_1^{(1)} = \phi_2^{(1)} \). Mode shape: \( \boldsymbol{\phi}^{(1)} = \begin{bmatrix}1\\1\end{bmatrix} \) — both masses move in the same direction with equal amplitude. This is a rigid-body mode (zero frequency, no elastic deformation).

Mode 2 (\( \omega_2 = \sqrt{2k/m} \)): \( (k-2k)\phi_1^{(2)} - k\phi_2^{(2)} = 0 \implies \phi_1^{(2)} = -\phi_2^{(2)} \). Mode shape: \( \boldsymbol{\phi}^{(2)} = \begin{bmatrix}1\\-1\end{bmatrix} \) — masses move in opposite directions, spring fully stretched/compressed. This is the elastic mode.

Example 7.2 — Two-Degree-of-Freedom System with Ground Springs.

System parameters: \( m_1 = 2 \) kg, \( m_2 = 1 \) kg, \( k_1 = 4000 \) N/m, \( k_2 = 2000 \) N/m, \( k_3 = 0 \).

Matrices:

\[ \left[ M \right] = \begin{bmatrix} 2 & 0 \\ 0 & 1 \end{bmatrix}, \quad \left[ K \right] = \begin{bmatrix} 6000 & -2000 \\ -2000 & 2000 \end{bmatrix} \]

Characteristic equation (let \( \lambda = \omega^2 \)):

\[ \det\begin{bmatrix} 6000 - 2\lambda & -2000 \\ -2000 & 2000 - \lambda \end{bmatrix} = (6000-2\lambda)(2000-\lambda) - 4{,}000{,}000 = 0 \]\[ 12{,}000{,}000 - 6000\lambda - 4000\lambda + 2\lambda^2 - 4{,}000{,}000 = 0 \]\[ 2\lambda^2 - 10000\lambda + 8{,}000{,}000 = 0 \implies \lambda^2 - 5000\lambda + 4{,}000{,}000 = 0 \]\[ \lambda = \frac{5000 \pm \sqrt{25{,}000{,}000 - 16{,}000{,}000}}{2} = \frac{5000 \pm 3000}{2} \]

\( \lambda_1 = 1000 \implies \omega_1 = 31.62 \) rad/s \( \approx 5.03 \) Hz

\( \lambda_2 = 4000 \implies \omega_2 = 63.25 \) rad/s \( \approx 10.07 \) Hz

Mode shapes: for \( \lambda_1 = 1000 \): \( (6000-2000)\phi_1 - 2000\phi_2 = 0 \implies 4000\phi_1 = 2000\phi_2 \implies \phi_2 = 2\phi_1 \). So \( \boldsymbol{\phi}^{(1)} = \begin{bmatrix}1\\2\end{bmatrix} \).

For \( \lambda_2 = 4000 \): \( (6000-8000)\phi_1 - 2000\phi_2 = 0 \implies -2000\phi_1 = 2000\phi_2 \implies \phi_2 = -\phi_1 \). So \( \boldsymbol{\phi}^{(2)} = \begin{bmatrix}1\\-1\end{bmatrix} \).

In mode 1 (lower frequency), mass 2 moves twice as far as mass 1, in the same direction. In mode 2, the masses move in opposite directions with equal amplitude.

7.3 Orthogonality of Mode Shapes

The orthogonality of mode shapes is the theoretical foundation of modal analysis. It guarantees that the transformation to modal coordinates decouples the equations of motion.

Orthogonality of Mode Shapes

Let \( \boldsymbol{\phi}^{(r)} \) and \( \boldsymbol{\phi}^{(s)} \) be mode shapes corresponding to distinct natural frequencies \( \omega_r \neq \omega_s \). Then:

\[ \left(\boldsymbol{\phi}^{(s)}\right)^T \left[ M \right] \boldsymbol{\phi}^{(r)} = 0 \quad (r \neq s) \]\[ \left(\boldsymbol{\phi}^{(s)}\right)^T \left[ K \right] \boldsymbol{\phi}^{(r)} = 0 \quad (r \neq s) \]

For \( r = s \): \( \left(\boldsymbol{\phi}^{(r)}\right)^T \left[ M \right] \boldsymbol{\phi}^{(r)} = m_r \) (modal mass) and \( \left(\boldsymbol{\phi}^{(r)}\right)^T \left[ K \right] \boldsymbol{\phi}^{(r)} = k_r = \omega_r^2 m_r \) (modal stiffness).

Proof of Orthogonality.

From the eigenvalue problem, mode \( r \) satisfies:

\[ \left[ K \right]\boldsymbol{\phi}^{(r)} = \omega_r^2 \left[ M \right]\boldsymbol{\phi}^{(r)} \quad \cdots (1) \]

and mode \( s \) satisfies:

\[ \left[ K \right]\boldsymbol{\phi}^{(s)} = \omega_s^2 \left[ M \right]\boldsymbol{\phi}^{(s)} \quad \cdots (2) \]

Premultiply (1) by \( \left(\boldsymbol{\phi}^{(s)}\right)^T \):

\[ \left(\boldsymbol{\phi}^{(s)}\right)^T \left[ K \right]\boldsymbol{\phi}^{(r)} = \omega_r^2 \left(\boldsymbol{\phi}^{(s)}\right)^T \left[ M \right]\boldsymbol{\phi}^{(r)} \quad \cdots (3) \]

Premultiply (2) by \( \left(\boldsymbol{\phi}^{(r)}\right)^T \):

\[ \left(\boldsymbol{\phi}^{(r)}\right)^T \left[ K \right]\boldsymbol{\phi}^{(s)} = \omega_s^2 \left(\boldsymbol{\phi}^{(r)}\right)^T \left[ M \right]\boldsymbol{\phi}^{(s)} \quad \cdots (4) \]

Since \( [M] \) and \( [K] \) are symmetric: \( \left(\boldsymbol{\phi}^{(r)}\right)^T \left[ K \right]\boldsymbol{\phi}^{(s)} = \left(\boldsymbol{\phi}^{(s)}\right)^T \left[ K \right]\boldsymbol{\phi}^{(r)} \) and similarly for \( [M] \). Therefore taking the transpose of (4) and subtracting from (3):

\[ 0 = (\omega_r^2 - \omega_s^2)\left(\boldsymbol{\phi}^{(s)}\right)^T \left[ M \right]\boldsymbol{\phi}^{(r)} \]

Since \( \omega_r \neq \omega_s \), we conclude \( \left(\boldsymbol{\phi}^{(s)}\right)^T \left[ M \right]\boldsymbol{\phi}^{(r)} = 0 \). Substituting back into (3): \( \left(\boldsymbol{\phi}^{(s)}\right)^T \left[ K \right]\boldsymbol{\phi}^{(r)} = 0 \). \( \square \)

The physical meaning of orthogonality is profound. The mass-orthogonality condition says that mode \( r \) imparts no kinetic energy (at unit velocity) to mode \( s \): the modes carry energy independently. Similarly, stiffness-orthogonality means mode \( r \) stores no strain energy in the deformation pattern of mode \( s \). Consequently, energy injected into one mode stays in that mode — modes do not exchange energy (in the linear, undamped case). This is why structures ring at multiple frequencies simultaneously after being struck: each mode rings independently. Damping slowly drains each mode’s energy, also independently (provided damping is also orthogonal — the so-called proportional or Rayleigh damping condition).

7.4 Modal Matrix and Coordinate Transformation

Collect the mode shapes as columns of the modal matrix \( \left[ \Phi \right] = \left[\boldsymbol{\phi}^{(1)} \, \boldsymbol{\phi}^{(2)} \, \cdots \, \boldsymbol{\phi}^{(N)}\right] \).

Define the modal coordinate vector \( \mathbf{q}(t) \) through the transformation:

\[ \mathbf{x}(t) = \left[ \Phi \right]\mathbf{q}(t) = \sum_{r=1}^{N} \boldsymbol{\phi}^{(r)} q_r(t) \]

Substitute into \( [M]\ddot{\mathbf{x}} + [K]\mathbf{x} = \mathbf{F}(t) \) and premultiply by \( [\Phi]^T \):

\[ [\Phi]^T[M][\Phi]\ddot{\mathbf{q}} + [\Phi]^T[K][\Phi]\mathbf{q} = [\Phi]^T\mathbf{F} \]

By orthogonality, \( [\Phi]^T[M][\Phi] = [m_r] \) (diagonal matrix of modal masses) and \( [\Phi]^T[K][\Phi] = [k_r] = [\omega_r^2 m_r] \) (diagonal matrix of modal stiffnesses). The system decouples into \( N \) independent SDOF equations:

\[ m_r \ddot{q}_r + \omega_r^2 m_r q_r = f_r(t), \quad r = 1, 2, \ldots, N \]

where the modal force is \( f_r(t) = \left(\boldsymbol{\phi}^{(r)}\right)^T \mathbf{F}(t) \).

Each modal equation is an independent SDOF system that can be solved by the methods of Chapters 5 and 6. The physical response is then reconstructed by superposition: \( \mathbf{x}(t) = \sum_r \boldsymbol{\phi}^{(r)} q_r(t) \).

7.5 Free Response via Modal Superposition

For the undamped free-vibration problem (\( \mathbf{F} = \mathbf{0} \)), each modal equation gives:

\[ q_r(t) = A_r \cos\omega_r t + B_r \sin\omega_r t \]

The physical response:

\[ \mathbf{x}(t) = \sum_{r=1}^{N} \boldsymbol{\phi}^{(r)}\left(A_r \cos\omega_r t + B_r \sin\omega_r t\right) \]

The constants \( A_r \) and \( B_r \) are found from the initial conditions via the orthogonality relations:

\[ A_r = \frac{\left(\boldsymbol{\phi}^{(r)}\right)^T [M] \mathbf{x}(0)}{m_r}, \qquad B_r = \frac{\left(\boldsymbol{\phi}^{(r)}\right)^T [M] \dot{\mathbf{x}}(0)}{m_r \omega_r} \]

Example 7.3 — Free Response of a 2-DOF System.

For the system of Example 7.2 (\( m_1 = 2 \) kg, \( m_2 = 1 \) kg, \( \omega_1 = 31.62 \) rad/s, \( \omega_2 = 63.25 \) rad/s, \( \boldsymbol{\phi}^{(1)} = \begin{bmatrix}1\\2\end{bmatrix} \), \( \boldsymbol{\phi}^{(2)} = \begin{bmatrix}1\\-1\end{bmatrix} \)).

Initial conditions: \( \mathbf{x}(0) = \begin{bmatrix}0.01\\0\end{bmatrix} \) m, \( \dot{\mathbf{x}}(0) = \mathbf{0} \).

Modal masses:

\[ m_1 = (\boldsymbol{\phi}^{(1)})^T[M]\boldsymbol{\phi}^{(1)} = \begin{bmatrix}1&2\end{bmatrix}\begin{bmatrix}2&0\\0&1\end{bmatrix}\begin{bmatrix}1\\2\end{bmatrix} = 2 + 4 = 6 \text{ kg} \]\[ m_2 = \begin{bmatrix}1&-1\end{bmatrix}\begin{bmatrix}2&0\\0&1\end{bmatrix}\begin{bmatrix}1\\-1\end{bmatrix} = 2 + 1 = 3 \text{ kg} \]

Modal initial displacements:

\[ A_1 = \frac{(\boldsymbol{\phi}^{(1)})^T[M]\mathbf{x}(0)}{m_1} = \frac{\begin{bmatrix}1&2\end{bmatrix}\begin{bmatrix}2&0\\0&1\end{bmatrix}\begin{bmatrix}0.01\\0\end{bmatrix}}{6} = \frac{0.02}{6} = 0.00333 \text{ m} \]\[ A_2 = \frac{(\boldsymbol{\phi}^{(2)})^T[M]\mathbf{x}(0)}{m_2} = \frac{\begin{bmatrix}1&-1\end{bmatrix}\begin{bmatrix}2&0\\0&1\end{bmatrix}\begin{bmatrix}0.01\\0\end{bmatrix}}{3} = \frac{0.02}{3} = 0.00667 \text{ m} \]

All \( B_r = 0 \) (zero initial velocities). Physical response:

\[ x_1(t) = 1(0.00333)\cos(31.62t) + 1(0.00667)\cos(63.25t) \]\[ x_2(t) = 2(0.00333)\cos(31.62t) + (-1)(0.00667)\cos(63.25t) \]\[ x_1(t) = 3.33\cos(31.62t) + 6.67\cos(63.25t) \text{ mm} \]\[ x_2(t) = 6.67\cos(31.62t) - 6.67\cos(63.25t) \text{ mm} \]

The response is a superposition of both modes. Note that mass 2’s response is purely the beat between the two modes (equal coefficients with opposite signs in the second mode).

7.6 Forced Response via Modal Superposition

For forced vibration, each decoupled modal equation:

\[ m_r \ddot{q}_r + m_r\omega_r^2 q_r = f_r(t) \]

is solved by the methods of Chapter 6 (for harmonic \( \mathbf{F}(t) = \hat{\mathbf{F}} e^{i\omega t} \), each modal equation has a harmonic particular solution). Including modal damping \( \zeta_r \) (added phenomenologically):

\[ \ddot{q}_r + 2\zeta_r\omega_r\dot{q}_r + \omega_r^2 q_r = \frac{f_r(t)}{m_r} \]

For harmonic excitation \( \mathbf{F}(t) = \hat{\mathbf{F}}\sin\omega t \):

\[ q_r(t) = \frac{\hat{f}_r/m_r}{\sqrt{(\omega_r^2 - \omega^2)^2 + (2\zeta_r\omega_r\omega)^2}}\sin(\omega t - \phi_r) \]

where \( \hat{f}_r = (\boldsymbol{\phi}^{(r)})^T \hat{\mathbf{F}} \) and \( \phi_r = \arctan\frac{2\zeta_r\omega_r\omega}{\omega_r^2 - \omega^2} \).

The total physical response:

\[ \mathbf{x}(t) = \sum_{r=1}^{N} \boldsymbol{\phi}^{(r)} q_r(t) \]

This is the modal superposition method — the workhorse of structural dynamics and finite element analysis. For a structure with \( N \) DOF under broadband excitation, only the first few modes typically contribute significantly, allowing massive computational savings by truncating the modal sum.

Example 7.4 — Harmonic Forced Response of a 2-DOF System.

Using the system of Example 7.2, apply a harmonic force \( F_1(t) = 10\sin(20t) \) N to mass 1, \( F_2 = 0 \). Assume modal damping \( \zeta_1 = \zeta_2 = 0.05 \).

Modal forces:

\[ \hat{f}_1 = (\boldsymbol{\phi}^{(1)})^T\hat{\mathbf{F}} = \begin{bmatrix}1&2\end{bmatrix}\begin{bmatrix}10\\0\end{bmatrix} = 10 \text{ N} \]\[ \hat{f}_2 = (\boldsymbol{\phi}^{(2)})^T\hat{\mathbf{F}} = \begin{bmatrix}1&-1\end{bmatrix}\begin{bmatrix}10\\0\end{bmatrix} = 10 \text{ N} \]

Modal response amplitudes (\( \omega = 20 \) rad/s, \( \omega_1 = 31.62 \) rad/s, \( \omega_2 = 63.25 \) rad/s):

For mode 1: \( r_1 = 20/31.62 = 0.633 \)

\[ Q_1 = \frac{10/6}{\sqrt{(1-0.633^2)^2 + (2\times0.05\times0.633)^2}} = \frac{1.667}{\sqrt{(0.600)^2 + (0.0633)^2}} = \frac{1.667}{\sqrt{0.3604 + 0.00401}} = \frac{1.667}{0.604} = 2.76 \text{ mm} \]

For mode 2: \( r_2 = 20/63.25 = 0.316 \)

\[ Q_2 = \frac{10/3}{\sqrt{(1-0.316^2)^2 + (2\times0.05\times0.316)^2}} = \frac{3.333}{\sqrt{(0.900)^2 + (0.0316)^2}} = \frac{3.333}{0.901} = 3.70 \text{ mm} \]

Physical response amplitudes (approximately, ignoring phase):

\( X_1 \approx |1 \times 2.76 + 1 \times 3.70| \approx 6.46 \) mm (modes in phase at \( \omega < \omega_1 \))

\( X_2 \approx |2 \times 2.76 + (-1) \times 3.70| \approx 1.82 \) mm

(Exact amplitude requires proper phase accounting; this illustrates the modal contribution magnitudes.)

7.7 Modal Analysis: Physical Interpretation and Engineering Significance

Modal analysis is to multi-DOF vibration what Fourier series is to signal processing: it decomposes a complex response into a sum of simple, independent components (modes), each oscillating at its own natural frequency. The key engineering insights are:

  • Mode shapes reveal vulnerability: A mode shape shows which parts of a structure move most (antinodes) and which are stationary (nodes). Placing sensors at antinodes maximizes observability; placing actuators at antinodes maximizes controllability.
  • Modal truncation: In practice, structures with thousands of DOF are characterized by measuring only the first 10–50 modes. Higher modes have high natural frequencies, are hard to excite, and contribute little to low-frequency response. This is the basis of experimental modal analysis (EMA) using accelerometers and impact hammers.
  • Modal damping: Even though real damping is distributed and complex, each mode's decay rate is well characterized by a single modal damping ratio \( \zeta_r \), measured via half-power bandwidth or logarithmic decrement on that mode's frequency response peak.
  • Anti-resonance: The modal superposition sum can produce cancellation at specific frequencies — anti-resonances — where the response is anomalously small. Anti-resonances appear between resonance peaks on a frequency response function plot. They are exploited in tuned vibration absorbers (DVAs).
  • Tuned vibration absorber (DVA): Adding a secondary mass-spring system tuned to the primary system's resonant frequency introduces an anti-resonance exactly at that frequency in the primary system's response — the classic Den Hartog vibration absorber. The two new resonances bracket the original resonance, providing excellent isolation over a narrow frequency band.

Example 7.5 — Dynamic Vibration Absorber (Conceptual).

A primary machine (\( M = 100 \) kg, \( K = 10^6 \) N/m, natural frequency \( \omega_n = 100 \) rad/s) suffers large amplitude vibration at the operating frequency \( \omega_{op} = 100 \) rad/s (at resonance). Design a tuned absorber.

The absorber (secondary system, mass \( m_a \), spring \( k_a \)) is tuned so \( \omega_a = \sqrt{k_a/m_a} = \omega_{op} = 100 \) rad/s.

Choose \( m_a = 0.05M = 5 \) kg (5% mass ratio — practical range 1%–10%). Then \( k_a = m_a \omega_a^2 = 5(100)^2 = 50{,}000 \) N/m.

After adding the absorber, the combined 2-DOF system has two new natural frequencies:

\[ \omega_{1,2}^2 = \frac{\omega_n^2 + \omega_a^2(1+\mu)}{2} \mp \sqrt{\left(\frac{\omega_n^2-\omega_a^2(1+\mu)}{2}\right)^2 + \mu\omega_n^2\omega_a^2} \]

with mass ratio \( \mu = m_a/M = 0.05 \) and \( \omega_a = \omega_n \):

\[ \omega_{1,2}^2 = \frac{\omega_n^2(2 + \mu)}{2} \mp \frac{\omega_n^2}{2}\sqrt{(2+\mu)^2 - 4(1+\mu)} = \frac{\omega_n^2(2.05)}{2} \mp \frac{\omega_n^2}{2}\sqrt{0.2025} \]\[ \omega_{1,2} = \omega_n\sqrt{1.025 \mp 0.2252} \implies \omega_1 \approx 0.899\omega_n = 89.9 \text{ rad/s}, \quad \omega_2 \approx 1.118\omega_n = 111.8 \text{ rad/s} \]

At the original resonant frequency (now between the two new resonances), the primary mass has theoretically zero amplitude. The absorber mass vibrates significantly — it acts as a reactive force that cancels the excitation on the primary.


Summary and Interconnections

This course presents two interlocking perspectives on machine dynamics. Part 1 establishes the geometric and kinematic foundations: how mechanisms are built (Grübler, Grashof), how they move (loop closure, differentiation), what forces they transmit (Newton/D’Alembert), and how gears achieve speed transformation (involute geometry, planetary trains). Part 2 translates the resulting forces and motions into the vibrational language of natural frequencies, damping, and resonance — the universal framework for predicting and controlling dynamic response in all mechanical systems.

The thread connecting the two parts is the concept of unbalanced periodic forces. Part 1 computes the forces; Part 2 predicts the response to those forces. Together they equip the mechanical engineer to design machines that run smoothly, reliably, and safely throughout their operating speed range.

Key Formulas Reference

Part 1 — Mechanisms

\[ M = 3(L-1) - 2J_1 - J_2 \quad \text{(Grübler)} \]\[ s + l \leq p + q \quad \text{(Grashof criterion)} \]\[ \frac{\omega_S - \omega_C}{\omega_R - \omega_C} = -\frac{N_R}{N_S} \quad \text{(Planetary train)} \]

Part 2 — Vibrations

\[ \omega_n = \sqrt{\frac{k}{m}}, \quad \zeta = \frac{c}{2\sqrt{km}}, \quad \omega_d = \omega_n\sqrt{1-\zeta^2} \]\[ M(r,\zeta) = \frac{1}{\sqrt{(1-r^2)^2 + (2\zeta r)^2}}, \quad r = \frac{\omega}{\omega_n} \]\[ \delta = \frac{2\pi\zeta}{\sqrt{1-\zeta^2}}, \quad \left[K\right]\boldsymbol{\phi}^{(r)} = \omega_r^2\left[M\right]\boldsymbol{\phi}^{(r)} \]\[ (\boldsymbol{\phi}^{(s)})^T[M]\boldsymbol{\phi}^{(r)} = 0, \quad (\boldsymbol{\phi}^{(s)})^T[K]\boldsymbol{\phi}^{(r)} = 0 \quad (r \neq s) \]
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