AEE Mechanical Engineering Core · Chapter 2

Theory of Machines and Vibrations

What to remember

Mechanisms are counted by degrees of freedom: F = 3(n − 1) − 2j − h. A four-bar chain is Grashof (full rotation possible) if s + l ≤ p + q.
Gears, cams, belts, brakes, flywheels and governors are the working elements of power plants and machines. Each has standard formulas (module, speed fluctuation, belt tension ratio e^(μθ), and so on).
Vibration of a single-degree system is described by ωn = √(k/m), damping ratio ζ = c/cc, and transmissibility. Isolation works only when the frequency ratio exceeds √2.

1. Mechanisms and kinematics

A link is a rigid body. A kinematic pair joins two links. In a lower pair the contact is over a surface (revolute, prismatic, screw). In a higher pair the contact is along a line or point (cam and follower, gear teeth).
Kinematic chain: links joined to give constrained relative motion. A mechanism is a chain with one link fixed. A machine is a mechanism that transmits force and does useful work.
Grubler's criterion for a plane mechanism: F = 3(n − 1) − 2j − h, with n links, j lower pairs and h higher pairs. For a single-degree mechanism with lower pairs only: 3n − 2j − 4 = 0. A four-bar chain has n = 4, j = 4, F = 1.
Grashof's law: if s + l ≤ p + q (s = shortest, l = longest, p and q the others), at least one link can rotate fully.

Fixed link	Result for a Grashof four-bar
Link adjacent to the shortest	Crank-rocker
Shortest link	Double crank (drag link)
Link opposite to the shortest	Double rocker

Inversions of the single slider-crank chain: reciprocating engine (the cylinder frame is the fixed link), Whitworth quick-return and crank-and-slotted-lever mechanism (used in shapers), oscillating cylinder engine, and the hand pump.
Double slider-crank inversions: Scotch yoke, Oldham's coupling, elliptical trammel.
Instantaneous centre: the point of zero velocity between two bodies. For n links the number of centres is n(n − 1)/2. Kennedy's theorem: three bodies in relative motion have three instantaneous centres on a straight line.
Velocity and acceleration: V = ωr. Centripetal acceleration = ω²r (towards centre). Tangential acceleration = αr. Coriolis acceleration = 2ωv, present when a link slides on a rotating link.
Steering gear: the Ackermann gear satisfies the correct-steering condition approximately: cotφ − cotθ = c/b (track c, wheelbase b). The Davis gear satisfies it exactly at the cost of sliding pairs.

2. Gears and gear trains

Law of gearing: the common normal at the point of contact must always pass through the pitch point, so that the velocity ratio is constant.
Involute profile satisfies the law, and centre distance changes do not affect velocity ratio.
Module m = pitch circle diameter/number of teeth (mm). Circular pitch p = πm. Diametral pitch = T/D (inverse of module). A standard pressure angle is 20°.
Velocity ratio = ω₁/ω₂ = T₂/T₁. Centre distance = m(T₁ + T₂)/2.
Interference occurs when the tip of a gear tooth cuts the flank of the mating tooth below the base circle. Minimum teeth to avoid interference on a pinion meshing with a rack: 2/sin²φ, which is about 17 for a 20° pressure angle (so 18 is taken in practice). Remedies: undercutting, longer addendum on the pinion, higher pressure angle, profile shift.
Contact ratio is the average number of tooth pairs in contact. It must be more than 1 (preferably more than 1.2).
Simple train: the idler changes direction but not the ratio. Compound train: two gears on the same shaft; ratio = product of driven teeth over product of driver teeth.
Epicyclic (planetary) train: an arm carries gears that rotate about their own axes and also around the sun gear. Solved by the tabular method. Gear boxes and reduction units in turbines use planetary gears.

3. Cams and followers

Terms: base circle, prime circle, pitch curve, trace point, pressure angle (angle between follower motion and the normal to the cam surface).
Followers: knife-edge, roller, flat-faced, mushroom; reciprocating or oscillating.
Motion types: uniform velocity (infinite acceleration at ends), uniform acceleration and retardation (parabolic), simple harmonic motion (SHM), and cycloidal (no jerk at ends, smoothest).
For SHM with stroke h, cam angle θ for the outstroke and angular speed ω: maximum velocity = πhω/(2θ), maximum acceleration = π²hω²/(2θ²).

4. Flywheels, governors and balancing

Flywheel: stores energy and smooths the fluctuation of speed within one cycle. Coefficient of fluctuation of speed Cs = (ωmax − ωmin)/ωmean. Fluctuation of energy ΔE = I·ωmean²·Cs.
Governor: controls the mean speed when the load changes over a long period. Types: centrifugal (Watt, Porter, Proell, Hartnell) and inertia.

Governor	Feature
Watt	Simple, no central load; height h = g/ω²
Porter	Heavy central sleeve load, more sensitive
Proell	Porter with the balls fixed on extension links
Hartnell	Spring-loaded, used for high speeds

Sensitiveness = (N₂ − N₁)/N, the range of speed over the mean speed. An isochronous governor has zero range (not practical). Hunting is continuous fluctuation of speed from over-sensitivity.
Balancing of rotating masses: for static balance, ΣmR = 0 (force polygon closes). For dynamic balance, also Σ mRl = 0 (couple polygon closes).
Reciprocating mass: primary unbalanced force = mω²r cosθ. Secondary force = mω²r cos2θ/n, with n = l/r. Reciprocating forces can be balanced only partly by a rotating mass. A locomotive suffers hammer blow, variation of tractive effort and swaying couple.
Gyroscopic couple: C = Iωωp (I = moment of inertia of the spinning body, ω = spin speed, ωp = precession speed). It acts on ships, aeroplanes and vehicles while turning.

5. Friction elements: belts, clutches and brakes

Belt drive: T₁/T₂ = e^(μθ). For a V-belt: T₁/T₂ = e^(μθ/sin β), with β half the groove angle. Power = (T₁ − T₂)v. Centrifugal tension Tc = mv² (m = mass per metre). Maximum power is transmitted at v = √(T/3m), where T is the maximum allowed tension.
Slip reduces the velocity ratio. Creep is small relative motion due to elastic stretch.
Clutch torque: T = nμW·Rmean. Uniform pressure: Rmean = (2/3)(R³ − r³)/(R² − r²). Uniform wear: Rmean = (R + r)/2. Uniform wear gives the lower torque, and it is the design basis for a worn-in clutch. n is the number of friction surfaces.
Band brake: tension ratio T₁/T₂ = e^(μθ). Braking torque = (T₁ − T₂)r.

6. Vibrations

Free undamped vibration: natural frequency ωn = √(k/m) rad/s. fn = ωn/2π. With static deflection δ: ωn = √(g/δ).
Springs: in parallel k = k₁ + k₂. In series 1/k = 1/k₁ + 1/k₂.
Damped vibration: damping ratio ζ = c/cc, with critical damping cc = 2√(km) = 2mωn. If ζ < 1 the system is underdamped (oscillatory). If ζ = 1 it is critically damped. If ζ > 1 it is overdamped (no oscillation). Damped frequency ωd = ωn√(1 − ζ²).
Logarithmic decrement δ = ln(x₁/x₂) = 2πζ/√(1 − ζ²), about 2πζ for light damping.
Forced vibration: frequency ratio r = ω/ωn. Resonance occurs at r = 1. Magnification factor is large near r = 1 and falls for r above 1.
Transmissibility TR = force transmitted/force applied = √(1 + (2ζr)²)/√((1 − r²)² + (2ζr)²). Vibration isolation (TR < 1) needs r > √2. For r > √2, more damping increases the transmitted force.
Whirling (critical) speed of a shaft = √(k/m) = √(g/δ), where δ is the static deflection of the shaft. Operating above critical speed needs a flexible shaft and fast passage through resonance. Turbo-generator rotors in thermal and hydel stations are balanced and monitored for this reason.
Torsional vibration: ωn = √(kt/I), with kt = GJ/L. For a two-rotor system, the node lies where I₁l₁ = I₂l₂.
Dunkerley's method: 1/f² = 1/f₁² + 1/f₂² + …, a lower-bound estimate of the fundamental frequency. Rayleigh's method uses energy balance.

7. Worked examples

Example 1 (degrees of freedom). A mechanism with 6 links and 7 lower pairs: F = 3(5) − 2(7) = 1.

Example 2 (gears). Pinion 20 teeth, gear 60 teeth, module 5 mm. Ratio = 3. Centre distance = 5 × 80/2 = 200 mm.

Example 3 (flywheel). I = 10 kg·m², ωmean = 100 rad/s, Cs = 0.02. ΔE = 10 × 10⁴ × 0.02 = 2000 J.

Example 4 (vibration). m = 10 kg, k = 1000 N/m. ωn = √100 = 10 rad/s, fn = 10/2π ≈ 1.59 Hz. cc = 2√(1000 × 10) = 200 N·s/m. If c = 100, ζ = 0.5.

Example 5 (belt). T₁ = 1000 N, T₂ = 400 N, v = 10 m/s. Power = 600 × 10 = 6000 W = 6 kW.

Example 6 (gyroscope). I = 0.5 kg·m², ω = 100 rad/s, ωp = 2 rad/s. C = 0.5 × 100 × 2 = 100 N·m.

Exam traps

Flywheel controls fluctuation within a cycle. Governor controls mean speed over load changes.
Uniform wear gives a smaller clutch torque than uniform pressure.
Isolation of vibration needs r > √2, not r > 1.
Higher pair means line or point contact. A lower pair has surface contact.
Idler gear does not change the gear ratio, only the rotation direction.
Tension ratio e^(μθ) applies to a flat belt. V-belts use e^(μθ/sin β).
Coriolis acceleration is 2ωv. Do not confuse with centripetal acceleration ω²r.
Static balance needs force balance only. Dynamic balance needs force and couple balance.
Damping ratio ζ = c/cc, not c/m.

One-liners

1. Grubler's equation: F = 3(n − 1) − 2j − h.
2. Number of instantaneous centres is n(n − 1)/2.
3. Module = pitch diameter/number of teeth.
4. Minimum pinion teeth for 20° pressure angle without interference is about 17 or 18.
5. Cycloidal cam motion avoids jerk at ends.
6. The shaper uses a quick-return mechanism.
7. Hartnell governor is spring loaded.
8. Gyroscopic couple C = Iωωp.
9. Natural frequency ωn = √(k/m).
10. Critical damping cc = 2√(km).
11. Resonance occurs at frequency ratio 1.
12. Whirling speed of a shaft equals its natural frequency of transverse vibration.

Practice questions

In a lower kinematic pair, the contact between the two links is
1. over a surface
2. along a line only
3. at a point only
4. through a flexible element
Answer
A. over a surface
Revolute, prismatic and screw pairs are lower pairs with surface contact.
Grubler's criterion for the degrees of freedom of a plane mechanism is
1. F = 3n − 2j − h
2. F = 3(n − 1) − j − 2h
3. F = 2(n − 1) − 3j − h
4. F = 3(n − 1) − 2j − h
Answer
D. F = 3(n − 1) − 2j − h
n = links, j = lower pairs, h = higher pairs.
The quick-return motion of a shaper is obtained by
1. the Geneva mechanism
2. the crank-and-slotted-lever mechanism
3. Oldham's coupling
4. the Scotch yoke
Answer
B. the crank-and-slotted-lever mechanism
It is an inversion of the single slider-crank chain.
The law of gearing requires that the common normal at the point of contact of two teeth
1. is parallel to the line of centres
2. passes through the pitch point
3. passes through the base circle centre only
4. is tangent to the pitch circle
Answer
B. passes through the pitch point
This keeps the velocity ratio constant.
The standard pressure angle of a modern involute gear is
1. 30°
2. 45°
3. 10°
4. 20°
Answer
D. 20°
14.5° is older practice; 20° is the standard.
A flywheel is used to control
1. the direction of rotation
2. the mean speed under changing load
3. fluctuation of speed within one cycle
4. the number of cycles per minute
Answer
C. fluctuation of speed within one cycle
A flywheel stores and releases energy within each cycle.
A governor is used to control
1. the pressure of steam
2. speed fluctuation within one cycle
3. the torque of the flywheel
4. the mean speed of an engine under changing load
Answer
D. the mean speed of an engine under changing load
Governors act on the fuel or steam supply.
A system is critically damped when the damping ratio is
1. 0
2. 0.5
3. 1
4. 2
Answer
C. 1
ζ = c/cc = 1 gives the fastest return without oscillation.
Oldham's coupling is an inversion of the
1. double slider-crank chain
2. whitworth chain
3. four-bar chain
4. single slider-crank chain
Answer
A. double slider-crank chain
It connects two parallel shafts having a small offset.
At resonance of a forced undamped system, the frequency ratio ω/ωn is
1. 0
2. 1
3. 0.5
4. √2
Answer
B. 1
Resonance occurs when the forcing frequency equals the natural frequency.
Vibration isolation (transmissibility less than 1) is achieved only when the frequency ratio is greater than
1. √2
2. 1
3. 0.5
4. 2
Answer
A. √2
Transmissibility equals 1 at r = √2 regardless of damping.
The Coriolis component of acceleration is given by
1. αr
2. ω²r
3. ωv/2
4. 2ωv
Answer
D. 2ωv
It arises when a slider moves on a rotating link.
Kennedy's theorem states that for three bodies in relative plane motion, the three instantaneous centres
1. form an equilateral triangle
2. lie on a circle
3. coincide at one point
4. lie on a straight line
Answer
D. lie on a straight line
This helps locate unknown centres.
Which cam follower motion avoids jerk at the start and end of the stroke?
1. Uniform acceleration
2. Cycloidal motion
3. Uniform velocity
4. Simple harmonic motion
Answer
B. Cycloidal motion
Cycloidal motion has continuous acceleration at the ends.
A rotor is statically balanced when
1. the sum of couples is zero only
2. the vector sum of the centrifugal forces is zero
3. its mass is concentrated at the axis
4. it has an even number of masses
Answer
B. the vector sum of the centrifugal forces is zero
Static balance needs ΣmR = 0; dynamic balance also needs ΣmRl = 0.
For the same dimensions, torque capacity of a clutch by the uniform wear theory compared with the uniform pressure theory is
1. higher
2. zero
3. lower
4. equal
Answer
C. lower
(R + r)/2 is smaller than the uniform-pressure mean radius.
Dunkerley's method gives an estimate of the fundamental frequency which is
1. exact
2. an upper bound
3. always twice the true value
4. a lower bound
Answer
D. a lower bound
1/f² = Σ1/fi² underestimates the frequency slightly.
A mechanism has 5 links, 5 lower pairs and 1 higher pair. Its degrees of freedom are
1. 2
2. 0
3. 3
4. 1
Answer
D. 1
F = 3(5 − 1) − 2(5) − 1 = 12 − 10 − 1 = 1.
The number of instantaneous centres in a mechanism of 6 links is
1. 12
2. 6
3. 15
4. 30
Answer
C. 15
n(n − 1)/2 = 6 × 5/2 = 15.
A gear has a pitch circle diameter of 120 mm and 30 teeth. Its module is
1. 4 mm
2. 6 mm
3. 3.6 mm
4. 3 mm
Answer
A. 4 mm
m = D/T = 120/30 = 4 mm.
Two gears of module 4 mm have 20 and 40 teeth. Their centre distance is
1. 60 mm
2. 240 mm
3. 120 mm
4. 80 mm
Answer
C. 120 mm
a = m(T₁ + T₂)/2 = 4 × 60/2 = 120 mm.
The minimum whole number of teeth on a pinion (20° pressure angle) meshing with a rack to avoid interference is
1. 18
2. 22
3. 12
4. 14
Answer
A. 18
2/sin²20° = 17.1, so 18 teeth.
A flywheel of I = 20 kg·m² runs at a mean speed of 50 rad/s with Cs = 0.04. The fluctuation of energy is
1. 4000 J
2. 2000 J
3. 200 J
4. 1000 J
Answer
B. 2000 J
ΔE = I·ω²·Cs = 20 × 2500 × 0.04 = 2000 J.
A belt has tight side tension 1500 N, slack side tension 500 N and speed 12 m/s. The power transmitted is
1. 18 kW
2. 6 kW
3. 24 kW
4. 12 kW
Answer
D. 12 kW
P = (T₁ − T₂)v = 1000 × 12 = 12,000 W.
For a flat belt with μ = 0.3 and angle of lap π radians, the tension ratio T₁/T₂ is about
1. 1.57
2. 2.57
3. 3.14
4. 9.42
Answer
B. 2.57
e^(0.3π) = e^0.942 ≈ 2.57.
A belt of mass 0.5 kg per metre runs at 20 m/s. The centrifugal tension is
1. 100 N
2. 10 N
3. 200 N
4. 400 N
Answer
C. 200 N
Tc = mv² = 0.5 × 400 = 200 N.
The maximum tension in a belt is 300 N and its mass is 1 kg per metre. The speed for maximum power transmission is
1. 10 m/s
2. 17.3 m/s
3. 100 m/s
4. 30 m/s
Answer
A. 10 m/s
v = √(T/3m) = √(300/3) = 10 m/s.
A clutch with two friction surfaces has inner radius 60 mm and outer radius 100 mm, μ = 0.25 and axial force 1 kN. Using the uniform wear theory, the torque capacity is
1. 20 N·m
2. 80 N·m
3. 16 N·m
4. 40 N·m
Answer
D. 40 N·m
T = nμW(R + r)/2 = 2 × 0.25 × 1000 × 0.08 = 40 N·m.
A system with m = 10 kg, k = 1000 N/m and viscous damping c = 40 N·s/m has a damping ratio of
1. 0.4
2. 0.1
3. 2
4. 0.2
Answer
D. 0.2
cc = 2√(km) = 200; ζ = 40/200 = 0.2.
For light damping with ζ = 0.05, the logarithmic decrement is about
1. 0.05
2. 3.14
3. 0.314
4. 0.0314
Answer
C. 0.314
δ ≈ 2πζ = 2π × 0.05 = 0.314.
A disc with I = 2 kg·m² spins at 50 rad/s and precesses at 0.5 rad/s. The gyroscopic couple is
1. 5 N·m
2. 50 N·m
3. 25 N·m
4. 100 N·m
Answer
B. 50 N·m
C = Iωωp = 2 × 50 × 0.5 = 50 N·m.
A shaft has a static deflection of 25 mm under its own load. Taking g = 10 m/s², its critical speed is
1. 20 rad/s
2. 10 rad/s
3. 40 rad/s
4. 400 rad/s
Answer
A. 20 rad/s
ωc = √(g/δ) = √(10/0.025) = √400 = 20 rad/s.
Consider the statements about kinematic pairs. 1. A higher pair has line or point contact. 2. A cam and follower is a higher pair. Which is/are correct?
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
C. Both 1 and 2
Both are standard facts.
Consider the statements about an idler gear. 1. It changes the velocity ratio of the train. 2. It changes the direction of rotation of the driven gear. Which is/are correct?
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
B. 2 only
An idler does not change the ratio.
Consider the statements about damping. 1. An underdamped system oscillates with decaying amplitude. 2. An overdamped system oscillates continuously with constant amplitude. Which is/are correct?
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
A. 1 only
An overdamped system returns without oscillation.
Consider the statements about vibration isolation. 1. Transmissibility is less than 1 only when the frequency ratio exceeds √2. 2. For frequency ratio above √2, increasing damping reduces the transmitted force. Which is/are correct?
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
A. 1 only
In the isolation region, more damping increases the transmitted force.
Consider the statements. 1. A flywheel controls the mean speed under load changes. 2. A governor controls the speed fluctuation within a cycle. Which is/are correct?
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
D. Neither 1 nor 2
The roles are the other way round.
Consider the statements about balancing. 1. For static balance the force polygon of rotating masses must close. 2. For dynamic balance both the force polygon and the couple polygon must close. Which is/are correct?
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
C. Both 1 and 2
Dynamic balance needs both conditions.
Consider the statements about clutches. 1. In the uniform pressure theory the mean radius is (R + r)/2. 2. In the uniform wear theory the mean radius is (R + r)/2. Which is/are correct?
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
B. 2 only
Uniform pressure gives (2/3)(R³ − r³)/(R² − r²).
Consider the statements about governors. 1. A Porter governor has a heavy central sleeve load. 2. A Porter governor is a spring-loaded governor. Which is/are correct?
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
A. 1 only
Hartnell governor is the spring-loaded type.
Consider the statements about acceleration. 1. Coriolis acceleration equals 2ωv. 2. It occurs when a slider moves along a rotating link. Which is/are correct?
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
C. Both 1 and 2
Both are correct.
Consider the statements about instantaneous centres. 1. The number of centres in a mechanism of n links is n(n − 1). 2. Three bodies in relative motion have three centres forming a triangle. Which is/are correct?
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
D. Neither 1 nor 2
The number is n(n − 1)/2 and the three centres lie on a straight line.
Consider the statements about frequency estimates. 1. Dunkerley's method gives a lower bound for the fundamental frequency. 2. Rayleigh's method uses an energy balance. Which is/are correct?
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
C. Both 1 and 2
Both are correct.
Which pair of mechanism and use is correctly matched?
1. Scotch yoke – transmission of power between intersecting shafts
2. Oldham's coupling – connecting parallel shafts with a small offset
3. Whitworth mechanism – speed control of engines
4. Geneva mechanism – force balancing of rotors
Answer
B. Oldham's coupling – connecting parallel shafts with a small offset
Oldham's coupling is a double slider-crank inversion for offset parallel shafts.
Which pair of governor and type is correct?
1. Porter – spring loaded
2. Proell – inertia type
3. Watt – spring loaded
4. Hartnell – spring loaded
Answer
D. Hartnell – spring loaded
Hartnell has a compressed spring balancing centrifugal force.

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