UNIT 4: Dynamics of Machines (ME-503 C)

Based on RGPV Past Papers (2022–2025)

I. Governors

Function & Classification

• Function: Maintain constant mean speed of engine under varying load by regulating fuel/steam supply.

• Classification:

  • Centrifugal/Inertia Governors: Use rotating masses (balls) to sense speed changes.

  • Simple vs. Compound: Simple has one set of rotating arms; compound has two (e.g., Porter, Proell).

Watt Governor

• Construction: Two balls on arms, hinged to a vertical spindle; arms connected to a sleeve.

• Height Derivation:

  For equilibrium:

$$ mg = \frac{m \omega^2 h}{g} \cdot \frac{h}{2} \quad \Rightarrow \quad h = \frac{g}{\omega^2} $$

With $$\displaystyle \omega = \frac{2\pi N}{60} $$,

$$ h = \frac{895}{N^2} \text{ (meters, } N \text{ in rpm)} $$

• Proof: $$\displaystyle h \propto \frac{1}{N^2} $$ directly from above equation.

[!TIP]

Watt governor is insensitive at high speeds; used for low-speed engines.

Porter Governor

• Construction: Upper arms pivoted on spindle; lower arms connected to sleeve via links. Central load $ W $ on sleeve.

• Equilibrium:

$$ \frac{m \omega^2 r}{g} = \frac{W + mg}{2} \left( \frac{a}{b} \right) \cos\theta $$

where $ a, b $ are lower/upper arm lengths, $ \theta $ = inclination of upper arm.

• Speed Range with Friction:

  Friction $$\displaystyle F_f $$ equivalent to force at sleeve. Limiting inclinations $$\displaystyle \theta_1, \theta_2 $$ give:

$$ N_{\text{max}} = \sqrt{\frac{(W + mg)(a/b) \cos\theta_1}{2m r_1} \cdot \frac{g}{1 - F_f/(W+mg)}} $$

$$ N_{\text{min}} = \sqrt{\frac{(W + mg)(a/b) \cos\theta_2}{2m r_2} \cdot \frac{g}{1 + F_f/(W+mg)}} $$

Range $$\displaystyle = N_{\text{max}} - N_{\text{min}} $$.

Proell Governor

• Construction: Lower arms extended beyond pivot; extensions parallel to axis at min radius.

• Equilibrium Speed:

  For min radius $$\displaystyle r_{\text{min}} $$ (extensions parallel):

$$ N_{\text{min}} = \sqrt{\frac{2(W + mg)(a/b)}{m r_{\text{min}}} \cdot \frac{g}{\cos\theta}} $$

For max radius $$\displaystyle r_{\text{max}} $$: similar with $$\displaystyle r_{\text{max}} $$.

• Sensitiveness Comparison:

  Proell governor has greater sensitiveness than Porter because for same $ r $, $ \cos\theta $ is larger in Proell (lower arms pivoted away), giving larger speed change for same radius change.

Governor Characteristics

[!TIP]

Stability Diagram:

• Stable: $$\displaystyle F_c $$ vs $ r $ curve with positive slope.

• Unstable: Negative slope.

• Isochronous: Straight line through origin.

II. Flywheels and Energy Fluctuation

Turning Moment Diagram (Four-Stroke Cycle)

• Interpretation:

  • One power stroke per two revolutions (720°).

  • Diagram shows torque variation: high during expansion, negative during compression, pumping losses.

  • Area above mean torque line = net work output per cycle.

Fluctuation of Energy ($ \Delta E $)

• Definition: Difference between max and min kinetic energy of flywheel.

• Relation to Diagram: $ \Delta E $ = maximum area of one of the loops between torque curve and mean resistance line.

• Calculation:

  If diagram scale: $$\displaystyle 1 \text{ mm} = k_1 \text{ N-m} $$ vertically, $$\displaystyle 1 \text{ mm} = k_2 \text{ rad} $$ horizontally,

$$ \Delta E = A_{\text{max}} \times k_1 \times \frac{1}{k_2} \quad (\text{in N-m}) $$

where $$\displaystyle A_{\text{max}} $$ = max enclosed area in mm².

Fluctuation of Speed ($ \Delta N $)

• Definition: $$\displaystyle \Delta N = N_{\text{max}} - N_{\text{min}} $$.

• Coefficient of Fluctuation of Speed: $$\displaystyle \delta = \frac{\Delta N}{N_{\text{mean}}} $$.

Coefficients

• Coefficient of Fluctuation of Energy (CE): $$\displaystyle C_E = \frac{\Delta E}{E_{\text{mean}}} $$.

• Coefficient of Fluctuation of Speed (CS): $$\displaystyle C_S = \delta $$.

Flywheel Design

• Mass & Radius of Gyration:

$$ \Delta E = I \omega^2 \delta \quad \Rightarrow \quad I = \frac{\Delta E}{\omega^2 \delta} $$

where $$\displaystyle I = m k^2 $$, $ k $ = radius of gyration.

$$ m = \frac{\Delta E}{k^2 \omega^2 \delta} $$

• Given Speed Limits: $$\displaystyle \delta = \frac{\Delta N}{N_{\text{mean}}} $$.

[!TIP]

Always convert units: $$\displaystyle \omega = \frac{2\pi N}{60} $$ rad/s. Use consistent units (N, m, s).

III. Balancing of Engines

Fundamentals

• Primary Balancing: Balance inertia forces due to crank rotation (first harmonic).

• Secondary Balancing: Balance forces due to connecting rod obliquity (second harmonic).

• Partial Balancing: Only fraction $ c $ of reciprocating mass balanced to avoid excessive vertical forces in locomotives.

• In-line Engines:

  • Primary forces can be balanced by crankshaft phase angles (e.g., 180° for 2-cylinder, 120° for 3-cylinder).

  • Complete balance impossible for reciprocating masses because secondary forces and couples remain.

Reciprocating Mass Forces (Neglecting rod obliquity)

Let $ m $ = reciprocating mass, $ r $ = crank radius, $ l $ = connecting rod length, $ \theta $ = crank angle from IDC.

1. Inertia Force:

$$ F_i = m \omega^2 r \left( \cos\theta + \frac{r}{l} \cos 2\theta \right) $$

2. Piston Effort ($$\displaystyle F_p $$): Net force on piston.

$$ F_p = P \cdot \frac{\pi D^2}{4} - F_i $$

where $ P $ = pressure difference (cover end - piston end).

3. Side Thrust on cylinder wall:

$$ F_s = \frac{F_p}{\tan\phi} \approx F_p \cdot \frac{r \sin\theta}{l \cos\theta} $$

(for small $ \phi $, $$\displaystyle \tan\phi \approx \frac{r \sin\theta}{l \cos\theta} $$).

4. Thrust in Connecting Rod ($$\displaystyle F_c $$):

$$ F_c = \frac{F_p}{\cos\phi} \approx \frac{F_p}{\cos\left( \frac{r}{l} \sin\theta \right)} $$

5. Crank Effort (Torque on crank shaft):

$$ T = F_c \cdot r \cdot \sin(\theta + \phi) \approx F_c \cdot r \cdot \sin\theta $$

(for small $ \phi $).

Radial Engines (e.g., 3-cylinder at 120°)

• Primary Forces:

  $$\displaystyle F_{p1} = m \omega^2 r \cos\theta_i $$, $$\displaystyle \theta_i = 0°, 120°, 240° $$.

  Sum $$\displaystyle \sum \cos\theta_i = 0 $$ → balanced.

• Secondary Forces:

  $$\displaystyle F_{p2} = m \omega^2 \frac{r^2}{l} \cos 2\theta_i $$.

  $$\displaystyle 2\theta_i = 0°, 240°, 480°≡120° $$ → sum $$\displaystyle = 0 $$ → balanced.

Locomotive Engines (Uncoupled Two-Cylinder, Crank Angle $ \alpha $)

• Assumptions: Reciprocating mass $ m $ per cylinder, crank radius $ r $, speed $ \omega $.

• Balancing Fraction: $ c $ (fraction balanced by revolving mass).

• Unbalanced Vertical Force (Hammer Blow):

  For $$\displaystyle \alpha = 90° $$:

$$ F_v = m r \omega^2 \left[ (1-c) \cos\theta \pm c \sin\theta \right] $$

Maximum hammer blow $$\displaystyle = m r \omega^2 \sqrt{(1-c)^2 + c^2} $$.

• Swaying Couple (about vertical axis):

$$ C_s = m r \omega^2 \cdot d \left[ (1-c) \sin\theta \pm c \cos\theta \right] $$

where $ d $ = distance between cylinder centerlines.

Maximum $$\displaystyle = m r \omega^2 d \sqrt{(1-c)^2 + c^2} $$.

• Variation in Tractive Effort:

  Due to inertia forces, effective driving force varies:

$$ \Delta F_t = \frac{1}{R} \left[ \pm m r \omega^2 \left( (1-c) \cos\theta \pm c \sin\theta \right) \right] $$

where $ R $ = driving wheel radius.

Balancing Mass

• Revolving Mass: Place mass $$\displaystyle m_b $$ at radius $ r $ opposite crank: $$\displaystyle m_b r = m r $$.

• Reciprocating Mass: Balance fraction $ c $ at crank radius: $$\displaystyle m_b r = c m r $$.

• Residual Unbalance Force after partial balancing:

$$ F_{\text{res}} = (1-c) m r \omega^2 \cos\theta \quad \text{(for single cylinder)} $$

[!TIP]

For two-cylinder locomotive with $$\displaystyle \alpha = 90° $$, primary forces cannot be completely balanced without causing vertical forces. Balance $ c \approx 0.6 $ to limit hammer blow.

IV. Friction Clutches

Single Plate Clutch

• Uniform Pressure Assumption ($$\displaystyle p = \text{constant} $$):

  Axial force $$\displaystyle F = \pi p (r_o^2 - r_i^2) $$.

  Torque:

$$ T = \mu F \cdot \frac{2}{3} \cdot \frac{r_o^3 - r_i^3}{r_o^2 - r_i^2} $$

• Uniform Wear Assumption ($$\displaystyle p r = \text{constant} $$):

  $$\displaystyle p_i r_i = p_o r_o = c $$.

  Axial force: $$\displaystyle F = 2\pi c (r_o - r_i) $$.

  Torque:

$$ T = \mu F \cdot \frac{r_o + r_i}{2} $$

Average pressure: $$\displaystyle p_{\text{avg}} = \frac{F}{\pi (r_o^2 - r_i^2)} $$.

Conical Clutch

• Construction: Cone-shaped friction surface on shaft or hub; axial force $ F $.

• Pressure Distribution: Uniform wear assumed ($$\displaystyle p r = \text{constant} $$).

• Torque Expression:

  Normal force $$\displaystyle F_n = \frac{F}{\sin\alpha} $$ (α = cone half-angle).

  Mean radius $$\displaystyle R_m = \frac{r_o + r_i}{2} $$.

$$ T = \mu F_n \cdot 2\pi R_m \cdot R_m = \mu \frac{F}{\sin\alpha} \cdot \pi (r_o + r_i) R_m $$

Simplified:

$$ T = \mu F \cdot \frac{\pi (r_o + r_i)^2}{2 \sin\alpha} $$

Multi-plate Clutch (Brief)

• Multiple friction discs alternately attached to shaft and hub.

• Torque proportional to number of friction surfaces.

[!TIP]

In design problems, given $$\displaystyle P, N, \mu, p_{\text{max}} $$, and $$\displaystyle R/b = 4 $$, use uniform wear for conservative design.

V. Brakes

Band Brake (with Lever)

• Tensions: $$\displaystyle T_1 $$ (tight), $$\displaystyle T_2 $$ (slack), $$\displaystyle \frac{T_1}{T_2} = e^{\mu\theta} $$ (θ in radians).

• Torque: $$\displaystyle T_b = (T_1 - T_2) r $$.

• Lever Mechanics:

  If band attached at distances $ a $ (from fulcrum to $$\displaystyle T_1 $$) and $ b $ (from fulcrum to $$\displaystyle T_2 $$), effort $ E $ at lever end:

$$ E \cdot l = T_1 \cdot a - T_2 \cdot b $$

Solve for $$\displaystyle T_1, T_2 $$, then $$\displaystyle T_b $$.

Internal Expanding Brake

• Construction: Two shoes inside drum, operated by cam or wheel.

• Self-energizing: Friction force on leading shoe adds to normal force, increasing torque.

• Working: Cam rotates, pushes shoes outward against drum.

Double Shoe Brake

• Force Calculation: For braking torque $$\displaystyle T_b $$,

$$ T_b = 2 \mu F_n R_m \cdot \theta \quad (\text{θ in radians, for uniform pressure}) $$

or $$\displaystyle T_b = 2 \mu F_n R_m \cdot \sin\theta $$ for curved shoes.

• Shoe Width: From allowable bearing pressure $ p $:

$$ F_n = p \cdot (\text{projected area}) = p \cdot (b \cdot 2R_m \sin\theta) $$

Solve for $ b $.

VI. Bearings and Friction Loss

Conical Pivot Bearing

• Uniform Pressure:

  Load: $$\displaystyle W = \frac{\pi p}{\sin\alpha} (r_o^2 - r_i^2) $$.

  Torque: $$\displaystyle M = \frac{2\pi \mu p}{3 \sin\alpha} (r_o^3 - r_i^3) $$.

  Power loss: $$\displaystyle P = M \omega $$.

• Uniform Wear:

  $$\displaystyle p r = c $$.

  Load: $$\displaystyle W = \frac{2\pi c}{\sin\alpha} (r_o - r_i) $$.

  Torque: $$\displaystyle M = \frac{\pi \mu c}{\sin\alpha} (r_o^2 - r_i^2) $$.

Collar Bearing

• Uniform Pressure:

  $$\displaystyle W = \pi p (r_o^2 - r_i^2) $$,

  $$\displaystyle M = \frac{2}{3} \mu \pi p (r_o^3 - r_i^3) $$.

• Uniform Wear:

  $$\displaystyle W = 2\pi p_i r_i (r_o - r_i) $$,

  $$\displaystyle M = \pi \mu p_i r_i (r_o^2 - r_i^2) $$.

• Number of Collars:

  $$\displaystyle n = \frac{W}{W_{\text{per collar}}} $$, where $$\displaystyle W_{\text{per collar}} $$ from above.

[!TIP]

For conical pivot, cone angle is usually included angle; use half-angle in formulas.

VII. Dynamometers

Absorption vs. Transmission

Torsion Dynamometer

• Construction: Torque shaft with strain gauges or angular twist measurement.

• Working:

  Torque $ T $ measured from strain $ \epsilon $: $$\displaystyle T = \frac{\pi d^3}{16} \cdot \frac{E \epsilon}{2} $$ (for circular shaft).

  Or from angular twist $ \phi $: $$\displaystyle T = \frac{G J \phi}{L} $$.

• Power: $$\displaystyle P = T \omega $$.

Other Types

• Hydraulic (fluid friction), Eddy Current (magnetic), Prony Brake (friction band).

VIII. Kinematic and Dynamic Analysis

Four-Bar Mechanism

• Velocity Analysis (Instantaneous Center Method):

  $$\displaystyle \omega_2 = \frac{v_B}{AB} $$, then $$\displaystyle v_C = \omega_3 \cdot BC $$.

• Acceleration Analysis:

  $$\displaystyle a_B = \alpha_2 \times AB - \omega_2^2 AB $$, then $$\displaystyle a_C = a_B + \alpha_3 \times BC - \omega_3^2 BC $$.

• Example: Given $$\displaystyle AB=60\text{mm}, BC=CD=70\text{mm}, DA=120\text{mm}, \omega_{AB}=10\text{ rad/s} $$ at $$\displaystyle \angle DAB=60° $$.

  Use vector loop: $$\displaystyle \vec{r}_{AB} + \vec{r}_{BC} - \vec{r}_{CD} - \vec{r}_{DA} = 0 $$.

Cams (Offset Circular Cam)

• Construction: Circular disc radius $ R $, center offset $ e $ from camshaft axis. Flat-faced follower.

• Follower Displacement: $$\displaystyle s = e(1 - \cos\theta) $$.

• Acceleration:

$$ a = e \omega^2 \cos\theta $$

where $ \theta $ = rotation angle from start of lift.

• Lift-off Condition:

  Spring force $$\displaystyle F_s = k(s + s_0) $$, where $$\displaystyle s_0 $$ = initial compression.

  Lift-off when dynamic force $$\displaystyle F_d = m a $$ overcomes $$\displaystyle F_s $$:

$$ m e \omega^2 \cos\theta = k(s + s_0) $$

Solve for $ \omega $.

Dynamically Equivalent System

• Definition: Replace a rigid body with two masses $$\displaystyle m_1, m_2 $$ at a reference point such that:

  1. Total mass same: $$\displaystyle m_1 + m_2 = m $$.

  2. Kinetic energy same: $$\displaystyle \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 = \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2 $$.

  3. Masses placed along line of motion: $$\displaystyle m_1 k_1^2 + m_2 k_2^2 = I $$.

• Common Choice: $$\displaystyle m_1 = m $$, $$\displaystyle m_2 = I/k^2 $$ at reference point.

IX. Special Topics

Friction Circle

• Definition: In journal bearings, friction force acts along tangent to journal surface; resultant lies on circle of radius $$\displaystyle r_f $$.

• Radius Derivation:

  Friction angle $ \phi $, $$\displaystyle \tan\phi = \mu $$.

$$ r_f = r_j \sin\phi \approx r_j \mu \quad (\text{for small } \phi) $$

where $$\displaystyle r_j $$ = journal radius.

Locomotive-Specific Derivation

• Frictional Couple (Uncoupled Two-Cylinder Four-Stroke):

  Due to side thrust on cylinder walls, couple about vertical axis:

$$ C_f = \mu m g \cdot d \cdot \sin\theta \quad \text{(approx)} $$

where $ d $ = distance between cylinders.

Comparison Notes

[!TIP]

Exam Focus: Derive hammer blow and swaying couple for two-cylinder locomotive with crank angle $ \alpha $. Prove Proell more sensitive than Porter by comparing $$\displaystyle \frac{dN}{d\theta} $$.

Final Note: Always check units (convert rpm to rad/s, mm to m). Use consistent SI units. For governor problems, draw free-body diagrams. For balancing, resolve forces horizontally/vertically.