Mechatronics (ME-503 (A)) - Unit 3 Short Notes

UNIT 3: MECHANICS OF MACHINES (DYNAMICS & CONTROL)

I. GOVERNORS

Function & Classification

• Function: Automatically controls the speed of an engine by regulating the fuel/steam supply in response to load changes.

• Classification:

  • Centrifugal vs. Inertia: Based on driving agency (centrifugal force vs. inertia of masses).

  • Simple vs. Compound: Based on number of revolving masses (1 set vs. 2 sets).

  • Types: Watt (simple centrifugal), Porter (compound, loaded sleeve), Proell (compound, offset lower pivots).

Watt Governor

• Construction: Two hinged arms with balls at ends, connected to a sleeve on the spindle.

• Working: As speed increases, centrifugal force lifts balls & sleeve, closing the fuel valve via linkage.

• Derivation of Height:

  At equilibrium for one ball:

$$ F_c = m \omega^2 r = \frac{W}{\cos \theta} \quad \text{and} \quad h = l \cos \theta $$

$$ \therefore h = \frac{g}{\omega^2} \quad \text{or} \quad h \propto \frac{1}{N^2 $$

> **Exam Tip**: This is a **fundamental derivation**. Prove using force triangle or resolving forces vertically/horizontally.

Porter Governor

• Construction: Similar to Watt, but lower arms are attached to a central load (W₁) on the sleeve.

• Force Analysis & Speed Equation:

  Consider forces on one ball and its arm. From geometry:

$$ r = \frac{m \omega^2 h}{g} \cdot \frac{l_2}{l_1 + l_2} + \frac{W_1}{2m \omega^2} \cdot \frac{l_1}{l_1 + l_2} $$

where \( l_1 \) = upper arm length, \( l_2 \) = lower arm length.

• Effect of Friction: Friction at sleeve increases minimum speed and decreases maximum speed, reducing sensitivity.

• Numerical: Use force equations for limiting inclinations \( \theta_1, \theta_2 \) with friction force \( F_f \) added/subtracted appropriately.

Proell Governor

• Construction: Lower arm pivots are offset from the axis by distance \( d \). Extensions \( e \) from pivots to ball centers.

• Force Analysis & Speed Equation:

$$ r = \frac{m \omega^2 h}{g} \cdot \frac{l_2}{l_1 + l_2} + \frac{W_1}{2m \omega^2} \cdot \frac{l_1 + d}{l_1 + l_2} $$

(At min speed, extensions parallel to axis).

• Comparison with Porter: For same \( m, W_1, l_1, l_2 \), Proell has higher sensitivity (larger \( \Delta N \)) because the term \( (l_1 + d) > l_1 \).

• Proof: Sensitiveness (Proell > Porter):

  Sensitiveness \( S = \frac{\Delta N}{N} \). From speed equations, \( \frac{dr}{dN} \) is larger for Proell due to \( (l_1 + d) \) term, hence \( \Delta N \) larger for same \( \Delta r \).

Governor Characteristics & Performance

Controlling Force Diagram: Plot \( F_c \) (vertical) vs. \( r \) (horizontal). Stability condition derived from slope of curve vs. slope of line from origin to point.

II. FLYWHEELS

Function & Distinction

• Flywheel: Stores kinetic energy during power stroke, releases during other strokes → reduces speed fluctuation.

• Governor: Controls mean speed by regulating energy input → does not affect fluctuation magnitude.

Turning Moment Diagram

• Drawn for one complete cycle (e.g., 720° for 4-stroke engine).

• Horizontal axis: Crank angle. Vertical axis: Turning moment (torque) on crank.

• Mean torque line: Horizontal line such that area above = area below.

• Interpretation: Areas between torque curve and mean line represent net energy added or subtracted per cycle.

Fluctuation Concepts

Design & Analysis

• Energy Stored in Flywheel:

$$ \Delta E = \frac{1}{2} I (\omega_{\text{max}}^2 - \omega_{\text{min}}^2) \approx I \omega \Delta \omega \quad (\text{for small } \Delta \omega) $$

where \( I = m k^2 \) (mass \( m \), radius of gyration \( k \)).

• Numerical Procedure:

  1. Find \( \Delta E \) from turning moment diagram (scale conversion).

  2. Given \( N \) (mean speed), \( \Delta N \) (limit), compute \( \omega = \frac{2\pi N}{60} \), \( \Delta \omega = \frac{2\pi \Delta N}{60} \).

  3. \( I = \frac{\Delta E}{\omega \Delta \omega} \), then \( m = \frac{I}{k^2} \) or \( k = \sqrt{I/m} \).

III. BALANCING OF RECIPROCATING & ROTATING MASSES

Fundamental Concepts

• Primary vs. Secondary:

  • Primary: Forces due to simple harmonic motion (SHM) assumption (\( \cos \omega t \)).

  • Secondary: Forces due to connecting rod obliquity (\( \cos 2\omega t \)), significant when \( \frac{r}{l} > \frac{1}{4} \).

• Dynamically Equivalent System: A system of two masses (one rotating, one reciprocating) that produces same inertia forces & moments as the original body.

• Partial Balancing: Only part of reciprocating mass is balanced to avoid excessive vertical unbalanced force (hammer blow) in locomotives. Typically balance 50-70%.

Forces in Engines

For a horizontal engine, crank angle \( \theta \) from IDC:

1. Inertia Force:

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

\( m_r \) = reciprocating mass, \( r \) = crank radius, \( l \) = connecting rod length.

2. Piston Effort (\( F_P \)): Net force on piston.

$$ F_P = F_{\text{gas}} - F_I $$

3. Crank Effort (\( F_T \)): Tangential force on crank.

$$ F_T = F_P \tan \phi \approx F_P \frac{r \sin \theta}{l \sin \theta} = F_P \frac{r}{l} \sin \theta \quad (\phi = \text{crank angle}) $$

4. Thrust on Cylinder Walls (\( F_C \)):

$$ F_C = \frac{F_P}{\cos \phi} \approx F_P \frac{l}{r} \frac{1}{\cos \theta} $$

5. Thrust in Connecting Rod (\( F_R \)):

$$ F_R = \frac{F_P}{\cos \phi} \approx F_P \frac{l}{r} \sec \theta $$

Unbalanced Forces & Couples

• Hammer Blow (Unbalanced Vertical Force):

  For a single uncoupled cylinder:

$$ F_V = (1 - c) m_r \omega^2 r \cos \theta $$

where \( c \) = fraction of reciprocating mass balanced. Acts vertically at crankpin.

> **Effect**: Causes **dynamic load variation** on rails, leading to vibration.

• Swaying Couple (Unbalanced Couple about vertical axis):

  For uncoupled two-cylinder engine (crank angles \( \theta, \theta+\phi \)):

$$ \text{Couple} = (1 - c) m_r \omega^2 r h (\cos \theta \pm \cos(\theta+\phi)) \cdot \frac{s}{2} $$

\( h \) = height of center of gravity, \( s \) = distance between cylinder centerlines. Sign depends on cylinder arrangement.

• Variation in Tractive Effort:

  For locomotives, net force on wheel-rail contact. Fluctuates due to unbalanced inertia forces. Maximum variation \( \Delta T = (1-c) m_r \omega^2 r \).

Multi-cylinder Engine Balancing

Key Formulae for Locomotive (Two-cylinder, crank angle \( \phi \)):

1. Fraction to balance:

$$ c = 1 - \frac{F_{\text{max}}}{m_r \omega^2 r} $$

where \( F_{\text{max}} \) = permissible hammer blow.

2. Maximum Swaying Couple:

$$ C_{\text{max}} = (1-c) m_r \omega^2 r \cdot \frac{s}{2} \cdot h $$

3. Variation in Tractive Effort:

$$ \Delta T = (1-c) m_r \omega^2 r $$

IV. FRICTION DEVICES & BEARINGS

Friction Clutches

Single Plate Clutch

• Assumptions:

  • Uniform Pressure: \( p = \text{constant} = \frac{W}{\pi (R_o^2 - R_i^2)} \)

$$ T = \mu W R_m, \quad R_m = \frac{2}{3} \frac{R_o^3 - R_i^3}{R_o^2 - R_i^2} $$

*   **Uniform Wear**: \( p \propto \frac{1}{r} \Rightarrow p R = \text{constant} \)

$$ T = \frac{1}{2} \mu W (R_o + R_i), \quad R_m = \frac{R_o + R_i}{2} $$

• Numerical: Given \( P, N, \mu, p_{\text{max}} \), find \( R_o, R_i, b \) (face width). Use \( T = \frac{P}{\omega} \), \( W = p_{\text{avg}} \cdot \text{area} \).

Conical Clutch

• Torque Expression:

$$ T = \mu W R_m \csc \alpha $$

where \( \alpha \) = cone angle, \( R_m \) = mean radius of contact. For uniform pressure, \( R_m \) as above; for uniform wear, \( R_m = \frac{R_o + R_i}{2} \).

Friction Brakes

Band Brake (Simple & Differential)

• Tensions: \( \frac{T_1}{T_2} = e^{\mu \theta} \) (\( \theta \) in radians).

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

• Differential Band: Lever attached to both ends. \( T_1 \) and \( T_2 \) related by lever principle.

• Numerical: Apply \( T_1/T_2 \) ratio and moment equilibrium about fulcrum.

Internal Expanding Shoe Brake (Self-energizing)

• Construction: Two shoes inside drum. Leading shoe (rotation direction) is self-energizing.

• Force Analysis: For leading shoe, friction adds to actuating force. Solve for \( T_1, T_2 \) using:

  • Moment about shoe pivot: \( F \cdot l = (T_1 - T_2) r \)

  • Friction relation: \( T_1 = T_2 e^{\mu \theta} \)

• Torque: \( T_b = (T_1 - T_2) r \).

Double Shoe Brake

• Construction: Two fixed shoes (or two expanding). Forces symmetric.

• Numerical: Given \( T_b \), find spring force \( F \) (each shoe). Use \( T_b = 2 (T_1 - T_2) r \) and force equilibrium on each shoe.

Bearings

Conical Pivot Bearing

• Uniform Pressure Assumption:

$$ p = \frac{W}{\pi (R_o^2 - R_i^2)} $$

$$ \text{Power loss } P = \frac{\mu W \omega (R_o^3 - R_i^3)}{3 (R_o^2 - R_i^2)} $$

• Uniform Wear Assumption:

  \( p \propto 1/r \Rightarrow p = \frac{W}{2\pi (R_o^2 + R_i^2)} \cdot \frac{R_o R_i}{R_o R_i} \) (derived from \( p R = \text{const} \))

$$ P = \frac{\mu W \omega (R_o^2 + R_i^2)}{4 R_o} $$

• Numerical: Given \( W, \omega, \mu, p_{\text{max}} \), find \( R_o, R_i \) from pressure condition, then \( P \).

Friction Circle

• Definition: Circle with radius \( r_f = r \sin \phi \), where \( r \) = journal radius, \( \phi \) = angle of friction.

• Concept: For a journal bearing with friction, the reaction force \( R \) acts along the tangent to this circle. The frictional force \( F = \mu R \) acts along the tangent to the friction circle.

• Radius Derivation:

  From geometry, \( \sin \phi = \frac{r_f}{r} \Rightarrow r_f = r \sin \phi \).

  Application: Used in velocity analysis of mechanisms with friction (e.g., cams, bearings).

V. DYNAMOMETERS

Classification & Purpose

• Absorption Dynamometer: Absorbs engine power (e.g., Prony brake, hydraulic). Measures brake power.

• Transmission Dynamometer: Transmits power while measuring (e.g., epicyclic train, torsion dynamometer). Measures power transmitted.

• Torsion Dynamometer: Subclass of transmission type. Measures torque on a shaft via angular twist.

Torsion Dynamometer

• Construction: Shaft with strain gauges or epicyclic gear train with torque arm.

• Working Principle:

  1. Torque \( T \) on shaft causes torsional twist \( \theta \).

  2. Strain gauges measure strain \( \epsilon \propto \theta \).

  3. Or, epicyclic gear: Torque reaction on a lever arm of length \( l \) gives force \( F = T/l \).

• Power Calculation:

$$ P = T \cdot \omega = \frac{2\pi N T}{60} $$

where \( T \) from measured strain or lever force.

VI. KINEMATICS OF MECHANISMS

Four-Bar Mechanism (Velocity & Acceleration)

• Instantaneous Center (IC) Method:

  • Kennedy's Theorem: Three bodies in plane have three ICs lying on a straight line.

  • Velocity: \( v = \omega \cdot \text{distance to IC} \).

  • Acceleration: Use relative acceleration equation:

$$ \vec{a}_B = \vec{a}_A + \vec{a}_{BA}^t + \vec{a}_{BA}^n $$

    where \( \vec{a}_{BA}^t = \alpha_{BA} \times \vec{r}_{BA} \), \( \vec{a}_{BA}^n = -\omega_{BA}^2 \vec{r}_{BA} \).

• Numerical Steps:

  1. Draw configuration to scale.

  2. Locate all ICs (IC of coupler with ground, etc.).

  3. Given \( \omega_{AB} \), find \( v_B = \omega_{AB} \cdot AB \).

  4. \( v_C = v_B \cdot \frac{BC_{\text{IC}}}{BC} \) (using IC of coupler).

  5. \( \omega_{BC} = v_C / BC \), \( \omega_{CD} = v_C / CD \).

  6. For acceleration, use \( a_B = a_A + a_{B/A}^t + a_{B/A}^n \) (with \( a_A=0 \) if ground).

Cams (Dynamic Analysis)

• Radial Cam with Offset Center & Flat-Faced Follower:

  • Offset: Cam center \( O \) offset by \( e \) from follower line of action (passes through shaft axis \( C \)).

  • Follower Lift: \( s = e(1 - \cos \theta) \) for simple harmonic motion? No – depends on cam profile.

  • Acceleration Derivation:

    Let cam rotate with \( \omega \). Follower position \( y \) (vertical). Cam radius \( r(\theta) \).

    For flat follower in contact with cam surface:

$$ y = r(\theta) \cos \phi + e \sin \phi \quad (\phi = \text{pressure angle}) $$

    But simpler: Use geometry of offset circle.

    If follower line passes through \( C \), and cam center \( O \) offset by \( e \):

$$ y = e(1 - \cos \theta) \quad \text{(for constant base circle?)} $$

Incorrect – actual depends on cam contour.

    **Correct approach**: For a **circular cam** (not common), but typical problem: Cam is a **circular disc** of radius \( R \), center \( O \) offset by \( e \) from camshaft axis \( C \). Follower has flat horizontal surface, line of action vertical through \( C \).

    Then, at angle \( \theta \), vertical distance from \( C \) to cam contact point:

$$ y = R - \sqrt{R^2 - (e \sin \theta)^2} - e \cos \theta \quad \text{(approx for small } e/R) $$

    Acceleration: \( a = \frac{d^2 y}{dt^2} = \omega^2 \frac{d^2 y}{d\theta^2} \).

*   **Condition for Lift-off**:

    Follower loses contact when **normal force** becomes zero. This occurs when **required acceleration** exceeds spring force capacity:

$$ m a_{\text{down}} > k \cdot \text{compression} \quad \text{or} \quad m a_{\text{up}} < \text{spring force} $$

    Typically: \( a_{\text{max}} > \frac{F_{\text{spring, max}}}{m} \).

VII. LOCOMOTIVE MECHANICS (SPECIAL APPLICATIONS)

Uncoupled vs. Coupled Locomotive Engines

• Uncoupled: Two cylinders with independent cranks (usually 90° apart). Causes swaying couple and hammer blow.

• Coupled: Cranks connected via coupling rods → forces partially balanced between wheels.

Balancing of Locomotives

1. Fraction of Reciprocating Mass to Balance:

   To limit hammer blow \( F_V \leq F_{\text{allow}} \):

$$ c = 1 - \frac{F_{\text{allow}}}{m_r \omega^2 r} $$

2. Maximum Swaying Couple:

$$ C_{\text{max}} = (1-c) m_r \omega^2 r \cdot \frac{s}{2} \cdot h $$

where \( s \) = distance between cylinder centerlines, \( h \) = height of CG above rail.

3. Variation in Tractive Effort:

$$ \Delta T = (1-c) m_r \omega^2 r $$

(Same as hammer blow magnitude for two-cylinder uncoupled).

KEY DERIVATIONS & PROOFS TO MASTER

1. Watt Governor Height:

$$ h = \frac{g}{\omega^2} \quad \text{and} \quad h \propto \frac{1}{N^2} $$

*From force balance: \( \frac{W}{\cos \theta} = m \omega^2 r \), \( r = l \sin \theta \), \( h = l \cos \theta \). Eliminate \( \theta \).*

2. Frictional Couple for Uncoupled Two-Cylinder Engine:

   For two cylinders with cranks at \( \theta \) and \( \theta+\phi \), unbalanced vertical forces:

$$ F_{V1} = (1-c) m_r \omega^2 r \cos \theta, \quad F_{V2} = (1-c) m_r \omega^2 r \cos(\theta+\phi) $$

Couple about CG (distance \( s/2 \) apart):

$$ \text{Couple} = \frac{s}{2} [F_{V1} - F_{V2}] = (1-c) m_r \omega^2 r \cdot \frac{s}{2} [\cos \theta - \cos(\theta+\phi)] $$

3. Hammer Blow & Swaying Couple:

   • Hammer Blow: \( F_V = (1-c) m_r \omega^2 r \cos \theta \) (max at \( \theta=0 \)).

   • Swaying Couple: As above.

4. Radius of Friction Circle:

$$ r_f = r \sin \phi $$

*From right triangle: \( \sin \phi = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{r_f}{r} \).*

5. Controlling Force Diagram & Stability:

   • Stable: \( \frac{dF_c}{dr} > \frac{F_c}{r} \) (CF curve steeper than line from origin).

   • Unstable: \( \frac{dF_c}{dr} < \frac{F_c}{r} \).

   • Isochronous: \( \frac{dF_c}{dr} = \frac{F_c}{r} \) (straight line through origin).

6. Acceleration for Offset Cam with Flat Follower:

   For circular cam of radius \( R \), offset \( e \), shaft axis \( C \), contact point geometry:

$$ y = R - \sqrt{R^2 - (e \sin \theta)^2} - e \cos \theta $$

Acceleration: \( a = \omega^2 \left[ R e \cos \theta \left( \frac{1}{\sqrt{R^2 - e^2 \sin^2 \theta}} - 1 \right) + e \cos \theta \right] \) (simplified for small \( e/R \): \( a \approx \omega^2 e \cos \theta \)).

> **Lift-off condition**: \( a_{\text{max}} > \frac{F_{\text{spring}}}{m} \).

Exam Strategy:

1. Governors: Be exact in force diagrams for Porter/Proell. Friction changes speed range.

2. Flywheels: Always convert diagram areas to energy using given scales. Use \( \Delta E = I \omega \Delta \omega \) for small fluctuations.

3. Balancing: Clearly distinguish primary/secondary, hammer blow (vertical force), swaying couple (moment). For locomotives, \( c \) is chosen based on hammer blow limit.

4. Friction Devices: Know uniform pressure vs. uniform wear assumptions. For bearings, remember power loss formulas.

5. Kinematics: IC method for velocity; relative acceleration for acceleration. For cams, derive acceleration from geometry, then apply lift-off condition.

6. Numericals: Draw neat diagrams, list given data, state assumptions (e.g., uniform pressure), show formula substitution.