UNIT 2: DYNAMICS OF MACHINES (ME-503 A) - SHORT NOTES

I. KINEMATIC ANALYSIS OF MECHANISMS

Four-Bar Linkage

• Velocity Analysis

  • Relative velocity method:

    $$\displaystyle v_B = \omega_{AB} \times r_{AB} $$

    $$\displaystyle v_C = v_B + \omega_{BC} \times r_{BC} $$

    $$\displaystyle v_C = v_D + \omega_{CD} \times r_{CD} $$ (with $$\displaystyle v_D = 0 $$ for fixed link $AD$).

    Solve for $$\displaystyle \omega_{BC} $$ and $$\displaystyle \omega_{CD} $$.

  • Instant centers (Aronhold-Kennedy):

    Number of ICs = $$\displaystyle \frac{n(n-1)}{2} $$ for $n$ links.

    Velocity at any point: $$\displaystyle v = \omega \times r $$ from the IC.

• Acceleration Analysis

  • Coriolis acceleration: $$\displaystyle a_c = 2 \omega \times v_{rel} $$, present when a point has relative velocity on a rotating link.

  • Tangential and normal components:

    $$\displaystyle a = \alpha \times r - \omega^2 r $$ (for a point on a rotating link).

    $$\displaystyle a_P = a_O + \alpha \times r + \omega \times (\omega \times r) + 2\omega \times v_{rel} + a_{rel} $$ (general).

Cams

• Offset Circular Cam (follower line of action through camshaft axis)

  • Follower displacement $$\displaystyle s = e(1 - \cos\theta) $$ for knife-edge follower? For flat-faced follower, derive from geometry.

  • Follower acceleration: $$\displaystyle a = -e \omega^2 \cos\theta $$ (for simple case) or more generally from $s(\theta)$.

• Critical Cam Shaft Speed (lift condition)

  • At the point of lift, spring force $$\displaystyle F_s = kx $$ equals inertia force $$\displaystyle F_i = m a $$.

  • Solve $$\displaystyle k x = m e \omega^2 \cos\theta $$ for $\omega$ (critical speed).

[!TIP] Common Pitfall: In acceleration analysis, forget Coriolis term when sliding exists on a rotating link.

II. BALANCING OF ENGINES

Fundamentals

• Dynamically equivalent system: A two-mass system that has same mass, center of mass, and moment of inertia about the center of mass as the original system.

• Primary balancing: Balances forces at crank frequency $\omega$ (first harmonic). Achieved by arranging cranks so primary forces cancel.

• Secondary balancing: Balances forces at $2\omega$ (second harmonic). Requires more cranks or special arrangements.

Reciprocating Engines

• Partial balancing: Only a fraction $c$ ($$\displaystyle 0 < c < 1 $$) of reciprocating mass $$\displaystyle m_r $$ is balanced by revolving mass $$\displaystyle m_{rev} $$.

  $$\displaystyle c = 1 $$ gives full primary balance but causes hammer blow.

• Hammer blow: Unbalanced vertical force due to revolving mass at crank speed.

  For a single cylinder: $$\displaystyle F_v = m_{rev} r \omega^2 \sin\theta $$.

• Swaying couple: Unbalanced couple due to horizontal components of unbalanced forces in an uncoupled engine.

  For two-cylinder engine with cranks at $$\displaystyle 180^\circ $$: $$\displaystyle C = m_r r \omega^2 \frac{r}{l} \cos\theta \cdot d $$, where $d$ is distance between cylinders.

• Variation in tractive effort: For locomotives, tractive effort varies due to unbalanced horizontal forces.

Multi-Cylinder Engines

• Radial engines (cylinders spaced equally around crankshaft):

  • Primary forces: Balanced if cranks are equally spaced (e.g., $$\displaystyle 120^\circ $$ for three cylinders).

  • Secondary forces: May not be fully balanced.

• In-line engines:

  • Primary forces: Can be balanced by proper crank arrangement (e.g., $$\displaystyle 180^\circ $$ for even number of cylinders in four-stroke engines).

  • Secondary forces: Cannot be fully balanced because they are in phase for all cylinders.

Locomotive Engines

• Unbalanced forces and couples: For multi-cylinder arrangements, calculate resultant unbalanced force and couple at the driving wheels.

• Balancing fraction: To limit hammer blow, balance only a fraction $c$ of reciprocating mass:

  $$\displaystyle c = \frac{\text{allowable hammer blow}}{m_r r \omega^2} $$.

• Frictional couple (uncoupled two-cylinder engine):

  $$\displaystyle C_f = 2 \mu \frac{W}{2} \frac{r}{l} \cos\theta \cdot d $$, where $W$ is total weight on wheels, $d$ is distance between cylinders.

[!TIP] Exam Tip: For in-line engines, remember primary forces can be balanced but secondary cannot. Hammer blow is due to revolving mass, not reciprocating.

III. FLYWHEELS AND ENERGY FLUCTUATION

Turning Moment Diagram

• Interpretation:

  • Mean torque line: average torque over cycle.

  • Areas above mean: energy surplus (stored in flywheel).

  • Areas below mean: energy deficit (released from flywheel).

• Energy fluctuation:

  $$\displaystyle \Delta E = \text{maximum cumulative area above or below mean line} $$.

Coefficients

• Coefficient of fluctuation of energy:

  $$\displaystyle \delta_E = \frac{\Delta E}{E_{\text{mean}}} $$.

• Coefficient of fluctuation of speed:

  $$\displaystyle \delta_N = \frac{N_{\text{max}} - N_{\text{min}}}{N_{\text{mean}}} $$ (or sometimes $$\displaystyle \frac{N_{\text{max}} - N_{\text{min}}}{2 N_{\text{mean}}} $$).

Flywheel Design

• Mass and radius of gyration:

  $$\displaystyle \Delta E = I \omega_{\text{mean}}^2 \delta_N $$, where $$\displaystyle I = m k^2 $$.

  $$\displaystyle \boxed{I = \frac{\Delta E}{\omega_{\text{mean}}^2 \delta_N}} $$.

• Maximum and minimum speed:

  From energy equation:

  $$\displaystyle E_{\text{max}} = E_{\text{mean}} + \Delta E = \frac{1}{2} I \omega_{\text{max}}^2 $$,

  $$\displaystyle E_{\text{min}} = E_{\text{mean}} - \Delta E = \frac{1}{2} I \omega_{\text{min}}^2 $$.

• Energy storage and release: During power stroke (torque > mean), flywheel stores energy, speed increases; during other strokes, flywheel releases energy, speed decreases.

[!TIP] Common Pitfall: Confusing $$\displaystyle \delta_N $$ definition. Use $$\displaystyle \delta_N = (N_{\text{max}} - N_{\text{min}})/N_{\text{mean}} $$ unless specified otherwise.

IV. GOVERNORS

Function and Classification

• Purpose: Maintain constant engine speed by regulating fuel supply against load changes.

• Types: Centrifugal (Watt, Porter, Proell, Wilson-Hartnell), inertia, etc.

Watt Governor

• Derivation of height:

  For each ball (mass $m$) and sleeve (mass $M$):

  $$\displaystyle 2 T \cos\theta = M g $$, $$\displaystyle T \sin\theta = m \omega^2 r $$, $$\displaystyle r = l \sin\theta $$, $$\displaystyle h = l \cos\theta $$.

  Eliminate $T$ and $\theta$:

  $$\displaystyle \boxed{h = \frac{M g}{2 m \omega^2}} $$.

• Proof $$\displaystyle h \propto 1/N^2 $$:

  $$\displaystyle \omega = \frac{2\pi N}{60} $$, so $$\displaystyle h = \frac{900 M g}{2 m \pi^2 N^2} \propto \frac{1}{N^2} $$.

Porter Governor

• Construction: Upper arms (length $a$) hinged to spindle; lower arms (length $b$) hinged to sleeve; balls at junction.

• Speed range with friction:

  Include friction force $$\displaystyle F_f $$ in sleeve equilibrium.

  Speeds $$\displaystyle N_1 $$ and $$\displaystyle N_2 $$ correspond to limiting inclinations $$\displaystyle \theta_{\text{min}} $$ and $$\displaystyle \theta_{\text{max}} $$.

• Limiting inclinations: Arms constrained by stops at angles $$\displaystyle \theta_1 $$ and $$\displaystyle \theta_2 $$.

Proell Governor

• Construction: Upper arms hinged on axis; lower arms pivoted at distance $f$ from axis; extensions of length $e$ to balls.

• Minimum speed: When extensions are parallel to axis, radius $$\displaystyle r_{\text{min}} = f + e $$? Actually, if extensions parallel, ball radial distance equals pivot distance $f$? Depends on geometry. Derive from force equations.

• Equilibrium speed: For given configuration, solve force equations for $\omega$.

Governor Performance

• Sensitiveness:

  $$\displaystyle S = \frac{N_{\text{max}} - N_{\text{min}}}{N_{\text{mean}}} $$. Higher $S$ means larger speed variation for load change.

• Isochronism: $$\displaystyle N_{\text{max}} = N_{\text{min}} $$, constant speed. Requires controlling force $$\displaystyle F_c \propto r $$.

• Hunting: Oscillations about mean speed due to over-sensitivity.

• Stability: Governor returns to equilibrium after disturbance.

  Condition: $$\displaystyle \frac{dF_c}{dr} > 0 $$. On $$\displaystyle F_c $$ vs $r$ diagram, curve must lie above $$\displaystyle F_c = m \omega^2 r $$ and have positive slope.

• Comparison: Porter vs Proell:

  Proell more sensitive because centrifugal force acts directly on lower arms.

  Proof: For same dimensions, Proell has larger $$\displaystyle \frac{dN}{dr} $$.

• Coefficient of insensitiveness:

  $$\displaystyle C_i = \frac{N_2 - N_1}{N_2} $$, where $$\displaystyle N_1 $$, $$\displaystyle N_2 $$ are speeds at max and min radius.

[!TIP] Exam Tip: For stability, remember $$\displaystyle dF_c/dr > 0 $$. Isochronous governor has $$\displaystyle F_c \propto r $$.

V. FRICTION DEVICES

Friction Circle

• Concept: When a journal rotates, friction force acts tangentially at contact point. The resultant force on bearing lies along the line of centers. The friction circle has radius $$\displaystyle r_f = r \sin\phi $$, where $$\displaystyle \phi = \tan^{-1}\mu $$.

• Radius expression:

  $$\displaystyle \boxed{r_f = r \sin\phi} $$.

Clutches

• Single Plate Clutch

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

    $$\displaystyle T = \frac{2}{3} \mu W \frac{r_2^3 - r_1^3}{r_2^2 - r_1^2} $$,

    $$\displaystyle r_m = \frac{2}{3} \frac{r_2^3 - r_1^3}{r_2^2 - r_1^2} $$.

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

    $$\displaystyle T = \mu W \frac{r_2 + r_1}{2} $$,

    $$\displaystyle r_m = \frac{r_2 + r_1}{2} $$.

• Conical Clutch

  • Working: Conical surfaces in contact; axial force $W$.

  • Torque (uniform wear assumption common):

    $$\displaystyle T = \mu W r_m \cot(\alpha/2) $$,

    where $\alpha$ is cone angle, $$\displaystyle r_m $$ mean radius.

Brakes

• Band Brake

  • Simple: One end fixed to fulcrum, other to lever.

    $$\displaystyle T_1 = T_2 e^{\mu\theta} $$, $$\displaystyle T_{\text{brake}} = (T_1 - T_2) r $$.

    Lever: $$\displaystyle P \cdot L = T_1 \cdot x $$ → $$\displaystyle T_{\text{brake}} = \frac{P L r}{x} (1 - e^{-\mu\theta}) $$.

  • Differential: Band attached to lever at two points; mechanical advantage differs.

• Internal Expanding Brake

  • Construction: Two shoes (leading and trailing) pushed outward by cam.

    Leading shoe self-energizing; trailing not.

  • Torque per shoe (approx):

    Leading: $$\displaystyle T = \mu W R (\tan\alpha + \alpha) $$;

    Trailing: $$\displaystyle T = \mu W R (\tan\alpha - \alpha) $$,

    where $2\alpha$ is contact angle, $R$ mean radius.

• Double Shoe Brake

  • Two shoes, spring force $$\displaystyle F_s $$ applies pressure.

  Braking torque $$\displaystyle T = 2 \mu W_{\text{shoe}} R_{\text{eff}} $$.

  Shoe width $b$ from pressure limit: $$\displaystyle p_{\text{max}} = \frac{W_{\text{shoe}}}{b \cdot \text{arc length}} \leq p_{\text{allow}} $$.

Bearings

• Conical Pivot Bearing

  • Uniform pressure:

    $$\displaystyle T = \frac{2 \mu W}{3 \cos(\alpha/2)} \frac{r_2^3 - r_1^3}{r_2^2 - r_1^2} $$,

    Power loss $$\displaystyle P = T \omega $$.

  • Uniform wear:

    $$\displaystyle T = \frac{\mu W (r_2 + r_1)}{2 \cos(\alpha/2)} $$.

• Collar Bearing

  • Uniform pressure:

    $$\displaystyle T = \frac{2}{3} \mu W \frac{r_2^3 - r_1^3}{r_2^2 - r_1^2} $$.

  • Uniform wear:

    $$\displaystyle T = \frac{1}{2} \mu W (r_2 + r_1) $$.

  • Number of collars: For uniform pressure, $$\displaystyle n \geq \frac{W}{p_{\text{allow}} \pi (r_2^2 - r_1^2)} $$.

[!TIP] Common Pitfall: In conical clutch, the cone angle affects the axial force–torque relationship. Remember the $\cot(\alpha/2)$ factor.

VI. DYNAMOMETERS

Classification

• Absorption dynamometers: Absorb and dissipate power as heat.

  Examples: Prony brake (friction on pulley), rope brake (weights on rope), eddy current (magnetic drag).

• Transmission dynamometers: Transmit power while measuring.

  Examples: Epicyclic train dynamometer, torsion dynamometer.

Torsion Dynamometer

• Working principle: Measure angle of twist $\theta$ on a shaft of known length $L$, polar moment $J$, and modulus $G$.

  Torque $$\displaystyle T = \frac{J G}{L} \theta $$.

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

Absorption vs Transmission

• Absorption: Power is wasted as heat; used for engine testing.

• Transmission: Power delivered to load; used for in-line measurement.

VII. SPECIAL TOPICS (FROM PAST PAPERS)

Locomotive Dynamics Integration

• Combined effect of balancing, hammer blow, and tractive effort:

  Balancing reduces hammer blow but may introduce unbalanced couples.

  Tractive effort variation affects locomotive’s pulling capacity.

  Design compromise: choose balancing fraction $c$ to limit hammer blow while accepting some tractive effort variation.

Engine Force Calculations

• Piston effort (force on piston):

  $$\displaystyle F_p = p A - m_i a $$,

  where $p$ = pressure, $A$ = piston area, $$\displaystyle m_i $$ = inertia mass, $a$ = piston acceleration.

• Inertia force:

  $$\displaystyle F_i = m_i a = m_i r \omega^2 \left( \cos\theta + \frac{r}{l} \cos2\theta \right) $$ (approximation for $r/l \ll 1$).

• Thrust on cylinder walls (side thrust):

  $$\displaystyle F_{\text{side}} = F_{\text{rod}} \sin\phi \approx F_{\text{rod}} \frac{r \sin\theta}{l} $$.

• Crank effort (torque on crank):

  $$\displaystyle T_{\text{crank}} = F_p \cdot r \left( \sin\theta + \frac{r}{2l} \sin2\theta \right) $$ (approx).

• Pressure difference and force:

  Given pressures $$\displaystyle p_{\text{cover}} $$, $$\displaystyle p_{\text{piston}} $$, net force $$\displaystyle F_{\text{net}} = (p_{\text{cover}} - p_{\text{piston}}) A_{\text{piston}} $$.

  Then piston effort $$\displaystyle F_p = F_{\text{net}} - F_i $$.

[!TIP] Exam Tip: In engine force problems, always draw the free-body diagram of the piston and connecting rod. Use $\theta$ measured from inner dead center (IDC).