UNIT 1: Dynamics of Machine – Short Notes

I. Kinematics and Dynamics of Mechanisms

Four-Bar Chain Analysis

Velocity Analysis by Instantaneous Center (IC) Method

• Instantaneous Center (IC): Point in a body with zero instantaneous velocity.

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

• Steps:

  1. Fix the frame (link AD). Identify all moving links.

  2. Locate ICs:

     • $$\displaystyle I_{12} $$: Between driver (AB) and frame (AD) → at A.

     • $$\displaystyle I_{23} $$: Between coupler (BC) and driver (AB) → intersection of AB and BC extended.

     • $$\displaystyle I_{34} $$: Between follower (CD) and frame (AD) → at D.

     • $$\displaystyle I_{24} $$: Between coupler (BC) and follower (CD) → intersection of BC and CD extended.

  3. Apply velocity relation: $$\displaystyle v_B = \omega_{AB} \times AB $$. Direction perpendicular to AB.

  4. Velocity of any point on a link: $$\displaystyle v = \omega \times \text{distance from IC} $$.

  5. Angular velocity: $$\displaystyle \omega_{BC} = \frac{v_B}{I_{23}B} $$, $$\displaystyle \omega_{CD} = \frac{v_C}{I_{34}C} $$.

[!TIP] Common Pitfall: Sign of angular velocity (CW/CCW) depends on direction of velocity vector relative to link. Use right-hand rule.

Acceleration Analysis

• For a point on a rotating link: $$\displaystyle \vec{a} = \vec{a}_O + \vec{a}_{rel} + \vec{a}_c $$

  • $$\displaystyle \vec{a}_O $$: Acceleration of IC or reference point.

  • $$\displaystyle \vec{a}_{rel} $$: Relative acceleration along link (tangential: $\alpha \times r$, radial: $$\displaystyle \omega^2 r $$).

  • $$\displaystyle \vec{a}_c $$: Coriolis acceleration ($$\displaystyle 2\omega \times v_{rel} $$) if point has relative velocity.

• For four-bar, often use IC method for velocity first, then differentiate or use relative acceleration equations.

Inertia Forces in Reciprocating Engines

Consider horizontal engine with crank radius $r$, connecting rod length $l$, mass of reciprocating parts $m$, crank angle $\theta$ from IDC, crank speed $\omega$.

Key Forces:

1. Inertia Force (FI): $$\displaystyle F_I = m \cdot a $$ (opposite to acceleration).

   • Acceleration of piston: $$\displaystyle a = \omega^2 r \left( \cos\theta + \frac{\cos 2\theta}{n} \right) $$ where $$\displaystyle n = l/r $$.

   • $$\displaystyle F_I = m \omega^2 r \left( \cos\theta + \frac{\cos 2\theta}{n} \right) $$ (to the left if $\theta$ measured from IDC with crank rotating CW).

2. Piston Effort (Fp): Net force on piston due to gas pressure and inertia.

   • $$\displaystyle F_p = P \cdot A - F_I - F_f $$ (where $P$ = pressure, $A$ = piston area, $$\displaystyle F_f $$ = friction).

   • If pressure difference between cylinder ends given: $$\displaystyle F_p = (P_{cover} - P_{piston}) \cdot A_{piston} - F_I $$.

3. Crank Effort (Fc): Component of force on crank along tangential direction.

   • $$\displaystyle F_c = \frac{F_p}{\sin\phi} $$ where $\phi$ = angle of connecting rod with line of stroke.

   • Approx: $$\displaystyle \sin\phi \approx \frac{r}{l}\sin\theta $$ → $$\displaystyle F_c \approx F_p \cdot \frac{n}{\sin\theta} $$.

4. Thrust on Cylinder Walls (Fw): $$\displaystyle F_w = F_p \cdot \tan\phi \approx F_p \cdot \frac{r}{l}\cos\theta $$.

5. Thrust in Connecting Rod (Fr): $$\displaystyle F_r = \frac{F_p}{\cos\phi} $$.

[!TIP] Exam Focus: "Difference of pressure between cylinder ends" means $$\displaystyle P_{cover} - P_{piston} $$ directly multiplies piston area to give net gas force.

Dynamically Equivalent System

• Definition: A system of masses (one or two) arranged such that:

  1. Total mass equals original mass.

  2. Center of mass (CG) position unchanged.

  3. Mass moment of inertia about CG unchanged.

• Two-Mass System: Place masses $$\displaystyle m_1 $$ and $$\displaystyle m_2 $$ at distances $$\displaystyle l_1 $$ and $$\displaystyle l_2 $$ from CG such that:

  • $$\displaystyle m_1 + m_2 = m $$

  • $$\displaystyle m_1 l_1 = m_2 l_2 $$ (CG condition)

  • $$\displaystyle m_1 l_1^2 + m_2 l_2^2 = I_G $$ (original $$\displaystyle I_G $$).

• Often choose $$\displaystyle l_1 + l_2 = b $$ (distance between masses) for convenience.

II. Flywheels and Energy Management

Purpose and Difference from Governor

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

• Governor: Controls mean speed by varying fuel supply → maintains constant average speed under load change.

Fluctuation of Energy and Speed

• Fluctuation of Energy ($\Delta E$): Maximum variation of kinetic energy in flywheel during cycle.

  • $$\displaystyle \Delta E = \text{Maximum energy} - \text{Minimum energy} $$.

• Fluctuation of Speed ($\Delta N$): $$\displaystyle N_{max} - N_{min} $$.

• Coefficient of Fluctuation of Energy ($$\displaystyle K_E $$): $$\displaystyle K_E = \frac{\Delta E}{\text{Mean kinetic energy}} $$.

• Coefficient of Fluctuation of Speed ($$\displaystyle K_N $$): $$\displaystyle K_N = \frac{\Delta N}{N_{mean}} $$.

Turning Moment Diagram

• Graph of crank torque (or piston effort × crank radius) vs. crank angle.

• Four-Stroke Engine: One power stroke per 2 revolutions → diagram repeats every 720°.

• Interpretation:

  • Area above mean torque line: Excess energy → stored in flywheel (speed increases).

  • Area below mean torque line: Deficiency → flywheel releases energy (speed decreases).

  • Net area over cycle = 0 (mean torque × 720° = area above = area below).

Design Calculations

• Energy Fluctuation: $$\displaystyle \Delta E = I \omega_{mean}^2 \cdot K_N $$ (approx for small $$\displaystyle K_N $$).

  • $I$ = mass moment of inertia of flywheel.

  • $$\displaystyle \omega_{mean} = \frac{2\pi N_{mean}}{60} $$.

• From turning moment diagram:

  • Let scales: $$\displaystyle 1 \text{ mm} = S_T \text{ N-m} $$ (torque), $$\displaystyle 1 \text{ mm} = S_\theta \text{ rad} $$ (angle).

  • Area $$\displaystyle A_i $$ (mm²) → actual area = $$\displaystyle A_i \cdot S_T \cdot S_\theta $$ (N-m·rad = N-m).

  • $$\displaystyle \Delta E = \text{Maximum cumulative deviation from mean} $$.

• Mass and Radius of Gyration:

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

  • $$\displaystyle m = \frac{\rho V}{?} $$ but usually given or found from $I$.

  • $$\displaystyle \boxed{I = \frac{\Delta E}{\omega_{mean}^2 \cdot K_N}} $$

• Speed Limits:

  • $$\displaystyle \omega_{max} = \omega_{mean} \left(1 + \frac{K_N}{2}\right) $$, $$\displaystyle \omega_{min} = \omega_{mean} \left(1 - \frac{K_N}{2}\right) $$ (approx).

[!TIP] In numericals, find cumulative areas from diagram, take maximum deviation as $\Delta E$ in N-m. Convert rpm to rad/s: $$\displaystyle \omega = \frac{2\pi N}{60} $$.

III. Governors

Function and Classification

• Function: Maintain constant engine speed by regulating fuel supply as load varies.

• Classification:

  • Centrifugal Governors: Rotating masses (balls) fly out with speed increase → sleeve rises → reduces fuel.

  • Inertia Governors: Use pendulums or flyweights with phase lag (e.g., Wilson-Hartnell).

Watt Governor

• Construction: Central spindle, two arms with balls at ends, sleeve on spindle.

• Working: As speed ↑, balls fly out, arms rise, sleeve lifts → linkage closes throttle.

• Derivation of Height (h):

  • Forces on one ball: weight $$\displaystyle W = mg $$, centrifugal force $$\displaystyle F_c = m \omega^2 r $$.

  • Tension $T$ in arm resolves: $$\displaystyle T \cos\theta = mg $$, $$\displaystyle T \sin\theta = m \omega^2 r $$.

  • $$\displaystyle \tan\theta = \frac{\omega^2 r}{g} $$.

  • From geometry: $$\displaystyle h = \frac{a}{\tan\theta} $$ where $a$ = distance from spindle axis to ball pivot.

  • $$\displaystyle \boxed{h = \frac{g}{\omega^2} \cdot \frac{a}{r}} $$ but $r$ varies. For small $\theta$, $r \approx a \sin\theta \approx a \theta$, $h \approx a \cos\theta \approx a$.

  • Inversely proportional to $$\displaystyle \omega^2 $$: $$\displaystyle h \propto \frac{1}{\omega^2} $$ (since $r$ changes slowly compared to $$\displaystyle \omega^2 $$).

Porter Governor

• Construction: Similar to Watt but with upper arms (pivoted on sleeve) and lower arms (pivoted on spindle). Central load $W$ on sleeve.

• Equilibrium:

  • Consider forces on one ball and lower arm.

  • Let $m$ = mass of each ball, $W$ = central load + friction (if any).

  • $$\displaystyle T_1 $$ tension in upper arm, $$\displaystyle T_2 $$ in lower arm.

  • Resolving vertically: $$\displaystyle 2T_1 \cos\theta + W = 2mg $$ (if $W$ includes sleeve weight).

  • Resolving horizontally: $$\displaystyle 2T_1 \sin\theta = 2m \omega^2 r $$.

  • From geometry: $$\displaystyle r = h \tan\theta $$ (if lower arm length $b$ and pivot distance $a$ from axis: $$\displaystyle r = a + b \sin\theta $$, $$\displaystyle h = b \cos\theta $$).

  • Eliminate tensions: $$\displaystyle \frac{\sin\theta}{\cos\theta} = \frac{m \omega^2 r}{mg - W/2} $$.

  • $$\displaystyle \boxed{\omega^2 = \frac{g(m - W/(2))}{r} \cdot \frac{\tan\theta}{?}} $$ Actually:

    • $$\displaystyle \tan\theta = \frac{m \omega^2 r}{mg - W/2} $$.

    • But $r$ and $\theta$ related by geometry.

  • Range of Speed with Friction:

    • Friction $$\displaystyle F_f $$ acts on sleeve. For equilibrium:

      • Rising: $$\displaystyle W + F_f $$ effective.

      • Falling: $$\displaystyle W - F_f $$ effective.

    • Speeds: $$\displaystyle \omega_{max} $$ for max $\theta$ (with $$\displaystyle W-F_f $$), $$\displaystyle \omega_{min} $$ for min $\theta$ (with $$\displaystyle W+F_f $$).

    • $$\displaystyle \omega^2 \propto \frac{1}{r} $$ approx for small $\theta$ variation.

Proell Governor

• Construction: Upper arms hinged on spindle, lower arms pivoted offset from axis at distance $e$. Ball extensions of length $c$ parallel to axis at min radius.

• Equilibrium:

  • Let $r$ = radius of ball path, $h$ = lift of sleeve.

  • Geometry: $$\displaystyle r = e + c \cos\phi + b \sin\phi $$, $$\displaystyle h = c \sin\phi + b \cos\phi $$ (where $\phi$ = angle of lower arm from vertical).

  • Forces: Similar to Porter but with offset.

  • $$\displaystyle \tan\phi = \frac{m \omega^2 r}{mg - W/(2)} $$ (if $W$ central load).

  • For minimum speed (min $r$): extensions parallel → $$\displaystyle \phi=0 $$? Actually at min radius, extensions are parallel to axis → $\phi$ such that $c \cos\phi$ part gives min $r$.

  • Solve simultaneously geometry and force equation for $\omega$.

Governor Characteristics

• Sensitiveness (S): $$\displaystyle S = \frac{\text{Range of speed}}{\text{Mean speed}} = \frac{\omega_{max} - \omega_{min}}{\omega_{mean}} $$.

  • High sensitiveness → large speed variation for small load change (not good).

• Isochronism: Governor maintains constant speed for all loads (within limits). $$\displaystyle S = 0 $$ ideally. Requires $$\displaystyle h \propto \frac{1}{\omega^2} $$ exactly.

• Hunting: Oscillations of governor about equilibrium position due to over-sensitiveness → speed fluctuations.

• Stability: Governor returns to equilibrium position after disturbance. Requires $$\displaystyle \frac{dF}{dr} > 0 $$ (controlling force increases with radius).

• Controlling Force Diagram: Plot $$\displaystyle F_c = m \omega^2 r $$ vs $r$.

  • Stable: $$\displaystyle F_c $$ curve steeper than $mg$ line → $$\displaystyle \frac{dF_c}{dr} > \frac{d(mg)}{dr} = 0 $$.

  • Unstable: $$\displaystyle F_c $$ curve less steep.

  • Isochronous: $$\displaystyle F_c $$ curve parallel to $mg$ line (constant $\omega$).

• Condition for Stability: $$\displaystyle \frac{d}{dr}(m\omega^2 r) > 0 $$ → $$\displaystyle \omega^2 > \frac{g}{r \tan\theta} \cdot \frac{d(r\tan\theta)}{dr} $$ (from force equation).

Coefficient of Insensitiveness

• $$\displaystyle \delta = \frac{\omega_{max} - \omega_{min}}{\omega_{mean}} $$ (same as sensitiveness). Sometimes defined as $1/S$.

Comparison of Porter and Proell Governors

• Proell has greater sensitiveness because:

  • For same $r$, $\omega$ is lower in Proell due to offset $e$ increasing effective moment arm of centrifugal force.

  • From equilibrium: $$\displaystyle \tan\phi = \frac{m \omega^2 r}{mg - W/2} $$.

  • In Proell, $$\displaystyle r = e + b \sin\phi + c \cos\phi $$ → for given $\phi$, $r$ larger → $$\displaystyle \omega^2 $$ smaller for same $m,g,W$.

  • Hence for same speed range, Proell has larger $\Delta \omega$? Actually sensitiveness $$\displaystyle S = \frac{\Delta \omega}{\omega_{mean}} $$ is higher because $$\displaystyle \omega_{min} $$ is lower and $$\displaystyle \omega_{max} $$ higher for same $h$ variation.

IV. Balancing of Reciprocating Engines

Primary and Secondary Balancing

• Primary Balance: Balance forces due to first order ($\omega$) acceleration.

  • Reciprocating mass acceleration: $$\displaystyle a_p = \omega^2 r (\cos\theta + \frac{\cos 2\theta}{n}) $$.

  • Primary: $$\displaystyle \omega^2 r \cos\theta $$ (in line with crank).

  • Secondary: $$\displaystyle \omega^2 r \frac{\cos 2\theta}{n} $$ (due to obliquity).

• Secondary Balance: Balance second order forces (often neglected for $$\displaystyle n>4 $$).

Partial Balancing of Reciprocating Masses

• Only part of reciprocating mass is balanced by revolving mass because:

  • Balancing all reciprocating mass with revolving mass would introduce vertical unbalanced force (hammer blow) at high speed due to $$\displaystyle m \omega^2 r \cos\theta $$ component.

  • In locomotives, hammer blow causes rail wear and derailment risk.

  • Typically balance 50-70% of reciprocating mass.

Unbalanced Forces and Couples in Locomotives (Uncoupled Two-Cylinder)

• Assumptions: Two cylinders, cranks at $$\displaystyle 90^\circ $$, same $$\displaystyle m_r $$ (recip mass), same $r$.

• Primary Unbalanced Force (Horizontal):

  • $$\displaystyle F_{p} = m_r \omega^2 r (\cos\theta + \cos(\theta + 90^\circ)) = m_r \omega^2 r (\cos\theta - \sin\theta) $$.

  • Magnitude: $$\displaystyle F_p = \sqrt{2} m_r \omega^2 r $$, angle $$\displaystyle 45^\circ $$ to horizontal.

• Swaying Couple (about vertical axis):

  • Couple due to primary forces separated by distance $d$ (cylinder center distance).

  • $$\displaystyle C_{sway} = F_{p1} \cdot \frac{d}{2} - F_{p2} \cdot \frac{d}{2} = \frac{d}{2} (F_{p1} - F_{p2}) $$.

  • $$\displaystyle F_{p1} = m_r \omega^2 r \cos\theta $$, $$\displaystyle F_{p2} = m_r \omega^2 r \cos(\theta+90) = -m_r \omega^2 r \sin\theta $$.

  • $$\displaystyle \boxed{C_{sway} = \frac{m_r \omega^2 r d}{2} (\cos\theta + \sin\theta)} $$.

• Hammer Blow (Unbalanced Vertical Force):

  • Secondary unbalanced force: $$\displaystyle F_s = m_r \omega^2 r \frac{\cos 2\theta}{n} $$.

  • For two cylinders $$\displaystyle 90^\circ $$ apart: $$\displaystyle F_{s1} = m_r \omega^2 r \frac{\cos 2\theta}{n} $$, $$\displaystyle F_{s2} = m_r \omega^2 r \frac{\cos 2(\theta+90)}{n} = -m_r \omega^2 r \frac{\sin 2\theta}{n} $$.

  • Vertical resultant: $$\displaystyle F_v = F_{s1} + F_{s2} = \frac{m_r \omega^2 r}{n} (\cos 2\theta - \sin 2\theta) $$.

  • Magnitude: $$\displaystyle F_v = \frac{\sqrt{2} m_r \omega^2 r}{n} $$.

  • This is the hammer blow.

• Variation in Tractive Effort:

  • Tractive effort $$\displaystyle T_e = \text{mean} + \text{fluctuation due to } F_p \sin\theta $$? Actually:

  • Crank effort $$\displaystyle F_c = \frac{F_p}{\sin\phi} \approx n F_p $$ (for small $\phi$).

  • $$\displaystyle F_p $$ has primary component $$\displaystyle m_r \omega^2 r \cos\theta $$ (if balanced partially, only unbalanced part).

  • Fluctuation in $$\displaystyle F_c $$ → variation in tractive effort.

Multi-Cylinder Engines

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

  • Primary forces: $$\displaystyle \sum m_r \omega^2 r \cos(\theta + \alpha_i) $$. For equal $$\displaystyle m_r,r $$, and $$\displaystyle \alpha_i = 0^\circ, 120^\circ, 240^\circ $$ → sum = 0.

  • Primary completely balanced.

  • Secondary: $$\displaystyle \sum m_r \omega^2 r \frac{\cos 2(\theta+\alpha_i)}{n} $$ → angles $$\displaystyle 0^\circ, 240^\circ, 480^\circ(=120^\circ) $$ → also sum = 0.

  • Secondary also balanced for 3-cylinder radial.

• In-line Engines:

  • Cylinders in line, cranks at various angles.

  • Can balance primary forces completely by choosing crank angles (e.g., 4-cylinder: 0°, 180°, 180°, 0°? Actually for even number, can balance primary but not secondary completely).

  • Impossible to balance completely both primary and secondary simultaneously for in-line engines with more than 2 cylinders because secondary forces have different phase relationships.

Locomotive Balancing (Two-Cylinder, Cranks at 90°)

• Let fraction $c$ of reciprocating mass $$\displaystyle m_r $$ balanced by revolving mass $$\displaystyle m_r' $$ at radius $r$.

• Balancing mass: $$\displaystyle m_b = c m_r $$ placed opposite crank.

• Unbalanced Primary Force (Horizontal):

  • Balanced part: $$\displaystyle c m_r \omega^2 r \cos\theta $$ (from revolving mass) cancels $$\displaystyle c m_r \omega^2 r \cos\theta $$ (reciprocating primary).

  • Unbalanced: $$\displaystyle (1-c) m_r \omega^2 r \cos\theta $$ from cylinder 1 + $$\displaystyle (1-c) m_r \omega^2 r \cos(\theta+90) = -(1-c) m_r \omega^2 r \sin\theta $$ from cylinder 2.

  • Resultant: $$\displaystyle F_{pu} = (1-c) m_r \omega^2 r \sqrt{\cos^2\theta + \sin^2\theta} = (1-c) m_r \omega^2 r $$ (constant magnitude? Actually vector sum magnitude = $$\displaystyle (1-c) m_r \omega^2 r \sqrt{\cos^2\theta + \sin^2\theta} = (1-c) m_r \omega^2 r $$ at angle $\theta$? Wait: $$\displaystyle F_x = (1-c) m_r \omega^2 r \cos\theta $$, $$\displaystyle F_y = -(1-c) m_r \omega^2 r \sin\theta $$ → magnitude = $$\displaystyle (1-c) m_r \omega^2 r $$, direction $\theta$ below horizontal? Actually angle $-\theta$ from x-axis. So magnitude constant but direction rotates.

• Hammer Blow (Vertical Unbalanced Secondary Force):

  • Secondary from reciprocating: $$\displaystyle \frac{(1-c) m_r \omega^2 r}{n} (\cos 2\theta - \sin 2\theta) $$? Actually each cylinder secondary: $$\displaystyle \frac{m_r \omega^2 r}{n} \cos 2(\theta+\alpha_i) $$.

  • For cylinder 1 ($$\displaystyle \alpha=0 $$): $$\displaystyle \frac{m_r \omega^2 r}{n} \cos 2\theta $$.

  • Cylinder 2 ($$\displaystyle \alpha=90 $$): $$\displaystyle \frac{m_r \omega^2 r}{n} \cos 2(\theta+90) = -\frac{m_r \omega^2 r}{n} \sin 2\theta $$.

  • Unbalanced vertical (since both have vertical components? Actually these forces are along cylinder axis (horizontal for horizontal engine). Hammer blow is vertical component due to connecting rod angularity? Wait: In horizontal locomotive, cylinders are horizontal, so primary and secondary forces are horizontal. Hammer blow is vertical force from unbalanced rotating masses? Actually in locomotives, hammer blow is due to unbalanced rotating masses (wheels, cranks) because centrifugal force $$\displaystyle m \omega^2 r $$ is vertical if rotation plane is horizontal. But here we are balancing reciprocating with revolving mass. The revolving mass $$\displaystyle c m_r $$ at radius $r$ on crank gives centrifugal force $$\displaystyle c m_r \omega^2 r $$ vertically (if crank rotates in horizontal plane). This is the main hammer blow.

  • So hammer blow = $$\displaystyle c m_r \omega^2 r $$ (magnitude, vertical).

  • Constraint: $$\displaystyle c m_r \omega^2 r \leq \text{ allowable} $$.

• Swaying Couple:

  • Due to unbalanced primary horizontal forces separated by $d$ (wheelbase? Actually distance between cylinder center lines).

  • $$\displaystyle C_{sway} = \frac{d}{2} [F_{p1} - F_{p2}] = \frac{d}{2} (1-c) m_r \omega^2 r (\cos\theta + \sin\theta) $$.

  • Max magnitude: $$\displaystyle C_{sway,max} = \frac{\sqrt{2}}{2} d (1-c) m_r \omega^2 r $$.

• Variation in Tractive Effort:

  • Tractive effort variation due to unbalanced primary forces? Actually tractive effort is tangential force on wheel. Unbalanced horizontal force affects wheel load distribution but not directly tractive effort? In exam context, variation in tractive effort often means fluctuation in crank effort due to pressure and inertia. But for balancing, we consider unbalanced forces on frame.

  • Typically asked: "variation in tractive effort" means the net horizontal force on locomotive frame from cylinders varies with $\theta$, causing "surge".

V. Friction in Machine Elements

Friction Circle

• Definition: Circle of radius $r \sin\phi$ (where $r$ = journal radius, $\phi$ = angle of friction) representing the locus of the resultant reaction force direction in a journal bearing.

• Significance: In journal bearings, the friction force $$\displaystyle F = \mu N $$ acts tangentially. The resultant reaction $R$ makes angle $\phi$ with normal. For pure rotation, the point of application of $R$ must lie within the friction circle to avoid sliding.

• Radius: $$\displaystyle r_f = r \sin\phi \approx r \mu $$ (for small $\phi$).

[!TIP] Used in analysis of bearings with rotating shaft: if resultant force vector from center of journal falls within friction circle, motion is possible without slip.

Clutches

Conical Clutch

• Working: Axial force $W$ presses conical surfaces. Friction transmits torque.

• Torque Transmission:

  • Uniform Pressure: $$\displaystyle p = \frac{W \sin\alpha}{\pi (r_2^2 - r_1^2)} $$.

    • Torque: $$\displaystyle T = \mu W \cdot \frac{r_2^3 - r_3^3}{3(r_2^2 - r_1^2)} \cdot \csc\alpha $$.

  • Uniform Wear: $$\displaystyle p r = \text{constant} $$ → $$\displaystyle p = \frac{W}{\pi (r_2^2 - r_1^2)} \cdot \frac{2}{\frac{1}{r_1} + \frac{1}{r_2}} $$? Actually:

    • $$\displaystyle p = \frac{W}{2\pi (r_2 - r_1) b} $$? Wait for cone: area = $$\displaystyle \pi (r_2 + r_1) \cdot \frac{(r_2 - r_1)}{\sin\alpha} $$.

    • Uniform wear: $p \propto 1/r$ → $$\displaystyle p = \frac{C}{r} $$.

    • $$\displaystyle W = \int p \cdot 2\pi r \cdot dr / \sin\alpha = \frac{2\pi C}{\sin\alpha} (r_2 - r_1) $$ → $$\displaystyle C = \frac{W \sin\alpha}{2\pi (r_2 - r_1)} $$.

    • Torque: $$\displaystyle T = \int \mu p \cdot r \cdot 2\pi r dr / \sin\alpha = \frac{2\pi \mu C}{\sin\alpha} \int_{r_1}^{r_2} r dr = \frac{\pi \mu C}{\sin\alpha} (r_2^2 - r_1^2) $$.

    • Substitute $C$: $$\displaystyle \boxed{T = \frac{\mu W}{2} \cdot \frac{r_2^2 - r_1^2}{r_2 - r_1} \cdot \csc\alpha = \frac{\mu W}{2} (r_2 + r_1) \csc\alpha} $$.

  • Mean Radius: $$\displaystyle R_m = \frac{2}{3} \cdot \frac{r_2^3 - r_1^3}{r_2^2 - r_1^2} $$ (uniform pressure) or $$\displaystyle \frac{r_2 + r_1}{2} $$ (uniform wear).

Plate Clutch (Single Dry Plate)

• Working: Axial force $W$ on friction lining (inner radius $$\displaystyle r_1 $$, outer $$\displaystyle r_2 $$).

• Uniform Wear Assumption (more realistic as pressure adjusts):

  • $$\displaystyle p r = \text{constant} $$.

  • $$\displaystyle W = \int_{r_1}^{r_2} p \cdot 2\pi r dr = 2\pi C \int_{r_1}^{r_2} dr = 2\pi C (r_2 - r_1) $$ → $$\displaystyle C = \frac{W}{2\pi (r_2 - r_1)} $$.

  • Torque: $$\displaystyle T = \int \mu p \cdot r \cdot 2\pi r dr = 2\pi \mu C \int_{r_1}^{r_2} r dr = \pi \mu C (r_2^2 - r_1^2) $$.

  • $$\displaystyle \boxed{T = \frac{\mu W}{2} \cdot \frac{r_2^2 - r_1^2}{r_2 - r_1} = \frac{\mu W}{2} (r_2 + r_1)} $$.

  • Mean Radius: $$\displaystyle R_m = \frac{r_2 + r_1}{2} $$.

• Design from Power & Speed:

  • $$\displaystyle P = \frac{2\pi N T}{60} $$ → $$\displaystyle T = \frac{60P}{2\pi N} $$.

  • Given $W$ limit, pressure $$\displaystyle p_{max} $$ limit.

  • From $$\displaystyle T = \mu W R_m $$ (using uniform wear approx) and $$\displaystyle W = 2\pi p_{avg} (r_2 - r_1) R_m $$? Actually:

    • $$\displaystyle W = 2\pi C (r_2 - r_1) $$, $$\displaystyle C = p_{max} r_2 $$? Not directly.

    • Better: $$\displaystyle p_{max} $$ at $$\displaystyle r_1 $$ (since $p \propto 1/r$). $$\displaystyle p_{max} = \frac{C}{r_1} $$ → $$\displaystyle C = p_{max} r_1 $$.

    • Then $$\displaystyle W = 2\pi p_{max} r_1 (r_2 - r_1) $$.

    • $$\displaystyle T = \pi \mu p_{max} r_1 (r_2^2 - r_1^2) = \pi \mu p_{max} r_1 (r_2 - r_1)(r_2 + r_1) $$.

    • Given ratio $$\displaystyle R_m / b = 4 $$ where $$\displaystyle b = r_2 - r_1 $$, $$\displaystyle R_m = (r_2+r_1)/2 $$.

    • Solve for $$\displaystyle r_1, r_2 $$.

Brakes

Internal Expanding Brake

• Working: Brake shoes inside drum. Hydraulic or mechanical actuation pushes shoes outward against drum.

• Torque: For two shoes, $$\displaystyle T = 2 \mu W R $$ where $W$ = force on one shoe, $R$ = effective radius (mean of $$\displaystyle r_1,r_2 $$).

• Self-energizing: One shoe helps increase force on other.

Band Brake

• Simple Band: Band around drum, one end fixed, other pulled by lever.

• Torque: $$\displaystyle T = (T_1 - T_2) R $$, where $$\displaystyle T_1 $$ = tight side tension, $$\displaystyle T_2 $$ = slack side.

  • $$\displaystyle T_1 = T_2 e^{\mu \theta} $$ ($\theta$ = angle of embrace in radians).

  • Lever: $$\displaystyle F \cdot l = T_1 r_1 - T_2 r_2 $$ (if lever fulcrum at one end, $F$ effort, $$\displaystyle r_1,r_2 $$ lever arms).

  • Solve for $$\displaystyle T_1, T_2 $$.

Double Shoe Brake

• Two shoes opposite each other, spring applies force, lever releases.

• Spring Force: For given torque $T$, each shoe contributes $T/2$.

  • $$\displaystyle T/2 = \mu W R $$ → $$\displaystyle W = \frac{T}{2\mu R} $$.

• Brake Shoe Width: Pressure $$\displaystyle p \leq p_{max} $$.

  • For uniform wear: $p \propto 1/r$ → max pressure at inner radius $$\displaystyle r_1 $$.

  • $$\displaystyle W = \int_{r_1}^{r_2} p \cdot 2\pi r dr \cdot b $$? Actually force on shoe: $$\displaystyle W = \int p \cdot b \cdot r d\theta $$? For shoe with angle $\theta$:

    • $$\displaystyle W = b \int_{r_1}^{r_2} p \cdot r d\theta $$? Better: Pressure acts on area $$\displaystyle b \cdot (r_2 - r_1) $$? Actually shoe contact area is rectangular: width $b$, length along radius $$\displaystyle (r_2 - r_1) $$? But pressure distribution: $$\displaystyle p = \frac{C}{r} $$.

    • $$\displaystyle W = \int_{r_1}^{r_2} p \cdot b \cdot r d\theta $$? For full $$\displaystyle 360^\circ $$? No, shoe covers angle $\alpha$.

    • Actually: $$\displaystyle W = \int_{\text{contact area}} p \, dA $$. For shoe with radial width $$\displaystyle b_r = r_2 - r_1 $$ and axial width $b$ (along shaft), and angle $\alpha$:

      • $$\displaystyle dA = b \cdot r d\theta $$? If $b$ is axial width, then area element = $b \cdot (r d\theta)$? That's for cylindrical surface. But shoe is flat? Typically brake shoe is curved to match drum. So area = $b \times \text{arc length}$? Actually pressure is normal to drum surface. For small $\alpha$, arc length ≈ $r \alpha$, but pressure varies with $r$.

      • Standard: $$\displaystyle W = \frac{1}{2} b \alpha (p_{max} r_1 + p_{min} r_2) $$? Not exactly.

      • For uniform wear: $$\displaystyle p = \frac{C}{r} $$, $$\displaystyle W = \int_{r_1}^{r_2} \int_{0}^{\alpha} \frac{C}{r} \cdot b \cdot r d\theta dr = b \alpha C \int_{r_1}^{r_2} dr = b \alpha C (r_2 - r_1) $$.

      • And $$\displaystyle p_{max} = \frac{C}{r_1} $$ → $$\displaystyle C = p_{max} r_1 $$.

      • So $$\displaystyle W = b \alpha p_{max} r_1 (r_2 - r_1) $$.

      • Also $$\displaystyle T/2 = \mu W R_m $$ with $$\displaystyle R_m = \frac{r_2^2 - r_1^2}{2(r_2 - r_1)} = \frac{r_2 + r_1}{2} $$ (uniform wear).

      • Given $\alpha$ (contact angle), solve for $b$.

Bearings

Conical Pivot Bearing

• Geometry: Cone angle $2\alpha$ (sometimes given as $\alpha$). Shaft diameter $d$, outer radius $R$, inner radius $r$ (if truncated).

• Power Loss in Friction:

  • Uniform Pressure: $$\displaystyle p = \text{constant} $$.

    • Load: $$\displaystyle W = \int p \cdot 2\pi r \cdot \frac{dr}{\sin\alpha} = \frac{2\pi p}{\sin\alpha} \int_{r}^{R} r dr = \frac{\pi p}{\sin\alpha} (R^2 - r^2) $$.

    • Friction force: $$\displaystyle dF = \mu p \cdot 2\pi r dr / \sin\alpha $$.

    • Torque: $$\displaystyle dT = dF \cdot r = \frac{2\pi \mu p}{\sin\alpha} r^2 dr $$.

    • $$\displaystyle T = \frac{2\pi \mu p}{\sin\alpha} \int_{r}^{R} r^2 dr = \frac{2\pi \mu p}{3\sin\alpha} (R^3 - r^3) $$.

    • Substitute $p$ from $W$: $$\displaystyle T = \frac{2\mu W}{3} \cdot \frac{R^3 - r^3}{R^2 - r^2} $$.

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

  • Uniform Wear: $$\displaystyle p r = \text{constant} $$ → $$\displaystyle p = \frac{C}{r} $$.

    • $$\displaystyle W = \int \frac{C}{r} \cdot 2\pi r dr / \sin\alpha = \frac{2\pi C}{\sin\alpha} (R - r) $$ → $$\displaystyle C = \frac{W \sin\alpha}{2\pi (R - r)} $$.

    • $$\displaystyle T = \int \mu \frac{C}{r} \cdot 2\pi r dr / \sin\alpha \cdot r = \frac{2\pi \mu C}{\sin\alpha} \int_{r}^{R} r dr = \frac{\pi \mu C}{\sin\alpha} (R^2 - r^2) $$.

    • $$\displaystyle T = \frac{\mu W}{2} \cdot \frac{R^2 - r^2}{R - r} = \frac{\mu W}{2} (R + r) $$.

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

• Design for Pressure Limit:

  • Given $$\displaystyle p_{max} $$, $W$, $\alpha$, find $R,r$.

  • For uniform pressure: $$\displaystyle p_{max} = \frac{W \sin\alpha}{\pi (R^2 - r^2)} $$.

  • If $$\displaystyle R = k r $$ (e.g., $$\displaystyle R=2r $$), solve.

Collar Bearing

• Power Absorption (Uniform Pressure):

  • Shaft diameter $d$, collar external radius $R$, internal $r$ (usually $r \approx d/2$).

  • Number of collars $n$.

  • Load: $$\displaystyle W = n \cdot p \cdot \pi (R^2 - r^2) $$.

  • Torque: $$\displaystyle T = n \cdot \mu p \cdot \frac{2\pi}{3} (R^3 - r^3) $$? Actually for collar (flat annular):

    • $$\displaystyle dT = \mu p \cdot 2\pi r dr \cdot r = 2\pi \mu p r^2 dr $$.

    • $$\displaystyle T = 2\pi \mu p \int_{r}^{R} r^2 dr = \frac{2\pi \mu p}{3} (R^3 - r^3) $$.

    • For $n$ collars: $$\displaystyle T = n \cdot \frac{2\pi \mu p}{3} (R^3 - r^3) $$.

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

• Number of Collars:

  • Given $W$, $$\displaystyle p_{max} $$, $R,r$, find $$\displaystyle n = \frac{W}{\pi p_{max} (R^2 - r^2)} $$.

VI. Dynamometers

Classification

• Absorption Dynamometers: Absorb entire engine power as heat (e.g., prony brake, rope brake).

• Transmission Dynamometers: Measure power while transmitting to load (e.g., torsion dynamometer, epicyclic train).

Torsion Dynamometers

• Working Principle: Measure torque via angular twist in a shaft.

  • Shaft with known $J$ (polar moment) and $G$ (modulus).

  • Torque $$\displaystyle T = \frac{G J \theta}{L} $$ where $\theta$ = angle of twist (radians), $L$ = length over which measured.

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

• Example: Torque shaft with strain gauges or dial gauge for $\theta$.

VII. Cam Dynamics

Offset Cam with Spring-Loaded Follower

• Offset Cam: Cam center $O$ offset by $e$ from camshaft axis $C$.

• Follower: Flat-faced, vertical line of action through $C$.

• Derivation of Follower Acceleration:

  • Cam radius $R$, offset $e$, cam rotates $\theta$.

  • Follower displacement $s$: distance from cam center to follower contact point along line of action.

  • Geometry: $$\displaystyle s = R \cos\phi + e \sin\phi $$, where $\phi$ = angle between line $OC$ and vertical? Actually if cam rotates $\theta$, the point of contact angle from vertical? Let $\psi$ be angle from vertical to line joining cam center to contact.

    • $$\displaystyle \tan\psi = \frac{e \sin\theta}{R - e \cos\theta} $$? Better:

    • Coordinates: Cam center at $(e \sin\theta, e \cos\theta)$ if axis at origin and $\theta$ from vertical? Assume $$\displaystyle \theta=0 $$ when offset along x-axis? Standard: Cam center at $(e \cos\theta, e \sin\theta)$? Let's define: Cam axis at $C(0,0)$, cam center $O(e,0)$ at $$\displaystyle \theta=0 $$. As cam rotates $\theta$, $O(e\cos\theta, e\sin\theta)$.

    • Follower contact point $P$ on cam periphery. Line of action vertical through $C(0,0)$.

    • $P$ has coordinates: $$\displaystyle x_P = e\cos\theta + R \cos(\theta + \alpha) $$, $$\displaystyle y_P = e\sin\theta + R \sin(\theta + \alpha) $$, where $\alpha$ is angle from radial line.

    • But since line of action is vertical ($$\displaystyle x=0 $$), set $$\displaystyle x_P=0 $$: $$\displaystyle e\cos\theta + R \cos(\theta+\alpha)=0 $$.

    • Solve for $\alpha$: $$\displaystyle \cos(\theta+\alpha) = -\frac{e}{R}\cos\theta $$.

    • Then $$\displaystyle s = y_P = e\sin\theta + R \sin(\theta+\alpha) $$.

    • Use $$\displaystyle \sin(\theta+\alpha) = \pm \sqrt{1 - \cos^2(\theta+\alpha)} $$ but sign depends.

    • Alternatively, from geometry: $$\displaystyle s = R - \sqrt{R^2 - e^2 \sin^2\theta} + e \cos\theta $$? Actually for flat-faced follower with offset, the displacement is: $$\displaystyle s = e(1 - \cos\theta) + R(1 - \cos\phi) $$? Not straightforward.

  • Simpler Approach: Consider virtual crank of length $e$? Actually offset cam equivalent to eccentric cam.

  • Standard result for acceleration:

    • $$\displaystyle s = e \cos\theta + \sqrt{R^2 - e^2 \sin^2\theta} $$ (if follower pushes when cam rotates).

    • Then differentiate twice:

      • $$\displaystyle v = \frac{ds}{dt} = \frac{ds}{d\theta} \omega $$.

      • $$\displaystyle a = \frac{d^2s}{d\theta^2} \omega^2 $$.

    • Compute $$\displaystyle \frac{ds}{d\theta} = -e\sin\theta + \frac{1}{2} (R^2 - e^2 \sin^2\theta)^{-1/2} \cdot (-2e^2 \sin\theta \cos\theta) = -e\sin\theta - \frac{e^2 \sin\theta \cos\theta}{\sqrt{R^2 - e^2 \sin^2\theta}} $$.

    • $$\displaystyle \frac{d^2s}{d\theta^2} = -e\cos\theta - e^2 \left[ \frac{ \cos^2\theta - \sin^2\theta }{\sqrt{R^2 - e^2 \sin^2\theta}} + \frac{ \sin^2\theta \cos^2\theta }{(R^2 - e^2 \sin^2\theta)^{3/2}} \right] $$? Messy.

  • Exam Focus: Derive expression for acceleration in terms of $\theta$, $e$, $R$, $\omega$. Often given as:

    • $$\displaystyle a = \omega^2 \left[ e \cos\theta + \frac{e^2}{R} \cos 2\theta \right] $$ for small $e/R$? Actually approximate:

      • $$\displaystyle \sqrt{R^2 - e^2 \sin^2\theta} \approx R - \frac{e^2 \sin^2\theta}{2R} $$.

      • Then $$\displaystyle s \approx e\cos\theta + R - \frac{e^2 \sin^2\theta}{2R} $$.

      • $$\displaystyle a \approx \omega^2 \left( -e\cos\theta - \frac{e^2}{R} \cos 2\theta \right) $$? Differentiate twice:

        • $$\displaystyle s \approx R + e\cos\theta - \frac{e^2}{2R} \sin^2\theta = R + e\cos\theta - \frac{e^2}{4R} (1 - \cos 2\theta) $$.

        • $$\displaystyle \frac{ds}{d\theta} \approx -e\sin\theta + \frac{e^2}{2R} \sin 2\theta $$.

        • $$\displaystyle \frac{d^2s}{d\theta^2} \approx -e\cos\theta + \frac{e^2}{R} \cos 2\theta $$.

        • So $$\displaystyle a = \omega^2 \frac{d^2s}{d\theta^2} \approx \omega^2 \left( -e\cos\theta + \frac{e^2}{R} \cos 2\theta \right) $$.

      • But sign depends on direction. Usually take magnitude.

Critical Speed for Lift

• Condition: Follower loses contact when spring force becomes zero or negative (i.e., cam acceleration > gravitational + spring reaction).

• For spring-loaded follower with mass $m$, spring stiffness $k$, preload $$\displaystyle F_0 $$:

  • Contact force $$\displaystyle N = F_0 + k s - m a $$ (if $a$ downward positive?).

  • Lift occurs when $$\displaystyle N=0 $$ → $$\displaystyle a = \frac{F_0 + k s}{m} $$.

  • Given $s(\theta)$, find maximum $a$ (magnitude). Set $$\displaystyle \omega^2 \cdot \text{max}\left|\frac{d^2s}{d\theta^2}\right| = \frac{F_0 + k s}{m} $$.

  • Solve for $$\displaystyle \omega_{crit} $$.

• Example from Nov 2022: Cam disc diameter 75 mm, offset 25 mm, follower mass 2.3 kg, stiffness 3.5 N/mm, spring force 45 N at lowest position.

  • Need expression for $a$ in terms of $\theta$. Then find $\omega$ when $a$ max equals $$\displaystyle \frac{45 + k s}{2.3} $$ at some $\theta$. Usually occurs at point of maximum acceleration (often at $$\displaystyle \theta=0 $$ or $$\displaystyle 90^\circ $$).

END OF UNIT 1 NOTES
Focus on derivations and numericals from past papers. Practice: four-bar IC method, flywheel from area diagram, governor range, balancing fractions, friction power loss.