THEORY OF MACHINES (ME-403) - Unit 3 Short Notes

UNIT 3 - THEORY OF MACHINES: EXAM-FOCUSED SHORT NOTES

Based on exhaustive analysis of RGPV past papers (Jun 2025, Dec 2024, Jun 2024, Nov 2023, Jun 2022).

1. FUNDAMENTALS OF MECHANISMS & KINEMATIC ANALYSIS

1.1 Basic Definitions & Classifications

• Mechanism: A set of links arranged to transmit motion and force. It is an assembly of bodies with constrained motion.

• Machine: A mechanism that transmits or modifies energy to perform useful work.

• Kinematic Chain: An assembly of links connected by kinematic pairs.

• Link/Joint: A rigid body. Joint (Pair) is connection between links.

  • Lower Pair: Surface contact (e.g., revolute, prismatic).

  • Higher Pair: Point/line contact (e.g., cam-follower, gear teeth).

• Kinematic Diagram: Schematic representation showing links and joints, not concerned with shape/size.

[!TIP] Exam Focus: Differentiating mechanism vs. machine is a very common 7-mark question. Use examples: a crank-slider (mechanism) vs. an internal combustion engine (machine).

1.2 Degree of Freedom (Mobility)

• Definition: Number of independent inputs required to define the configuration of a mechanism.

• Kutzbach's Criterion for Plane Mechanisms:

$$ \boxed{M = 3(N - 1) - 2J_1 - J_2} $$

Where:

*   $M$ = Mobility (DoF)

*   $N$ = Number of links (including frame)

*   $$\displaystyle J_1 $$ = Number of lower pairs (1 DoF each)

*   $$\displaystyle J_2 $$ = Number of higher pairs (2 DoF each)

• Gruebler's Equation: For simple chain with only revolute joints ($$\displaystyle J_1 = J $$, $$\displaystyle J_2 = 0 $$):

$$ \boxed{M = 3(N - 1) - 2J} $$

**Limitation:** Fails for **redundant constraints** (over-constrained mechanisms like parallel linkages).

• Redundant Degree of Freedom: Extra DoF due to special geometry (e.g., when all four links of a 4-bar are equal, it has 1 DoF but Gruebler gives 0). Requires special analysis.

[!TIP] Common Pitfall: Always count the frame as a link. For a standard 4-bar: $$\displaystyle N=4, J=4 \Rightarrow M=3(4-1)-2(4)=1 $$.

1.3 Grashof's Law for Four-Bar Linkages

• Statement: For a 4-bar chain with link lengths $s$ (shortest), $l$ (longest), $p, q$ (intermediate):

$$ \boxed{s + l < p + q} \quad \text{(Grashof condition)} $$

• Proof Idea: Based on continuous rotation possibility of the shortest link.

• Classification:

  | Condition | Shortest Link | Fixed Link | Mechanism Type | Inversions (Crank possibilities) | | :--- | :--- | :--- | :--- | :--- | | $$\displaystyle s+l < p+q $$ | Crank | Adjacent to s | Crank-Rocker | 1 (s or p/q) | | $$\displaystyle s+l < p+q $$ | Crank | Opposite to s | Double-Crank (Drag-link) | 2 (s & p/q) | | $$\displaystyle s+l = p+q $$ | Any | Any | Change Point | Special case | | $$\displaystyle s+l > p+q $$ | None | Any | Double-Rocker | 0 |

[!TIP] Exam Pattern: Given 4 link lengths, first apply Grashof's inequality. Then, based on which link is fixed, determine the type. If fixed link is longest, it's always Double-Rocker (even if Grashof satisfied).

1.4 Inversions of Mechanisms

• Definition: Obtained by fixing a different link as frame in a kinematic chain.

• Four-Bar Chain Inversions:

  1. 1st Inversion (Fix shortest): Crank-Rocker (e.g., Pumping unit).

  2. 2nd Inversion (Fix longest): Double-Rocker (e.g., Pantograph).

  3. 3rd Inversion (Fix one intermediate): Double-Crank (e.g., Oscillating engine).

  4. 4th Inversion (Fix other intermediate): Double-Crank (e.g., Watt's indicator).

• Slider-Crank Chain Inversions:

  1. Reciprocating Engine: Fix frame (1st). Crank → piston motion.

  2. Oscillating Engine: Fix connecting rod (3rd). Crank → oscillating cylinder.

  3. Whitworth Quick Return: Fix crank (2nd). Slider → oscillating tool.

  4. Hand Pump/Pneumatic: Fix piston (4th). Crank → oscillating handle.

• Double Slider-Crank Chain Inversions:

  1. Oldham's Coupling: Fix frame. Two sliders → parallel shafts with angular displacement.

  2. Scotch Yoke: Fix one slider. Crank → reciprocating yoke.

  3. Beam Engine: Fix crank. Two sliders → oscillating beam.

[!TIP] Memory Aid: Link inversions to applications. Whitworth = Quick return (fix crank). Oldham's = Coupling (fix frame). Reciprocating engine = Fix frame.

1.5 Instantaneous Center (IC) Method

• Definition: Point in a body where velocity is zero instantaneously.

• Types:

  • Fixed IC ($$\displaystyle I_{12} $$): On the frame, velocity zero for both bodies.

  • Moving IC ($$\displaystyle I_{13} $$): On a moving link, velocity zero w.r.t. another moving link.

  • Instantaneous Axis of Rotation: IC for a rigid body in plane motion.

• Kennedy's Theorem (Three ICs in a Line): For three rigid bodies (1,2,3) in plane motion, the three ICs ($$\displaystyle I_{12}, I_{23}, I_{31} $$) are collinear.

• Procedure to Locate ICs:

  1. Identify all links.

  2. Fixed ICs are at revolute joints (pin connections).

  3. For slider, IC is at infinity along the direction of motion.

  4. Use Kennedy's theorem to find remaining ICs at intersections of lines joining known ICs.

• Velocity Analysis:

$$ v = \omega \times r \quad \text{or} \quad v_B = \frac{v_A}{I_{12}A} \times I_{12}B $$

Velocity of any point = (Angular velocity) × (Perpendicular distance from IC).

[!TIP] Key Skill: For slider-crank, IC $$\displaystyle I_{12} $$ at crank pivot, $$\displaystyle I_{23} $$ at connecting rod-piston pin, $$\displaystyle I_{13} $$ at infinity (vertical). Use Kennedy to find $$\displaystyle I_{23} $$ relative to frame.

1.6 Velocity & Acceleration Analysis

• Relative Motion Method (Graphical):

  • Velocity Diagram: $$\displaystyle v_B = v_A + v_{BA} $$. $$\displaystyle v_{BA} $$ ⊥ AB.

  • Acceleration Diagram: $$\displaystyle a_B = a_A + a_{BA}^n + a_{BA}^t $$.

    • $$\displaystyle a_{BA}^n = \omega_{BA}^2 \cdot BA $$ (towards $$\displaystyle I_{23} $$).

    • $$\displaystyle a_{BA}^t = \alpha_{BA} \cdot BA $$ (⊥ AB).

  • Coriolis Component: When a point on a rotating link has relative motion along the link.

$$ \boxed{a_c = 2 \omega \times v_{rel}} $$

    *   **Magnitude:** $$\displaystyle 2 \omega v_{rel} $$

    *   **Direction:** Rotate $$\displaystyle v_{rel} $$ by **+90°** in direction of $\omega$.

• Analytical Methods:

  • Complex Numbers (Complex Algebra): Represent position vectors as complex numbers. Differentiate for velocity/acceleration.

  • Loop Closure Equations: For a loop: $$\displaystyle \vec{r_1} + \vec{r_2} + ... = 0 $$. Differentiate for velocity/acceleration.

• Comparison:

  | Graphical | Analytical | | :--- | :--- | | Visual, intuitive | Precise, good for programming | | Error-prone for complex mechanisms | Handles many variables | | Good for single position | Good for entire cycle |

[!TIP] Coriolis is crucial for mechanisms with sliders on rotating links (e.g., Whitworth quick return, shaper). Always check: is there relative sliding along a rotating link?

2. KINEMATIC SYNTHESIS & DYNAMIC ANALYSIS

2.1 Kinematic Synthesis

• Type (Number) Synthesis: Determining number of links and joints to achieve desired DoF. Uses Gruebler's equation.

• Motion Analysis: Given mechanism → find motion (velocity/acceleration).

• Dimensional Synthesis: Given motion requirements → find link lengths.

  • Function Generation: Input-output relationship.

  • Path Generation: Point on coupler follows prescribed path.

  • Motion Generation: Entire body achieves prescribed positions/orientations.

• Freudenstein's Equation (Position Synthesis of 4-bar):

  For 4-bar with fixed link $$\displaystyle O_2O_4 $$, input $$\displaystyle \theta_2 $$, output $$\displaystyle \theta_4 $$:

$$ \boxed{K_1 \cos \theta_4 + K_2 \cos \theta_2 + K_3 \cos(\theta_2 - \theta_4) + K_4 = 0} $$

Where $$\displaystyle K_1, K_2, K_3, K_4 $$ are functions of link lengths ($a, b, c, d$).

Used for **two-position synthesis**. For three positions, use **Chebyshev spacing** and solve three equations.

2.2 Dynamic Analysis of Mechanisms

• D'Alembert's Principle: Convert dynamic problem to static by adding inertia forces.

$$ \sum (\vec{F} - m\vec{a}) = 0 \quad \text{or} \quad \sum (\vec{T} - J\vec{\alpha}) = 0 $$

**Significance:** Allows use of **static equilibrium equations** with **fictitious inertia forces/torques**.

• Dynamic Force Analysis Steps:

  1. Perform kinematic analysis (find $\omega, \alpha$ at given position).

  2. Calculate inertia forces ($$\displaystyle -m\vec{a}_G $$) and inertia torques ($$\displaystyle -J_G \vec{\alpha} $$) for each link.

  3. Apply D'Alembert's principle + external forces (e.g., piston load, driving torque).

  4. Solve for unknown reactions at joints using static equilibrium for each link (consider 2-force members).

• Application to Slider-Crank:

  • Link 1 (crank): Inertia torque $$\displaystyle -J_1 \alpha_1 $$, driving torque $$\displaystyle T_{in} $$.

  • Link 2 (connecting rod): Inertia force $$\displaystyle -m_2 a_{G2} $$, inertia torque $$\displaystyle -J_{G2} \alpha_2 $$.

  • Link 3 (piston): Inertia force $$\displaystyle -m_3 a_3 $$, external load $$\displaystyle F_{piston} $$.

  • Solve starting from piston (1 unknown reaction at pin).

• Kinematically-Determined vs. Statically-Determinate:

  • Kinematically Determinate: All motions known from geometry (e.g., 4-bar with known input $$\displaystyle \omega_2 $$).

  • Statically Determinate: Number of unknown reactions = number of equilibrium equations (3 per link). Often requires momentum method for indeterminate structures.

[!TIP] Exam Alert: Dynamic analysis problems often ask for input torque or bearing reactions. Start from the link with fewest unknowns (usually piston/slider).

3. CAMS & FOLLOWER SYSTEMS (HIGH FREQUENCY)

3.1 Classification of Cams & Followers

[!TIP] Roller follower is most common in machinery. Offset follower reduces pressure angle.

3.2 Fundamental Cam Terminology

• Base Circle ($$\displaystyle r_b $$): Smallest radius of cam profile. Foundation for design.

• Pitch Circle: Circle on which pitch point lies. Radius $$\displaystyle r_p = r_b + \frac{h}{2} $$ (for SHM).

• Pressure Angle ($\phi$): Angle between normal to pitch curve and direction of follower motion.

  • Effect: High $\phi$ → high lateral force → wear/jamming.

  • Limit: $$\displaystyle \phi_{max} \approx 30^\circ $$ (translating), $$\displaystyle 35^\circ $$ (oscillating).

• Undercutting: Occurs when cam radius of curvature < follower radius. Causes sharp corners and interference.

  • Prevention: Increase base circle, use roller follower, use inverted cam.

• Cam Profile Generation:

  • Direct Method: Draw follower positions for successive cam angles → join.

  • Inverse Method (Common): Draw displacement diagram → invert to get profile (using radius vector).

3.3 Follower Motion Curves & Cam Design

• Displacement Diagram: Plot $s$ (lift) vs. $\theta$ (cam rotation). Key intervals: Rise (R), Dwell (D), Return (R), Dwell (D).

• Common Motion Laws:

  | Law | Displacement $s$ | Velocity $v$ | Acceleration $a$ | Pros/Cons | | :--- | :--- | :--- | :--- | :--- | | Uniform Velocity | $$\displaystyle s = \frac{h}{\beta} \theta $$ | Constant | $\infty$ at start/end | Not used (shock) | | Uniform Acceleration (Parabolic) | $$\displaystyle s = \frac{h}{\beta^2} \theta^2 $$ (rise) | Linear | Constant | Finite $a$, but discontinuity at mid-point | | Simple Harmonic Motion (SHM) | $$\displaystyle s = \frac{h}{2} \left(1 - \cos \frac{\pi \theta}{\beta}\right) $$ | Cosine | Sine | Smooth, but finite jerk at ends | | Cycloidal Motion | $$\displaystyle s = \frac{h}{\pi} \left(\frac{\pi \theta}{\beta} - \sin \frac{\pi \theta}{\beta}\right) $$ | $1-\cos$ | $\sin$ | Best (zero $a$ at ends), smooth |

• Design Steps:

  1. Draw displacement diagram (choose motion law for each interval).

  2. Calculate max velocity/acceleration:

     • SHM: $$\displaystyle v_{max} = \frac{\pi h}{2\beta} $$, $$\displaystyle a_{max} = \frac{\pi^2 h}{\beta^2} \cdot \frac{N^2}{60} $$ (rad/s²).

     • Uniform Accel: $$\displaystyle v_{max} = \frac{2h}{\beta} $$, $$\displaystyle a_{max} = \frac{2h}{\beta^2} \cdot \frac{N^2}{60} $$.

  3. Draw velocity & acceleration diagrams (optional but good for check).

  4. Lay out cam profile:

     • Divide base circle into $\Delta \theta$ intervals.

     • For each $$\displaystyle \theta_i $$, find $$\displaystyle s_i $$ from diagram.

     • Draw radius vector of length $$\displaystyle r_b + s_i $$ at angle $$\displaystyle \theta_i $$.

     • Draw normal at pitch point → intersect radius vector → cam profile point.

• Tangent Cam (Symmetrical): For roller follower. Profile: straight flanks + circular nose.

  • Principal Dimensions: $$\displaystyle r_b $$, $$\displaystyle r_1 $$ (nose radius), $h$, $\phi$ (ascent angle).

  • Acceleration at key points:

    • Start/end of lift (flank-nose junction): $$\displaystyle a = \frac{2v^2}{r_1 \sin \phi} $$ (max).

    • Apex of nose: $$\displaystyle a = \frac{v^2}{r_1} $$.

[!TIP] Cam Design is a 14-mark question. Always: (1) Draw displacement diagram with proper scaling. (2) Label all angles (rise $$\displaystyle \beta_r $$, dwell $$\displaystyle \beta_d $$, etc.). (3) Calculate $$\displaystyle v_{max}, a_{max} $$ numerically for asked intervals. (4) For profile, show construction clearly (radius vectors, normals).

3.4 Special Cams

• Critical Path Motion: Motion where follower loses contact with cam due to inertia (especially at high speed).

  • Condition: $$\displaystyle a_{follower} > a_{cam} \cdot \sin\phi $$ (downward acceleration).

• Torque on Camshaft:

$$ T = F_r \cdot r_{eff} $$

Where $$\displaystyle F_r $$ is **radial force** (from pressure angle & spring force), $$\displaystyle r_{eff} $$ is **effective radius** (usually pitch circle radius). Varies with cam angle.

4. GEARS & GEAR TRAINS (HIGH FREQUENCY)

4.1 Fundamentals of Gearing

• Law of Gearing: For constant velocity ratio, the common normal at the point of contact must always pass through a fixed point (the pitch point) on the line of centers.

  • Derivation: For involute, normal is along generatrix (line of action), which is tangent to base circle → always passes through pitch point.

• Conjugate Profiles: Tooth profiles satisfying law of gearing. Involute is the most common.

• Involute Profile:

  • Generation: Unwinding a string from base circle.

  • Properties:

    1. Conjugacy: Satisfies law of gearing.

    2. Center Distance Variation: Velocity ratio remains constant even if center distance varies slightly (important for manufacturing tolerances).

    3. Pressure Angle Constant: $$\displaystyle \phi = \angle $$ between line of action and tangent.

4.2 Gear Terminology

• Module (m): $$\displaystyle m = \frac{\text{Pitch Diameter}}{\text{Number of teeth}} $$ (mm). Standard values.

• Pressure Angle ($\phi$): Standard $$\displaystyle 20^\circ $$ (now), older $$\displaystyle 14.5^\circ $$.

• Contact Ratio ($\epsilon$): Average number of teeth in contact. $$\displaystyle \epsilon > 1 $$ ensures smooth transmission. Typically 1.6–2.2.

4.3 Interference & Undercutting

• Interference: Occurs when addendum tip of driving gear (pinion) contacts addendum root of driven gear (outside the line of action). Results in non-conjugate action and wear.

• Minimum Teeth to Avoid Interference (Full-depth Involute):

$$ \boxed{z_{min} = \frac{2h_a^*}{\sin^2 \phi}} $$

Where $$\displaystyle h_a^* $$ = addendum coefficient (usually 1).

*   For $$\displaystyle \phi=20^\circ $$, $$\displaystyle z_{min} \approx 18 $$ (pinion). **Gear can have fewer teeth** if pinion has enough.

• Addendum Modification (Profile Shifting): Shift tool inward/outward during cutting.

  • Positive shift (+x): Increases addendum, reduces undercutting → allows <18 teeth pinion.

  • Negative shift (-x): Reduces addendum, increases contact ratio.

[!TIP] Interference Problem: Given $$\displaystyle z_1, z_2, \phi, m $$, check if $$\displaystyle z_1 < z_{min} $$. If yes, interference occurs. To avoid, either increase $$\displaystyle z_1 $$ or use profile shift.

4.4 Gear Trains

• Simple Gear Train: $$\displaystyle \frac{\omega_1}{\omega_n} = (-1)^n \frac{z_2 z_4 ...}{z_1 z_3 ...} $$. Disadvantage: Large size for high ratio.

• Compound Gear Train: Two gears on same shaft. Ratio = product of individual ratios. Compact.

• Epicyclic (Planetary) Gear Train:

  • Elements: Sun (central), Planet (carried by arm), Ring (annular, internal teeth).

  • Velocity Ratio Methods:

    1. Tabular (Relative Motion) Method:

       • Step 1: Assume arm fixed → find relative speeds.

       • Step 2: Add arm speed to all.

       • Formula: $$\displaystyle \frac{\omega_{sun} - \omega_{arm}}{\omega_{ring} - \omega_{arm}} = -\frac{Z_{ring}}{Z_{sun}} $$.

    2. Algebraic Method: Use $$\displaystyle \omega_{relative} $$ and sign convention (CW/CCW).

  • Special Advantages:

    • High power-to-weight ratio.

    • Coaxial shafts possible (sun & ring).

    • High velocity ratios in single stage.

    • Applications: Automatic transmissions, differentials, industrial drives, wind turbines.

[!TIP] Epicyclic is a 7-mark question. Tabular method is foolproof. Always: (1) Identify fixed element (if any). (2) Write speed relation when arm fixed. (3) Add arm speed. Example: If ring fixed, $$\displaystyle \frac{\omega_{sun} - \omega_{arm}}{0 - \omega_{arm}} = -\frac{Z_{ring}}{Z_{sun}} $$.

4.5 Gear Design Problems

• Least Number of Teeth for Given Ratio ($i$): For pinion ($$\displaystyle z_1 $$) and gear ($$\displaystyle z_2 = i z_1 $$), avoid interference on both:

$$ z_1 \ge z_{min} \quad \text{and} \quad z_2 \ge \frac{z_{min}}{1 - \frac{1}{i} \cdot \frac{2h_a^*}{\sin^2 \phi}} $$

• Path of Contact ($L$): Distance along line of action from start (addendum circle intersection) to end (other addendum circle intersection).

$$ L = \sqrt{r_{a1}^2 - r_{b1}^2} + \sqrt{r_{a2}^2 - r_{b2}^2} - (r_{b1} + r_{b2}) \sin \phi $$

• Arc of Contact ($\theta$): Angle subtended on pitch circle by path of contact: $$\displaystyle \theta = \frac{L}{r_p} $$.

• Contact Ratio: $$\displaystyle \epsilon = \frac{L}{p_c} $$.

• Sliding Velocity: At point of contact, $$\displaystyle v_s = (\omega_1 r_1 - \omega_2 r_2) \sin \phi $$. Max at start/end of contact.

5. BELT, ROPE & CHAIN DRIVES (HIGH FREQUENCY)

5.1 Belt Drives

• Types:

  • Flat Belt: Flexible, used for large center distances, high speed.

  • V-Belt: Trapezoidal, wedges in groove → higher friction, compact.

  • Open Drive: Pulleys rotate same direction.

  • Crossed Drive: Pulleys rotate opposite direction (stress reversal → lower power).

• Velocity Ratio (Considering Thickness $t$):

$$ \frac{N_2}{N_1} = \frac{D_1 + t}{D_2 + t} \quad (\text{Open}) \quad ; \quad \frac{N_2}{N_1} = \frac{D_1 + t}{D_2 + t} \quad (\text{Cross, same}) $$

*   **Slip ($s$):** Actual speed ratio < theoretical due to slip. $$\displaystyle s = \frac{N_1 - N_2'}{N_1} $$.

*   **Creep:** Local stretching of belt → similar effect.

• Power Transmission - Euler's Equation:

$$ \boxed{\frac{T_1}{T_2} = e^{\mu \theta}} $$

Where $\theta$ = **angle of lap** (radians) on **smaller pulley**.

*   **Open Drive:** $$\displaystyle \theta = \pi - 2\alpha $$, $$\displaystyle \sin \alpha = \frac{r_1 - r_2}{C} $$.

*   **Cross Drive:** $$\displaystyle \theta = \pi + 2\alpha $$.

• Maximum Power Transmission:

  • Power $$\displaystyle P = (T_1 - T_2) v $$.

  • Given max tension $$\displaystyle T_{max} $$ and tension ratio $$\displaystyle k = e^{\mu\theta} $$:

$$ T_1 = k T_2, \quad T_1 + T_2 = 2T_{max} \quad (\text{if centrifugal tension neglected}) $$

$$ \Rightarrow T_1 = \frac{2k}{k+1} T_{max}, \quad T_2 = \frac{2}{k+1} T_{max} $$

$$ P_{max} = \frac{2k}{k+1} T_{max} \cdot v \left(1 - \frac{1}{k}\right) = \frac{2(k-1)}{k+1} T_{max} v $$

*   **With Centrifugal Tension ($$\displaystyle T_c = m v^2 $$):**

    Effective tension $$\displaystyle T_1' = T_1 - T_c $$, $$\displaystyle T_2' = T_2 - T_c $$.

    $$\displaystyle P = (T_1' - T_2') v = (T_1 - T_2) v $$ (same form, but $$\displaystyle T_1, T_2 < T_{max} $$).

• Stress in Belt:

  • Open Drive: $$\displaystyle \sigma_{max} = \frac{T_1}{A} $$ (tight side).

  • Cross Drive: $$\displaystyle \sigma_{max} = \frac{T_1 + T_2}{A} $$ (both sides add).

  • With centrifugal tension: $$\displaystyle \sigma_{max} = \frac{T_1}{A} + \sigma_c $$, $$\displaystyle \sigma_c = \frac{m v^2}{A} $$.

[!TIP] Belt Problems are frequent. Always: (1) Find $\theta$ on smaller pulley. (2) Use Euler's $$\displaystyle T_1/T_2 = e^{\mu\theta} $$. (3) For max power, use $$\displaystyle T_1 = \frac{2k}{k+1}T_{max} $$. (4) For stress in cross drive, use $$\displaystyle T_1+T_2 $$.

5.2 Chain Drives

• Comparison with Belts:

  | Chain Drive | Belt Drive | | :--- | :--- | | No slip (positive drive) | Slip possible | | Compact, high load | Needs large tension | | Chordal action (velocity fluctuation) | Smooth | | Lubrication needed | Usually not | | Used for short center distances | Long distances |

• Velocity Ratio: $$\displaystyle N_1/N_2 = z_2/z_1 $$ (no slip).

• Chordal Action: Due to polygonal effect, pitch line velocity fluctuates. Pitch must be chosen to limit fluctuation.

5.3 Friction in Power Transmission

• Factors: $\mu$ (friction coefficient), $\theta$ (angle of lap), $$\displaystyle T_1, T_2 $$ (tensions), $v$ (speed), centrifugal force.

• Efficiency: $$\displaystyle \eta = \frac{T_1 - T_2}{T_1} = 1 - \frac{1}{k} $$ (without centrifugal). Decreases with $\theta$? No, $k$ increases with $\theta$, so $\eta$ increases with $\theta$.

6. FRICTION DEVICES: BRAKES & CLUTCHES (MEDIUM FREQUENCY)

6.1 Clutches

• Function: Connect/disconnect power transmission gradually (while both shafts rotating).

• Difference from Brake: Brake stops/retards a rotating member; Clutch engages two rotating members.

• Classification:

  • Friction Clutches: Plate (Single/Multi-plate), Cone, Centrifugal.

  • Positive Clutches: Jaw/Claw (no slip, for indexing).

• Plate Clutch:

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

$$ \boxed{T = \frac{2}{3} \mu p \pi (r_2^3 - r_1^3)} $$

*   **Uniform Wear:** Assumes $$\displaystyle p r = \text{constant} $$ (more realistic for old clutches).

$$ \boxed{T = \frac{1}{2} \mu p_{max} \pi (r_2^2 - r_1^2) \cdot \frac{r_2^3 - r_1^3}{r_2^2 - r_1^2}} = \frac{1}{2} \mu W (r_2 + r_1) \quad (\text{since } W = \pi p_{max} (r_2^2 - r_1^2)) $$

    **Simpler form:** $$\displaystyle T = \mu W R_f $$, where $$\displaystyle R_f = \frac{2}{3} \frac{r_2^3 - r_1^3}{r_2^2 - r_1^2} $$ (mean radius of friction).

• Cone Clutch:

$$ \boxed{T = \frac{1}{2} \mu W (r_1 + r_2) \csc \alpha} $$

Where $\alpha$ = **semi-cone angle**.

• Centrifugal Clutch:

  • Principle: At high $\omega$, shoes fly out → engage.

  • Torque: $$\displaystyle T = \mu W R \cdot n $$, where $$\displaystyle W = m (\omega^2 R - g \tan \alpha) $$ (centrifugal force minus spring force), $R$ = radius of engagement, $n$ = number of shoes.

[!TIP] Plate clutch with uniform wear is most common in exams. Remember: $$\displaystyle T = \frac{1}{2} \mu W (r_2 + r_1) $$. For max/min pressure, use $p \propto 1/r$.

6.2 Brakes

• Function: Stop/retard a rotating member by dissipating energy as heat.

• Classification: Block, Band & Block, Disc, Drum (Internal/External Expanding).

• Band & Block Brake:

  • Derivation: For one block, $$\displaystyle T_0 = T_n \left(\frac{1+\mu \tan \theta}{1-\mu \tan \theta}\right) $$.

  • For $n$ blocks (independent): $$\displaystyle \frac{T_0}{T_n} = \left(\frac{1+\mu \tan \theta}{1-\mu \tan \theta}\right)^n $$.

  • Analysis: Draw free body diagram of band. $$\displaystyle T_0 $$ (tight side) → blocks → $$\displaystyle T_n $$ (slack side). Use moment equilibrium about drum center.

• Simple Block Brake:

  • With shoe pivot: $$\displaystyle T = \mu P R \frac{a+b}{b} $$ (where $a$ = distance pivot to force, $b$ = pivot to drum center).

  • Without pivot (pivoted at center): $$\displaystyle T = \mu P R $$.

• Comparison Block vs. Band & Block: For same drum dia, Band & Block gives higher braking torque for same force $P$ because of self-energizing effect ($$\displaystyle T_0 > \mu P R $$).

6.3 Other Friction Devices

• Pivot & Collars:

  • Flat Pivot: $$\displaystyle T = \frac{2}{3} \mu W R $$ (uniform pressure).

  • Conical Pivot: $$\displaystyle T = \frac{2}{3} \mu W R \csc \alpha $$.

  • Collar Bearing: $$\displaystyle T = \frac{1}{2} \mu W (R_2 + R_1) $$ (uniform wear).

• Power Screws (Threaded Spindles):

  • Torque to Raise Load:

$$ \boxed{T_{raise} = W \frac{d_m}{2} \left( \frac{\tan \phi + \tan \lambda}{1 - \tan \phi \tan \lambda} \right) + \frac{W}{2} \mu_c (D_o^2 - d_c^2)/(d_m)} $$

    Where $\phi$ = friction angle ($$\displaystyle \tan^{-1}\mu $$), $\lambda$ = lead angle, $$\displaystyle \mu_c $$ = collar friction.

*   **Torque to Lower Load:**

$$ T_{lower} = W \frac{d_m}{2} \left( \frac{\tan \lambda - \tan \phi}{1 + \tan \phi \tan \lambda} \right) + \text{collar term} $$

*   **Efficiency:**

$$ \eta_{raise} = \frac{\tan \lambda}{\tan(\phi + \lambda)} \quad (\text{ignoring collar}) $$

*   **Self-locking:** $$\displaystyle \eta_{raise} < 0.5 $$ or $$\displaystyle \tan \lambda < \tan \phi $$ → load won't back-drive.

[!TIP] Power screw problems: Always separate thread torque and collar torque. For efficiency, use $$\displaystyle \eta = \frac{Work_{out}}{Work_{in}} = \frac{W \cdot \text{lead}}{2\pi T} $$.

7. VIBRATIONS & BALANCING (LOWER FREQUENCY BUT ASKED)

7.1 Free Vibrations

• Types & Natural Frequency ($$\displaystyle f_n $$):

  | Type | System | $$\displaystyle f_n $$ (Hz) | | :--- | :--- | :--- | | Longitudinal | Spring-mass (linear) | $$\displaystyle f_n = \frac{1}{2\pi} \sqrt{\frac{k}{m}} $$ | | Transverse | Shaft/beam with mass | $$\displaystyle f_n = \frac{1}{2\pi} \sqrt{\frac{k_{eq}}{m}} $$ (k from deflection) | | Torsional | Shaft with disc | $$\displaystyle f_n = \frac{1}{2\pi} \sqrt{\frac{C}{J}} $$ (C = torsional stiffness, J = polar moment) |

7.2 Balancing of Rotating Masses

• Single Plane (Static) Balancing: All masses in same plane. $$\displaystyle \sum m r = 0 $$, $$\displaystyle \sum m r \theta = 0 $$. Use polygon method or complex numbers.

• Two Plane (Dynamic) Balancing: Masses in different planes (along shaft).

  • Method: Choose two balancing planes (X, Y). For each original mass $$\displaystyle m_i $$ at radius $$\displaystyle r_i $$, angle $$\displaystyle \theta_i $$, distance $$\displaystyle a_i $$ from reference:

    • Moment (couple) polygon: $$\displaystyle \sum m_i r_i a_i $$ (in plane perpendicular to shaft).

    • Force polygon: $$\displaystyle \sum m_i r_i $$ (in plane of reference).

  • Solve for balancing masses $$\displaystyle m_X, m_Y $$ at given radii.

• Primary & Secondary Balancing (Reciprocating Engines):

  • Primary: Balance $$\displaystyle m \omega^2 r $$ (unbalanced force at engine frequency).

  • Secondary: Balance $$\displaystyle m \omega^2 r \frac{r}{l} \cos 2\theta $$ (due to piston acceleration approximation). Requires extra masses.

[!TIP] Balancing Problem: Given masses $$\displaystyle m_i $$, radii $$\displaystyle r_i $$, angles $$\displaystyle \theta_i $$, and plane positions $$\displaystyle x_i $$, find $$\displaystyle m_X, m_Y $$ at $$\displaystyle R_X, R_Y $$. Steps: (1) Compute $$\displaystyle m_i r_i $$ and $$\displaystyle m_i r_i x_i $$. (2) Draw force polygon → get $$\displaystyle m_X r_X + m_Y r_Y $$. (3) Draw couple polygon → get $$\displaystyle m_X r_X x_X + m_Y r_Y x_Y $$. (4) Solve.

8. SPECIAL MECHANISMS & APPLICATIONS

8.1 Quick Return Mechanisms

• Crank and Slotted Lever:

  DiagramCANVAS: Sketch: Fixed pivot O1, crank O1P rotating, slotted lever O2B with slot, slider P in slot. Driving stroke (slow) when crank rotates from θ1 to θ2, return stroke (fast) from θ2 to θ1. Label angles α (slow stroke angle), β (fast stroke angle).

  • Time Ratio (TR): $$\displaystyle \boxed{TR = \frac{\text{Time of cutting stroke}}{\text{Time of return stroke}} = \frac{\beta + \alpha}{\alpha}} $$

  • Design: Given $r$ (crank), $l$ (lever length), $d$ (distance O1O2), find $\alpha, \beta$ from geometry.

• Whitworth Quick Return:

  DiagramCANVAS: Sketch: Crank OA rotating, connecting rod AB, slotted lever O1C with slot, tool at D. Crank center O, lever pivot O1. Driving stroke when A moves from left to right, return when A moves right to left.

  • Time Ratio: $$\displaystyle \boxed{TR = \frac{360^\circ - \psi}{\psi}} $$, where $$\displaystyle \psi = 2 \cos^{-1} \left( \frac{r}{d} \right) $$ (if $$\displaystyle r < d $$).

  • Design Synthesis (from Jun 2025): Given return stroke $L$, TR, crank length $r$ → find $d$.

    • From geometry: $$\displaystyle L = 2 \sqrt{d^2 - r^2} $$.

    • From TR: $$\displaystyle \psi = \frac{360^\circ}{TR + 1} $$.

    • Then $$\displaystyle \cos(\psi/2) = r/d $$ → $$\displaystyle d = r / \cos(\psi/2) $$.

8.2 Tractor Link & Straight Line Generating Mechanisms

• Tractor Link: Inversion of 4-bar where coupler point traces approximate straight line (used in tractor front wheels).

• Straight Line Mechanisms: Watt's, Robert's, Peaucellier-Lipkin (exact straight line).

8.3 Power Transmission Elements Summary

[!TIP] Quick Return design is a synthesis problem. Remember: Whitworth uses $$\displaystyle \psi = 2\cos^{-1}(r/d) $$; Crank & Slotted Lever uses geometry of slot.

Final Exam Strategy:

1. Definitions first: Always start with clear definitions (Mechanism, DoF, Cam terms, etc.).

2. Formulas boxed: Key equations (Gruebler, Grashof, Euler, Torques) must be memorized.

3. Diagrams essential: For cams, gears, mechanisms, sketch neatly and label.

4. Step-by-step: For synthesis/analysis, show all steps (displacement diagram, velocity polygon, force equilibrium).

5. Units: Convert all to consistent units (mm, N, s, rad).

All the best! Focus on HIGH FREQUENCY topics (Cams, Gears, Belt Drives, Inversions, Quick Return) as they appear in every paper.