UNIT 5: THEORY OF MACHINES - SHORT NOTES

Based on RGPV Past Papers (Jun 2025, Dec 2024, Jun 2024, Jun 2023, Nov 2023, Jun 2022)

I. FUNDAMENTAL CONCEPTS OF MECHANISMS

Mechanisms vs. Machines

• Mechanism: A set of links designed to transmit motion and force in a predetermined manner. It is a motion transformer.

• Machine: A mechanism plus a power source and a useful work output. It transmits and modifies energy.

• Key Difference: All machines contain mechanisms, but not all mechanisms are complete machines (e.g., a door hinge is a mechanism, not a machine).

Degree of Freedom (DOF)

• Definition: The number of independent coordinates required to define the position of all links in a mechanism relative to a fixed frame.

• Significance: Determines if the mechanism can move as intended. DOF = 1 for a simple mechanism driven by one input.

• Kutzbach's Criterion (Plane Mechanisms):

$$DOF = 3(n - 1) - 2j - h$$

Where:

*   $n$ = total number of links (including fixed link)

*   $j$ = number of binary joints (lower pairs)

*   $h$ = number of higher pairs (point/line contact)

> [!TIP] For a 4-bar linkage (n=4, j=4, h=0), DOF = 3(3) - 2(4) = 1.

• Gruebler's Criterion:

$$DOF = 3(n - 1) - 2j$$

(Assumes no higher pairs and all joints are 1-DOF binary joints).

*   **Distinction**: Gruebler's equation is a special case of Kutzbach's where $$\displaystyle h=0 $$. Kutzbach's is more general.

Constrained Motion

• Completely Constrained Motion: Motion is possible in one direction only. Example: Piston in an engine cylinder (sliding pair).

• Incompletely Constrained Motion: Motion is possible in more than one direction. Example: Shaft in a journal bearing (can rotate and slide).

• Successfully Constrained Motion: Incompletely constrained motion made to have only one degree of freedom by the application of a special device or force. Example: Shaft in a step bearing (rotation prevented by key).

II. MECHANISM CLASSIFICATION, GRASHOF'S LAW & INVERSIONS

Grashof's Law

• Statement: For a 4-bar linkage with link lengths $S$ (shortest), $L$ (longest), $P$, $Q$ (intermediate), the condition for at least one link to make a full rotation is:

$$S + L < P + Q$$

• Interpretation & Classification:

  | Grashof Condition | Fixed Link | Resulting Mechanism | Example | | :--- | :--- | :--- | :--- | | $$\displaystyle S + L < P + Q $$ | Shortest ($S$) | Double Crank (both adjacent links rotate) | Coupling rod in locomotive | | $$\displaystyle S + L < P + Q $$ | Longest ($L$) | Crank-Rocker (one link rotates, other oscillates) | Steam engine mechanism | | $$\displaystyle S + L < P + Q $$ | Any other ($P$ or $Q$) | Double Rocker (both adjacent links oscillate) | Common in fixtures | | $$\displaystyle S + L > P + Q $$ | Any link | Change Point/Non-Grashof | No link can rotate fully | | $$\displaystyle S + L = P + Q $$ | Any link | Special Case (Parallelogram or Deltoid) | Pantograph, parallel motion |

Inversions of Mechanisms

• Concept: Fixing one of the links in a kinematic chain to form a mechanism. A 4-bar chain has 4 inversions.

• Inversions of Four-Bar Linkage:

  1. Fixed Link = Frame (Ground): Standard 4-bar linkage.

  2. Fixed Link = Crank: Oscillating Engine (output link oscillates).

  3. Fixed Link = Coupler: Watt's Indicator (output is straight line motion approx.).

  4. Fixed Link = Rocker: Double Crank Mechanism.

• Inversions of Slider-Crank Mechanism:

  1. Fixed Link = Frame (Crank fixed, slider moves): Reciprocating Engine/Pump (Standard).

  2. Fixed Link = Connecting Rod: Oscillating Engine (Crank oscillates, output is rotary).

  3. Fixed Link = Crank: Hand Pump/Screw Jack (Slider is input, crank is output).

  4. Fixed Link = Slider: Whitworth Quick Return (Crank is input, connecting rod oscillates).

• Inversions of Double Slider Crank Mechanism:

  1. Fixed Link = Frame: Elliptical Trammel (Generates elliptical motion).

  2. Fixed Link = One Slider: Scotch Yoke (Converts rotary to reciprocating).

  3. Fixed Link = Other Slider: Oldham's Coupling (Connects parallel shafts with lateral shift).

  4. Fixed Link = Coupler: Double Slider Crank (Rarely used).

Quick Return Mechanisms

• Whitworth Quick Return:

  • Construction: Slotted lever ($$\displaystyle O_2B $$) pivoted at $$\displaystyle O_2 $$, crank ($$\displaystyle O_1P $$) connected to lever via sliding block $P$ in slot. Link $CD$ is the tool.

  • Working: Forward stroke (cutting) occurs during slower crank rotation through angle $\beta$. Return stroke (non-cutting) occurs during faster rotation through $(360° - \beta)$.

  • Time Ratio Derivation:

$$\text{Time Ratio } (k) = \frac{\text{Cutting Time}}{\text{Return Time}} = \frac{\beta}{360° - \beta}$$

    From geometry of triangle $$\displaystyle O_1O_2P $$:

$$\cos\left(\frac{\beta}{2}\right) = \frac{d^2 + r^2 - R^2}{2 d r}$$

    where $r$ = crank length, $R$ = lever length $$\displaystyle O_2B $$, $d$ = distance $$\displaystyle O_1O_2 $$.

> [!TIP] $$\displaystyle k > 1 $$ for quick return. Design given $k$, $r$, find $d$ or $R$.

• Crank and Slotted Lever Mechanism:

  • Construction: Similar to Whitworth but crank drives the slotted lever directly (no sliding block).

  • Time Ratio Derivation:

    Let crank $$\displaystyle O_1A $$ rotate at constant $\omega$. Angle turned for forward stroke = $\beta$, for return = $360° - \beta$.

$$\tan\phi = \frac{r \sin\theta}{d + r \cos\theta}$$

(from geometry)

    Time ratio $$\displaystyle k = \frac{\beta}{360° - \beta} $$ can be found by solving for $\theta$ when $\phi$ is max/min.

III. KINEMATIC ANALYSIS OF MECHANISMS

Instantaneous Center (IC) Method

• Definition: The point in a body which has zero instantaneous velocity relative to the fixed link.

• Types:

  • Fixed IC: Lies on fixed link (e.g., $$\displaystyle I_{12} $$ for crank in slider-crank).

  • Moving IC: Lies on neither link (e.g., $$\displaystyle I_{23} $$ between coupler and crank).

  • Permanent IC: Lies on the same point in both links (e.g., $$\displaystyle I_{34} $$ for slider and ground).

• Kennedy's Theorem: For three bodies (1,2,3) in plane motion, the three ICs ($$\displaystyle I_{12} $$, $$\displaystyle I_{23} $$, $$\displaystyle I_{13} $$) are collinear.

• Location of ICs in Slider-Crank (Crank AB, Connecting Rod BC, Slider C, Ground 1):

  • $$\displaystyle I_{12} $$: At $O$ (pivot of crank).

  • $$\displaystyle I_{23} $$: Intersection of lines perpendicular to AB at $B$ and BC at $C$.

  • $$\displaystyle I_{13} $$: At infinity (slider moves linearly).

  • $$\displaystyle I_{24} $$ (if considering follower): At slider path intersection.

  [!TIP] Velocity of any point = $\omega \times$ (distance from IC).

Velocity & Acceleration Analysis

• Absolute Motion: Motion relative to fixed frame.

• Relative Motion: Motion of a point relative to another moving point.

• Velocity Diagram:

  • $$\displaystyle \vec{v}_{BA} = \vec{v}_B - \vec{v}_A $$ (relative velocity of B w.r.t A).

  • Construction: $$\displaystyle v_A $$ known, $$\displaystyle v_B $$ perpendicular to AB (if rotating about A), $$\displaystyle v_{BA} $$ perpendicular to AB.

• Acceleration Diagram:

  • $$\displaystyle \vec{a}_B = \vec{a}_A + \vec{a}_{BA}^t + \vec{a}_{BA}^c $$

  • $$\displaystyle \vec{a}_{BA}^t = \alpha_{AB} \times \vec{BA} $$ (tangential, $\perp$ to AB)

  • $$\displaystyle \vec{a}_{BA}^c = \omega_{AB}^2 \times \vec{BA} $$ (centripetal, $\parallel$ to AB towards A)

Coriolis Component of Acceleration

• Occurs when: A point on a link has both rotational motion and a sliding motion relative to another link.

• Magnitude: $$\displaystyle a_c = 2 \omega v_{rel} $$

• Direction: Perpendicular to the relative velocity vector $$\displaystyle v_{rel} $$. Rotated $90°$ in the direction of $\omega$.

• Derivation Sketch: For a point $P$ on a rotating link with $\omega$, sliding along the link with $$\displaystyle v_{rel} $$:

$$\vec{a}_P = \vec{a}_O + \vec{a}_{PO}^t + \vec{a}_{PO}^c + \vec{a}_c$$

where $$\displaystyle \vec{a}_c = 2 \vec{\omega} \times \vec{v}_{rel} $$.

D'Alembert's Principle

• Statement: The sum of the differences between the forces acting on a system of particles and the time derivatives of the momenta of those particles is zero. For rigid bodies, it introduces inertia forces ($-m\vec{a}$) and inertia torques ($-I\alpha$).

• Significance: Converts a dynamic problem into a static equilibrium problem. Allows use of static force analysis techniques for accelerating mechanisms.

• Application to Slider-Crank: Apply inertia forces at each link's center of mass. For crank: inertia force $$\displaystyle -m_2 a_{G2} $$ and inertia torque $$\displaystyle -I_2 \alpha_2 $$. For connecting rod: resolve inertia force along and perpendicular to its axis. For slider: inertia force $$\displaystyle -m_3 a_3 $$. Then take moments about pins to find bearing reactions or input torque.

IV. KINEMATIC SYNTHESIS OF LINKAGES

Number Synthesis vs. Motion Analysis

• Number Synthesis: Determines the number of links, joints, and DOF required to achieve a specified motion task. Uses Gruebler's criterion. "How many links and joints are needed?"

• Motion Analysis: Given a mechanism (links/joints known), determines its motion (position, velocity, acceleration). "How does this mechanism move?"

Analytical & Graphical Methods

• Analytical Method: Uses coordinate geometry and loop-closure equations to solve for unknown link lengths/angles.

  • Advantages: Precise, suitable for computer-aided synthesis.

  • Limitations: Complex equations, multiple solutions, difficult for higher-order synthesis.

• Graphical Method: Uses geometric constructions (e.g., Chebyshev's spacing, precision points) to determine link dimensions.

  • Advantages: Intuitive, good for understanding motion.

  • Limitations: Less accurate, tedious for many precision points.

Freudenstein's Equation

• Application: Synthesizes a 4-bar linkage to satisfy one position with specified input and output angular velocities (or accelerations).

• Equation:

$$K_1 \cos\theta_4 + K_2 \cos\theta_2 + K_3 = \cos(\theta_2 - \theta_4)$$

Where $$\displaystyle \theta_2 $$ = input angle, $$\displaystyle \theta_4 $$ = output angle.

$$K_1 = \frac{d}{a}, \quad K_2 = \frac{d}{c}, \quad K_3 = \frac{a^2 - b^2 + c^2 + d^2}{2ac}$$

with $a, b, c, d$ = lengths of crank, coupler, rocker, fixed link respectively.

• For given $$\displaystyle \theta_2, \theta_4, \omega_2, \omega_4 $$, differentiate Freudenstein's equation w.r.t time to get:

$$-K_1 \sin\theta_4 \cdot \omega_4 - K_2 \sin\theta_2 \cdot \omega_2 = \sin(\theta_2 - \theta_4)(\omega_2 - \omega_4)$$

Solve simultaneously with original equation for $$\displaystyle K_1, K_2 $$. Then use $$\displaystyle K_3 $$ equation to find link ratios.

V. CAMS AND FOLLOWERS (HIGH FREQUENCY)

Classification

• By Cam Shape:

  • Radial/Disc Cam: Follower moves perpendicular to cam axis.

    DiagramSEARCH: radial disc cam

  • Cylindrical Cam: Follower moves parallel to cam axis (groove cam).

  • Plate Cam: Flat plate with contoured edge.

• By Follower Type:

  • Knife-edge: Sharp point. High wear, used for low-speed.

  • Roller: Reduces friction. Most common.

  • Flat-faced: Large contact area, used for heavy loads.

  • Mushroom/Spherical: Self-aligning.

Follower Motion Curves (Displacement Diagrams)

For lift $h$, cam angle $\theta$, total angle for motion $\beta$.

• Simple Harmonic Motion (SHM):

$$s = \frac{h}{2}\left(1 - \cos\frac{\pi\theta}{\beta}\right)$$

$$v_{max} = \frac{\pi h \omega}{2\beta}, \quad a_{max} = \frac{\pi^2 h \omega^2}{2\beta^2}$$

at $$\displaystyle \theta = \beta/2 $$.

• Uniform Velocity:

$$s = \frac{h\theta}{\beta}$$

$$v = \text{constant} = \frac{h\omega}{\beta}, \quad a = 0 \text{ (infinite at start/end - not practical)}$$

• Uniform Acceleration & Deceleration (with dwells):

  • Ascent/Descent split equally into accel ($\beta/2$) and decel ($\beta/2$).

  • For ascent: $0 \le \theta \le \beta/2$: $$\displaystyle s = \frac{2h}{\beta^2} \theta^2 $$

  • $\beta/2 \le \theta \le \beta$: $$\displaystyle s = h - \frac{2h}{\beta^2} (\beta - \theta)^2 $$

  • $$\displaystyle v_{max} = \frac{2h\omega}{\beta} $$ at midpoint.

  • $$\displaystyle a_{max} = \frac{4h\omega^2}{\beta^2} $$ constant during accel/decel.

Cam Design Parameters

• Base Circle: Smallest circle that can be drawn tangent to the cam profile. Radius $$\displaystyle r_b $$.

• Pitch Circle: Circle on which the follower motion is referenced. Radius $$\displaystyle r_p = r_b + \frac{h}{2} $$ for knife-edge.

• Pressure Angle ($\phi$): Angle between the direction of follower motion and the normal to the cam profile at the point of contact.

  • Significance: High $\phi$ increases follower side thrust and wear.

  • Maximum Permissible: $30°$ for translating followers, $35°$ for oscillating (often $25°-30°$ for high speed).

• Stroke of Follower: Maximum upward or downward movement ($h$).

Cam Profile Construction (Radial Cam, Knife-Edge Follower)

1. Draw base circle of radius $$\displaystyle r_b $$.

2. Divide cam rotation ($360°$) and follower displacement diagram into same number of divisions.

3. From each division angle on base circle, draw radial line.

4. Along each radial line, mark displacement $s$ from base circle. These are pitch points.

5. Join pitch points with a smooth curve → pitch curve.

6. Draw offset curve (normal to pitch curve) at distance equal to knife-edge radius (≈0) → cam profile.

[!TIP] For roller follower, pitch curve is offset by roller radius $$\displaystyle r_r $$ to get prime circle ($$\displaystyle r_p = r_b + r_r $$). Cam profile is drawn tangent to roller circles at pitch points.

[!TIP] For flat-faced follower, draw normal to pitch curve at contact point and mark follower width along it.

Undercutting

• Cause: When the cam radius is too small, the pitch curve has a concave portion. The follower may lose contact or the cam profile becomes impossible to manufacture (undercut).

• Effect: Reduces strength, causes jerky motion.

• Prevention: Increase base circle radius, use roller follower (larger prime circle), or use a different motion curve.

Critical Path Motion & Torque on Cam Shaft

• Critical Path Motion: The motion (usually acceleration) that causes the maximum force on the follower, hence maximum torque on cam shaft.

• Torque Calculation:

$$T_{cam} = F_{follower} \times r_p \cdot \sin\phi$$

where $$\displaystyle F_{follower} $$ includes inertia force ($m a$) and any external load. Maximum torque occurs at peak acceleration or when pressure angle is max.

VI. GEARS AND GEAR TRAINS (HIGH FREQUENCY)

Law of Gearing

• Statement: For two gears in mesh, the common normal at the point of contact must always pass through a fixed point on the line of centers (the pitch point) to achieve a constant velocity ratio.

• Derivation:

  Let gears 1 and 2 have centers $$\displaystyle O_1, O_2 $$, radii $$\displaystyle r_1, r_2 $$. Point of contact $P$.

  Velocity of $P$ on gear 1: $$\displaystyle v_P = \omega_1 r_1 $$ (perpendicular to $$\displaystyle O_1P $$).

  Velocity of $P$ on gear 2: $$\displaystyle v_P = \omega_2 r_2 $$ (perpendicular to $$\displaystyle O_2P $$).

  For no slip, these velocities must be equal and opposite along the common normal.

$$\frac{\omega_1}{\omega_2} = \frac{r_1}{r_2} = \text{constant}$$

This requires that the common normal at $P$ passes through the pitch point $C$ on $$\displaystyle O_1O_2 $$ where $$\displaystyle O_1C / O_2C = r_1/r_2 $$.

• Conjugate Gears: Gear profiles that satisfy the law of gearing. Involute profile is the most common conjugate profile because it satisfies the law for all positions of contact along the line of action.

Involute Gear Profile

• Generation: Unwinding a taut string from a base circle. The path traced by the string's end is the involute.

• Properties:

  • Satisfies law of gearing (common normal is tangent to base circles).

  • Center distance can vary slightly without changing velocity ratio (interchangeability).

  • Pressure angle $\phi$ is constant (angle between line of action and tangent to pitch circle).

• Standard Pressure Angle: $20°$ (formerly $14.5°$).

Spur Gears (Involute, Full-Depth Teeth)

• Key Parameters (Module $m$, $N$ = number of teeth):

  • Pitch Circle Diameter: $$\displaystyle D = m N $$

  • Base Circle Diameter: $$\displaystyle D_b = D \cos\phi $$

  • Addendum: $$\displaystyle a = m $$ (standard)

  • Dedendum: $$\displaystyle b = 1.25m $$ (standard)

  • Whole Depth: $$\displaystyle h = 2.25m $$

• Arc of Contact ($$\displaystyle \epsilon_\alpha $$): Length of arc on pitch circle through which a tooth remains in contact.

$$\epsilon_\alpha = \frac{\text{Path of contact}}{\cos\phi}$$

*   **Path of Contact**: Distance along line of action from start of engagement (addendum circle of driver) to end (addendum circle of driven).

$$\text{Path of contact} = \sqrt{r_{a1}^2 - r_{b1}^2} + \sqrt{r_{a2}^2 - r_{b2}^2} - (r_{b1} + r_{b2})\sin\phi$$

where $$\displaystyle r_a = r + a $$, $$\displaystyle r_b = r \cos\phi $$.

• Contact Ratio ($\epsilon$): Average number of teeth in contact.

$$\epsilon = \frac{\text{Arc of contact}}{\text{Circular pitch}} = \frac{\text{Path of contact}}{\pi m \cos\phi}$$

*   Must be $$\displaystyle > 1 $$ for continuous transmission. Typical: $$\displaystyle 1.2 < \epsilon < 2.0 $$.

• Interference:

  • Cause: When the addendum tip of the driven gear tooth undercuts the flank of the driving gear tooth before engagement ends. Happens when the number of teeth on the smaller gear (pinion) is too low.

  • Minimum Teeth to Avoid Interference (standard full-depth, $$\displaystyle a=m $$):

$$N_{min} = \frac{2a}{\sin^2\phi} \left( \sqrt{1 + \frac{h_a}{a} \sin^2\phi} - 1 \right) \approx \frac{2a}{\sin^2\phi}$$

    For $$\displaystyle \phi=20° $$, $$\displaystyle a=m $$: $$\displaystyle N_{min} \approx 18 $$ (pinion).

*   **Prevention**: **Addendum Modification (Crowning)**: Reduce pinion addendum and increase gear addendum. Increases $$\displaystyle N_{min} $$ effectively.

Classification of Gears

Gear Trains

• Simple Gear Train: One gear per shaft. Speed ratio = $$\displaystyle -\frac{N_2}{N_1} $$ (negative for opposite rotation).

• Compound Gear Train: Two gears on same intermediate shaft. Speed ratio = product of individual ratios.

• Epicyclic (Planetary) Gear Train:

  • Construction: Sun gear (center), planet gears (meshing with sun and ring), carrier (holds planets), ring gear (internal teeth).

  • Fundamental Equation (for one planet, ignoring sign):

$$\frac{\omega_s - \omega_a}{\omega_p - \omega_a} = (-1)^n \frac{N_p}{N_s}$$

    where $$\displaystyle \omega_s, \omega_p, \omega_a $$ = angular velocities of sun, planet, arm. $n$ = number of idlers (0 or 1).

*   **Tabular Method**: More reliable for complex trains. Write $$\displaystyle \omega_{relative\ to\ arm} = \omega - \omega_a $$ for each shaft. Then apply simple train formula to relative speeds.

*   **Special Advantages**:

    1.  High speed reduction in compact space.

    2.  Co-axial input/output possible.

    3.  Multiple outputs possible from one input.

    4.  Can combine motions (addition/subtraction of speeds).

VII. POWER TRANSMISSION SYSTEMS

Belt Drives

• Flat Belts: Used for large center distances, high speed. Friction coefficient lower.

• V-Belts: Wedge action increases normal force → higher friction. More compact, no slip.

• Open vs Cross Belt Drive:

  • Open: Pulleys rotate in same direction. Angle of contact on smaller pulley: $$\displaystyle \theta = 180° - 2\alpha $$, where $$\displaystyle \alpha = \sin^{-1}\left(\frac{D-d}{2C}\right) $$.

  • Cross: Pulleys rotate in opposite directions. Angle of contact on each pulley: $$\displaystyle \theta = 180° + 2\alpha $$.

  • Tension Ratio:

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

(Euler's formula, $\theta$ in radians)

    where $$\displaystyle T_1 $$ = tight side tension, $$\displaystyle T_2 $$ = slack side tension.

• Slip vs Creep:

  • Slip: Relative motion between belt and pulley due to insufficient friction (global, $$\displaystyle S\% = \frac{v_1 - v_2}{v_1} \times 100 $$).

  • Creep: Local stretching of belt as it passes from slack to tight side (elastic deformation). Always occurs.

• Power Transmission:

$$P = (T_1 - T_2) v$$

**Condition for Maximum Power**:

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

Maximum $$\displaystyle P_{max} = T_2 v (e^{\mu\theta} - 1) $$ occurs when $$\displaystyle T_2 = \frac{T}{e^{\mu\theta} + 1} $$, $$\displaystyle T_1 = \frac{T e^{\mu\theta}}{e^{\mu\theta} + 1} $$, where $$\displaystyle T = T_1 + T_2 $$ (maximum allowable tension).

Also, centrifugal tension $$\displaystyle T_c = m v^2 $$ reduces effective tension. For max power considering $$\displaystyle T_c $$, $$\displaystyle T_1 - T_2 = \frac{T}{3} $$ when $$\displaystyle \mu\theta = 1 $$.

• Stress in Belt:

$$\sigma_{max} = \frac{T_1}{b t} \quad \text{(Open drive)}$$

For **cross drive**, due to twist, effective tension is $$\displaystyle T_1 + T_2 $$? Actually, stress calculation uses same formula but with $$\displaystyle T_1 $$ from tension ratio. However, cross belt experiences additional stress due to bending? Standard approach: compute $$\displaystyle T_1, T_2 $$ from Euler's formula with appropriate $\theta$, then $$\displaystyle \sigma_{max} = T_1/(b t) $$.

Clutches

• Classification:

  • Plate Clutch: Single plate, multi-plate (increases torque for given size).

  • Cone Clutch: Conical surfaces, self-centering.

  • Centrifugal Clutch: Engages at high speed via centrifugal force.

• Torque Transmission Derivation:

  • Cone Clutch:

    Consider elemental ring at radius $r$, width $dr$.

    Normal force $$\displaystyle dN = p \cdot 2\pi r \cdot \frac{dr}{\sin\alpha} $$ ($\alpha$ = cone angle).

    Friction force $$\displaystyle dF = \mu dN $$.

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

    Uniform Pressure ($$\displaystyle p = constant $$):

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

    Axial force $$\displaystyle W = \frac{\pi p}{\sin\alpha} (r_2^2 - r_1^2) $$.

    Eliminate $p$: 

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

*   **Plate Clutch**:

    *   **Uniform Pressure** ($$\displaystyle p = constant $$):

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

    *   **Uniform Wear** ($$\displaystyle p r = constant = c $$):

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

        (Derivation: $$\displaystyle dN = c \cdot 2\pi dr $$, $$\displaystyle dT = \mu c \cdot 2\pi r dr $$, integrate; $$\displaystyle W = \int dN = 2\pi c (r_2 - r_1) $$).

Brakes

• Band and Block Brake:

  • Construction: Flexible band with wooden blocks, wrapped on drum. One end fixed, other pulled by lever.

  • Tension Ratio Derivation:

    Consider one block subtending angle $2\theta$ at center. Tensions $$\displaystyle T_1 $$ (tight), $$\displaystyle T_2 $$ (slack) on either side.

    Equilibrium of block: $$\displaystyle T_1 - T_2 = \mu N $$, $$\displaystyle N = T_1 + T_2 $$ (resolving radially, assuming $\theta$ small).

$$\frac{T_1}{T_2} = \frac{1 + \mu\tan\theta}{1 - \mu\tan\theta}$$

    For $n$ blocks, ratio multiplies: $$\displaystyle \left(\frac{T_1}{T_2}\right)^n $$.

*   **Braking Torque**: $$\displaystyle T_b = (T_1 - T_2) R $$, where $R$ = drum radius.

• Disc & Drum Brakes: Disc brakes have better heat dissipation, less fade. Drum brakes self-energizing (self-servo).

Friction Devices

• Pivot & Collars:

  • Torque for uniform pressure: $$\displaystyle T = \frac{\mu W}{2} \frac{r_2^3 - r_1^3}{r_2^2 - r_1^2} $$ (similar to cone clutch).

  • Torque for uniform wear: $$\displaystyle T = \frac{\mu W}{2} (r_2 + r_1) $$.

  • Power Loss: $$\displaystyle P_{loss} = T \omega $$.

• Power Screws (Traverse Screws):

  • Torque to Raise Load:

$$T_{raise} = W \cdot \frac{d_m}{2} \left( \tan\lambda + \tan\phi \right)$$

    where $$\displaystyle d_m $$ = mean diameter, $\lambda$ = helix angle ($$\displaystyle \tan\lambda = \frac{p}{\pi d_m} $$), $\phi$ = friction angle ($$\displaystyle \tan\phi = \mu $$).

*   **Torque to Lower Load**:

$$T_{lower} = W \cdot \frac{d_m}{2} \left( \tan\lambda - \tan\phi \right)$$

*   **Efficiency**:

$$\eta_{raise} = \frac{\tan\lambda}{\tan(\lambda + \phi)}$$

$$\eta_{lower} = \frac{\tan\lambda}{\tan(\lambda - \phi)}$$

*   **Self-locking**: Occurs when $$\displaystyle \tan\lambda < \tan\phi $$ → $$\displaystyle T_{lower} > 0 $$ (load won't back-drive).

VIII. VIBRATIONS AND BALANCING (HIGH FREQUENCY)

Free Vibrations

• Longitudinal Vibration: Particles move along the axis of the member (e.g., spring-mass system along a rod). Frequency $$\displaystyle f = \frac{1}{2\pi}\sqrt{\frac{k}{m}} $$.

• Transverse Vibration: Particles move perpendicular to the axis (e.g., beam with mass, string). Frequency depends on boundary conditions and EI.

• Torsional Vibration: Particles rotate about the axis (twisting). Frequency $$\displaystyle f = \frac{1}{2\pi}\sqrt{\frac{GJ}{I \cdot l}} $$ for uniform shaft.

Balancing of Rotating Masses

• Static Balancing: Masses in a single plane. Condition: $$\displaystyle \Sigma m r = 0 $$ (vector sum of centrifugal forces zero). Required when mass centers lie in one plane.

• Dynamic Balancing: Masses in multiple planes. Conditions:

  1. $$\displaystyle \Sigma m r = 0 $$ (force balance)

  2. $$\displaystyle \Sigma m r d = 0 $$ (moment balance about any plane), where $d$ = distance from reference plane.

  • Necessary when mass centers are in different axial planes (couples exist).

• Two-Plane Balancing (e.g., four masses in different planes):

  • Choose two balancing planes (e.g., $X$ and $Y$).

  • Write force polygon equations: $$\displaystyle \Sigma m r \cos\theta = M_x r_x \cos\theta_x + M_y r_y \cos\theta_y $$, $$\displaystyle \Sigma m r \sin\theta = M_x r_x \sin\theta_x + M_y r_y \sin\theta_y $$.

  • Write moment equations about one plane (e.g., $X$): $$\displaystyle \Sigma m r d = M_y r_y d_{yx} $$.

  • Solve for $$\displaystyle M_x, M_y $$ and their angles.

  [!TIP] For given masses $$\displaystyle m_i $$ at radii $$\displaystyle r_i $$ and angles $$\displaystyle \theta_i $$, in planes at distances $$\displaystyle d_i $$ from a reference, with balance masses $$\displaystyle M_1, M_2 $$ at $$\displaystyle (r_1, \theta_1) $$ in plane 1 and $$\displaystyle (r_2, \theta_2) $$ in plane 2, set up:

  • $$\displaystyle \Sigma m_i r_i \cos\theta_i = M_1 r_1 \cos\theta_1 + M_2 r_2 \cos\theta_2 $$

  • $$\displaystyle \Sigma m_i r_i \sin\theta_i = M_1 r_1 \sin\theta_1 + M_2 r_2 \sin\theta_2 $$

  • $$\displaystyle \Sigma m_i r_i d_i = M_2 r_2 d_{2} $$ (taking moments about plane 1, where $$\displaystyle d_2 $$ = distance between planes 1 and 2).

  Solve sequentially.

\boxed{\text{END OF UNIT 5 SHORT NOTES}}