UNIT 2: THEORY OF MACHINES - EXAM-FOCUSED SHORT NOTES

I. FUNDAMENTALS OF MECHANISMS & KINEMATICS

1.1 Definitions & Basic Concepts

• Machine: A combination of resistant bodies with relative motion to transmit force and do work. Example: Engine, lathe.

• Mechanism: A kinematic chain with one link fixed, used to transmit motion only. Example: Slider-crank, four-bar.

• Kinematic Link: A resistant body that forms part of a machine and has relative motion with respect to another link.

• Kinematic Pair: Two links in contact with relative motion. Classified by:

  • Nature of Contact: Lower pair (surface contact, e.g., pin joint), Higher pair (line/point contact, e.g., cam-follower, gear teeth).

  • Relative Motion: Sliding pair, Turning pair, Rolling pair, Helical pair.

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

• Degree of Freedom (DoF) / Mobility: The number of independent inputs required to completely specify the mechanism's configuration.

  • Kutzbach's Criterion (Plane Mechanisms):

$$ \boxed{DoF = 3(n - 1) - 2j_1 - j_2} $$

    *   `n` = Total number of links (including frame).

    *   `j₁` = Number of pairs with 1 DoF (lower pairs).

    *   `j₂` = Number of pairs with 2 DoF (higher pairs).

*   **Gruebler's Criterion (Plane Mechanisms, all lower pairs):**

$$ \boxed{DoF = 3(n - 1) - 2j} \quad (\text{where } j_2 = 0) $$

    *   Often written as: `DoF = 3(n-1) - 2j` for `j` total binary joints.

    > [!TIP] **Common Pitfall:** Forgetting to include the **fixed link (frame)** in `n`. A mechanism must have **DoF = 1** to be a useful mechanism.

• Constrained Motion: Motion with restrictions.

  • Completely Constrained: Unique motion for given input (e.g., piston in cylinder).

  • Partially Constrained: More than one motion possible for a given input (e.g., shaft in footstep bearing - can rotate and float).

  • Incompletely Constrained: Less motion than intended (e.g., square bar in square hole - can't rotate).

1.2 Four-Bar Linkage & Grashof's Law

• Grashof's Law: For a four-bar chain with links of lengths s (shortest), l (longest), p, q:

  If s + l < p + q, then at least one link can rotate continuously relative to the others.

  If s + l > p + q, then no link can make a full rotation (double rocker only).

  If s + l = p + q, it's a change point mechanism (special case).

• Classification (based on which link is fixed):

  | Fixed Link | Grashof Condition (s+l < p+q) | Mechanism Type | | :--- | :--- | :--- | | Shortest (s) | Yes | Double Crank (Drag Link) | | Longest (l) | Yes | Crank Rocker | | Any other (p or q) | Yes | Double Rocker | | Any | No (s+l > p+q) | Double Rocker |

  Example (Dec 2024): Given O1O2=50mm, O1A=62mm, AB=37mm, O2B=68mm. Sorted: s=37 (AB), l=68 (O2B), p=50, q=62. s+l = 105, p+q=112. Since 105 < 112, Grashof satisfied.

  • If AB (s) is fixed → Double Rocker.

  • If AB (s) is crank → Not possible as fixed link must be longest or a coupler for crank-rocker/double-crank.

1.3 Inversions of Four-Bar Mechanism

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

  • 1st Inversion: Fix link AD (ground). → Crank-Rocker/Double-Crank/Double-Rocker (standard four-bar).

  • 2nd Inversion: Fix link AB (crank). → Watt's Indicator Mechanism (used in engines to indicate piston motion).

  • 3rd Inversion: Fix link BC (coupler). → Quick Return Mechanisms (e.g., Whitworth, Crank & Slotted Lever).

  • 4th Inversion: Fix link CD (rocker). → Hoecken's Straight Line Motion (approximate straight line).

1.4 Special Mechanisms & Quick Return Motions

• Whitworth Quick Return Mechanism (3rd Inversion of 4-bar):

  • Construction: Crank OA (driver) → Slotted lever O1C (rocker) → Slider R (follower).

  • Time Ratio (TR): TR = (Time of cutting stroke) / (Time of return stroke).

  • Key Formula:

$$ \boxed{TR = \frac{180^\circ + \psi}{180^\circ - \psi}} $$

    where `ψ` = angle through which slotted lever oscillates (cutting stroke angle).

*   **Design:** Given `TR`, `stroke (L)`, `crank length (r)`. Find distance `O1O2` (between crank center & slotted lever pivot) and `O1C` (lever length).

    > **Steps:** Use geometry in extreme positions to relate `L`, `r`, `O1O2`, `O1C`, and `ψ`.

• Crank and Slotted Lever Mechanism (Also 3rd Inversion):

  • Similar to Whitworth but the slotted lever is the driver and crank is the follower. Used in shaping machines.

• Double Slider Crank Mechanism:

  • Chain: Two sliding pairs and two turning pairs.

  • Inversions:

    1. 1st Inversion (Fixed link 1): Scotch Yoke Mechanism (Link 2 is yoke, Link 4 is crank). Converts rotary to reciprocating motion with simple harmonic displacement.

    2. 2nd Inversion (Fixed link 2): Oldham's Coupling (Link 1 & 3 are shafts, Link 4 is intermediate member with tongues). Transmits motion between non-collinear shafts with constant velocity ratio.

    3. 3rd Inversion (Fixed link 3): Hand Pump / Elliptical Trammel (Used for drawing ellipses).

II. KINEMATIC ANALYSIS (Velocity & Acceleration)

2.1 Relative Motion & Reference Frames

• Absolute Motion: Motion observed from a fixed (inertial) reference frame.

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

• Fundamental Equation: For two points A and B on a rigid body:

$$ \vec{v}_B = \vec{v}_A + \vec{v}_{B/A} \quad \text{(Velocity)} $$

$$ \vec{a}_B = \vec{a}_A + \vec{a}_{B/A} + \vec{\alpha} \times \vec{r}_{B/A} + \vec{\omega} \times (\vec{\omega} \times \vec{r}_{B/A}) \quad \text{(Acceleration)} $$

*   `v_B/A` = Relative velocity of B w.r.t A (due to rotation).

*   `α × r_B/A` = **Tangential acceleration**.

*   `ω × (ω × r_B/A)` = **Centripetal (normal) acceleration**.

2.2 Instantaneous Center (IC) Method

• Definition: The point in a plane which is instantaneously at rest (zero velocity) relative to the fixed plane.

• Significance: Velocity of any point on a rigid body is perpendicular to the line joining that point to the IC, and magnitude v = ω * r.

• Location of ICs (Kennedy's Theorem): For three bodies (1,2,3) in plane motion, the three ICs (I₁₂, I₁₃, I₂₃) are collinear.

• Procedure for Velocity Analysis:

  1. Identify all instantaneous centers in the mechanism.

  2. Known velocity of one point (e.g., crank pin) → Find ω_crank = v / r.

  3. Velocity of any other point = ω_crank * (distance from that point to IC of the link containing the known point).

  Example (Slider-Crank): IC of connecting rod (link 3) is at the intersection of extensions of crank (link 2) and slider path (link 1). Velocity of piston = ω_crank * (IC to piston distance).

2.3 Analytical & Graphical Methods

• Complex Algebra / Loop Closure Equation: For a loop 1-2-3-...-1:

$$ \sum \vec{r} = 0 \quad \Rightarrow \quad \sum (r e^{i\theta}) = 0 $$

Separate into real & imaginary parts to solve for unknown positions, velocities (`\dot{\theta}`), accelerations (`\ddot{\theta}`).

• Graphical Velocity Polygon:

  1. Draw velocity diagram to scale.

  2. Use v = ωr for known rotations.

  3. Use v_B = v_A + v_{B/A} (perpendicular to AB).

  4. Complete the polygon to find unknown velocities.

• Graphical Acceleration Polygon:

  1. Draw acceleration diagram.

  2. Components: a_t = αr (⊥ to link), a_n = ω²r (towards IC).

  3. Use a_B = a_A + a_{B/A} + α×r + ω×(ω×r).

  4. Construct using Coriolis component if point moves on a moving link.

2.4 Dynamic Analysis: D'Alembert's Principle

• Statement: The inertia forces (and torques) acting on a body, together with the external forces (and torques), form a system in dynamic equilibrium.

$$ \vec{F} - m\vec{a} = 0 \quad \text{and} \quad \vec{T} - I\vec{\alpha} = 0 $$

where `-ma` is the **inertia force** and `-Iα` is the **inertia torque**.

• Application to Slider-Crank:

  1. For each link (crank, connecting rod, piston), draw free body diagram.

  2. Introduce inertia force F_i = -m a_G at C.G. and inertia torque T_i = -I_G α (or -I_O α if rotating about fixed axis).

  3. Apply static equilibrium equations (ΣF_x=0, ΣF_y=0, ΣM=0) to solve for unknown forces/torques.

  Key: Inertia force acts opposite to acceleration of C.G. For a rotating link about a fixed axis, inertia torque acts opposite to angular acceleration.

2.5 Motion Synthesis vs. Analysis

• Motion Analysis: Given mechanism geometry & input motion → Find output motion (velocity, acceleration, forces).

• Motion Synthesis: Given prescribed output motion → Find mechanism geometry (link lengths, angles).

  • Number Synthesis: Determine n, j, DoF (type of mechanism needed).

  • Kinematic Synthesis:

    • Function Generation: Input-output relationship (angle-angle).

    • Path Generation: Point on output link follows a prescribed path.

    • Motion Generation: Output link achieves prescribed positions, velocities, accelerations.

• Freudenstein's Equation (Four-Bar Function Generation):

  For prescribed angles θ₂ (input) and θ₄ (output):

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

where:

$$ K_1 = \frac{d_2}{a}, \; K_2 = \frac{d_2}{a}, \; K_3 = \frac{a^2 - b^2 + c^2 + d^2}{2ac}, \; K_4 = \frac{b^2 + c^2 - d^2}{2ac} $$

*   `a, b, c, d` are link lengths (crank, coupler, rocker, fixed).

*   For **three positions**, solve three equations for `K₁, K₂, K₃`.

*   For **velocity/acceleration synthesis**, differentiate Freudenstein's equation w.r.t. time.

III. CAMS & FOLLOWER SYSTEMS

3.1 Classification of Cams & Followers

• By Cam Shape:

  • Disc/Rotary: Most common. Cam rotates about a fixed axis.

  • Cylindrical: Cam is a cylinder, follower moves parallel to cam axis.

  • Linear/Translating: Cam translates.

• By Follower Type:

  • Knife Edge: Sharp point. High wear, undercutting likely.

  • Roller: Reduces friction & wear. Increases cam size.

  • Flat-Faced/Mushroom: Large contact area, good for high-speed, heavy loads. Pressure angle must be small.

• By Follower Motion:

  • Radial: Line of stroke passes through cam axis.

  • Offset: Line of stroke does not pass through cam axis. Causes swinging motion.

• By Constraint:

  • Rise: Follower moves away from cam.

  • Fall/Dwell: Follower stationary or returns.

  • Dwell: Period of no motion.

3.2 Cam Terminology & Geometry

• Base Circle: Smallest circle that can be drawn tangent to the cam profile. Radius = R_b.

• Pitch Circle: Circle on which follower motion is defined (for roller follower, it's the circle through roller centers). Radius = R_p.

• Prime Circle: For flat-faced followers, the circle on which the pitch point lies. R_p = R_b + h/2 (h = lift).

• Pressure Angle (φ): Angle between the direction of follower motion and the normal to the pitch curve at the point of contact.

  • Critical for design: High φ → high lateral force → follower jamming. Limit: φ_max ≤ 30° (industry), ≤ 35° (high-speed).

• Stroke/Lift (h): Maximum distance the follower moves from its lowest to highest position.

• Angle of Action: Cam rotation angle for a specific motion (rise, fall, dwell).

• Undercutting: Occurs when the pitch curve radius becomes too small during design, causing the cam profile to have a cusp or concave portion where the follower cannot be manufactured. Prevented by: Increasing base circle radius, using roller follower, or using a different motion curve.

3.3 Follower Motion Curves & Design

• Standard Displacement Diagrams (θ vs. s):

  | Motion | Displacement (s) | Max Velocity (v_max) | Max Acceleration (a_max) | Characteristics | | :--- | :--- | :--- | :--- | :--- | | UVM (Uniform Velocity) | Linear | h / (θ_r/ω) | Infinite at start/end | Not used (shock) | | UAM (Uniform Accel/Decel) | Parabolic (2 halves) | 2h / (θ_r/ω) | 4hω² / θ_r² | Constant a, discontinuity in a at junction | | SHM (Simple Harmonic) | s = (h/2)[1 - cos(πθ/θ_r)] | πhω / (2θ_r) | (π²hω²)/(2θ_r²) | Smooth v & a, but high a at mid-point | | Cycloidal | s = h[θ/θ_r - (1/(2π)) sin(2πθ/θ_r)] | 2πhω / θ_r | 2π²hω² / θ_r² | Best: Zero a at start/end, smooth curves. |

  • θ_r = angle for rise/fall (radians), ω = cam angular speed (rad/s).

  • For descent (fall), same formulas apply but h is negative.

• Cam Profile Drawing (Radial Follower, Knife-Edge):

  1. Draw base circle (radius R_b).

  2. Draw s-axis diagram (displacement vs. cam angle) on a prime circle of radius R_p = R_b + h_max (for knife-edge, R_p = R_b).

  3. Divide the s-axis into intervals corresponding to motion segments (rise, dwell, fall, dwell).

  4. Transfer displacement s from s-axis diagram to radial lines on the prime circle.

  5. Draw offset circle of radius s from each radial line.

  6. The envelope of these offset circles is the cam profile.

  For Roller Follower: Use pitch circle (R_p = R_b + roller radius). The offset is along the normal to the pitch curve, not radial.

3.4 Critical Path Motion & Torque on Camshaft

• Critical Path Motion: The follower motion that requires the maximum force from the cam, typically during the rise when accelerating against spring force and inertia.

• Torque on Camshaft:

$$ T_{cam} = F_{follower} \times (R_p \sin \phi) $$

where `F_follower` is the total force on follower (spring force + inertia force + friction).

*   Torque is **maximum** when `φ` is maximum (usually at the beginning/end of rise/fall).

*   **Design Implication:** Camshaft must be sized for this maximum torque.

IV. GEAR TRAINS & GEAR GEOMETRY

4.1 Fundamentals of Gearing

• Law of Gearing: "The common normal at the point of contact of two gear teeth must always pass through a fixed point on the line of centers, called the pitch point, to give a constant velocity ratio."

  • Proof: For constant ω₁/ω₂, the instantaneous center of rotation must be on the line of centers. This IC is the pitch point P. The common normal must pass through P.

  • Conjugate Gears: Gear pairs that satisfy the law of gearing. Involute profile is the most common conjugate profile.

• Involute Gear Tooth Profile:

  • Generation: Unwinding a taut string from a base circle.

  • Properties:

    1. Constant velocity ratio (satisfies law of gearing).

    2. Interference: Occurs when the addendum circle of the pinion extends beyond the limit of contact on the gear, causing non-involute contact. Prevented by: Increasing number of teeth on pinion (N_p ≥ N_min), using profile shift (addendum modification).

    3. Undercutting: To avoid interference, standard full-depth involute gears with less than minimum teeth are cut with a undercut at the root, weakening the tooth.

• Pressure Angle (φ): Standard values: 14.5° (old), 20° (modern, better efficiency, less undercut).

  • Base Circle Radius: R_b = R_p cos φ (where R_p is pitch circle radius).

4.2 Gear Terminology & Calculations

• Module (m): m = D / N (mm). Standardized (e.g., 1, 2, 2.5, ... 50 mm).

• Circular Pitch (p_c): p_c = πm (distance along pitch circle between corresponding points on adjacent teeth).

• Diametral Pitch (P_d): P_d = N / D (teeth per inch) - Imperial unit.

• Addendum (a): Radial distance from pitch circle to top of tooth. a = m (standard full-depth).

• Dedendum (b): Radial distance from pitch circle to root of tooth. b ≈ 1.25m.

• Tooth Thickness: t = p_c / 2 = πm/2 (on pitch circle).

• Path of Contact (L): Length of common normal between start of contact (at addendum circle of driver) and end of contact (at addendum circle of driven).

• Arc of Contact (Arc): Length of arc on pitch circle between start and end of contact.

$$ \text{Arc} = \sqrt{r_{a1}^2 - r_{b1}^2} + \sqrt{r_{a2}^2 - r_{b2}^2} - (r_{b1} \tan \phi - r_{b2} \tan \phi) $$

where `r_a`, `r_b` are addendum and base circle radii.

• Contact Ratio (ε): Average number of teeth in contact.

$$ \boxed{\varepsilon = \frac{\text{Path of Contact}}{\text{Circular Pitch}} = \frac{L}{p_c}} $$

*   `ε > 1` ensures smooth transmission (at least one pair always in contact).

*   **Minimum requirement:** `ε ≥ 1.2` for smooth operation.

• Interference & Minimum Teeth:

  • Minimum Teeth on Pinion to Avoid Interference:

$$ \boxed{N_{p(min)} = \frac{2a}{\sqrt{1 - (N_2 \sin^2 \phi / N_1^2)} - 1}} \quad \text{or approximate:} \quad N_{p(min)} = \frac{2a}{\sin^2 \phi} $$

    where `a = addendum / module = 1` (standard).

*   For `φ=20°`, `N_p(min) ≈ 17` (standard full-depth). With **profile shift**, `N_p(min)` can be reduced.

4.3 Classification of Gears

4.4 Gear Trains

• Simple Gear Train: Series of gears. Speed ratio:

$$ \frac{\omega_1}{\omega_n} = (-1)^{n-1} \frac{T_2 T_4 ... T_n}{T_1 T_3 ... T_{n-1}} $$

*   `(-1)^{n-1}` indicates direction reversal for odd number of idlers.

• Compound Gear Train: Each shaft carries two gears. Speed ratio:

$$ \frac{\omega_1}{\omega_4} = \frac{T_2 T_4}{T_1 T_3} \quad \text{(for 2-stage)} $$

*   Used for **higher reductions** in less space.

• Epicyclic (Planetary) Gear Train:

  • Elements: Sun gear (S), Planet gears (P) on a carrier/arm (A), Ring gear (R) (internal teeth).

  • Key: At least one element is fixed or has known motion.

  • Tabular (Arm/Stationary) Method:

    1. Fix the arm (make it stationary). Convert epicyclic to simple train.

    2. Calculate speed ratios assuming arm fixed: ω_S / ω_R = -T_R / T_S.

    3. Add arm speed ω_A to all speeds (since we "unfixed" it).

    4. Apply constraints (e.g., ω_R=0 for fixed ring) to solve.

    Example (Dec 2024): Ring fixed (ω_R=0), ω_S known, find ω_A. From step 2: (ω_S - ω_A)/(ω_R - ω_A) = -T_R/T_S. Substitute ω_R=0 and solve for ω_A.

  • Advantages: High reduction in compact space, coaxial shafts, multiple outputs possible.

  • Synthesis: Given gear ratio i = ω_S/ω_A (with ring fixed), find teeth numbers.

    • Condition: (N_S + 2N_P) = N_R (for meshing).

    • From velocity ratio: i = 1 + N_R/N_S.

    • Choose N_S, then N_R = N_S(i-1), N_P = (N_R - N_S)/2.

V. POWER TRANSMISSION ELEMENTS

5.1 Belt Drives

• Types: Flat, V-Belt (wedge effect, higher μ), Rope.

• Open vs. Cross Belt: Cross belt reverses direction, needs twist.

• Kinematics:

  • Velocity Ratio (Ideal, no slip): N₁D₁ = N₂D₂ or ω₁D₁ = ω₂D₂.

  • Considering Belt Thickness: N₁(D₁ + t) = N₂(D₂ + t).

  • Slip (s%): Actual speed reduction. N₂(actual) = N₁(D₁/D₂)(1 - s/100).

  • Creep: Local deformation at contact, similar effect to slip.

• Analysis (Flat Belt):

  • Euler's Equation (Tension Ratio):

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

    *   `T₁` = Tight side tension, `T₂` = Slack side tension.

    *   `μ` = coefficient of friction.

    *   `θ` = angle of lap (radians) on smaller pulley.

    *   For **cross belt**, `θ = π + 2α` (α = sin⁻¹((D₁-D₂)/(2C))).

*   **Power Transmitted:**

$$ \boxed{P = (T_1 - T_2) \times v} $$

    where `v = π D₁ N₁ / 60` (m/s).

*   **Maximum Power Transmission:**

    *   Condition: `T₁ / T₂ = e^{μθ}` and `T₁ = σ_max × b × t` (σ_max = max allowable stress).

    *   **Optimum Tension Ratio for Max Power:** `T₁/T₂ = 3` (when `T₁ = 3T₂` and `T₁ = σ_max bt`).

    *   **Max Power:** `P_max = (2/3) T₁ v` (when `T₁=3T₂`).

*   **Stress in Belt:**

    1.  **Tension Stress:** `σ_t = T / (b t)`.

    2.  **Centrifugal Tension:** `T_c = m v²` (m = mass/length).

    3.  **Total Maximum Tension:** `T₁ = T (tight) + T_c`.

    4.  **Effective Tension for Power:** `T₁ - T₂ = T (e^{μθ} - 1) / e^{μθ}` (where `T` is initial tension).

*   **Numerical Approach:** Given `P`, `v`, `μ`, `θ`, `σ_max`, `b`, `t`:

    1.  Find `T₁ - T₂ = P/v`.

    2.  Use `T₁/T₂ = e^{μθ}` to find `T₁` and `T₂`.

    3.  Check `T₁ ≤ σ_max b t` (open) or `T₁ + T_c ≤ σ_max b t` (cross, considering centrifugal tension adds to both sides? Actually, centrifugal tension acts radially, but for stress calculation on belt section, it's `T + T_c`).

    4.  For **cross belt**, centrifugal tension `T_c` adds to **both** `T₁` and `T₂`? Actually, centrifugal tension is uniform along belt and acts outward, so the **net tension** due to centrifugal effect is `T_c` on top of the transmitted tension. So max total tension = `T₁ + T_c` (tight side), min = `T₂ + T_c` (slack side). Stress check on tight side: `(T₁ + T_c) / (b t) ≤ σ_max`.

5.2 Friction Devices

• Clutch vs. Brake:

  • Clutch: Connects/disconnects power transmission (input & output shafts). Engages gradually.

  • Brake: Stops/retards a moving member (usually one shaft). Engages abruptly.

• Plate Clutch (Single/Multi-plate):

  • Uniform Pressure Theory (New): Pressure p constant. T = μ p π (r_o² - r_i²) n (n = number of friction surfaces).

  • Uniform Wear Theory (Worn): p r = constant. T = (2/3) μ W (r_o³ - r_i³)/(r_o² - r_i²).

  • Given W (axial force), find T: Use appropriate theory. Often uniform wear is more realistic.

  • Example (Jun 2025): W=4kN, r_i=50mm, r_o=100mm, μ=0.3. Uniform wear:

$$ T = \frac{2}{3} \mu W \frac{r_o^3 - r_i^3}{r_o^2 - r_i^2} = \frac{2}{3} \times 0.3 \times 4000 \times \frac{100^3 - 50^3}{100^2 - 50^2} = 800 \times \frac{875000}{7500} = 800 \times 116.67 = 93333.33 \text{ Nmm} = 93.33 \text{ Nm} $$

    Pressure: `p_max = W / (π(r_o² - r_i²))` (uniform pressure), `p_min = (r_i/r_o) p_max` (uniform wear).

• Band & Block Brake:

  • Derivation of Tension Ratio:

    Consider one block subtending angle 2θ at drum center. Equilibrium of block:

    T₁ = T₂ + μ N (horizontal), N = T₁ sin θ + T₂ sin θ (vertical).

    Solving: T₁/T₂ = (1+μ tan θ)/(1-μ tan θ).

    For n identical blocks:

$$ \boxed{\frac{T_0}{T_n} = \left(\frac{1+\mu \tan \theta}{1-\mu \tan \theta}\right)^n} $$

    *   `T₀` = tension on tight side (where operator applies force), `T_n` = tension on slack side.

*   **Braking Torque:** `T_brake = (T₀ - T_n) × R` (R = drum radius).

*   **Self-energizing:** If the **friction force aids** the applied force (common in **band & block** when `θ` is large). Condition: `T₀` applied on slack side? Actually, for self-energizing, the friction should help tighten the band. In **simple band brake** with lever, if the band is on the **same side** as the lever pivot, it's self-energizing.

• Power Screws:

  • Torque to Raise Load:

$$ \boxed{T_{raise} = \frac{W d_m}{2} \left( \frac{l + \pi \mu d_m}{\pi d_m - \mu l} \right)} $$

    where `d_m` = mean diameter, `l` = lead (pitch for single start), `μ` = friction coefficient.

*   **Torque to Lower Load:**

$$ T_{lower} = \frac{W d_m}{2} \left( \frac{\pi \mu d_m - l}{\pi d_m + \mu l} \right) $$

*   **Efficiency (raising):**

$$ \eta = \frac{\text{Work output}}{\text{Work input}} = \frac{W l}{2\pi T_{raise}} = \frac{l}{l + \pi \mu d_m} \times \frac{\pi d_m - \mu l}{\pi d_m} $$

    *   **Maximum Efficiency:** Occurs when `μ = l/(π d_m)`. Then `η_max = (1 - sin φ)/(1 + sin φ)` where `φ` = friction angle (`tan φ = μ`).

*   **Collar Friction:** Additional torque if load rests on a collar of mean radius `R_c`: `T_collar = μ_f W R_c` (μ_f = collar friction coeff.).

5.3 Pivots & Collars

• Friction Torque for Flat Pivot (Uniform Wear):

$$ T = \frac{2}{3} \mu W R $$

where `R` = outer radius of pivot (assuming uniform wear, pressure `p ∝ 1/r`).

• Friction Torque for Conical Pivot:

$$ T = \frac{2}{3} \mu W R \csc \alpha $$

where `α` = semi-angle of cone.

• Friction Torque for Spherical Pivot: Similar to flat pivot but with mean radius.

VI. VIBRATIONS & BALANCING

6.1 Free Vibrations

• Fundamental Equation (1-DOF, Undamped):

$$ \boxed{m \frac{d^2x}{dt^2} + k x = 0} $$

*   Solution: `x = A sin(ω_n t + φ)`

*   **Natural Frequency:** `ω_n = √(k/m)` (rad/s), `f_n = ω_n/(2π)` (Hz).

• Types:

  • Longitudinal: Motion along axis (spring-mass).

  • Transverse: Motion perpendicular to axis (beam, shaft).

  • Torsional: Angular oscillation (shaft with disc): J d²θ/dt² + k_t θ = 0, ω_n = √(k_t/J).

6.2 Balancing of Rotating Masses

• Single Plane (Static) Balancing:

  • For masses rotating in same plane.

  • Condition: Σ m r = 0 (vector sum of centrifugal forces = 0).

  • Graphical: Polygon of m r vectors. Find balancing mass M at radius R such that M R closes the polygon.

  • Analytical: Resolve into x and y components:

$$ \sum (m_i r_i \cos \theta_i) + M R \cos \phi = 0 $$

$$ \sum (m_i r_i \sin \theta_i) + M R \sin \phi = 0 $$

    Solve for `M R` and `φ`.

• Two Plane (Dynamic) Balancing:

  • For masses in different planes along the shaft.

  • Unbalance: Causes couple (moment) in addition to force.

  • Procedure: Place trial masses in two arbitrary planes (e.g., at both ends). Measure m r effect at each bearing (or use ω & α from vibration). Solve for balancing masses M₁, M₂ in chosen planes.

  • Vector Method: Use complex numbers or polygons for m r and m r l (moment about a reference plane).

  • Example (Nov 2023, Jun 2022): Given masses M₁..M₄ at radii r₁..r₄ and angles θ₁..θ₃ in planes at distances l₁, l₂, l₃ from reference. Balancing masses in planes X, Y at distances d₁, d₂ from reference. Set up equations:

    • Force balance: Σ (m_i r_i e^{iθ_i}) + M_X R_X e^{iφ_X} + M_Y R_Y e^{iφ_Y} = 0

    • Moment balance about ref: Σ (m_i r_i l_i e^{iθ_i}) + M_X R_X d₁ e^{iφ_X} + M_Y R_Y d₂ e^{iφ_Y} = 0

    Solve for M_X R_X and M_Y R_Y vectors, then magnitudes & angles.

6.3 Coriolis Acceleration

• Definition: Additional acceleration component experienced by a particle moving on a rotating link.

• Derivation:

  For a point P on a rotating link OA with ω, and P moving radially with velocity u:

$$ \vec{a}_P = \vec{a}_O + \vec{a}_{P/O,rel} + \vec{a}_{t} + \vec{a}_{c} + \boxed{\vec{a}_c} $$

where:

*   `a_t = α × r` (tangential)

*   `a_c = ω × (ω × r)` (centripetal)

*   **Coriolis component:** `\vec{a}_c = 2 \vec{\omega} \times \vec{u}`

• Magnitude: a_c = 2 ω u

• Direction: Perpendicular to u. Leading the radial velocity u by 90° in the direction of rotation.

• Application: Slider-crank with oscillating piston (piston has both radial and tangential components relative to crank pin).

HIGH-FREQUENCY TOPICS: QUICK RECAP