3.3 Rolling-Element & Journal Bearings, Power Screws, Clutches, Brakes & Flywheels

Rolling-Element Bearings, Hydrodynamic Journal Bearings, Clutches & Brakes

Machine assemblies require specialized mechanical components to support rotating shafts, regulate speed, absorb kinetic energy, and smooth torque fluctuations. This section provides complete coverage of four major machine elements tested on the PE Mechanical exam: hydrodynamic journal bearings, rolling-element bearings ($L_{10}$ fatigue life), friction clutches and brakes (disc, band, drum), and flywheel inertia sizing and rim stresses.

+---------------------------------------------------------------------------------------------------+
|                         POWER TRANSMISSION COMPONENT ARCHITECTURE                                 |
|                                                                                                   |
|   [JOURNAL BEARINGS]       [ROLLING BEARINGS]       [CLUTCHES & BRAKES]      [FLYWHEELS]          |
|   - Hydrodynamic Film      - Ball / Roller Elements - Uniform Wear / Press   - Energy Storage     |
|   - Sommerfeld Number (S)  - Basic Dynamic Load (C) - Disc / Drum / Band     - Fluctuation (C_s)  |
|   - Petroff Friction       - L10 Life (p=3, 10/3)   - Capstan: T1/T2=e^(mu*t)- Inertia Sizing     |
|   - Thermal Dissipation    - Eq Load: P = X*Fr+Y*Fa - Thermal Dissipation    - Rim Hoop Stress    |
+---------------------------------------------------------------------------------------------------+

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1. Hydrodynamic Lubrication & Journal Bearings

In a hydrodynamic journal bearing, shaft rotation drags viscous lubricant into a converging clearance wedge between the journal (shaft) and the bushing, generating sufficient hydrodynamic pressure to support external radial loads without solid metal-to-metal contact.

+---------------------------------------------------------------------------------------------------+
|                             HYDRODYNAMIC JOURNAL BEARING GEOMETRY                                 |
|                                                                                                   |
|                                    +-----------------------+                                      |
|                                  /                           \                                    |
|                                /       STATIONARY BUSHING      \                                  |
|                               |            (Radius R)           |                                 |
|                               |        +---------------+        |                                 |
|                               |       /                 \       |                                 |
|                               |      /  ROTATING JOURNAL \      |  Radial Clearance: c = R - r    |
|                               |     |     (Radius r)      |     |  Eccentricity: e                |
|                               |      \                   /      |  Min Film: h_0 = c - e          |
|                               |       \  w (Rotation)   /       |  Eccentricity Ratio: eps = e/c  |
|                                \       +---------------+       /                                  |
|                                  \                           /                                    |
|                                    +-----------------------+                                      |
+---------------------------------------------------------------------------------------------------+

The Stribeck Curve & Lubrication Regimes

The lubrication regime is governed by the bearing characteristic parameter $\frac{\mu N'}{P}$:

1. Boundary Lubrication (Low $\mu N'/P$): Solid asperity contact carries the load ($f \approx 0.08 - 0.15$). High wear occurs during startup/shutdown.
2. Mixed-Film Lubrication: Intermediate regime with partial asperity contact and partial hydrodynamic pressure.
3. Hydrodynamic (Full-Film) Lubrication (High $\mu N'/P$): A continuous fluid film separates surfaces completely ($f \approx 0.001 - 0.005$). Zero mechanical wear.

1. Petroff's Equation (Concentric Journal Friction)

For a lightly loaded, high-speed concentric journal bearing ($e \approx 0$), Petroff's equation gives the frictional torque $T_f$ and coefficient of friction $f$:

$T_f = \frac{4 \pi^2 \mu r^3 l N'}{c}$

$f = 2 \pi^2 \left( \frac{r}{c} \right) \left( \frac{\mu N'}{P} \right)$

Where:

• $\mu = \text{Dynamic viscosity (}\text{Pa}\cdot\text{s in SI; } \text{reyns} = \text{lbf}\cdot\text{s/in}^2 \text{ in US Customary, with } 1\text{ reyn} = 6.895 \times 10^6\text{ cP)}$
• $r = \text{Journal radius}$, $l = \text{Bearing axial length}$, $c = \text{Radial clearance} = R - r$
• $N' = \text{Rotational speed in rev/s} = N / 60$
• $P = \text{Projected bearing pressure} = \frac{W}{2 r l} = \frac{W}{d l}$

2. The Sommerfeld Number ($S$)

The dimensionless Sommerfeld number (Bearing Characteristic Number) is the master design parameter for hydrodynamic journal bearings:

$S = \left( \frac{r}{c} \right)^2 \left( \frac{\mu N'}{P} \right)$

Using Raimondi-Boyd charts with $S$ and length-to-diameter ratio ($l/d$):

• Minimum film thickness: $h_0 = c (1 - \epsilon)$, where $\epsilon = e/c$ is the eccentricity ratio.
• Friction variable: $\left(\frac{r}{c}\right) f \implies f$.
• Power Loss: $H_{\text{loss}} = 2\pi N' T_f = f W (2\pi r N')$.
• Lubricant Temperature Rise: $\Delta T = \frac{H_{\text{loss}}}{\dot{m} c_p}$, where $\dot{m} = \rho Q$ is the lubricant mass flow rate.

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2. Rolling-Element Bearings & $L_{10}$ Fatigue Life

Rolling contact bearings replace sliding friction with rolling elements (balls or rollers). Bearing life is limited by subsurface shear fatigue spalling.

+---------------------------------------------------------------------------------------------------+
|                                 ROLLING-ELEMENT BEARING TYPES                                     |
|                                                                                                   |
|   DEEP GROOVE BALL:        CYLINDRICAL ROLLER:      TAPERED ROLLER:          SPHERICAL ROLLER:    |
|   - Radial + moderate axial- High radial capacity   - Heavy combined loads   - Self-aligning      |
|   - Low friction / high spd- Zero thrust capacity   - Radial + high thrust   - Heavy radial loads |
|   - Life exponent p = 3    - Life exponent p = 10/3 - Used in preloaded pairs- Angular tolerance  |
+---------------------------------------------------------------------------------------------------+

Equivalent Radial Dynamic Load ($P_e$)

When a bearing experiences simultaneous radial load $F_r$ and axial thrust load $F_a$:

$P_e = X V F_r + Y F_a$

Where:

• $V = \text{Rotation factor (}1.0\text{ for rotating inner ring; } 1.2\text{ for rotating outer ring)}$
• $X = \text{Radial factor}$ and $Y = \text{Thrust factor}$ (from manufacturer tables based on $F_a / (V F_r) > e$)
• If $F_a / (V F_r) \le e$, then $X = 1.0$ and $Y = 0$, so $P_e = V F_r$.

$L_{10}$ Rating Life Sizing Equations

The basic rating life $L_{10}$ (the operating life completed or exceeded by $90%$ of identical bearings before first evidence of fatigue) is:

$L_{10} = \left( \frac{C}{P_e} \right)^p \times 10^6 \quad (\text{revolutions})$

$L_{10h} = \frac{10^6}{60 N} \left( \frac{C}{P_e} \right)^p \quad (\text{operating hours})$

Where:

• $C = \text{Basic dynamic load rating (load for } 10^6 \text{ revolutions)}$
• $P_e = \text{Equivalent radial dynamic load}$
• $N = \text{Rotational speed in rpm}$
• $p = 3.0 \text{ for ball bearings}$; $p = \frac{10}{3} \approx 3.333 \text{ for roller bearings}$

Reliability and Adjusted Life ($L_{na}$)

For reliability levels other than $90%$ ($R > 0.90$):

$L_{na} = a_1 a_2 a_3 L_{10}$

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3. Clutches & Disc Brakes: Uniform Wear vs. Uniform Pressure

Friction clutches and disc brakes transmit torque through mechanical contact over annular friction surfaces ($r_i \le r \le r_o$).

+---------------------------------------------------------------------------------------------------+
|                             ANNULAR DISC CLUTCH / BRAKE GEOMETRY                                  |
|                                                                                                   |
|                                    +-----------------------+                                      |
|                                  /   Outer Radius (r_o)      \                                    |
|                                /       +---------------+       \                                  |
|                               |       /  Inner Rad (r_i)\       |  Friction Surfaces: N           |
|                               |      |   (Bore)          |      |  Friction Coeff: mu             |
|                               |       \                 /       |  Max Pressure: p_max            |
|                                \       +---------------+       /   Axial Clamping Force: F        |
|                                  \                           /                                    |
|                                    +-----------------------+                                      |
+---------------------------------------------------------------------------------------------------+

Model 1: Uniform Wear Model (Broken-In Surfaces)

Wear is assumed proportional to frictional work ($PV = \text{const} \implies p \cdot r = C = p_{\max} r_i$). Maximum pressure occurs at inner radius $r_i$. This is the standard, conservative model for design:

$F = 2 \pi p_{\max} r_i (r_o - r_i)$

$T = N \mu F r_{\text{avg}} = N \mu F \left( \frac{r_o + r_i}{2} \right) = \pi N \mu p_{\max} r_i (r_o^2 - r_i^2)$

Where $N = \text{Number of active friction interface pairs}$ (for $M$ driving plates and $K$ driven plates, $N = M + K - 1$).

Model 2: Uniform Pressure Model (Brand New Surfaces)

Assumes rigid, perfectly flat new plates with uniform pressure distribution ($p = p_{\max} = \text{const}$):

$F = \pi p_{\max} (r_o^2 - r_i^2)$

$T = \frac{2}{3} N \mu F \left( \frac{r_o^3 - r_i^3}{r_o^2 - r_i^2} \right) = \frac{2}{3} \pi N \mu p_{\max} (r_o^3 - r_i^3)$

[!IMPORTANT] Optimal Disc Geometry Ratio: To maximize torque capacity for a given maximum outer radius $r_o$ under the uniform wear model, set $\frac{d}{dr_i}[r_i (r_o^2 - r_i^2)] = r_o^2 - 3 r_i^2 = 0 \implies \frac{r_i}{r_o} = \frac{1}{\sqrt{3}} \approx 0.577$.

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4. Band Brakes & Drum Brakes

+---------------------------------------------------------------------------------------------------+
|                                    BAND BRAKE MECHANICS                                           |
|                                                                                                   |
|                                       T_1 (Tight Side Tension)                                    |
|                                         ^                                                         |
|                                         |  Wrap Angle (theta rad)                                 |
|                                   +-----+-----+                                                   |
|                                 /               \                                                 |
|                                |  DRUM ROTATION  | ---> Torque T = (T_1 - T_2) * r                |
|                                 \               /                                                 |
|                                   +-----+-----+                                                   |
|                                         |                                                         |
|                                         v                                                         |
|                                       T_2 (Slack Side Tension)                                    |
|                                                                                                   |
|   CAPSTAN FORMULA: T_1 / T_2 = e^(mu * theta)   ===>   T_1 = T_2 * e^(mu * theta)                 |
+---------------------------------------------------------------------------------------------------+

Band Brake Governing Equations

• Capstan Friction Tension Ratio: $\frac{T_1}{T_2} = e^{\mu \theta}$ (Where $\theta$ is the band wrap angle in radians, and $T_1$ is tight side tension opposing drum rotation).
• Braking Torque Transmitted: $T_{\text{brake}} = (T_1 - T_2) r = T_2 (e^{\mu \theta} - 1) r$
• Actuating Force: Solved from lever moment equilibrium $\sum M_{\text{pivot}} = 0$.
• Self-Locking Phenomenon: In a differential band brake, if friction assists the actuating force such that $F_{\text{actuate}} \le 0$, the brake locks automatically upon contact without external operator force.

Brake Energy Dissipation & Temperature Rise

When absorbing kinetic energy $\Delta E_k$ from moving mass $m$ or rotating inertia $I$:

$\Delta E_k = \frac{1}{2} m (v_1^2 - v_2^2) + \frac{1}{2} I (\omega_1^2 - \omega_2^2)$

$\Delta T = \frac{\Delta E_k}{m_{\text{drum}} c_p}$

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5. Flywheel Energy Storage & Rim Stresses

Flywheels smooth cyclic torque variations in reciprocating machinery by storing kinetic energy during excess torque phases and delivering it during peak demands.

+---------------------------------------------------------------------------------------------------+
|                                 FLYWHEEL CRANK-TORQUE DIAGRAM                                     |
|                                                                                                   |
|   Torque                                                                                          |
|     ^             +---------------+ (Peak Torque)                                                 |
|     |            /                 \                                                              |
|     |           /    Excess Work    \                                                             |
|     |          /    (+Delta E)       \                                                            |
|   T_mean =====+=======================+============================= (Mean Resisting Load)        |
|     |          \                     /                                                            |
|     |           \   Deficit Work    /                                                             |
|     |            \   (-Delta E)    /                                                              |
|     +-------------+---------------+--------------------------> Crank Angle Theta (rad)           |
+---------------------------------------------------------------------------------------------------+

1. Coefficient of Speed Fluctuation ($C_s$)

$C_s = \frac{\omega_1 - \omega_2}{\omega_{\text{avg}}} = \frac{N_1 - N_2}{N_{\text{avg}}}$

Where $\omega_{\text{avg}} = \frac{\omega_1 + \omega_2}{2}$ (Standard values: $C_s = 0.01 - 0.02$ for generators; $C_s = 0.05 - 0.10$ for punch presses).

2. Sizing Required Mass Moment of Inertia ($I$)

The maximum energy change $\Delta E$ is:

$\Delta E = \frac{1}{2} I (\omega_1^2 - \omega_2^2) = I C_s \omega_{\text{avg}}^2 \implies I = \frac{\Delta E}{C_s \omega_{\text{avg}}^2}$

3. Flywheel Rim Geometry and Hoop Stress

For a rim-type flywheel where mass is concentrated in a rim of mean radius $R$ ($I \approx m_{\text{rim}} R^2$):

$m_{\text{rim}} = \frac{I}{R^2} = \rho (2\pi R b t)$

Centrifugal acceleration induces tensile hoop stress $\sigma_t$ in the thin rim:

$\sigma_t = \rho v^2 = \rho (\omega R)^2$

Where $\rho$ is mass density ($\text{kg/m}^3$ or $\text{slug/in}^3$) and $v = \omega R$ is rim linear speed.

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6. Power Screws & Threaded Linear Actuation

The Machine Design and Materials specification groups power screws with belts, chains, clutches, and brakes under Power Transmission. A power screw converts input rotation and torque into linear motion and force — jack screws, C-clamps, vises, valve stems, and machine-tool lead screws are the classic exam applications.

Thread Geometry Terminology

• Pitch ($p$): axial distance between adjacent thread crests.
• Lead ($l$): axial advance per screw revolution. Single-start thread: $l = p$; double-start: $l = 2p$.
• Helix angle ($\lambda$): $\tan\lambda = l / (\pi d_m)$, measured at the mean (pitch) thread diameter $d_m$.
• Thread forms: square threads maximize efficiency and strength per turn; Acme (29° included angle) threads are cheaper to machine and self-centering; buttress threads carry heavy loads in one direction; ball screws replace sliding friction with rolling contact.

Torque to Raise and Lower the Load

Raising a load $F$ against thread friction requires input torque:

$T_R = \frac{F\, d_m}{2}\left(\frac{l + \pi f d_m}{\pi d_m - f l}\right) + T_{\text{collar}}, \qquad T_{\text{collar}} = \frac{F f_c d_c}{2}$

where $f$ is the thread coefficient of friction and the collar term accounts for any thrust-bearing surface of mean diameter $d_c$. The lowering expression flips the friction terms in sign — but a self-locking screw still needs torque to pay friction going down.

Friction Angle, Efficiency & the Self-Locking Condition

Define the friction angle $\phi = \tan^{-1} f$. Thread efficiency is:

$e = \frac{F l}{2 \pi T_R} = \frac{\tan\lambda}{\tan(\lambda + \phi)}$

• Self-locking condition: $\phi > \lambda$ (efficiency below 50%). A self-locking screw holds the applied load with zero sustaining input torque — jacks do not back-drive.
• Ball screws reach $e \approx 90%$ and are therefore back-drivable; vertical axes need a holding brake or counterweight.
• Long slender screws loaded in compression must also clear column buckling (Euler/Johnson criteria from Section 2.4) — thread strength alone is not sufficient.

Worked Mini-Example

A single-start square-thread jack screw has $d_m = 20\text{ mm}$, $p = l = 5\text{ mm}$, $\mu = 0.10$ (collar friction neglected), lifting $F = 10\text{ kN}$:

• $\lambda = \tan^{-1}(5 / (20\pi)) = 4.55°$; $\phi = \tan^{-1}(0.10) = 5.71°$.
• $\phi > \lambda$ → self-locking ✓ (the jack holds the 10 kN with the handle released).
• $T_R = \frac{F d_m}{2}\tan(\lambda + \phi) = \frac{10{,}000 \times 0.020}{2} \times \tan(10.26°) = 100 \times 0.181 \approx 18.1\text{ N·m}$.
• $e = \tan(4.55°) / \tan(10.26°) = 0.0796 / 0.181 \approx 0.44$ (44%).

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7. Step-by-Step Worked Problem: Rolling Bearing Life & Flywheel Sizing

Problem Statement

1. A cylindrical roller bearing ($C = 54\text{ kN}$, $p = 10/3$) operates at $1500\text{ rpm}$ under a purely radial load $F_r = 9.0\text{ kN}$ ($V = 1.0$). Compute the $L_{10h}$ rating life in operating hours.
2. A punch press requires energy absorption $\Delta E = 3600\text{ J}$ per stroke. Mean speed is $240\text{ rpm}$, and speed fluctuation must not exceed $C_s = 0.04$. Sizing a steel rim flywheel ($\rho = 7850\text{ kg/m}^3$) with mean radius $R = 0.60\text{ m}$, determine the required mass moment of inertia $I$, required rim mass $m_{\text{rim}}$, and the resulting rim hoop stress $\sigma_t$.

Step-by-Step Solution

Part 1: Roller Bearing Life Calculation

• Equivalent dynamic load: $P_e = F_r = 9.0\text{ kN}$.
• Compute $L_{10}$ in millions of revolutions (for roller bearings, $p = 10/3 = 3.333$): $L_{10} = \left( \frac{C}{P_e} \right)^{10/3} = \left( \frac{54}{9.0} \right)^{3.333} = (6.0)^{3.333} = 390.1\text{ million revs}$
• Convert to operating hours at $N = 1500\text{ rpm}$: $L_{10h} = \frac{L_{10} \times 10^6}{60 N} = \frac{390.1 \times 10^6}{60 \times 1500} = \frac{3.901 \times 10^8}{9.0 \times 10^4} = \mathbf{4,334\text{ hours}}$

Part 2: Flywheel Sizing Calculation

• Mean angular velocity: $\omega_{\text{avg}} = \frac{2\pi (240)}{60} = 25.133\text{ rad/s}$.
• Required mass moment of inertia: $I = \frac{\Delta E}{C_s \omega_{\text{avg}}^2} = \frac{3600\text{ J}}{0.04 \times (25.133)^2} = \frac{3600}{0.04 \times 631.65} = \frac{3600}{25.266} = \mathbf{142.48\text{ kg}\cdot\text{m}^2}$
• Required rim mass (at $R = 0.60\text{ m}$): $m_{\text{rim}} = \frac{I}{R^2} = \frac{142.48}{(0.60)^2} = \frac{142.48}{0.36} = \mathbf{395.8\text{ kg}}$
• Rim linear speed: $v = \omega R = (25.133)(0.60) = 15.08\text{ m/s}$.
• Rim hoop tensile stress: $\sigma_t = \rho v^2 = (7850\text{ kg/m}^3) \times (15.08\text{ m/s})^2 = 7850 \times 227.4 = 1.785 \times 10^6\text{ Pa} = \mathbf{1.79\text{ MPa}}$ (Well within allowable limit for cast iron/steel, typically $\sigma_{\text{allow}} \approx 30 - 50\text{ MPa}$).

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8. Exam Tips & Common Traps

[!TIP] Speedy Exam Rules:

• In band brake capstan calculations ($e^{\mu\theta}$), wrap angle $\theta$ must always be in radians ($180^\circ = \pi\text{ rad}$). Degrees will give massive exponential errors.
• For rolling bearings, double-check whether the problem specifies a ball bearing ($p=3$) or a roller bearing ($p=10/3$). Doubling load on a ball bearing reduces life by $2^3 = 8\times$, while on a roller bearing it reduces life by $2^{3.333} = 10.08\times$.

[!WARNING] Common Traps:

• Clutch Pressure Model Trap: Unless a problem explicitly specifies "brand new, rigid flat plates," always use the uniform wear model. Uniform pressure overestimates capacity and only applies before initial wear occurs.
• Flywheel Percentage Fluctuation: If $C_s$ is given as $3%$, use $0.03$ in equations, not $0.3$.