Mechanical Engineering Design · Part 4 — Special Topics

Clutches, Brakes, Couplings, and Flywheels

Devices that start, stop, store, and smooth rotation: the static analysis of friction brakes and clutches (rim, band, disk, cone), the uniform-wear and uniform-pressure torque models, the energy and temperature-rise of a braking cycle, friction materials, couplings, and flywheel sizing by the coefficient of speed fluctuation.

Learning objectives

Distinguish devices that connect, stop, accommodate misalignment, and smooth rotational speed.
Construct friction-device free-body diagrams with consistent motion and friction directions.
Select uniform-wear, uniform-pressure, capstan, energy, temperature, and flywheel models.
Calculate torque capacity, actuation demand, dissipated energy, and required rotational inertia.
Separate static torque capacity from thermal duty, fatigue life, wear, and controllability.
Use supplier data, standards, tests, and iteration to make auditable component selections.

16–1Static Analysis of Clutches and Brakes

A Common Analytical Frame

A clutch synchronizes rotating members, a brake transfers torque to a fixed frame, a coupling joins shafts during normal service, and a flywheel exchanges stored kinetic energy with a machine. Recognition matters because torque equilibrium alone cannot size thermal mass, and an energy calculation alone cannot establish static holding capacity. Friction devices require a pressure model, a free-body diagram, and a heat-flow model appropriate to the duty cycle.

Self-Energizing and Self-Locking

Friction opposes relative slip at the interface, not necessarily the absolute motion of either isolated member. A shoe is self-energizing when its friction moment assists actuation; reversing drum rotation can make the same shoe self-deenergizing. If friction sustains contact without applied force, the geometry is self-locking and may grab unpredictably. Draw the assumed rotation and friction arrow before writing moments.

Physical meaning

Friction torque controls angular acceleration through \(\sum T=I\alpha\), while slip energy becomes heat. Higher torque shortens a stop but does not erase the kinetic energy that must be absorbed; it releases approximately the same energy in less time and may raise flash temperature.

Problem-solving strategy

Identify whether the target is holding torque, dynamic stopping, synchronization, or speed smoothing.
Isolate each rotating member and mark positive rotation, relative slip, friction, and reactions.
Select the pressure distribution and obtain friction data at the expected pressure, speed, and temperature.
Use force and moment equilibrium for capacity, then angular momentum or energy for transient duty.
Check pressure, rubbing speed, energy per event, average heat rejection, wear, and fail-safe behavior.
Iterate geometry and actuation, then verify the selected hardware with current supplier data and tests.

16–2Internal Expanding Rim Clutches and Brakes

The Drum Brake

An internal-expanding shoe (the automotive drum brake) pivots at one end while an actuating force at the other presses the lining against the drum's inner surface. The pressure is taken proportional to \(\sin\theta\) (zero at the heel pivot, maximum near the toe). Integrating the pressure and friction over the lining arc gives the moment of the normal forces \(M_N\), the moment of the friction forces \(M_f\), and the actuating force — from which the braking torque follows.

This sine pressure idealization assumes a rigid shoe and drum, known contact arc, and a pivot geometry for which pressure vanishes at the heel. It is a static model; lining compliance, local hot spots, wear, and transient drum expansion require test or finite-element evidence.

Pressure Distribution & Braking Torque

\[ p = p_a\,\frac{\sin\theta}{\sin\theta_a} \qquad\qquad T = \frac{f\,p_a\,b\,r}{\sin\theta_a}\int_{\theta_1}^{\theta_2}\sin\theta\,(r - a\cos\theta)\,d\theta \]Eqs. (16–2, 16–6)

Actuating Force from Shoe Moments

\[ F = \frac{M_N - M_f}{c} \quad(\text{self-energizing}) \qquad F = \frac{M_N + M_f}{c} \quad(\text{self-deenergizing}) \]Eqs. (16–3, 16–4)

16–3External Contracting Rim Clutches and Brakes

Outside the Drum

External-contracting shoes press inward on the drum's outer surface (common in industrial and crane brakes). The same moment analysis applies, but the friction-moment sign flips relative to the internal case, so the self-energizing and self-deenergizing roles of the two shoes swap. Symmetric pivoted shoes can be arranged so the friction moments cancel the pin reactions.

Because rotation direction changes which shoe leads, analyze forward and reverse operation separately. A design that has low pedal force in one direction may demand much more force in the other or become sensitive to friction variation.

16–4Band-Type Clutches and Brakes

The Capstan Relation

A flexible band wrapped around a drum develops a tension ratio governed by the belt-friction (capstan) equation \(P_1/P_2 = e^{f\phi}\), where \(P_1\) is the tight-side tension, \(P_2\) the slack-side, \(f\) the friction coefficient, and \(\phi\) the wrap angle. The braking torque is \((P_1 - P_2)\) times the drum radius. The pressure is proportional to the local tension, so the maximum pressure occurs at the tight end (toe).

The capstan relation assumes a flexible band, uniform friction coefficient, impending or steady slip, and wrap angle \(\phi\) in radians. It predicts tension ratio, not band strength; pin loads, fatigue at attachments, drum heating, lining pressure, and release clearance remain separate.

Band Tension Ratio & Torque

\[ \frac{P_1}{P_2} = e^{f\phi} \qquad\qquad T = (P_1 - P_2)\,\frac{D}{2} \qquad\qquad p_a = \frac{2 P_1}{b\,D} \]Eqs. (16–18)–(16–20)

EXAMPLE 1 Band-Brake Capacity and Peak Pressure

A stationary band brake wraps \(210^\circ\) around a \(400\ \text{mm}\)-diameter drum. At impending slip, the dry lining coefficient is \(f=0.28\), slack-side tension is \(700\ \text{N}\), and band width is \(50\ \text{mm}\). Determine tight-side tension, braking torque, and maximum nominal lining pressure.

Recognition strategy

A flexible band with known wrap, friction, and one end tension signals the capstan relation. Once both tensions are known, their difference supplies drum torque and tight-side tension supplies the peak nominal pressure.

Given information

\(\phi=210^\circ=3.6652\ \text{rad}\), \(f=0.28\), \(P_2=700\ \text{N}\).
\(D=0.400\ \text{m}\), \(b=0.050\ \text{m}\); \(P_1\) denotes the tight-side tension.
Impending clockwise drum slip is assumed, with uniform stated friction and a flexible band.

Model and interpretation

Isolate the drum. Tangential band forces oppose clockwise motion, and the net resisting torque is reported as a positive magnitude. Pressure from the elementary capstan model rises toward the tight end, where the nominal maximum is evaluated.

Step-by-step solution

1. Convert wrap angle. The exponential requires radians: \(\phi=210(\pi/180)=3.6652\ \text{rad}\), giving the dimensionless exponent \(f\phi=(0.28)(3.6652)=1.0263\).

2. Find tight-side tension. \(P_1/P_2=e^{1.0263}=2.790\), so \(P_1=(700\ \text{N})(2.790)=1953\ \text{N}\).

3. Calculate braking torque. \(T=(P_1-P_2)D/2=(1953-700)(0.400/2)=250.6\ \text{N}\!\cdot\!\text{m}\), acting counterclockwise against the assumed drum slip.

4. Calculate peak nominal pressure. \(p_a=2P_1/(bD)=2(1953)/[(0.050)(0.400)]=195300\ \text{Pa}=195\ \text{kPa}\).

Verification

Units: The tension ratio is dimensionless; force times radius gives \(\text{N}\!\cdot\!\text{m}\), and \(\text{N}/(\text{m}\!\cdot\!\text{m})=\text{Pa}\) for lining pressure.

Signs and direction: \(P_1>P_2\) as required for a tight side. The reported torque is a resisting magnitude; on a signed clockwise-positive drum equation it would enter as \(-250.6\ \text{N}\!\cdot\!\text{m}\).

Plausibility: With \(f\to0\), the exponential ratio approaches one and torque approaches zero. Here the moderate wrap produces a reasonable ratio of \(2.79\), and direct substitution recovers the specified slack tension.

Final result: \(\boxed{P_1=1.95\ \text{kN},\quad T_B=251\ \text{N}\!\cdot\!\text{m},\quad p_a=195\ \text{kPa}}\).

Engineering interpretation

These are static nominal demands, not allowable strengths. The band, anchors, drum, and lining must be checked using qualified fatigue, pressure, wear, temperature, and rubbing-speed data, including friction scatter and both rotation directions.

16–5Frictional-Contact Axial Clutches

Two Models: Uniform Wear vs. Uniform Pressure

A disk (plate) clutch transmits torque across annular friction faces moved axially together. Two limiting models bracket reality. Uniform wear (after break-in) assumes \(pr=\text{constant}\), so pressure is greatest at the inner radius. Uniform pressure approximates a new, rigid, accurately aligned face. Neither model is universally conservative for every output: select it from the surface condition, state whether \(N\) counts active friction interfaces, and compare local pressure and thermal duty as well as torque.

Uniform Wear — Force and Torque

\[ F = \frac{\pi\,p_a\,d}{2}(D - d) \qquad\qquad T = \frac{F\,f}{4}(D + d) \]Eqs. (16–23, 16–24)

Uniform Pressure — Torque

\[ T = \frac{F\,f}{3}\,\frac{D^3 - d^3}{D^2 - d^2} \]Eq. (16–26)

EXAMPLE 2 Broken-In Multi-Surface Clutch

A broken-in dry disk clutch has outer diameter \(D=280\ \text{mm}\), inner diameter \(d=160\ \text{mm}\), four active friction interfaces, axial clamp force \(F=8.00\ \text{kN}\), and supplied hot friction coefficient \(f=0.32\). Find uniform-wear torque capacity and maximum pressure, then quantify the torque difference if uniform pressure were incorrectly assumed.

Recognition strategy

The phrase broken-in selects uniform wear. Clamp force is common to each active interface, while total torque is the per-interface torque multiplied by the explicitly stated interface count. The alternate model is evaluated only as a sensitivity comparison.

Given information

\(D=0.280\ \text{m}\), \(d=0.160\ \text{m}\), \(N=4\), \(F=8000\ \text{N}\), and \(f=0.32\).
Uniform wear is the operating model; uniform pressure is a comparison for the same force.
Faces are flat, axial force is uniform among interfaces, and torque values are static capacities.

Model and interpretation

Use an annular friction-face free body. Positive transmitted torque opposes relative slip. Under uniform wear the effective radius is \((D+d)/4\), and \(p_a\) is the inner-edge maximum, not an area-average pressure.

Step-by-step solution

1. Establish effective radius. For uniform wear, \(r_e=(D+d)/4=(0.280+0.160)/4=0.110\ \text{m}\), which lies between the physical radii \(0.080\ \text{m}\) and \(0.140\ \text{m}\).

2. Calculate total uniform-wear torque. \(T_w=NFfr_e=(4)(8000)(0.32)(0.110)=1126.4\ \text{N}\!\cdot\!\text{m}\).

3. Recover maximum pressure. \(p_a=2F/[\pi d(D-d)]=2(8000)/[\pi(0.160)(0.120)] =265300\ \text{Pa}=265\ \text{kPa}\).

4. Compare the alternate model. \(T_p=NFf(D^3-d^3)/[3(D^2-d^2)] =1154.3\ \text{N}\!\cdot\!\text{m}\), which is \((1154.3/1126.4-1)100=2.48\%\) higher.

Verification

Units: Force times the dimensionless friction coefficient and metre radius gives \(\text{N}\!\cdot\!\text{m}\). The pressure expression has newtons over square metres and therefore produces pascals.

Signs and direction: Capacity is stated as a positive magnitude. On either rotating member's free body, friction torque acts opposite that member's relative slip; the mating member receives an equal and opposite torque.

Plausibility: The effective radius \(0.110\ \text{m}\) lies inside the annulus, and \(T_w=4(8000)(0.32)(0.110)\) independently reproduces the result. The small model difference is credible for this radius ratio.

Final result: \(\boxed{T_w=1.126\ \text{kN}\!\cdot\!\text{m},\quad p_a=265\ \text{kPa}; T_p=1.154\ \text{kN}\!\cdot\!\text{m}\text{ is }2.48\%\text{ higher}}\).

Engineering interpretation

Model choice changes this torque by only \(2.48\%\), but that does not establish an adequate design margin. The hot minimum friction value, clamp-force tolerance, plate flatness, spline fatigue, engagement energy, wear allowance, and supplier pressure limit still govern release.

16–6Disk Brakes

Caliper-and-Rotor Brakes

A disk brake (caliper) presses friction pads against a rotor over an annular sector rather than a full annulus, eliminating the self-energizing variability of drum brakes and dissipating heat well. The analysis uses an effective radius at which the total friction force acts; for the uniform-wear assumption it is the mean of the inner and outer pad radii. The torque is the friction force times this effective radius, summed over the pads.

Do not confuse piston force, caliper clamp force, and the normal force on one pad. State which force appears in the selected equation. Rotor stress, pad taper, hydraulic pressure limits, boiling margin, and parking-brake holding are separate checks.

Disk-Brake Effective Radius & Torque

\[ r_e = \frac{r_o + r_i}{2}\ (\text{uniform wear}) \qquad\qquad T = f\,F\,r_e\,(\text{per pad pair}) \]Eq. (16–33 region)

16–7Cone Clutches and Brakes

The Wedging Cone

A cone clutch uses a conical friction surface at half-angle \(\alpha\); the wedging action multiplies the normal force for a given axial actuating force, so a smaller force produces a given torque (at the cost of engagement smoothness). Using the uniform-wear model, the torque is the plate-clutch result divided by \(\sin\alpha\), and the axial engaging force is correspondingly reduced.

Small cone angles reduce required axial force but increase sensitivity to tolerance, wear, and sticking. Manufacturing must control runout and cone-angle match, and release travel must remain adequate throughout the wear allowance.

Cone-Clutch Torque (uniform wear)

\[ T = \frac{F\,f}{4\sin\alpha}(D + d) \]Eq. (16–43)

16–8Energy Considerations

Heat from Slippage

During clutch engagement or braking, the relative slip between the two members dissipates kinetic energy as heat. Modeling two inertias \(I_1, I_2\) at speeds \(\omega_1, \omega_2\) connected through the friction device, the total energy dissipated depends only on the inertias and the speed difference — not on the friction torque, which only sets how fast (and how long) the heat is released. This energy fixes the thermal duty the device must survive.

The relation assumes no external work during synchronization and constant inertias referred to a common shaft. External motor torque, load torque, changing ratio, or translational masses must be included when present. For a brake stopping one inertia, it reduces to \(E=\tfrac12 I\omega^2\).

Energy Dissipated in Engagement

\[ E = \frac{I_1 I_2\,(\omega_1 - \omega_2)^2}{2\,(I_1 + I_2)} \]Eq. (16–52)

16–9Temperature Rise

Estimating the Heat Pulse

A first estimate of the bulk temperature rise per engagement treats the dissipated energy as absorbed by the mass of the clutch/brake parts at their specific heat: \(\Delta T=E/(c_p m)\). This lumped, adiabatic pulse estimate requires a supplied heat partition, effective heated mass, and temperature-appropriate specific heat. Repeated cycles also require an average and transient cooling model. Bulk temperature can be acceptable while interface flash temperature causes fade or cracking.

Bulk Temperature Rise per Engagement

\[ \Delta T = \frac{E}{C_p\,m} \]Eq. (16–58)

EXAMPLE 3 Screen a Repeating Brake Duty

A production indexer brakes a referred inertia \(I=18.0\ \text{kg}\!\cdot\!\text{m}^2\) from \(900\ \text{rev/min}\) to rest in \(4.00\ \text{s}\), once every \(25.0\ \text{s}\). Neglect external work during each stop. A rotor absorbs \(85\%\) of the heat, has effective mass \(24.0\ \text{kg}\) and supplied \(c_p=460\ \text{J/(kg}\!\cdot\!\text{K)}\). Its validated cooling rating at the temperature limit is \(3.20\ \text{kW}\), and hot torque capacity is \(500\ \text{N}\!\cdot\!\text{m}\). Screen torque, pulse rise, and average cooling.

Recognition strategy

Stop time calls for angular dynamics and torque capacity; speed and inertia determine event energy; heat partition and thermal mass determine an adiabatic pulse; cycle time converts event energy to average cooling duty. All three constraints must pass.

Given information

\(I=18.0\ \text{kg}\!\cdot\!\text{m}^2\), \(n_0=900\ \text{rev/min}\), \(t_s=4.00\ \text{s}\), and cycle \(t_c=25.0\ \text{s}\).
Rotor fraction \(\chi=0.85\), \(m=24.0\ \text{kg}\), and \(c_p=460\ \text{J/(kg}\!\cdot\!\text{K)}\).
Supplier-validated limits: \(\dot Q_{\mathrm{cap}}=3.20\ \text{kW}\) and \(T_{\mathrm{cap,hot}}=500\ \text{N}\!\cdot\!\text{m}\).

Model and interpretation

Take initial rotation positive and constant brake torque opposite rotation. The brake alone removes kinetic energy, deceleration is uniform, the rotor receives the specified energy fraction, and the pulse is adiabatic. Average cooling is a separate periodic steady-duty check.

Step-by-step solution

1. Convert speed and find torque. \(\omega_0=900(2\pi/60)=94.248\ \text{rad/s}\). Thus \(\alpha=(0-\omega_0)/4.00=-23.562\ \text{rad/s}^2\) and brake-torque magnitude is \(I|\alpha|=424.1\ \text{N}\!\cdot\!\text{m}\).

2. Calculate event energy. \(E=\tfrac12I\omega_0^2=\tfrac12(18.0)(94.248)^2=79944\ \text{J}=79.94\ \text{kJ}\). With no external work, this is the total heat generated per stop.

3. Estimate rotor pulse rise. \(E_r=\chi E=(0.85)(79944)=67952\ \text{J}\), so \(\Delta T_r=E_r/(mc_p)=67952/[(24.0)(460)]=6.16\ \text{K}\) per stop.

4. Screen repeated duty. Rotor average heat is \(\overline{\dot Q}_r=67952/25.0=2.718\ \text{kW}\), below \(3.20\ \text{kW}\) by \(0.482\ \text{kW}\). Hot torque margin is \(500/424.1=1.179\).

Verification

Units: \((\text{kg}\!\cdot\!\text{m}^2)(\text{rad/s}^2)=\text{N}\!\cdot\!\text{m}\); \(I\omega^2\) is joules, and \(\text{J}/[\text{kg}(\text{J/(kg}\!\cdot\!\text{K)})]=\text{K}\).

Signs and direction: Angular acceleration and brake torque are negative under the positive-rotation convention. Capacity and heat are compared as positive magnitudes; a positive margin indicates the supplied limit exceeds demand.

Plausibility: Constant torque does work \(T(\omega_0t_s/2)=(424.1)(188.50)=79.94\ \text{kJ}\), independently matching the kinetic-energy loss. The rotor fraction is below one and its average heat is below total event heat per cycle.

Final result: \(\boxed{T_B=424\ \text{N}\!\cdot\!\text{m},\quad E=79.94\ \text{kJ},\quad \Delta T_r=6.16\ \text{K},\quad \overline{\dot Q}_r=2.72\ \text{kW};\ \text{screen passes}}\).

Engineering interpretation

The margins are modest, so this is not a released design. Confirm worst-case inertia, voltage or pneumatic-force variation, hot minimum friction, emergency-stop energy, flash temperature, thermal transients, wear life, guarding, and the supplier's allowable duty by test.

16–10Friction Materials

Lining Selection

A good friction material has a high and stable coefficient of friction, resistance to fade and wear at temperature, mechanical strength, and compatibility with the mating surface. Common linings: molded and woven organics, sintered metal (metallic-ceramic), cermets, and carbon composites for severe duty. Key tabulated properties are the friction coefficient, the maximum allowable pressure, the maximum operating temperature, and the maximum rubbing velocity. Asbestos has been phased out for non-asbestos organic and metallic formulations.

Use supplier curves or test data for the actual mating material, surface finish, lubricant, pressure, sliding speed, and temperature. A single nominal coefficient is not a strength value and should not be used as both a minimum-capacity value and a maximum-actuation value. Check wear, corrosion, environmental restrictions, availability, and inspection criteria.

16–11Miscellaneous Clutches and Couplings

Specialized Devices

Beyond friction clutches: a square-jaw clutch gives positive (non-slip) engagement; an overrunning (one-way) clutch transmits torque in one direction and freewheels in the other (ratchets, sprags, rollers); a centrifugal clutch engages automatically above a threshold speed. Couplings join shafts: rigid couplings demand precise alignment; flexible couplings (elastomeric, gear, disk, Oldham) accommodate misalignment and damp shock; universal joints connect intersecting shafts.

Selection begins with continuous and peak torque, speed, starts, torsional stiffness, backlash, shaft sizes, environment, and quantified angular, parallel, and axial misalignment. A flexible coupling accommodates limited misalignment but does not excuse poor alignment. Verify hub-shaft connection, key or spline stresses, balance grade, guard requirements, fatigue, and manufacturer derating.

16–12Flywheels

Smoothing Speed with Stored Energy

A flywheel absorbs energy when the driving torque exceeds the load and releases it when the load exceeds the drive, smoothing the speed of machines with fluctuating torque (presses, engines). The coefficient of speed fluctuation \(C_s = (\omega_2 - \omega_1)/\omega\) quantifies the allowable speed swing about the mean. The required flywheel inertia follows from the energy swing \(E_2 - E_1\) over one cycle and the permissible \(C_s\).

The energy swing is the maximum-minus-minimum cumulative area under the net-torque-versus-angle curve, not the total absolute area. The equation assumes a small speed fluctuation about the mean. A rim-dominated flywheel may efficiently provide inertia, but rim stress, burst containment, balance, shaft critical speed, bearing load, fatigue, and guarding can govern.

Coefficient of Speed Fluctuation

\[ C_s = \frac{\omega_2 - \omega_1}{\omega}, \qquad \omega = \frac{\omega_1 + \omega_2}{2} \]Eq. (16–62)

Required Flywheel Inertia

\[ E_2 - E_1 = C_s\,I\,\omega^2 \quad\Rightarrow\quad I = \frac{E_2 - E_1}{C_s\,\omega^2} \]Eqs. (16–64, 16–65)

Physical meaning

For fixed allowable speed fluctuation, required inertia falls with the square of mean speed. That attractive scaling is limited by centrifugal stress and rotor dynamics, so increasing speed is a system-level design change rather than a free reduction in mass.

Chapter recap

Start every friction analysis with relative motion, a free body, and a justified pressure model.
Static torque, dynamic energy, temperature, wear, and fatigue are different checks and may govern independently.
Uniform-wear and uniform-pressure disk models require explicit surface condition and interface count.
Coupling selection includes alignment, stiffness, balance, connection strength, environment, and supplier derating.
Flywheel inertia follows from energy fluctuation, while stress, balance, critical speed, and containment limit speed.

Before moving on: you should be able to choose a friction model, assign force directions, compute torque and heat, distinguish demand from rated capacity, and list the standards, catalog data, tolerances, controls, and tests needed before release.