10.4 Clutches, Brakes & Power Screws

10.4 Clutches, Brakes & Power Screws

Friction clutches and brakes start, stop and hold machinery by converting kinetic energy into heat at a sliding interface, and power screws convert torque into large axial forces. All three depend on the friction coefficient and on how the geometry lets friction either help or oppose the applied force.

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Friction Clutches & Brakes

Friction clutches and brakes control the transmission of torque through dry or lubricated friction interfaces.

Plate and Multi-Disk Clutches

Consider annular friction disks with inner radius $r_i$ and outer radius $r_o$ clamped together by axial actuating force $F$ across $N_p$ active friction interfaces ($N_p = N_{\text{driving}} + N_{\text{driven}} - 1$). Torque capacity is calculated under two classic assumptions:

1. Uniform Wear Theory (Worn-In Steady State)

• Premise: Wear rate is proportional to frictional work, which is the product of local pressure and sliding velocity: $\text{Wear} \propto p \cdot v \propto p \cdot r = C$ (constant). Once broken in, wear distributes uniformly across the disk surface, forcing pressure to vary inversely with radius: $p(r) = C/r$. Peak pressure $p_{\max}$ occurs at the inner radius $r_i$.
• Clamping Force ($F$):

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

• Torque Capacity ($T$):

$$T = \frac{1}{2} \mu F (r_o + r_i) N_p = \mu F r_m N_p$$

where the mean friction radius is:

$$r_m = \frac{r_o + r_i}{2}$$

2. Uniform Pressure Theory (New, Rigid Disks)

• Premise: Assumes brand-new, perfectly flat and rigid friction plates where normal clamping pressure is uniform across the entire face: $p = \text{const} = p_{\max}$.
• Clamping Force ($F$):

$$F = p \pi (r_o^2 - r_i^2)$$

• Torque Capacity ($T$):

$$T = \frac{2}{3} \mu F \left( \frac{r_o^3 - r_i^3}{r_o^2 - r_i^2} \right) N_p$$

Band Brakes & Self-Locking

A flexible friction band wrapped around a rotating brake drum of radius $r$ through wrap angle $\theta$ (in radians) obeys the capstan tension relationship:

$$\frac{F_1}{F_2} = e^{f \theta}$$

where $F_1$ is tight-side band tension and $F_2$ is slack-side band tension. Braking torque is:

$$T = (F_1 - F_2) r = F_2 (e^{f\theta} - 1) r$$

Depending on which band end connects to the operating lever relative to the lever pivot, the frictional drag of the rotating drum either assists the operator (self-energizing) or opposes the operator:

• Self-Energizing: Drum rotation pulls the band toward the applied lever force, multiplying braking torque for a given hand force.
• Self-Locking: Occurs when the self-energizing moment equals or exceeds the restoring moment about the pivot. The required actuating force drops to zero or becomes negative ($F_{\text{act}} \le 0$). Once engaged, the brake grabs and locks automatically without external force; it cannot be disengaged until rotation reverses.

Block (Shoe) Brakes

A short block pressed against a drum of radius $r$ with normal force $N$ produces friction force $fN$ and braking torque $T = f N r$. When the shoe is carried on a pivoted lever, take moments about the pivot. Let the actuating force $F$ act at distance $c$ from the pivot, the normal force $N$ at distance $b$, and the friction force $fN$ along a line offset $a$ from the pivot:

$$F c = N b \mp f N a \quad \Longrightarrow \quad N = \frac{F c}{b \mp f a}$$

• Self-energizing (minus sign): drum rotation drags the shoe toward the drum, so less hand force is needed. The brake becomes self-locking when $b \le f a$, because the required $F$ drops to zero or below.
• De-energizing (plus sign): rotation in the other direction pushes the shoe away, so more force is needed.
• A double-block brake uses two opposed shoes so the radial forces on the shaft cancel; its torque is $T = 2 f N r$ for equal shoe forces.

Long shoes on drum brakes (internal expanding or external contracting) do not have uniform pressure; their analysis assumes pressure proportional to $\sin\theta$ around the drum, as tabulated in machine-design texts.

Mini-example: A double-block brake presses each shoe with $N = 4.0\,\text{kN}$ against a $250\,\text{mm}$-radius drum with $f = 0.30$, giving $T = 2(0.30)(4{,}000)(0.250) = 600\,\text{N}\cdot\text{m}$. If one lever has $c = 900\,\text{mm}$, $b = 300\,\text{mm}$ and $a = 50\,\text{mm}$, the self-energizing shoe needs $F = N(b - fa)/c = 4{,}000(300 - 15)/900 = 1{,}267\,\text{N}$, while the de-energizing direction needs $4{,}000(315)/900 = 1{,}400\,\text{N}$. Self-locking would require $b \le fa = 15\,\text{mm}$, so this brake is safe from grabbing.

Disc (Caliper) Brakes

A caliper squeezes pads against both faces of a rotor with clamping force $F$. For annular pads the effective friction radius follows the same two models as plate clutches: $r_e = (r_o + r_i)/2$ for uniform wear and $r_e = \tfrac{2}{3}(r_o^3 - r_i^3)/(r_o^2 - r_i^2)$ for uniform pressure. With one pad on each face,

$$T = 2 f F r_e$$

Disc brakes resist fade better than drums because the exposed rotor sheds heat quickly and is not self-energizing, so torque stays nearly proportional to clamp force as friction changes.

Thermal Energy Dissipation & Brake Fade

During a single stopping cycle, mechanical kinetic energy ($E_k = \frac{1}{2} I \omega^2$ or $\frac{1}{2} m v^2$) converts directly into thermal energy in the brake drum/disk:

$$\Delta T = \frac{E_k}{m_b \cdot c_p}$$

where $m_b$ is the mass of the brake rotor/drum and $c_p$ is its specific heat capacity. If repetitive cycling causes interface temperature to exceed $250^\circ - 300^\circ\text{C}$, the friction coefficient drops sharply—a catastrophic phenomenon known as brake fade.

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Power Screws: Square & Acme Threads

Power screws convert rotational torque into high-force linear motion for jacks, presses, machine tool lead screws, and tensile testing frames.

Thread Geometry

• Square Threads: Thread flank angle $\alpha = 0^\circ$. Delivers maximum mechanical efficiency (no radial bursting component), but cannot be adjusted for wear and is difficult to cut with split nuts.
• Acme Threads: Standardized with a $29^\circ$ included angle (flank angle $\alpha = 14.5^\circ$, $\sec\alpha = \sec 14.5^\circ = 1.03294$). Permits split nuts to take up wear and provides higher shear strength at the root.
• Mean Diameter ($d_m$): For nominal major diameter $d$ and thread pitch $p$:

$$d_m = d - \frac{p}{2}$$

• Lead ($l$): Axial advance per one revolution. For a single-start screw, $l = p$; for a double-start screw, $l = 2p$.
• Lead Angle ($\lambda$):

$$\tan\lambda = \frac{l}{\pi d_m}$$

Torque to Raise Load ($T_R$)

Raising an axial compressive load $F$ requires overcoming thread friction on the inclined plane plus collar thrust bearing friction:

$$T_R = \frac{F d_m}{2} \left( \frac{l + \pi f d_m \sec\alpha}{\pi d_m - f l \sec\alpha} \right) + \frac{F f_c d_c}{2}$$

where $f$ is the coefficient of thread friction, $f_c$ is the collar friction coefficient, and $d_c$ is the mean diameter of the thrust collar washer.

Torque to Lower Load ($T_L$)

When lowering the load, the axial force assists downward motion, reversing the lead term:

$$T_L = \frac{F d_m}{2} \left( \frac{\pi f d_m \sec\alpha - l}{\pi d_m + f l \sec\alpha} \right) + \frac{F f_c d_c}{2}$$

The Self-Locking Condition

A power screw is self-locking if torque $T_L$ remains strictly positive ($T_L > 0$) without collar friction ($f_c = 0$). This ensures the screw holds its load statically without backdriving or spinning downward:

$$\pi f d_m \sec\alpha - l > 0 \iff \tan\lambda \le f \sec\alpha$$

• If $\tan\lambda \le f \sec\alpha$, the screw is self-locking.
• If $\tan\lambda > f \sec\alpha$, the screw is overhauling (backdriveable); an external brake is required to prevent the load from driving the screw.

Mechanical Lifting Efficiency ($\eta$)

The overall mechanical efficiency of a power screw compares ideal work to actual input work:

$$\eta = \frac{F \cdot l}{2\pi T_R}$$

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Worked Engineering Example: Acme Power Screw Jack Analysis

Problem Statement

A heavy-duty screw jack utilizes a single-start Acme thread to raise a vertical compressive load of $F = 25\,\text{kN}$ ($25,000\,\text{N}$). Design specifications:

• Major screw diameter: $d = 36\,\text{mm}$
• Pitch: $p = 6\,\text{mm}$ (single-start $\implies$ lead $l = 6\,\text{mm} = 0.006\,\text{m}$)
• Flank angle: $\alpha = 14.5^\circ$ ($\sec\alpha = 1.03294$)
• Thrust collar mean diameter: $d_c = 50\,\text{mm} = 0.050\,\text{m}$
• Coefficient of thread friction: $f = 0.14$
• Coefficient of collar friction: $f_c = 0.10$

Determine:

1. The mean thread diameter $d_m$ and lead angle $\lambda$.
2. Whether the screw is self-locking.
3. The total torque required to raise the load ($T_R$).
4. The total torque required to lower the load ($T_L$).
5. The overall mechanical lifting efficiency $\eta$.

Step-by-Step Solution

Step 1: Geometric Parameters

Compute mean thread diameter:

$$d_m = d - \frac{p}{2} = 36\,\text{mm} - 3\,\text{mm} = 33\,\text{mm} = 0.033\,\text{m}$$

Compute lead angle tangent $\tan\lambda$ and angle $\lambda$:

$$\tan\lambda = \frac{l}{\pi d_m} = \frac{0.006\,\text{m}}{\pi (0.033\,\text{m})} = \frac{0.006}{0.103673} = 0.057875$$ $$\lambda = \arctan(0.057875) = 3.313^\circ$$

Step 2: Verify the Self-Locking Condition

Evaluate the self-locking threshold $f \sec\alpha$:

$$f \sec\alpha = 0.14 \times \sec(14.5^\circ) = 0.14 \times 1.03294 = 0.14461$$

Comparing:

$$\tan\lambda = 0.05788 < f \sec\alpha = 0.14461$$

Because $\tan\lambda < f \sec\alpha$, the screw is positively self-locking. The load cannot backdrive the screw.

Step 3: Compute Raising Torque ($T_R$)

Compute thread torque $T_t$:

$$T_t = \frac{F d_m}{2} \left( \frac{l + \pi f d_m \sec\alpha}{\pi d_m - f l \sec\alpha} \right)$$

Evaluate numerator and denominator:

• Numerator: $l + \pi f d_m \sec\alpha = 0.006 + \pi(0.14)(0.033)(1.03294) = 0.006 + 0.014992 = 0.020992\,\text{m}$
• Denominator: $\pi d_m - f l \sec\alpha = \pi(0.033) - (0.14)(0.006)(1.03294) = 0.103673 - 0.000868 = 0.102805\,\text{m}$

$$\frac{\text{Numerator}}{\text{Denominator}} = \frac{0.020992}{0.102805} = 0.20419$$ $$T_t = \frac{(25,000\,\text{N})(0.033\,\text{m})}{2} \times 0.20419 = 412.5 \times 0.20419 = 84.23\,\text{N}\cdot\text{m}$$

Compute collar friction torque $T_c$:

$$T_c = \frac{F f_c d_c}{2} = \frac{(25,000\,\text{N})(0.10)(0.050\,\text{m})}{2} = 62.50\,\text{N}\cdot\text{m}$$

Total raising torque:

$$T_R = T_t + T_c = 84.23 + 62.50 = 146.73\,\text{N}\cdot\text{m} \approx 146.7\,\text{N}\cdot\text{m}$$

Step 4: Compute Lowering Torque ($T_L$)

Compute thread lowering torque $T_{t,\text{lower}}$:

$$T_{t,\text{lower}} = \frac{F d_m}{2} \left( \frac{\pi f d_m \sec\alpha - l}{\pi d_m + f l \sec\alpha} \right)$$

• Numerator: $\pi f d_m \sec\alpha - l = 0.014992 - 0.006 = 0.008992\,\text{m}$
• Denominator: $\pi d_m + f l \sec\alpha = 0.103673 + 0.000868 = 0.104541\,\text{m}$

$$\frac{\text{Numerator}}{\text{Denominator}} = \frac{0.008992}{0.104541} = 0.08601$$ $$T_{t,\text{lower}} = 412.5 \times 0.08601 = 35.48\,\text{N}\cdot\text{m}$$

Total lowering torque (collar friction opposes motion):

$$T_L = T_{t,\text{lower}} + T_c = 35.48 + 62.50 = 97.98\,\text{N}\cdot\text{m} \approx 98.0\,\text{N}\cdot\text{m}$$

(A positive lowering torque of $98.0\,\text{N}\cdot\text{m}$ confirms that manual effort must be exerted to lower the jack).

Step 5: Compute Overall Mechanical Efficiency ($\eta$)

$$\eta = \frac{F \cdot l}{2\pi T_R} = \frac{(25,000\,\text{N})(0.006\,\text{m})}{2\pi (146.73\,\text{N}\cdot\text{m})} = \frac{150\,\text{J}}{921.93\,\text{J}} = 0.1627 \approx 16.3\%$$

The thread alone has an efficiency of $\eta_{\text{thread}} = 150 / (2\pi \times 84.23) = 28.3\%$; collar friction reduces the net jack efficiency to $16.3\%$.