Design of drum and disc brakes

Vehicle braking energy and torque, disc and drum brakes, lever-operated block brakes with self-energising and self-locking, long shoes and band brakes, and temperature rise, with a car disc-brake and a block-brake example.

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Why it matters

Brakes turn the vehicle's kinetic energy into heat, and they must do so repeatedly without fading, locking or overheating the fluid. Designing a brake means finding the torque each wheel needs, the clamp or actuating force that produces it, and the temperature rise of the drum or disc. Lever-operated block brakes and band brakes are also standard GATE problems because they show self-energising and self-locking clearly.

Key ideas

Energy and torque: stopping a vehicle of mass m from speed v absorbs ½·m·v² (plus rotating-mass and gradient terms). For a deceleration a, the total braking force at the tyres is m·a, shared between axles according to the dynamic load transfer during braking (typically 60–75 % at the front of a car). The brake torque per wheel is that wheel's force times the rolling radius.

Disc brakes: a caliper squeezes two pads onto the two faces of a rotating disc, so there are two friction surfaces. Torque T = 2·μ·W·R_f, with R_f the effective pad radius. Discs are exposed to air, so they cool well and resist fade; ventilated discs have internal vanes. Their torque is proportional to μ, so it is stable (no self-energising), and they shed water quickly.

Drum brakes: shoes are pushed outward against the inside of a rotating drum. A leading shoe is pulled into the drum by friction (self-energising), so it gives more torque for the same actuating force; a trailing shoe is pushed away and gives less. Drum brakes are compact, cheap and suit parking brakes, but they fade more because heat is trapped and the drum expands away from the shoes.

Block (shoe) brake with a lever: a block pressed on a drum of radius r with normal force N gives friction μ·N and torque T = μ·N·r. Taking moments about the lever fulcrum: if the friction force's moment about the fulcrum helps the applied force, the brake is self-energising, N = P·l/(b − μ·a); if b ≤ μ·a, no applied force is needed and the brake is self-locking (undesirable, it grabs). For the opposite rotation, N = P·l/(b + μ·a). Here l is the lever arm of P, b the arm of N, and a the distance of the friction force's line from the fulcrum.

Long shoes: when the contact angle 2θ exceeds about 60°, pressure is not uniform and the equivalent friction coefficient is μ′ = 4μ·sin θ/(2θ + sin 2θ).

Band brake: a flexible band wrapped through angle θ: T₁/T₂ = e^(μθ), and braking torque (T₁ − T₂)·r.

Heat and fade: temperature rise of a disc or drum for one stop ≈ energy absorbed / (m·c), if all the heat stays in it. High temperatures reduce lining friction (fade) and can boil brake fluid. Data books give limits on lining pressure and on the power absorbed per unit lining area.

Formulas

Kinetic energy: E = ½·m·v² Braking force: F_b = m·a; per wheel torque T = F_wheel·R_w Disc brake: T = 2·μ·W·R_f Block brake: T = μ·N·r Lever: N = P·l / (b ∓ μ·a); self-locking if b ≤ μ·a Long shoe: μ′ = 4μ·sin θ / (2θ + sin 2θ) Band brake: T₁/T₂ = e^(μ·θ), T = (T₁ − T₂)·r Temperature rise: ΔT = E / (m_d·c)

Symbols: m = vehicle mass (kg); v = speed (m/s); a = deceleration (m/s²); F_b = braking force (N); R_w = rolling radius (m); μ = friction coefficient (–); W = pad clamp force per pad (N); R_f = effective pad radius (m); N = normal force on block (N); r = drum radius (m); P = applied lever force (N); l, b, a = moment arms of P, N and friction force about the fulcrum (m); 2θ = shoe contact angle (rad); T₁, T₂ = tight- and slack-side band tensions (N); m_d = disc or drum mass (kg); c = specific heat (J/kg·K); ΔT = temperature rise (K).

Worked examples

Example 1 (standard). A 1200 kg car brakes from 90 km/h (25 m/s) to rest at 7 m/s². The front axle provides 70 % of the braking. Rolling radius 0.3 m. Each front disc has pads at R_f = 0.11 m, μ = 0.4, and the disc mass is 6 kg (c = 460 J/kg·K). Find the clamp force per caliper and the temperature rise of a front disc.

  1. Total braking force = 1200 × 7 = 8400 N; per front wheel = 0.7 × 8400 / 2 = 2940 N.
  2. Torque per front wheel = 2940 × 0.3 = 882 N·m.
  3. Clamp force: W = T/(2·μ·R_f) = 882 / (2 × 0.4 × 0.11) = 10.0 kN.
  4. Energy: E = ½ × 1200 × 25² = 375 kJ; per front disc = 0.7 × 375/2 = 131.25 kJ.
  5. ΔT = 131 250 / (6 × 460) = 47.6 K for one stop, assuming all heat stays in the disc.

Example 2 (GATE level). A lever-operated block brake acts on a drum of 200 mm radius; μ = 0.3. The applied force P = 500 N acts 1000 mm from the fulcrum, the block is 400 mm from the fulcrum, and the line of the friction force is 50 mm from the fulcrum. Find the braking torque for both directions of rotation and check for self-locking.

  1. Rotation in which friction assists P: N = 500 × 1000 / (400 − 0.3 × 50) = 500 000/385 = 1298.7 N; T = 0.3 × 1298.7 × 0.2 = 77.9 N·m.
  2. Opposite rotation: N = 500 000 / (400 + 15) = 1204.8 N; T = 0.3 × 1204.8 × 0.2 = 72.3 N·m.
  3. Self-locking needs b ≤ μ·a, i.e. 400 ≤ 15 mm, which is false, so the brake is not self-locking.

Common mistakes

  • Using one friction surface for a disc brake; the pads grip both faces.
  • Using the drum or disc outer radius instead of the pad effective radius.
  • Getting the sign of μ·a wrong: decide from the figure whether the friction moment helps or opposes the applied force.
  • Using degrees instead of radians in e^(μθ) or in the long-shoe formula.
  • Assuming the rear brakes share braking equally with the front; load transfers forward when braking.
  • Treating self-locking as desirable in a service brake.

For GATE ME

Expect: braking torque of block, band and band-and-block brakes; lever force for a required torque; self-energising and self-locking conditions; energy absorbed and temperature rise; and stopping force or distance. Practise free-body diagrams of the lever for both rotation directions.

Quick check

  1. A car of 1000 kg stops from 20 m/s. Energy absorbed?
  2. Disc brake: μ = 0.35, W = 5 kN, R_f = 0.1 m. Torque?
  3. Band brake: μ = 0.3, θ = 270°. Find T₁/T₂.
  4. When is a lever block brake self-locking?
  5. Why do disc brakes fade less than drum brakes?

Answers: 1. 200 kJ. 2. 2 × 0.35 × 5000 × 0.1 = 350 N·m. 3. e^(0.3 × 4.712) = 4.11. 4. When b ≤ μ·a, so the friction moment alone holds the shoe on. 5. The disc is exposed to cooling air, and its expansion does not move it away from the pads.

Interview questions

All Design of Machine and Automotive Elements interview questions

Try answering each one aloud before you open it.

  1. 1.What is a drum brake and how does it work?Concept

    A drum brake is a type of brake that uses friction caused by a set of shoes or pads that press outward against a rotating cylinder-shaped part called a drum. When the brake pedal is pressed, hydraulic fluid is forced into the wheel cylinder, pushing the brake shoes against the drum. This friction slows down the rotation of the wheel, thereby stopping the vehicle.

  2. 2.Explain the working principle of a disc brake.Concept

    A disc brake works by using calipers to squeeze pairs of pads against a disc or rotor to create friction. This action slows the rotation of a shaft, such as a vehicle axle, either to reduce its rotational speed or to hold it stationary. The friction between the pads and the disc converts kinetic energy into thermal energy, which is dissipated into the atmosphere.

  3. 3.What are the main differences between drum brakes and disc brakes?Concept

    Drum brakes are enclosed and use brake shoes that press outward against a drum, while disc brakes are open and use brake pads that squeeze against a rotor. Disc brakes generally provide better stopping performance, especially in wet conditions, and are more resistant to brake fade. Drum brakes, however, are typically less expensive and can provide more braking force in a smaller package, making them suitable for rear-wheel applications.

  4. 4.Why are disc brakes preferred over drum brakes in high-performance vehicles?Application

    Disc brakes are preferred in high-performance vehicles because they offer better heat dissipation, which reduces the risk of brake fade during repeated or sustained braking. They also provide more consistent braking performance in various conditions, including wet weather, due to their open design that allows water to be expelled more easily.

  5. 5.What happens if the brake fluid in a hydraulic brake system is not maintained properly?Application

    If the brake fluid is not maintained properly, it can absorb moisture from the air, leading to a lower boiling point. This can cause vapor lock, where the fluid boils and creates gas bubbles, leading to a spongy brake pedal and reduced braking efficiency. Contaminated fluid can also corrode brake components, leading to leaks and brake failure.

  6. 6.Why is it important to have a ventilated disc in a disc brake system?Application

    A ventilated disc is important because it helps dissipate heat more effectively than a solid disc. The ventilation allows air to flow through the disc, cooling it down and reducing the risk of overheating and brake fade. This is particularly important in high-performance or heavy-duty applications where brakes are subjected to high thermal loads.

  7. 7.What materials are commonly used for brake pads and why?Application

    Brake pads are commonly made from materials such as semi-metallic, ceramic, and organic compounds. Semi-metallic pads are durable and provide good heat transfer, making them suitable for high-performance applications. Ceramic pads offer quieter operation and produce less dust, while organic pads are softer and provide a smoother braking experience but may wear out faster.

  8. 8.Calculate the braking force required to stop a vehicle with a mass of 1500 kg traveling at 20 m/s within a distance of 50 meters.Numerical

    To calculate the braking force, use the work-energy principle: Work done = Change in kinetic energy. Initial kinetic energy = 0.5 * m * v^2 = 0.5 * 1500 * 20^2 = 300,000 J. Final kinetic energy = 0 J (since the vehicle stops). Work done = Force * distance = 300,000 J. Therefore, Force = 300,000 J / 50 m = 6000 N.

  9. 9.A disc brake rotor has a diameter of 0.3 meters and a thickness of 0.02 meters. If the rotor is made of cast iron with a density of 7200 kg/m³, calculate its mass.Numerical

    Treating the rotor as a solid disc, its volume is V = π·(d/2)²·t = π × 0.15² × 0.02 = 1.414 × 10⁻³ m³. Its mass is m = ρ·V = 7200 × 1.414 × 10⁻³ ≈ 10.2 kg. A real rotor is lighter because of its centre bore, hat section and ventilation vanes; the mass matters because it sets the temperature rise per stop, ΔT = E/(m·c).

  10. 10.Explain the role of the master cylinder in a hydraulic brake system.Concept

    The master cylinder is a critical component in a hydraulic brake system. It converts the mechanical force from the brake pedal into hydraulic pressure. When the brake pedal is pressed, the master cylinder pushes brake fluid through the brake lines to the wheel cylinders or calipers, which then apply the brakes. It ensures that the force applied by the driver is transmitted efficiently to the brakes.

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