Drum and disc brakes

Why it matters

The brake at each wheel is where the vehicle's kinetic energy is turned into heat. Whether it is a drum or a disc decides how much torque a given line pressure produces, how stable that torque is as the linings heat up or get wet, and how quickly the heat escapes. Modern cars use discs at the front and discs or drums at the rear; trucks and buses still use large drums widely, though air-operated discs are spreading.

Key ideas

Drum brake. A cast-iron drum turns with the wheel. Two curved shoes lined with friction material sit on a fixed back plate and are pushed outward against the inside of the drum by a hydraulic wheel cylinder (cars) or an S-cam (air brakes). Return springs pull the shoes back; an adjuster takes up lining wear.

• Leading shoe — the shoe whose tip is pushed in the direction of drum rotation. Friction drags it harder into the drum, so its friction moment assists the actuating force: self-energising.
• Trailing shoe — friction tends to push it away from the drum: de-energising. A leading–trailing brake gives the same torque in both directions of rotation.
• Two-leading-shoe (two wheel cylinders) — more torque forward, much less in reverse. Duo-servo — the first shoe's friction force is passed to the second, giving very high gain; used for parking brakes.
• Brake factor = total drum friction force ÷ actuating force. A high brake factor means low pedal effort but high sensitivity to changes in μ: as linings heat and μ falls, torque drops sharply (fade). The drum also expands when hot, so the pedal travels further.
• Advantages: cheap, high brake factor, enclosed (dust and mud kept out), easy to add a parking brake. Disadvantages: poor heat dissipation, fade, water and dirt trapped inside, more parts, needs adjustment.

Disc brake. A disc (rotor) turns with the wheel; a caliper straddles it and hydraulic pistons press friction pads on both faces.

• Fixed (opposed-piston) caliper — pistons on both sides; rigid, used on performance and heavy vehicles.
• Floating (sliding) caliper — piston(s) on the inboard side only; the caliper slides on guide pins and pulls the outer pad against the disc. Lighter, cheaper, more compact; most cars.
• Ventilated discs have radial vanes between two faces to pump cooling air; solid discs are used at the rear and on small cars. Material: grey cast iron; carbon–ceramic on some high-performance cars.
• Advantages: friction surfaces exposed to air (better cooling, less fade), brake factor ≈ 2μ (torque nearly proportional to pressure, stable), self-cleaning and quick recovery when wet, self-adjusting through piston-seal rollback, easy pad change. Disadvantages: no self-energising — needs higher clamping forces and usually a booster; more exposed to dirt; parking brake more complex.

Fade, glazing and heat. Repeated hard stops raise lining temperature; above the material's working range μ drops (fade). The brakes must store the energy of a stop and then shed it to air, so disc mass, ventilation and airflow matter.

Formulas

F_c = p · A_p

• F_c = clamping (piston) force (N), p = line pressure (Pa), A_p = piston area (m²); for a floating caliper with one piston, this force acts on both pads.

T_disc = n · μ · F_c · r_e

• T_disc = brake torque (N·m), n = number of friction faces (2 for a normal caliper), μ = pad–disc friction coefficient, r_e = effective radius (m).

r_e = (r_o + r_i) / 2 (uniform wear), r_e = (2/3)·(r_o³ − r_i³) / (r_o² − r_i²) (uniform pressure)

• r_o, r_i = outer and inner radii of the pad's contact (m). Uniform wear is the usual assumption for worn-in pads.

F_b = T / R

• F_b = braking force at the tyre (N), R = tyre rolling radius (m).

N_lead = P·a / (c − μ·d), N_trail = P·a / (c + μ·d)

• N = normal force on a short, pivoted shoe (N), P = actuating force (N), a = moment arm of P about the shoe pivot (m), c = moment arm of the normal force about the pivot (m), d = moment arm of the friction force about the pivot (m). Short-shoe (concentrated force) model; if c ≤ μ·d the leading shoe would self-lock.

T_drum = μ·(N_lead + N_trail)·r

• r = drum radius (m).

E = ½·m·v², ΔT = E_disc / (m_d · c_p)

• E = kinetic energy of the vehicle (J), ΔT = temperature rise of one disc (K) if it absorbs E_disc in a single stop with no cooling, m_d = disc mass (kg), c_p = specific heat (J/kg·K; about 460 for cast iron).

Worked examples

Example 1 (standard) — disc brake torque. A floating caliper has one 54 mm piston. Line pressure is 6 MPa, μ = 0.38, the pads contact between radii 90 mm and 140 mm, and the tyre rolling radius is 0.30 m. Assuming uniform wear, find the brake torque and the braking force at the tyre.

1. A_p = π × 0.054² / 4 = 2.290 × 10⁻³ m².
2. F_c = p · A_p = 6 × 10⁶ × 2.290 × 10⁻³ = 13 740 N.
3. r_e = (r_o + r_i)/2 = (0.14 + 0.09)/2 = 0.115 m.
4. T_disc = n·μ·F_c·r_e = 2 × 0.38 × 13 740 × 0.115 = 1201 N·m.
5. F_b = T / R = 1201 / 0.30 = 4003 N. T ≈ 1.20 kN·m, F_b ≈ 4.0 kN per wheel.

Example 2 (GATE level) — leading and trailing shoes. Each shoe of a drum brake is pivoted at one end and pushed at the other by P = 1000 N. About the pivot, P acts at 0.20 m, the resultant normal force at 0.10 m and the friction force at 0.08 m. μ = 0.35 and the drum radius is 0.12 m. Find the normal force on each shoe and the total brake torque.

1. Leading shoe (friction moment assists P): N_lead = P·a/(c − μ·d) = 1000 × 0.20 / (0.10 − 0.35 × 0.08) = 200 / 0.072 = 2778 N.
2. Trailing shoe (friction moment opposes P): N_trail = P·a/(c + μ·d) = 200 / 0.128 = 1563 N.
3. Torques: leading 0.35 × 2778 × 0.12 = 116.7 N·m; trailing 0.35 × 1563 × 0.12 = 65.6 N·m.
4. T_drum = 116.7 + 65.6 = 182.3 N·m. N_lead ≈ 2.78 kN, N_trail ≈ 1.56 kN, T ≈ 182 N·m — the leading shoe gives 1.78 times the torque of the trailing shoe from the same actuating force.

Example 3 — temperature rise in one stop. A 1200 kg car stops from 100 km/h; the front axle takes 70 % of the braking, shared by two 6 kg cast-iron discs (c_p = 460 J/kg·K). E = ½ × 1200 × 27.78² = 463 kJ; per front disc 0.35 × 463 = 162 kJ; ΔT = 162 000 / (6 × 460) = 58.7 K per stop, before cooling — which is why repeated stops cause fade.

Common mistakes

• Using one friction face instead of two for a caliper (n = 2).
• Thinking the caliper force equals the braking force at the tyre — convert through μ, r_e and R.
• Swapping the signs for leading and trailing shoes: the leading shoe has (c − μd) in the denominator.
• Calling a high brake factor purely good; it also makes the brake sensitive to μ and prone to fade.
• Saying drum brakes are always rear-only: heavy trucks use drums on all axles.
• Forgetting to convert piston diameter in mm to m² area.

For GATE ME

This topic maps directly onto machine-design brake questions: short-shoe and long-shoe drum brakes, self-energising and self-locking conditions, band and block brakes, disc brake torque with uniform pressure or uniform wear, and energy and temperature rise. Practise taking moments about the shoe pivot with the correct direction of friction.

Quick check

1. Which shoe is self-energising?
2. How many friction faces does a normal caliper brake have?
3. What is the uniform-wear effective radius for pads between 80 mm and 120 mm?
4. Why does a disc recover quickly when wet?
5. What is the self-locking condition for a leading shoe in the short-shoe model?

Answers: 1. The leading shoe. 2. Two. 3. 100 mm. 4. Centrifugal action and pad wiping throw water off its exposed faces. 5. c ≤ μ·d.