Couplings, clutches and brakes

Why it matters

Couplings join motors to pumps and gearboxes; clutches let a machine start, stop and change gear without stopping the prime mover; brakes absorb kinetic energy safely in hoists, vehicles and machine tools. Clutches and brakes are both friction devices, so the same pressure and wear assumptions give their torque capacity, and these calculations are standard design and GATE problems.

Key ideas

Couplings.

• Rigid couplings (sleeve/muff, split-muff, flange) need accurately aligned shafts. In a flange coupling, torque passes through bolts in shear (fitted bolts) or by friction between the flanges.
• Flexible couplings (bushed-pin, jaw with elastomer, gear, Oldham, universal joint) tolerate some angular, parallel or axial misalignment and damp shock. The Oldham coupling handles parallel offset; the Hooke's (universal) joint handles angular misalignment but gives a fluctuating output speed unless used in pairs.

Friction clutches. Disc (single- or multi-plate), cone and centrifugal clutches transmit torque by friction between surfaces pressed together by an axial force W. Two assumptions give the torque:

• Uniform pressure: valid for new, well-fitted surfaces. Gives a slightly larger friction radius.
• Uniform wear: after running in, wear ∝ p·v is constant, so p·r = constant; the maximum pressure is at the inner radius. Gives smaller (conservative) torque and is used for design. Each pair of contacting surfaces counts once in n. A single-plate automobile clutch has the plate faced on both sides, so n = 2. For a multi-plate clutch, n = (number of plates in contact) − 1.

Cone clutch. The cone (semi-angle α) multiplies the normal force: torque T = μ·W·R_m/sin α, so a small axial force gives large torque, but too small an α makes the cone stick (α is kept above about 12°).

Brakes. A brake converts kinetic or potential energy into heat.

• Block (shoe) brake: a block pressed on a drum; friction force μ·R_n acts at the drum radius. Depending on where the fulcrum is relative to the friction line, the friction moment can help (self-energising) or oppose the applied force. If it alone can hold the lever, the brake is self-locking, which is usually undesirable.
• Band brake: a band wrapped round a drum; tensions follow the belt relation T₁/T₂ = e^(μθ), and the braking torque is (T₁ − T₂)·r. In a simple band brake, the lever force depends on which end (tight or slack) is attached away from the fulcrum, so the direction of drum rotation matters. A differential band brake can be made self-locking.
• Internal expanding shoe (drum) and disc brakes in vehicles; disc brakes dissipate heat better and fade less.

Energy and heat. The energy a brake absorbs (change of kinetic and potential energy) appears as heat; its rate decides the temperature rise and the lining wear, often checked through a p·v limit from a data book.

Formulas

Disc clutch torque: T = n·μ·W·R_f — T (N·m), n pairs of friction surfaces, μ coefficient of friction, W axial force (N), R_f friction radius (m).

Uniform pressure: R_f = (2/3)·(r_o³ − r_i³)/(r_o² − r_i²), W = π·p·(r_o² − r_i²).

Uniform wear: R_f = (r_o + r_i)/2, W = 2π·p_max·r_i·(r_o − r_i) (p_max at r_i, in Pa).

Cone clutch: T = μ·W·R_f/sin α.

Band brake: T₁/T₂ = e^(μθ) (θ in rad); braking torque T_B = (T₁ − T₂)·r.

Block brake: T_B = μ·R_n·r (R_n normal force on the block).

Power: P = 2π·N·T/60 (W, N in rev/min).

Flange coupling with n fitted bolts of diameter d_b on pitch-circle diameter D: T = n·(π/4)·d_b²·τ·(D/2).

Worked examples

Example 1 (standard). A single-plate clutch, faced on both sides, has outer and inner diameters 300 mm and 200 mm, μ = 0.3 and an allowable maximum pressure of 0.1 MPa. Assuming uniform wear, find the axial force, the torque and the power at 1500 rev/min.

1. r_o = 150 mm, r_i = 100 mm, n = 2.
2. Axial force: W = 2π·p_max·r_i·(r_o − r_i) = 2π × 0.1 × 100 × 50 = 3142 N (MPa × mm² = N).
3. Friction radius: R_f = (150 + 100)/2 = 125 mm.
4. Torque: T = n·μ·W·R_f = 2 × 0.3 × 3142 × 0.125 = 235.6 N·m.
5. Power: P = 2π × 1500 × 235.6/60 = 37.0 kW.
6. W ≈ 3.14 kN, T ≈ 236 N·m, P ≈ 37 kW. (Uniform pressure would give R_f = 126.7 mm, slightly higher.)

Example 2 (GATE level). A simple band brake has a drum of 500 mm diameter, a wrap angle of 270° and μ = 0.25. It must provide a braking torque of 300 N·m. The tight end of the band is attached to the fulcrum of the lever; the slack end is attached 100 mm from the fulcrum; the operating force acts 500 mm from the fulcrum. Find the band tensions and the operating force.

1. θ = 270° = 4.712 rad; T₁/T₂ = e^(0.25 × 4.712) = e^1.178 = 3.248.
2. T₁ − T₂ = T_B/r = 300/0.25 = 1200 N.
3. T₂ = 1200/(3.248 − 1) = 533.8 N; T₁ = 3.248 × 533.8 = 1733.8 N.
4. The tight end passes through the fulcrum (no moment). Moments about the fulcrum: P × 500 = T₂ × 100.
5. P = 533.8 × 100/500 = 106.8 N.
6. T₁ ≈ 1734 N, T₂ ≈ 534 N, P ≈ 107 N. If the drum turned the other way, the slack and tight ends would swap and P would be 1734 × 100/500 ≈ 347 N.

Common mistakes

• Using n = 1 for a single-plate clutch faced on both sides.
• Using uniform-pressure results for design of a worn clutch; uniform wear is the safe assumption.
• Forgetting sin α in the cone clutch formula, or using the full cone angle.
• Using θ in degrees in e^(μθ).
• Taking the braking torque as a lever force times lever length; braking torque is the friction force times the drum radius.
• Ignoring the direction of drum rotation in band and block brakes.

For GATE PI

Expect torque capacity of disc and cone clutches under uniform wear or pressure, number of friction surfaces in multi-plate clutches, band tensions and lever force in band brakes, block-brake self-locking conditions, and bolt sizing of flange couplings. Practise drawing the lever free-body diagram and labelling tight and slack sides first.

Quick check

1. A clutch has μ = 0.4, W = 1500 N, R_f = 0.2 m and one pair of surfaces. Torque?
2. Which assumption gives the larger torque for the same W: uniform pressure or uniform wear?
3. A band brake has T₁/T₂ = e^(μθ) with μ = 0.3 and θ = π. What is T₁/T₂?
4. Which coupling connects parallel shafts with a small lateral offset?

Answers: 1. 120 N·m; 2. uniform pressure; 3. 2.57; 4. the Oldham coupling.