Keys and couplings

Types of keys and couplings, key design in shear and crushing, and rigid flange coupling checks, with key-length and flange-coupling bolt examples.

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Why it matters

A gear, pulley, flywheel or propeller-shaft flange is useless unless it is locked to its shaft. Keys, splines and couplings carry the full engine torque through very small areas, so they are classic weak links. They are also deliberately made the weaker part in some designs, so that a cheap key shears before an expensive shaft or gear is damaged.

Key ideas

Keys sit partly in a keyway in the shaft and partly in the hub.

  • Sunk keys (rectangular, square, parallel or taper): the commonest. For a standard proportion, width b ≈ d/4 and height h ≈ 2b/3 (rectangular) or h = b (square); take exact sizes from the standard for the shaft diameter.
  • Feather key: a parallel key fixed to the shaft or hub that lets the hub slide axially (gear-shifting sleeves).
  • Woodruff key: a semicircular disc in a curved seat. It can rock in its seat and self-align with the hub keyway, so it suits tapered shafts and light torque (e.g. small pulleys and alternator drives), but its deep seat weakens the shaft.
  • Saddle and round keys: friction or light-duty use only.
  • Splines: several integral keys; used where torque is high or the hub slides (gearbox main shaft, propeller-shaft slip joint).

Key failure modes:

  1. Shear across the key on the plane between shaft and hub: area b·l.
  2. Crushing (bearing) on the side faces: area (h/2)·l, half the height in each part. For a square key (h = b) the two strengths are equal when σ_c = 2τ. Making the key as strong in shear as the shaft gives, for b = d/4 and the same τ, l = π·d/2 ≈ 1.57·d.

Keyways weaken the shaft. The stress concentration at the keyway root and the loss of section are allowed for by a strength-reduction factor from the data book or by a higher factor of safety.

Couplings join two shafts end to end.

  • Rigid: muff (sleeve), split-muff (clamp) and flange couplings. They need accurate alignment and cannot absorb shock.
  • Flexible: bushed-pin (rubber bushes on pins), jaw/elastomer, gear and disc couplings, accommodating small angular, parallel and axial misalignment and damping shocks.
  • Universal (Hooke's) joint: for large angles between shafts, as in propeller shafts.

Flange coupling design checks: shaft in torsion; key in shear and crushing; hub (a hollow shaft of outer diameter D_h ≈ 2d); flange in shear at the hub junction; bolts in shear at the pitch circle (and crushing on the flange thickness).

Formulas

Shaft: T = π·d³·τ / 16 Key in shear: T = τ·b·l·(d/2) Key in crushing: T = σ_c·(h/2)·l·(d/2) Equal shear strength of key and shaft (b = d/4): l = π·d / 2 Flange bolts in shear: T = n·(π/4)·d₁²·τ_b·(D/2) Flange at hub (shear): T = π·D_h²·t_f·τ_f / 2 Torque from power: T = 60·P / (2π·N)

Symbols: T = torque (N·mm); d = shaft diameter (mm); τ = permissible shear stress (MPa); b, h, l = key width, height and length (mm); σ_c = permissible crushing stress (MPa); n = number of bolts; d₁ = bolt shank diameter (mm); τ_b = bolt shear stress (MPa); D = bolt pitch-circle diameter (mm); D_h = hub diameter (mm); t_f = flange thickness (mm); τ_f = flange shear stress (MPa); P = power (W); N = speed (rpm).

Worked examples

Example 1 (standard). A 40 mm shaft transmits 400 N·m through a rectangular key 12 mm wide and 8 mm high. Permissible shear stress 60 MPa, crushing stress 120 MPa. Find the key length.

  1. Shear: l = 2T/(τ·b·d) = 2 × 400 000 / (60 × 12 × 40) = 27.8 mm.
  2. Crushing: l = 4T/(σ_c·h·d) = 4 × 400 000 / (120 × 8 × 40) = 41.7 mm.
  3. Crushing governs. Take l = 45 mm.

Example 2 (GATE level). A flange coupling connects two 50 mm shafts transmitting 15 kW at 200 rpm. Six bolts sit on a 150 mm pitch circle; permissible bolt shear stress = 40 MPa. The hub diameter is 100 mm and the flange is 25 mm thick. Find the bolt diameter and the flange shear stress.

  1. T = 60 × 15 000 / (2π × 200) = 716.2 N·m = 716 200 N·mm.
  2. Bolts: d₁² = 8T/(n·π·τ_b·D) = 8 × 716 200 / (6 × π × 40 × 150) = 50.7 mm², so d₁ = 7.12 mm. Choose M10 fitted bolts (shank larger than 7.12 mm).
  3. Flange shear at the hub: τ_f = 2T/(π·D_h²·t_f) = 2 × 716 200 / (π × 100² × 25) = 1.82 MPa, far below the usual cast-iron limit, so the flange is safe in shear.

Common mistakes

  • Using b·h as the shear area of a key. The key shears over b·l; it crushes over (h/2)·l.
  • Forgetting the d/2 lever arm, or using the full shaft diameter as the arm.
  • Answering "it depends on the shaft radius" when the radius is given: the force on the key acts at d/2.
  • Computing the coupling's capacity with the bolt circle diameter as if it were the radius.
  • Mixing N·m and N·mm (1 N·m = 1000 N·mm).
  • Choosing a rigid coupling where the shafts cannot be held in alignment.

For GATE ME

Expect: key length or width from shear and crushing; torque capacity of a key or shaft; equal-strength conditions for keys; bolt size in a flange coupling; and type identification (feather, Woodruff, splines, flexible couplings). Practise converting power and speed into torque first.

Quick check

  1. A key 10 mm wide and 60 mm long on a 50 mm shaft, τ = 50 MPa. What torque can it carry in shear?
  2. Which key allows the hub to slide along the shaft?
  3. For a square key, what relation between σ_c and τ makes shear and crushing strengths equal?
  4. Why is a Woodruff key used on tapered shafts?
  5. Which coupling type would you use between a gearbox and a rear axle with a large varying angle?

Answers: 1. 50 × 10 × 60 × 25 = 750 000 N·mm = 750 N·m. 2. Feather key (or splines). 3. σ_c = 2τ. 4. It rocks in its seat and aligns itself with the hub keyway. 5. A universal (Hooke's) joint.

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 key in the context of machine design, and what is its primary function?Concept

    A key is a machine element used to connect a rotating machine element to a shaft. Its primary function is to transmit torque from the shaft to the rotating element, such as a gear or pulley, ensuring that they rotate together without slipping.

  2. 2.Explain the different types of keys used in mechanical design.Concept

    Sunk keys (rectangular or square, parallel or taper) are the general-purpose choice; their width is about a quarter of the shaft diameter. A feather key is a parallel key fixed to one part that lets the hub slide axially, as in gear-shifting. A Woodruff key is a semicircular disc that rocks in its seat and aligns itself with the hub, which suits tapered shafts and light torque, though its deep seat weakens the shaft. Saddle and round keys rely on friction and are for light duty, and splines are multiple integral keys used for high torque or sliding hubs.

  3. 3.What is a coupling, and why is it used in mechanical systems?Concept

    A coupling connects two shafts end to end so that power can be transmitted, and lets shafts made or supplied separately (motor and pump, gearbox and propeller shaft) be joined and disconnected. Rigid couplings (muff, clamp, flange) need accurate alignment and transmit everything, including shock. Flexible couplings (bushed-pin, elastomer, gear, disc) also accommodate small angular, parallel and axial misalignment and absorb shock and vibration. For large shaft angles a universal joint is used instead.

  4. 4.Explain the difference between rigid and flexible couplings.Concept

    Rigid couplings are used when precise alignment of the two shafts is required, as they do not allow for any misalignment. Flexible couplings, on the other hand, can accommodate some degree of misalignment and are used in applications where slight shaft misalignment is unavoidable or where vibration damping is needed.

  5. 5.Why are Woodruff keys used, and what is their drawback?Application

    A Woodruff key is a semicircular disc that sits in a matching curved seat in the shaft. Because it can rock in its seat, it aligns itself with the hub keyway during assembly, which is especially useful on tapered shaft ends. It is cheap to make and hold in place. Its drawback is the deep seat, which removes a lot of shaft section and raises the stress concentration, so it is used for light torque and small shafts.

  6. 6.What could happen if a key is improperly fitted in a shaft and hub assembly?Application

    If a key is improperly fitted, it can lead to several issues such as slippage between the shaft and the hub, uneven distribution of load, and increased wear and tear. This can ultimately result in mechanical failure, reduced efficiency, and potential damage to the machine components.

  7. 7.Why might a flexible coupling be used in a drive system with misaligned shafts?Application

    A flexible coupling is used in a drive system with misaligned shafts because it can accommodate angular, parallel, and axial misalignments. This flexibility helps in reducing stress on the shafts and bearings, minimizing vibration, and preventing premature wear or failure of the components.

  8. 8.A rectangular key 10 mm wide, 8 mm high and 50 mm long is fitted on a 40 mm shaft. If the permissible shear stress in the key is 40 MPa, what torque can it transmit in shear?Numerical

    The key shears on the plane between shaft and hub, of area b × l = 10 × 50 = 500 mm². The shear force acts at the shaft surface, r = d/2 = 20 mm. T = τ·b·l·(d/2) = 40 × 500 × 20 = 400 000 N·mm = 400 N·m. The key should also be checked in crushing on its side faces, over the area (h/2)·l.

  9. 9.A coupling is used to connect two shafts with a misalignment of 2 degrees. What type of coupling would you recommend and why?Application

    Two degrees of angular misalignment is more than a rigid coupling can tolerate and at the upper end of what many elastomeric or bushed-pin couplings are rated for, so the first step is to check the coupling manufacturer's rated angle. A gear coupling or a disc coupling rated for that angle, or a universal (Hooke's) joint, would be suitable; a universal joint is the usual choice when the angle is larger or varies, as in a propeller shaft. The aim is to keep the reaction loads on the shafts and bearings low.

  10. 10.Determine the maximum torque a 12 mm square key, 50 mm long, can transmit in shear on a 50 mm diameter shaft if the permissible shear stress is 50 MPa.Numerical

    The shear area is b × l = 12 × 50 = 600 mm² and the force acts at the shaft radius, 25 mm. T = τ·b·l·(d/2) = 50 × 600 × 25 = 750 000 N·mm = 750 N·m. For a square key with permissible crushing stress twice the shear stress, the crushing capacity would be the same, so shear is a fair estimate here.

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